tìm x
a, x - 3 = \(2\frac{1}{4}\)-\(\frac{1}{2}\)
b, x : 17,6 = 100 - 63,2
giúp mik vs T.T
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a:=>x^2-1-x=2x-1
=>x^2-x-1=2x-1
=>x^2-3x=0
=>x=0(loại) hoặc x=3(nhận)
b:=>x+2=0 hoặc 5-3x=0
=>x=-2 hoặc x=5/3
c:=>20(1-2x)+6x=9(x-5)-24
=>20-40x+6x=9x-45-24
=>-34x+20=9x-69
=>-43x=-89
=>x=89/43
d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3
=>2x^2+4x-19=-2x+7
=>2x^2+6x-26=0
=>x^2+3x-13=0
=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)
e: =>(2x-3)(2x-3-x-1)=0
=>(2x-3)(x-4)=0
=>x=4 hoặc x=3/2
\(\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x^2+x}\)
\(\Leftrightarrow\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x\left(x+1\right)}\)
ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)
Ta có : `(x-1)/x -1/(x+1) =(2x-1)/(x(x+1))`
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}-\dfrac{x}{x\left(x+1\right)}=\dfrac{2x-1}{x\left(x+1\right)}\)
`=> x^2 +x -x-1 -x-2x+1=0`
`<=> x^2 -3x =0`
`<=> x(x-3)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=3\end{matrix}\right.\)
__
`(x+2)(5-3x)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\5-3x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\3x=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{5}{3}\end{matrix}\right.\)
__
\(\dfrac{5\left(1-2x\right)}{3}+\dfrac{x}{2}=\dfrac{3\left(x-5\right)}{4}-2\)
\(\Leftrightarrow\dfrac{20\left(1-2x\right)}{12}+\dfrac{6x}{12}=\dfrac{9\left(x-5\right)}{12}-\dfrac{24}{12}\)
`<=> 2x- 40x + 6x = 9x - 45 -24`
`<=> 2x- 40x + 6x-9x + 45 +24=0`
`<=>-41x+69=0`
`<=>-41x=-69`
`<=> x=69/41`
a) \(\left|x+\frac{1}{5}\right|-4=-2\)
=) \(\left|x+\frac{1}{5}\right|=-2+4=2\)
=) \(x+\frac{1}{5}=2\)hoặc \(x+\frac{1}{5}=-2\)
=) \(x=2-\frac{1}{5}=\frac{9}{5}\); =) \(x=\left(-2\right)-\frac{1}{5}=\frac{-11}{5}\)
Vậy \(x=\left\{\frac{9}{5},\frac{-11}{5}\right\}\)
b)\(2x-\frac{1}{5}=\frac{6}{5}x-\frac{1}{2}\)
=) \(2x-\frac{6}{5}x=\frac{-1}{2}+\frac{1}{5}\)
=) \(x.\left(2-\frac{6}{5}\right)=\frac{-3}{10}\)
=) \(x.\frac{4}{5}=\frac{-3}{10}\)
=) \(x=\frac{-3}{10}:\frac{4}{5}\)
=) \(x=\frac{-3}{8}\)
c) \(\left(x-3\right)^{x+2}-\left(x-3\right)^{x+8}=0\)
=) \(\left(x-3\right)^{x+2}.\left(1-6\right)=0\)
=) \(\left(x-3\right)^{x+2}=0:\left(1-6\right)=0\)
Mà chỉ có \(0^x=0\)
=) \(x-3=0\)
=) \(x=0+3\)
=) \(x=3\)
a,
\(\left|x+\frac{1}{5}\right|-4=-2\)
\(\Rightarrow\left|x+\frac{1}{5}\right|=2\)
\(\Rightarrow\hept{\begin{cases}x+\frac{1}{5}=2\\x+\frac{1}{5}=-2\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{9}{5}\\x=-\frac{11}{5}\end{cases}}\)
b,
\(2x-\frac{1}{5}=\frac{6}{5}x-\frac{1}{2}\)
\(\Rightarrow2x-\frac{6}{5}x=-\frac{1}{2}+\frac{1}{5}\)
\(\Rightarrow\frac{4}{5}x=-\frac{3}{10}\Leftrightarrow x=-\frac{3}{8}\)
c,
\(\left[x-3\right]^{x+2}-\left[x-3\right]^{x+8}=0\)
=> [x-3]x + 2 = [x-3]x+8
=> x + 2 = x + 8
=> x không tồn tại
7/4.x+3/2=-4/5
7/4.x=-4/5-3/2
7/4.x=-23/10
x=-23/10:7/4
x=-46/35
vậy x=-46/35
1/4+3/4.x=3/4
1.x=3/4
x=3/4:1
x=3/4
vậy x=3/4
x.(1/4+1/5)-(1/7+1/8)=0
x.9/20-15/56=0
x.51/280=0
x=0:51/280
x=0
vậy x=0
3/35-(3/5+x)=2/7
(3/5+x)=3/35-2/7
(3/35+x)=-1/5
x=-1/5-3/5
x=-4/5
vậy x=-4/5
\(a,1\frac{3}{4}.x+1\frac{1}{2}=\frac{4}{5}\)
\(\frac{7}{4}.x=\frac{4}{5}-\frac{3}{2}\)
\(\frac{7}{4}.x=\frac{-7}{10}\)
\(x=\frac{-7}{10}:\frac{7}{4}\)
\(x=\frac{-2}{5}\)
\(b,\frac{1}{4}+\frac{3}{4}.x=\frac{3}{4}\)
\(\frac{3}{4}.x=\frac{3}{4}-\frac{1}{4}\)
\(\frac{3}{4}.x=\frac{1}{2}\)
\(x=\frac{1}{2}:\frac{3}{4}\)
\(x=\frac{2}{3}\)
\(c,x.\left(\frac{1}{4}+\frac{1}{5}\right)-\left(\frac{1}{7}+\frac{1}{8}\right)=0\)
\(x.\frac{9}{20}-\frac{15}{56}=0\)
\(x.\frac{9}{20}=\frac{15}{56}\)
\(x=\frac{15}{56}:\frac{9}{20}\)
\(x=\frac{25}{42}\)
\(d,\frac{3}{35}-\left(\frac{3}{5}+x\right)=\frac{2}{7}\)
\(\frac{3}{5}+x=\frac{3}{35}-\frac{2}{7}\)
\(\frac{3}{5}+x=\frac{-1}{5}\)
\(x=\frac{-1}{5}-\frac{3}{5}\)
\(x=\frac{-4}{5}\)
Học tốt
ui mk bó tay vì chưa hok đến lóp 9!!! ^^
54746767765858578758788974686865876546456475675685785
\(bn\)\(xem\)\(lai\)\(giup\)\(mk\)\(cho\)\(\frac{x+522}{7}\)\(neu\)\(thay\)\(bang\)\(\frac{x+552}{7}\)\(thi\)\(dug\)\(hon\)
thế thì bạn giải thử xem cô t ra đề thế mà ừ thì cứ cho là x + 552 cx đc
A, số nguyên âm lớn nhất là -1
=> 10-x=-1
X= 10 - (-1)
X= 11
B, |x-1| = |-4|
TH1: x-1=-4
X = -4 + 1
X = -3
TH2: x-1 = 4
X = 4+1
X = 5
Mình làm vật thôi ^_^ chúc bạn học tốt
Đức™
🤟🏿🤟🏿🤟🏿🤟🏿🤟🏿🌹
a/ 10-x là số nguyên âm lớn nhất => 10-x=-1
=>x=11
b/ I x-1 I =I-4I
=> Ix-1I=4 =>x-1=4 hoặc x-1=-4
=> x= 5 hoặc x= -3
c/(4-61-19x).(-4)2 =0 => -57-19x=0
=> 19x=-57 => x=-3
d/ 5x/-12=1/4+3/-2
=> 5x/-12=5/-12
=> 5x=5 => x=1
a) \(\left|x+\frac{1}{2}\right|=\frac{1}{3}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+\frac{1}{2}=\frac{1}{3}\\x+\frac{1}{2}=-\frac{1}{3}\end{cases}}\) \(\Leftrightarrow\)\(\orbr{\begin{cases}x=-\frac{1}{6}\\x=-\frac{5}{6}\end{cases}}\)
Vậy....
b) \(\left|x-\frac{1}{2}\right|=\frac{1}{3}-\frac{1}{2}\)
\(\Leftrightarrow\)\(\left|x-\frac{1}{2}\right|=-\frac{1}{6}\) vô lí do \(\left|a\right|\ge0\)
Vậy pt vô nghiệm
c) \(\left|x+\frac{1}{3}\right|-4=-1\)
\(\Leftrightarrow\)\(\left|x+\frac{1}{3}\right|=3\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+\frac{1}{3}=3\\x+\frac{1}{3}=-3\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{8}{3}\\x=-\frac{10}{3}\end{cases}}\)
Vậy..
d) \(\left|x-\frac{1}{5}\right|+\frac{1}{3}=\frac{1}{4}-\left|-\frac{3}{2}\right|\)
\(\Leftrightarrow\)\(\left|x-\frac{1}{5}\right|+\frac{1}{3}=-\frac{5}{4}\)
\(\Leftrightarrow\)\(\left|x-\frac{1}{5}\right|=-\frac{19}{12}\)vô lí do \(\left|a\right|\ge0\)với mọi a
Vậy pt vô nghiệm
e) \(\left|x-\frac{5}{2}\right|=\frac{4}{3}-\left(\frac{2}{3}-\frac{1}{2}\right)\)
\(\Leftrightarrow\)\(\left|x-\frac{5}{2}\right|=\frac{7}{6}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-\frac{5}{2}=\frac{7}{6}\\x-\frac{5}{2}=-\frac{7}{6}\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=3\frac{2}{3}\\x=\frac{4}{3}\end{cases}}\)
Vậy...
a)x-3=9/4-1/2
x-3=9/4-2/4
x-3=7/4
x=7/4+3
x=7/4+12/4
x=19/4
b)x:17,6=100-63,2
x:17,6=36,8
x=36,8x17,6
x=647,68
câu a bn viết đề bài rồi làm nha