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a) Điều kiện: x≥0x≥0
√16x=816x=8⇔(√16x)2=82⇔(16x)2=82 ⇔16x=64⇔16x=64 ⇔x=6416⇔x=4⇔x=6416⇔x=4 (thỏa mãn điều kiện)
Vậy x=4x=4.
Cách khác:
√16x=8⇔√16.√x=8⇔4√x=8⇔√x=2⇔x=22⇔x=416x=8⇔16.x=8⇔4x=8⇔x=2⇔x=22⇔x=4
b) Điều kiện: 4x≥0⇔x≥04x≥0⇔x≥0
√4x=√54x=5 ⇔(√4x)2=(√5)2⇔4x=5⇔x=54⇔(4x)2=(5)2⇔4x=5⇔x=54 (thỏa mãn điều kiện)
Vậy x=54x=54.
c) Điều kiện: 9(x−1)≥0⇔x−1≥0⇔x≥19(x−1)≥0⇔x−1≥0⇔x≥1
√9(x−1)=219(x−1)=21⇔3√x−1=21⇔3x−1=21⇔√x−1=7⇔x−1=7 ⇔x−1=49⇔x=50⇔x−1=49⇔x=50 (thỏa mãn điều kiện)
Vậy x=50x=50.
Cách khác:
√9(x−1)=21⇔9(x−1)=212⇔9(x−1)=441⇔x−1=49⇔x=509(x−1)=21⇔9(x−1)=212⇔9(x−1)=441⇔x−1=49⇔x=50
d) Điều kiện: x∈Rx∈R (vì 4.(1−x)2≥04.(1−x)2≥0 với mọi x)x)
√4(1−x)2−6=04(1−x)2−6=0⇔2√(1−x)2=6⇔2(1−x)2=6 ⇔|1−x|=3⇔|1−x|=3 ⇔[1−x=31−x=−3⇔[1−x=31−x=−3 ⇔[x=−2x=4⇔[x=−2x=4
Vậy x=−2;x=4.
a, \(\sqrt{16x}=8\Leftrightarrow4\sqrt{x}=8\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)
b, \(\sqrt{4x}=\sqrt{5}\)ĐK : x \(\ge0\)
bình phương 2 vế ta được : \(4x=5\Leftrightarrow x=\frac{5}{4}\)
c, \(\sqrt{9\left(x-1\right)}=21\Leftrightarrow3\sqrt{x-1}=21\Leftrightarrow\sqrt{x-1}=7\)
bình phương 2 vế ta được : \(x-1=49\Leftrightarrow x=50\)
d, \(\sqrt{4\left(1-x\right)^2}-6=0\Leftrightarrow2\left|1-x\right|=6\Leftrightarrow\left|1-x\right|=3\)
TH1 : \(1-x=3\Leftrightarrow x=-2\)
TH2 : \(1-x=-3\Leftrightarrow x=4\)
a) ( x - 3)4 + ( x - 5)4 = 82
Đặt : x - 4 = a , ta có :
( a + 1)4 + ( a - 1)4 = 82
⇔ a4 + 4a3 + 6a2 + 4a + 1 + a4 - 4a3 + 6a2 - 4a + 1 = 82
⇔ 2a4 + 12a2 - 80 = 0
⇔ 2( a4 + 6a2 - 40) = 0
⇔ a4 - 4a2 + 10a2 - 40 = 0
⇔ a2( a2 - 4) + 10( a2 - 4) = 0
⇔ ( a2 - 4)( a2 + 10) = 0
Do : a2 + 10 > 0
⇒ a2 - 4 = 0
⇔ a = + - 2
+) Với : a = 2 , ta có :
x - 4 = 2
⇔ x = 6
+) Với : a = -2 , ta có :
x - 4 = -2
⇔ x = 2
KL.....
b) ( n - 6)( n - 5)( n - 4)( n - 3) = 5.6.7.8
⇔ ( n - 6)( n - 3)( n - 5)( n - 4) = 1680
⇔ ( n2 - 9n + 18)( n2 - 9n + 20) = 1680
Đặt : n2 - 9n + 19 = t , ta có :
( t - 1)( t + 1) = 1680
⇔ t2 - 1 = 1680
⇔ t2 - 412 = 0
⇔ ( t - 41)( t + 41) = 0
⇔ t = 41 hoặc t = - 41
+) Với : t = 41 , ta có :
n2 - 9n + 19 = 41
⇔ n2 - 9n - 22 = 0
⇔ n2 + 2n - 11n - 22 = 0
⇔ n( n + 2) - 11( n + 2) = 0
⇔ ( n + 2)( n - 11) = 0
⇔ n = - 2 hoặc n = 11
+) Với : t = -41 ( giải tương tự )
@Giáo Viên Hoc24.vn
@Giáo Viên Hoc24h
@Giáo Viên
@giáo viên chuyên
@Akai Haruma
a) Ta có: \(2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20\)
\(\Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)
\(\Leftrightarrow4\sqrt{x-3}=20\)
\(\Leftrightarrow x-3=25\)
hay x=28
b) Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)
\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)
\(\Leftrightarrow2\sqrt{x+2}=6\)
\(\Leftrightarrow x+2=9\)
hay x=7
a)
`4(x-2)^2 =4`
`<=>(x-2)^2 =1`
`<=>x-2=1` hoặc `x-2=-1`
`<=>x=3` hoặc `x=1`
b)
`5(x^2 -6x+9)=5`
`<=>(x-3)^2 =1`
`<=>x-3=1`hoặc `x-3=-1`
`<=>x=4` hoặc `x=2`
c)
`4x^2 +4x+1=0`
`<=>(2x+1)^2 =0`
`<=>2x+1=0`
`<=>x=-1/2`
d)
`9x^2 +6x+1=2`
`<=>(3x+1)^2 =2`
\(< =>\left[{}\begin{matrix}3x+1=\sqrt{2}\\3x+1=-\sqrt{2}\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{\sqrt{2}-1}{3}\\x=\dfrac{-\sqrt{2}-1}{3}\end{matrix}\right.\)
a: Khi x=16 thì \(A=\dfrac{6}{16-3\cdot4}=\dfrac{6}{4}=\dfrac{3}{2}\)
b: P=A:B
\(=\dfrac{6}{\sqrt{x}\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{6}{\sqrt{x}\left(\sqrt{x}-3\right)}\cdot\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{6}\)
\(=\dfrac{\sqrt{x}+3}{\sqrt{x}}\)
c: \(P-1=\dfrac{\sqrt{x}+3-\sqrt{x}}{\sqrt{x}}=\dfrac{3}{\sqrt{x}}>0\)
=>P>1
\(f,\sqrt{x^2-25}-\sqrt{x-5}=0\)
=> \(\sqrt{x^2-25}=\sqrt{x-5}\)
=>\(x^2-25=x-5\)
=>\(x^2-x=25-5=20\)
=>( đến đoạn này mình xin chịu )
\(a,\sqrt{16x}=8\)
=>\(16x=8^2\)
=>\(16x=64\)
=>\(x=64:16=4\)
Vậy \(x\in\left\{4\right\}\)
\(b,\sqrt{x^2}=2x-1\)
=>\(x=2x-1\)
=>\(2x-x=1\)
=>\(x=1\)
Vậy \(x\in\left\{1\right\}\)
\(c,\sqrt{9.\left(x-1\right)}=21\)
=>\(9.\left(x-1\right)=21^2=441\)
=> \(x-1=441:9=49\)
=>\(x=49+1=50\)
Vậy \(x\in\left\{50\right\}\)
\(d,\sqrt{4\left(1-x\right)^2}-6=0\)
=>\(\sqrt{4\left(1-x\right)^2}=0+6=6\)
=> \(4\left(1-x\right)^2=6^2=36\)
=>\(\left(1-x\right)^2=36:4=9\)
=>\(1-x=\sqrt{9}=3\)
=>\(x=1-3=-2\)
Vậy \(x\in\left\{-2\right\}\)
\(g,\sqrt{9\left(2-3x\right)^2}=6\)
=> \(9.\left(2-3x\right)^2=6^2=36\)
=> \(\left(2-3x\right)^2=36:9=4\)
=> \(2-3x=\sqrt{4}=2\)
=>\(3x=2-2=0\)
=>\(x=0:3=0\)
Vậy \(x\in\left\{0\right\}\)
( còn các bài còn lại mình sẽ nghĩ tiếp , HS6-7 làm bài )
a) \(\sqrt{x^4}=2\)( ĐK x ∈ R )
⇔ \(\sqrt{\left(x^2\right)^2}=2\)
⇔ \(\left|x^2\right|=2\)
⇔ \(\orbr{\begin{cases}x^2=2\\x^2=-2\left(loai\right)\end{cases}}\)
⇔ x2 - 2 = 0
⇔ ( x - √2 )( x + √2 ) = 0
⇔ x - √2 = 0 hoặc x + √2 = 0
⇔ x = ±√2
b) \(3\sqrt{x+1}-8=0\)( ĐK x ≥ -1 )
⇔ \(3\sqrt{x+1}=8\)
⇔ \(\sqrt{x+1}=\frac{8}{3}\)
⇔ \(x+1=\frac{64}{9}\)
⇔ \(x=\frac{55}{9}\)( tm )
c) \(2\sqrt{x-3}+\sqrt{25x-75}=14\)( ĐK x ≥ 3 ) ( Vầy hợp lí hơn á )
⇔ \(2\sqrt{x-3}+\sqrt{5^2\left(x-3\right)}=14\)
⇔ \(2\sqrt{x-3}+5\sqrt{x-3}=14\)
⇔ \(7\sqrt{x-3}=14\)
⇔ \(\sqrt{x-3}=2\)
⇔ \(x-3=4\)
⇔ \(x=7\)( tm )
d) \(\sqrt{\left(3x-1\right)^2}=5\)( ĐK x ∈ R )
⇔ \(\left|3x-1\right|=5\)
⇔ \(\orbr{\begin{cases}3x-1=5\\3x-1=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{4}{3}\end{cases}}\)
e) \(\sqrt{x^2+4x+4}-6=0\)( ĐK x ∈ R )
⇔ \(\sqrt{\left(x+2\right)^2}=6\)
⇔ \(\left|x+2\right|=6\)
⇔ \(\orbr{\begin{cases}x+2=6\\x+2=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-8\end{cases}}\)
\(a)\)\(\sqrt{x^4}=2\)\(\Leftrightarrow\)\(x^2=2\)\(\Rightarrow\)\(\orbr{\begin{cases}x=\sqrt{2}\\x=-\sqrt{2}\end{cases}}\)
Vậy \(x=\sqrt{2}\)\(hoặc\)\(x=-\sqrt{2}\)
\(b)\)\(ĐK:x\ge0\)
\(3\sqrt{x+1}-8=0\)\(\Leftrightarrow\)\(3\sqrt{x}=8\)\(\Leftrightarrow\)\(\sqrt{x}=\frac{8}{3}\)\(\Leftrightarrow\)\(x=(\frac{8}{3})^2\)\(\Leftrightarrow\)\(x=\frac{64}{9}\)\((TM)\)
Vậy \(x=\frac{64}{9}\)
\(d)\)\(\sqrt{(3x-1)^2}=5\)\(\Leftrightarrow\)\(|3x-1|=5\)\((1)\)
- Nếu \(x\ge\frac{1}{3}\)thì \(\left(1\right)\Leftrightarrow3x-1=5\)\(\Leftrightarrow\)\(3x=6\)\(\Leftrightarrow\)\(x=2\)\(\left(TM\right)\)
- Nếu \(x< \frac{1}{3}\)thì \((1)\Leftrightarrow-\left(3x-1\right)=5\)\(\Leftrightarrow\)\(3x-1=-5\)\(\Leftrightarrow\)\(3x=-5+1\)\(\Leftrightarrow\)\(3x=-4\)\(\Leftrightarrow\)\(x=\frac{-4}{3}\left(TM\right)\)
Vậy \(x\in\hept{2;\frac{-4}{3}}\)
- \(e)\)\(\sqrt{x^2+4x+4}-6=0\)\(\Leftrightarrow\)\(\sqrt{(x+2)^2}=6\)\(\Leftrightarrow\)\(|x+2|=6\)\(\left(2\right)\)
-Nếu \(x\ge-2\)thì \(\left(2\right)\Leftrightarrow x+2=6\Leftrightarrow x=4(TM)\)
-Nếu \(x< -2\)thì \(\left(2\right)\Leftrightarrow-\left(x+2\right)=6\Leftrightarrow x+2=-6\Leftrightarrow x=-8\left(TM\right)\)
Vậy \(x=4;x=-8\)
1) \(\sqrt[]{9\left(x-1\right)}=21\)
\(\Leftrightarrow9\left(x-1\right)=21^2\)
\(\Leftrightarrow9\left(x-1\right)=441\)
\(\Leftrightarrow x-1=49\Leftrightarrow x=50\)
2) \(\sqrt[]{1-x}+\sqrt[]{4-4x}-\dfrac{1}{3}\sqrt[]{16-16x}+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}+\sqrt[]{4\left(1-x\right)}-\dfrac{1}{3}\sqrt[]{16\left(1-x\right)}+5=0\)
\(\)\(\Leftrightarrow\sqrt[]{1-x}+2\sqrt[]{1-x}-\dfrac{4}{3}\sqrt[]{1-x}+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}\left(1+3-\dfrac{4}{3}\right)+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}.\dfrac{8}{3}=-5\)
\(\Leftrightarrow\sqrt[]{1-x}=-\dfrac{15}{8}\)
mà \(\sqrt[]{1-x}\ge0\)
\(\Leftrightarrow pt.vô.nghiệm\)
3) \(\sqrt[]{2x}-\sqrt[]{50}=0\)
\(\Leftrightarrow\sqrt[]{2x}=\sqrt[]{50}\)
\(\Leftrightarrow2x=50\Leftrightarrow x=25\)
1) \(\sqrt{9\left(x-1\right)}=21\) (ĐK: \(x\ge1\))
\(\Leftrightarrow3\sqrt{x-1}=21\)
\(\Leftrightarrow\sqrt{x-1}=7\)
\(\Leftrightarrow x-1=49\)
\(\Leftrightarrow x=49+1\)
\(\Leftrightarrow x=50\left(tm\right)\)
2) \(\sqrt{1-x}+\sqrt{4-4x}-\dfrac{1}{3}\sqrt{16-16x}+5=0\) (ĐK: \(x\le1\))
\(\Leftrightarrow\sqrt{1-x}+2\sqrt{1-x}-\dfrac{4}{3}\sqrt{1-x}+5=0\)
\(\Leftrightarrow\dfrac{5}{3}\sqrt{1-x}+5=0\)
\(\Leftrightarrow\dfrac{5}{3}\sqrt{1-x}=-5\) (vô lý)
Phương trình vô nghiệm
3) \(\sqrt{2x}-\sqrt{50}=0\) (ĐK: \(x\ge0\))
\(\Leftrightarrow\sqrt{2x}=\sqrt{50}\)
\(\Leftrightarrow2x=50\)
\(\Leftrightarrow x=\dfrac{50}{2}\)
\(\Leftrightarrow x=25\left(tm\right)\)
4) \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\left(ĐK:x\ge-\dfrac{1}{2}\right)\\2x+1=-6\left(ĐK:x< -\dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\left(tm\right)\\x=-\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)
5) \(\sqrt{\left(x-3\right)^2}=3-x\)
\(\Leftrightarrow\left|x-3\right|=3-x\)
\(\Leftrightarrow x-3=3-x\)
\(\Leftrightarrow x+x=3+3\)
\(\Leftrightarrow x=\dfrac{6}{2}\)
\(\Leftrightarrow x=3\)
`đk:x ne +-3,x ne -2`
`B=(21/(x^2-9)-(x-4)/(3-x)-(x-1)/(3+x)):(1-1/(x+3))`
`=(21/(x^2-9)+(x-4)/(x-3)-(x-1)/(x+3)):((x+3-1)/(x+3))`
`=((21+x^2-x-12-x^2+4x-3)/((x-3)(x+3))):(x+2)/(x+3)`
`=(3x+6)/((x-3)(x+3))*(x+3)/(x+2)`
`=(3x+6)/((x-3)(x+2))`
`=3/(x-3)`
`b)|2x+1|=5`
`<=>` \(\left[ \begin{array}{l}2x=4\\2x=-6\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=2(tm)\\x=-3(l)\end{array} \right.\)
`=>B=3/(2-3)=-3`
`c)B=-3/5`
`<=>3/(x-3)=3/(-5)`
`<=>x-3=-5`
`<=>x=-2(l)`
`d)B<0`
`<=>3/(x-3)<0`
Mà `3>0`
`=>x-3<0<=>x<3`
a) đk: \(x\ne\pm3\)
\(B=\left[\dfrac{21}{\left(x-3\right)\left(x+3\right)}+\dfrac{x-4}{x-3}-\dfrac{x-1}{x+3}\right]:\left(\dfrac{x+3-1}{x+3}\right)\)
= \(\left[\dfrac{21+\left(x-4\right)\left(x+3\right)-\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right]:\dfrac{x+2}{x+3}\)
= \(\dfrac{21+x^2-x-12-x^2+4x-3}{\left(x-3\right)\left(x+3\right)}.\dfrac{x+3}{x+2}\)
= \(\dfrac{3x+6}{\left(x-3\right)\left(x+3\right)}.\dfrac{x+3}{x+2}=\dfrac{3}{x-3}\)
b) Để \(\left|2x+1\right|=5\)
<=> \(\left[{}\begin{matrix}2x+1=5< =>x=2\left(c\right)\\2x+1=-5< =>x=-3\left(l\right)\end{matrix}\right.\)
Thay x = 2, ta có;
B = \(\dfrac{3}{2-3}=-3\)
c) Để B = \(\dfrac{-3}{5}\)
<=> \(\dfrac{3}{x-3}=\dfrac{-3}{5}\)
<=> x - 3 = -5
<=> x = -2
d) Để B < 0
<=> \(\dfrac{3}{x-3}< 0\)
<=> x - 3 < 0
<=> x < 3
a) √16x = 8 (điều kiện: x ≥ 0)
⇔ 16x = 82 ⇔ 16x = 64 ⇔ x = 4
(Hoặc: √16x = 8 ⇔ √16.√x = 8
⇔ 4√x = 8 ⇔ √x = 2 ⇔ x = 4)
b) điều kiện: x ≥ 0
c) điều kiện: x - 1 ≥ 0 ⇔ x ≥ 1 (*)
x = 50 thỏa mãn điều kiện (*) nên x = 50 là nghiệm của phương trình.
d) Vì (1 - x)2 ≥ 0 ∀x nên phương trình xác định với mọi giá trị của x.
- Khi 1 – x ≥ 0 ⇔ x ≤ 1
Ta có: 2|1 – x| = 6 ⇔ 2(1 – x) = 6 ⇔ 2(1 – x) = 6
⇔ –2x = 4 ⇔ x = –2 (nhận)
- Khi 1 – x < 0 ⇔ x > 1
Ta có: 2|1 – x| = 6 ⇔ 2[– (1 – x)] = 6
⇔ x – 1 = 3 ⇔ x = 4 (nhận)
Vậy phương trình có hai nghiệm: x = - 2; x = 4