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a) \(\left(2.x-1\right)^6=\left(2.x-1\right)^8\)
\(\Leftrightarrow\left(2.x-1\right)^8-\left(2.x-1\right)^6=0\)
\(\Leftrightarrow\left(2x-1\right)^6.\left[\left(2x-1\right)-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^6=0\\\left(2x-1\right)-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x-1=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=1\end{matrix}\right.\)
Vậy : \(x\in\left\{\frac{1}{2},1\right\}\)
b) \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
\(\Leftrightarrow\left(x-1\right)^{x+4}-\left(x-1\right)^{x+2}=0\)
\(\Leftrightarrow\left(x-1\right)^{x+2}.\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^{x+2}=0\\\left(x-1\right)^2-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x-1=1\\x-1=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)
Vậy : \(x\in\left\{0,1,2\right\}\)
Chúc học tốt nhé !!
a, Ta có : \(\left(2x-1\right)^4=16\)
=> \(\left(\left(2x-1\right)^2\right)^2-\left(2^2\right)^2=0\)
=> \(\left(\left(2x-1\right)^2-2^2\right)\left(\left(2x-1\right)^2+2^2\right)=0\)
=> \(\left(2x-1-2\right)\left(2x-1+2\right)\left(\left(2x-1\right)^2+2^2\right)=0\)
Mà \(\left(2x-1\right)^2+2^2>0\)
=> \(\left(2x-3\right)\left(2x+1\right)=0\)
=> \(\left[{}\begin{matrix}x=\frac{3}{2}\\x=-\frac{1}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{\frac{3}{2};-\frac{1}{2}\right\}\)
b, Ta có : \(\left(2x+1\right)^4=\left(2x+1\right)^6\)
=> \(\left(2x+1\right)^6-\left(2x+1\right)^4=0\)
=> \(\left(2x+1\right)^4\left(\left(2x+1\right)^2-1\right)=0\)
=> \(\left(2x+1\right)^4\left(2x+1-1\right)\left(2x+1+1\right)=0\)
=> \(2x\left(2x+1\right)^4\left(2x+2\right)=0\)
=> \(\left[{}\begin{matrix}2x=0\\2x+1=0\\2x+2=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=0\\x=-\frac{1}{2}\\x=-1\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{0;-1;-\frac{1}{2}\right\}\)
c, Ta có : \(\left|\left|x+3\right|-8\right|=20\)
TH1 : \(x+3\ge0\left(x\ge-3\right)\)
=> \(\left|x+3\right|=x+3\)
=> \(\left|x-5\right|=20\)
TH1.1 : \(x-5\ge0\left(x\ge5\right)\)
=> \(\left|x-5\right|=x-5=20\)
=> \(x=25\left(TM\right)\)
TH1.2 : \(x-5< 0\left(x< 5\right)\)
=> \(\left|x-5\right|=5-x=20\)
=> \(x=-15\) ( không thỏa mãn )
TH2 : \(x+3< 0\left(x< -3\right)\)
=> \(\left|x+3\right|=-x-3\)
=> \(\left|-x-11\right|=20\)
TH1.1 : \(-x-11\ge0\left(x\le-11\right)\)
=> \(\left|-x-11\right|=-x-11=20\)
=> \(x=-31\left(TM\right)\)
TH1.2 : \(-x-11< 0\left(x>-11\right)\)
=> \(\left|-x-11\right|=x+11=20\)
=> \(x=9\) ( không thỏa mãn )
Vậy phương trình có tập nghiệm là \(S=\left\{-31;25\right\}\)
a, ( 2x - 1 )4 = 16
=> 2x - 1 = 2 hoặc -2
TH1: 2x - 1 = 2
=> 2x = 2 + 1 = 3; => x = \(\frac{3}{2}\)
TH2: 2x - 1 = -2
=> 2x = -2 + 1 = -1; => x =- \(\frac{1}{2}\)
b, ( 2x + 1 )4 = ( 2x + 1 )6
=> ( 2x + 1 )4 - ( 2x + 1 )6 = 0
= ( 2x + 1 )4 - ( 2x - 1 )2 . ( 2x - 1 )4
= ( 2x + 1 )4 . [ 1 - ( 2x - 1 )2 ] = 0
Ta có ( 2x + 1 )4 và ( 2x - 1 )2 \(\ge\) 0 vì có số mũ chẵn
Ta có 2 TH
TH1: ( 2x - 1 )4 = 0
=> 2x - 1 = 0; => x = \(\frac{1}{2}\)
TH2: 1 - ( 2x - 1 )2 = 0; => ( 2x - 1 )2 = 1
=> 2x - 1 = 1; => x = 1
c, //x + 3/ - 8/ = 20
Ta có 2 TH, mỗi TH lại chia thành 2 TH nhỏ hơn
TH1: /x + 3/ - 8 = 20
=> /x + 3/ = 28
=> x + 3 = 28 hoặc -28
TH1 nhỏ: x + 3 = 28; => x = 25
TH2 nhỏ: x + 3 = -28; => x = -31
TH2: /x + 3/ - 8 = -20
=> /x + 3/ = -12; => TH này loại
=> x = 25; -31
a) \(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{2}\right)^3\)
\(\Rightarrow x-\frac{1}{2}=\frac{1}{2}\)
\(\Rightarrow x=1\)
b) \(\left(x+\frac{1}{2}\right)^2=\left(\frac{2}{3}\right)^2\)
\(\Rightarrow x+\frac{1}{2}=\frac{2}{3}\)
\(\Rightarrow x=\frac{1}{6}\)
\(\left|x+\dfrac{1}{1.5}\right|+\left|x+\dfrac{1}{5.9}\right|+\left|x+\dfrac{1}{9.14}\right|+...+\left|x+\dfrac{1}{397.401}\right|\ge0\)
\(\Rightarrow101x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow x+\dfrac{1}{1.5}+x+\dfrac{1}{5.9}+...+x+\dfrac{1}{397.401}=101x\)
\(\Rightarrow101x+\left(\dfrac{1}{1.5}+\dfrac{1}{5.9}+...+\dfrac{1}{397.401}\right)=x\)
\(\Rightarrow\dfrac{1}{4}\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+...+\dfrac{4}{397.401}\right)=x\)
\(\Rightarrow x=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+....+\dfrac{1}{397}-\dfrac{1}{401}\right)\)
\(\Rightarrow x=\dfrac{1}{4}\left(1-\dfrac{1}{401}\right)\)
\(\Rightarrow x=\dfrac{1}{4}.\dfrac{400}{401}\)
\(\Rightarrow x=\dfrac{100}{401}\)
a) \(\left(x-5\right)\left(x+8\right)-\left(x+4\right)\left(x-1\right)\)
\(=\left(x^2+3x-40\right)-\left(x^2+3x-4\right)\)
\(=x^2+3x-40-x^2-3x+4\)
\(=-36\)
b)\(x^4\left(x^2-1\right)\left(x^2+1\right)\)
\(=x^4\left(x^4-1\right)\)
\(=x^8-x^4\)
a. \(\left(x+5\right)^3=-64\)
\(\left(x+5\right)^3=\left(-4\right)^3\)
\(\Rightarrow x+5=-4\)
=> x = -9
b. \(|2x-5|=8\)
\(\left[{}\begin{matrix}2x-5=8\\2x-5=-8\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=13\\2x=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{13}{2}\\x=\dfrac{-3}{2}\end{matrix}\right.\)
c. \(\left|\dfrac{3}{4}x-\dfrac{1}{5}\right|=2\)
\(\left[{}\begin{matrix}\dfrac{3}{4}x-\dfrac{1}{5}=2\\\dfrac{3}{4}x-\dfrac{1}{5}=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{11}{5}\\\dfrac{3}{4}x=\dfrac{-9}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{44}{15}\\x=\dfrac{-12}{5}\end{matrix}\right.\)
d. \(\left|3x-6\right|=x+4\)
\(\left[{}\begin{matrix}3x-6=x+4\\3x-6=-x-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x-x=4+6\\3x+x=-4+6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=10\\4x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{1}{2}\end{matrix}\right.\)
e. \(\left|x-3\right|=2x+1\)
\(\left[{}\begin{matrix}x-3=2x+1\\x-3=-2x-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-2x=1+3\\x+2x=-1+3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}-x=4\\3x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{2}{3}\end{matrix}\right.\)
a/ (x - 1)6 = (x - 1)8
=> (x - 1)6 [1 - (x - 1)2] = 0
=> (x - 1)6 (1 - x2 + 2x - 1) = 0
=> (x - 1)6 (-x2 + 2x) = 0
=> x - 1 = 0 => x = 1
hoặc - x2 + 2x = 0 => x = 0 hoặc x = 2
Vậy x = 0, x = 1, x = 2