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Khách
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KH
22 tháng 9 2018
* Trả lời:
\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)
\(\Leftrightarrow-3+6x-4-12x=-5x+5\)
\(\Leftrightarrow6x-12x+5x=3+4+5\)
\(\Leftrightarrow x=12\)
\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)
\(\Leftrightarrow6x-15-6+24x=-3x+7\)
\(\Leftrightarrow6x+24x+3x=15+6+7\)
\(\Leftrightarrow33x=28\)
\(\Leftrightarrow x=\dfrac{28}{33}\)
\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)
\(\Leftrightarrow1-3x-6x+12=-4x-5\)
\(\Leftrightarrow-3x-6x+4x=-1-12-5\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)
\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)
\(\Leftrightarrow-x-5x=-7\)
\(\Leftrightarrow-6x=-7\)
\(\Leftrightarrow x=\dfrac{7}{6}\)
\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)
\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)
\(\Leftrightarrow-15x+3x=4\)
\(\Leftrightarrow-12x=4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
24 tháng 9 2021
\(\left(2x-1\right)^6=\left(2x-1\right)^8\\ \Rightarrow\left(2x-1\right)^6-\left(2x-1\right)^8=0\\ \Rightarrow\left(2x-1\right)^6\left[1-\left(2x-1\right)^2\right]=0\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\\left(2x-1\right)^2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\2x-1=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)
24 tháng 9 2021
\(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=-1\\2x-1=1\\2x-1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{1}{2}\end{matrix}\right.\)
25 tháng 8 2020
a) \(2x+\frac{3}{15}=\frac{7}{5}\)
=> \(2x=\frac{7}{5}-\frac{3}{15}=\frac{21}{15}-\frac{3}{15}=\frac{18}{15}\)
=> \(x=\frac{18}{15}:2=\frac{18}{15}\cdot\frac{1}{2}=\frac{9}{15}\cdot\frac{1}{1}=\frac{9}{15}\)
b) \(x-\frac{2}{9}=\frac{8}{3}\)
=> \(x=\frac{8}{3}+\frac{2}{9}\)
=> \(x=\frac{24}{9}+\frac{2}{9}=\frac{26}{9}\)
c) \(\frac{-8}{x}=\frac{-x}{18}\)
=> x(-x) = (-8).18
=> -x2 = -144
=> x2 = 144(bỏ dấu âm)
=> x = \(\pm\)12
d) \(\frac{2x+3}{6}=\frac{x-2}{5}\)
=> 5(2x + 3) = 6(x - 2)
=> 10x + 15 = 6x - 12
=> 10x + 15 - 6x + 12 = 0
=> 4x + 27 = 0
=> 4x = -27
=> x = -27/4
e) \(\frac{x+1}{22}=\frac{6}{x}\)
=> x(x + 1) = 132
=> x(x + 1) = 11.12
=> x = 11
f) \(\frac{2x-1}{2}=\frac{5}{x}\)
=> x(2x - 1) = 10
=> 2x2 - x = 10
=> 2x2 - x - 10 = 0
tới đây tự làm đi nhé
g) \(\frac{2x-1}{21}=\frac{3}{2x+1}\)
=> (2x - 1)(2x + 1) = 63
=> 4x2 - 1 = 63
=> 4x2 = 64
=> x2 = 16
=> x = \(\pm\)4
h) Tương tự
NC
25 tháng 8 2020
a) \(\frac{2x+3}{15}=\frac{7}{5}\Leftrightarrow10x+15=105\Leftrightarrow10x=90\Rightarrow x=9\)
b) \(\frac{x-2}{9}=\frac{8}{3}\Leftrightarrow3x-6=72\Leftrightarrow3x=78\Rightarrow x=26\)
c) \(\frac{-8}{x}=\frac{-x}{18}\Leftrightarrow x^2=144\Leftrightarrow\orbr{\begin{cases}x=12\\x=-12\end{cases}}\)
d) \(\frac{2x+3}{6}=\frac{x-2}{5}\Leftrightarrow10x+15=12x-12\Leftrightarrow2x=27\Rightarrow x=\frac{27}{2}\)
e) \(\frac{x+1}{22}=\frac{6}{x}\Leftrightarrow x^2+x-132=0\Leftrightarrow\left(x-11\right)\left(x+12\right)=0\Leftrightarrow\orbr{\begin{cases}x=11\\x=-12\end{cases}}\)
f) \(\frac{2x-1}{2}=\frac{5}{x}\Leftrightarrow2x^2-x-10=0\Leftrightarrow\left(x-2\right)\left(2x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{5}{2}\end{cases}}\)
g) \(\frac{2x-1}{21}=\frac{3}{2x+1}\Leftrightarrow4x^2=64\Leftrightarrow x^2=16\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
h) \(\frac{10x+5}{6}=\frac{5}{x+1}\Leftrightarrow10x^2+15x-25=0\Leftrightarrow5\left(x-1\right)\left(2x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)
MN
21 tháng 1 2017
Vì \(\left(2x-1\right)^6;\left(2x-1\right)^8\ge0,\forall x\)
\(\Rightarrow\left(2x-1\right)^6=\left(2x-1\right)^8\Leftrightarrow\left(2x-1\right)=0\Leftrightarrow x=\frac{1}{2}\)
21 tháng 9 2016
(2x - 1)6 = (2x - 1)8
=> (2x - 1)8 - (2x - 1)6 = 0
=> (2x - 1)6.[(2x - 1)2 - 1] = 0
=> \(\orbr{\begin{cases}\left(2x-1\right)^6=0\\\left(2x-1\right)^2-1=0\end{cases}}\)=> \(\orbr{\begin{cases}2x-1=0\\\left(2x-1\right)^2=1\end{cases}}\)=> \(\orbr{\begin{cases}2x=1\\2x-1\in\left\{1;-1\right\}\end{cases}}\)=> \(\orbr{\begin{cases}x=\frac{1}{2}\\2x\in\left\{2;0\right\}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x\in\left\{1;0\right\}\end{cases}}\)
Vậy \(x\in\left\{\frac{1}{2};1;0\right\}\)
TM
21 tháng 9 2016
\(\left(2x-1\right)^6=\left(2x-1\right)^8\)
=>\(\left(2x-1\right)^6-\left(2x-1\right)^8=0\)
=>\(\left(2x-1\right)^6\left[1-\left(2x-1\right)^2\right]=0\)
=>\(\left(2x-1\right)^6\left(1-2x+1\right)\left(1+2x-1\right)=0\)
=>\(\left(2x-1\right)^6\left(2-2x\right)2x=0\)
=>(2x-1)6=0 hoặc 2-2x=0 hoặc x=0
+)Với (2x-1)2=6 => x=\(\frac{1}{2}\)
+)Với 2-2x=0 => x=1
Vậy \(x\in\left\{0;\frac{1}{2};1\right\}\)
AN
1
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 3 2023
a.
ĐKXĐ: \(x\ne6\)
\(\dfrac{7}{x-6}=\dfrac{x-6}{7}\)
\(\Leftrightarrow\dfrac{49}{7\left(x-6\right)}=\dfrac{\left(x-6\right)^2}{7\left(x-6\right)}\)
\(\Rightarrow\left(x-6\right)^2=49=7^2\)
\(\Rightarrow\left[{}\begin{matrix}x-6=7\\x-6=-7\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=13\\x=-1\end{matrix}\right.\) (thỏa mãn)
b. ĐKXĐ: \(x\ne\dfrac{1}{2}\)
\(\dfrac{2x-1}{8}=\dfrac{-2}{1-2x}\)
\(\Leftrightarrow\dfrac{\left(2x-1\right)^2}{8\left(2x-1\right)}=\dfrac{16}{8\left(2x-1\right)}\)
\(\Rightarrow\left(2x-1\right)^2=16=4^2\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=4\\2x-1=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\) (thỏa mãn)
\(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\Rightarrow\left(2x-1\right)^8-\left(2x-1\right)^6=0\)
\(\Rightarrow\left(2x-1\right)^6\left[\left(2x-1\right)^2-1\right]=0\)
\(\Rightarrow\left(2x-1\right)^6\left(2x-1+1\right)\left(2x-1-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(2x-1\right)^6=0\\2x=0\\2x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=0\\x=1\end{matrix}\right.\)
Vậy......................
\(\left(2x-1\right)^6=\left(2x-1\right)^8\\ \Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\2x-1=1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)