4/5 x 3/x x 2/11 = 24/165
x = ?
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17 tháng 7 2016
1.2x^2+x-6=2x^2+4x-3x+6=(2x^2+4x)-(3x+6)=2x(x+2)-3(x+2)=(x+2)(2x-3)
2.x^3-9x^2+14x
=x*(x^2-9x+14)
=x*(x^2-7x-2x+14)
=x*((x^2-7x)-(2x-14))
=x*(x(x-7)-2(x-7))
=x*((x--7)(x-2))
=x*(x-7)(x-2)
PD
2
17 tháng 8 2021
a,
\(\left(x-6\right)^2=9\\ \Rightarrow\left[{}\begin{matrix}x-6=-3\\x-6=3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=9\end{matrix}\right.\)
b,
\(\left|x\right|=3\\ \Rightarrow\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)
c,
\(\left|x+5\right|=15\\ \Rightarrow\left[{}\begin{matrix}x+5=-15\\x+5=15\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-20\\x=10\end{matrix}\right.\)
d,
\(2^{x-1}=16\\ \Rightarrow2^{x-1}=2^4\\ \Rightarrow x-1=4\\ \Rightarrow x=5\)
e,
\(5^{x+1}=125\\ \Rightarrow5^{x+1}=5^3\\ \Rightarrow x+1=3\\ \Rightarrow x=2\)
17 tháng 8 2021
a: Ta có: \(\left(x-6\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=3\\x-6=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=3\end{matrix}\right.\)
b: Ta có: \(\left|x\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
c: Ta có: \(\left|x+5\right|=15\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=15\\x+5=-15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-20\end{matrix}\right.\)
PT
2
15 tháng 4 2021
Đặt x2 + 10x + 24 = y
pt đã cho trở thành ( y + 4x ).y - 165x2 = 0
<=> y2 + 4xy - 165x2 = 0
<=> y2 - 11xy + 15xy - 165x2 = 0
<=> y( y - 11x ) + 15x( y - 11x ) = 0
<=> ( y - 11x )( y + 15x ) = 0
=> ( x2 + 10x + 24 - 11x )( x2 + 10x + 24 + 15x ) = 0
<=> ( x2 - x + 24 )( x2 + 25x + 24 ) = 0
<=> ( x2 - x + 24 )( x2 + 24x + x + 24 ) = 0
<=> ( x2 - x + 24 )[ x( x + 24 ) + ( x + 24 ) ] = 0
<=> ( x2 - x + 24 )( x + 24 )( x + 1 ) = 0
Vì x2 - x + 24 > 0 ∀ x
nên pt <=> ( x + 24 )( x + 1 ) = 0 <=> x = -24 hoặc x = -1
Vậy ...
NC
16 tháng 4 2021
Đặt t = \(x^2+14x+24\)
\(\Rightarrow\)\(t\left(t-4x\right)-165x^{^2}=0\)
\(\Leftrightarrow t^2-4xt-165x^2=0\)
\(\Leftrightarrow t^2+11xt-15xt-165x^2=0\)
\(\Leftrightarrow t\left(t+11x\right)-15x\left(t+11x\right)=0\)
\(\Leftrightarrow\left(t+11x\right)\left(t-15x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}t+11x=0\\t-15x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}t=-11x\\t=15x\end{cases}}}\)
với t= -11x
\(\Rightarrow x^2+14x+24=-11x\)
\(\Leftrightarrow x^2+25x+24=0\)
\(\Leftrightarrow x^2+x+24x+24=0\)
\(\Leftrightarrow x\left(x+1\right)+24\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+24\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+24=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-24\end{cases}}}\)
với t=15x
\(\Rightarrow x^2+14x+24=15x\)
\(\Leftrightarrow x^2-x+24=0\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{95}{4}=0\)(Vô Lí)
vậy....
BB
3
BD
20 tháng 8 2023
\(a.\dfrac{1}{2}:3+x=\dfrac{14}{5}\)
\(\dfrac{1}{6}+x=\dfrac{14}{5}\)
\(=>x=\dfrac{79}{30}\)
\(b.\dfrac{8}{5}:x:\dfrac{7}{4}=\dfrac{11}{6}\)
\(\left(\dfrac{8}{5}\cdot\dfrac{4}{7}\right):x=\dfrac{11}{6}\)
\(\dfrac{32}{35}:x=\dfrac{11}{6}\)
\(x=\dfrac{192}{385}\)
\(c.\dfrac{24}{10}+x:\dfrac{3}{4}=\dfrac{11}{3}\)
\(x:\dfrac{3}{4}=\dfrac{11}{3}-\dfrac{24}{10}\)
\(x:\dfrac{3}{4}=\dfrac{38}{30}\)
\(=>x=\dfrac{19}{20}\)
HQ
Hà Quang Minh
Giáo viên
20 tháng 8 2023
\(a,\dfrac{1}{2}:3+x=\dfrac{14}{5}\\ \Leftrightarrow x+\dfrac{1}{6}=\dfrac{14}{5}\\ \Leftrightarrow x=\dfrac{79}{30}\\ b,\dfrac{8}{5}:x:\dfrac{7}{4}=\dfrac{11}{6}\\ \Leftrightarrow x=\dfrac{192}{385}\\ c,\dfrac{24}{10}+x:\dfrac{3}{4}=\dfrac{11}{3}\\ \Leftrightarrow\dfrac{4}{3}x=\dfrac{19}{15}\\ \Leftrightarrow x=\dfrac{19}{20}\)
NN
1
A
1
20 tháng 2 2020
(x^2 +24+14x) (x^2+24+10x) =165x^2
Đặt t = x^2 + 24+12x
(t-2x)(t+2x) = 165x^2
t^2 - 4x^2 =165x^2
t^2 = 169x^2
t = 13x hay t = -13x
Nếu t = 13x thì
x^2 +12x + 24= 13x
x^2 - x + 24 = 0 (Vô nghiệm vì vế trái > 0)
Nếu t = -13x thì:
x^2 +12x+24 = -13x
x^2 +25x +24=0
(x+1)(x+24) = 0
x + 1 =0 hay x+24 = 0
x = -1 hay x= -24
Vậy...
Học tốt!
DQ
2
31 tháng 7 2023
a: =>(x-2)^3*[(x-2)^8-1]=0
=>(x-2)(x-3)(x-1)=0
=>\(x\in\left\{2;3;1\right\}\)
b: (x-5)^24=(x-5)^9
=>\(\left(x-5\right)^9\cdot\left[\left(x-5\right)^{15}-1\right]=0\)
=>x-5=0 hoặc x-5=1
=>x=6 hoặc x=5
c: =>(x-5)^4*[(x-5)^21-1]=0
=>x-5=0 hoặc x-5=1
=>x=5 hoặc x=6
H9
HT.Phong (9A5)
CTVHS
31 tháng 7 2023
a) \(\left(x-2\right)^{11}=\left(x-2\right)^3\)
\(\Rightarrow\left(x-2\right)^{11}-\left(x-2\right)^3=0\)
\(\Rightarrow\left(x-2\right)^3\left[\left(x-2\right)^8-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-2\right)^3=0\\\left(x-2\right)^8-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\\left(x-2\right)^8=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x-2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
b) \(\left(x-5\right)^{24}=\left(x-5\right)^9\)
\(\Rightarrow\left(x-5\right)^{24}-\left(x-5\right)^9=0\)
\(\Rightarrow\left(x-5\right)^9\left[\left(x-5\right)^{15}-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^9=0\\\left(x-5\right)^{15}-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^{15}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x-5=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)
c) \(\left(x-5\right)^{25}=\left(x-5\right)^4\)
\(\Rightarrow\left(x-5\right)^{25}-\left(x-5\right)^4\)
\(\Rightarrow\left(x-5\right)^4\left[\left(x-5\right)^{21}-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^4=0\\\left(x-5\right)^{21}-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^{21}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x-5=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)
CD
2
16 tháng 4 2017
\(\frac{4}{5}\cdot\frac{3}{x}\cdot\frac{2}{11}=\frac{24}{165}\)
\(\frac{3}{x}\) \(=\frac{24}{165}:\frac{2}{11}:\frac{4}{5}\)
\(\frac{3}{x}\) \(=1\)
\(x\) \(=3\)
\(\frac{4}{3}:\frac{x}{5}=\frac{5}{13}\cdot\frac{4}{9}\)
\(\frac{4}{3}:\frac{x}{5}=\frac{20}{117}\)
\(\frac{x}{5}=\frac{4}{3}:\frac{20}{117}\)
\(\frac{x}{5}=\frac{39}{5}\)
\(x=39\)
\(\frac{4}{5}\)x\(\frac{3}{x}\)x\(\frac{2}{11}=\frac{24}{165}\)
\(\Rightarrow\frac{4}{5}\)x\(\frac{3}{x}\)x\(\frac{2}{11}=\frac{8}{55}\)
\(\Rightarrow\frac{4}{5}\)x\(\frac{3}{x}\)= \(\frac{8}{55}\):\(\frac{2}{11}\)
\(\Rightarrow\frac{4}{5}\)x\(\frac{3}{x}\)=\(\frac{4}{5}\)
\(\Rightarrow\frac{3}{x}\)=\(\frac{4}{5}:\frac{4}{5}\)
\(\Rightarrow\frac{3}{x}=1\)
\(\Rightarrow\frac{3}{x}=\frac{3}{3}\)
\(\Rightarrow x=3\)