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a) \(\left(2x-1\right)^{10}=\left(1-2x\right)^5\)
\(\Rightarrow\left(2x-1\right)^2=1-2x\)
\(\Rightarrow4x^2-4x+1=1-2x\)
\(\Rightarrow4x^2-4x=-2x\)
\(\Rightarrow2x^2-2x=-x\)
\(\Rightarrow2x^2-x=0\)
\(\Rightarrow x.\left(2x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\2x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy ...
b) \(\left(3x-1\right)^{15}=\left(1-3x\right)^8\)
\(\Rightarrow\left(3x-1\right)^{15}-\left(1-3x\right)^8=0\)
\(\Rightarrow\left(3x-1\right)^{15}-\left(-\left(3x-1\right)\right)^8=0\)
\(\Rightarrow\left(3x-1\right)^{15}-\left(3x-1\right)^8=0\)
\(\Rightarrow\left(3x-1\right)^8.\left(\left(3x-1\right)^7-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(3x-1\right)^8=0\\\left(3x-1\right)^7-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy ....
c) Tự lm
a, \(\left|x+1\right|-4=0\)
\(\Rightarrow\left|x+1\right|=4\)
+) Xét \(x\ge-1\) có:
\(x+1=4\Rightarrow x=3\) ( t/m )
+) Xét x < -1 có:
\(-x-1=4\Rightarrow x=-5\) ( t/m )
Vậy x = 3 hoặc x = -5
b, tương tự
c, \(3x+\left|2x\right|=5x\)
\(\Rightarrow\left|2x\right|=2x\)
+) Xét \(x\ge0\)
\(\Rightarrow2x=2x\Rightarrow x\in R\forall x>0\)
+) Xét x < 0
\(\Rightarrow-2x=2x\Rightarrow x=0\)
Vậy \(x\ge0\)
d, \(\left|2x+1\right|=\left|x-3\right|\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=x-3\\2x+1=3-x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy...
a) \(\left|0,5x-2\right|-\left|x+\frac{1}{3}\right|=0\)
=> \(\left|0,5x-2\right|=\left|x+\frac{1}{3}\right|\)
=> \(\orbr{\begin{cases}0,5x-2=x+\frac{1}{3}\\0,5x-2=-x-\frac{1}{3}\end{cases}}\)
=> \(\orbr{\begin{cases}-0,5x=\frac{7}{3}\\1,5x=\frac{5}{3}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{14}{3}\\x=\frac{10}{9}\end{cases}}\)
b) \(2x-\left|x+1\right|=\frac{1}{2}\)
=> \(\left|x+1\right|=2x-\frac{1}{2}\) (Đk: \(2x-\frac{1}{2}\ge0\) <=> \(x\ge\frac{1}{4}\))
=> \(\orbr{\begin{cases}x+1=2x-\frac{1}{2}\\x+1=\frac{1}{2}-2x\end{cases}}\)
=> \(\orbr{\begin{cases}-x=-\frac{3}{2}\\3x=-\frac{1}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{6}\end{cases}}\)
a: \(\Leftrightarrow\left\{{}\begin{matrix}x>=-2\\\left(3x+8+2x+4\right)\left(3x+8-2x-4\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-2\\\left(5x+12\right)\left(x+4\right)=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
b: \(\Leftrightarrow\left|4x+2\right|=x+15\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-15\\\left(4x+2+x+15\right)\left(4x+2-x-15\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-15\\\left(5x+17\right)\left(3x-13\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{17}{5};\dfrac{13}{3}\right\}\)
c: =>3x+7>=0
hay x>=-7/3
d: =>|2x-5|=-2x+5
=>2x-5<=0
hay x<=5/2
a) Ta có \(|5\left(2x+3\right)\ge0\)
\(|2\left(2x+3\right)|\ge0\)
\(|2x+3|\ge0\)
\(\Rightarrow|5\left(2x+3\right)|+|\left(2x+3\right)|+|2x+3|\ge0\)
\(\Rightarrow5\left(2x+3\right)+2\left(2x+3\right)+2x+3=16\)
\(\Rightarrow10x+15+4x+6+2x+3=16\)
\(\Rightarrow\left(10x+4x+2x\right)+\left(15+6+3\right)=16\)
\(\Rightarrow16x+24=16\)
\(\Rightarrow24=16x-16\)
\(\Rightarrow24=x\)
Vậy x=24
a, \(\frac{3}{5}\left(2x-\frac{1}{3}\right)+\frac{4}{15}=\frac{12}{30}\)
\(\Leftrightarrow\frac{3}{5}\left(2x-\frac{1}{3}\right)=\frac{2}{15}\)
\(\Leftrightarrow2x-\frac{1}{3}=\frac{2}{9}\)
\(\Leftrightarrow2x=\frac{5}{9}\)
\(\Leftrightarrow x=\frac{5}{18}\)
b,\(\left(-0,2\right)^x=\frac{1}{25}\)
\(\Leftrightarrow\left(\frac{-1}{5}\right)^x=\left(\frac{-1}{5}\right)^2\)
\(\Leftrightarrow x=2\)
c,\(\left|x-1\right|-\frac{3}{12}=\left(-\frac{1}{2}\right)^2\)
\(\Leftrightarrow\left|x-1\right|-\frac{3}{12}=\frac{1}{4}\)
\(\Leftrightarrow\left|x-1\right|=\frac{1}{2}\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=\frac{1}{2}\\x-1=-\frac{1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{1}{2}\end{cases}}\)
\(a,\frac{3}{5}\left(2x-\frac{1}{3}\right)=\frac{12}{30}-\frac{4}{15}\)
\(\frac{3}{5}\left(2x-\frac{1}{3}\right)=\frac{2}{15}\)
\(2x-\frac{1}{3}=\frac{2}{9}\)
\(x=\frac{5}{18}\)
\(b,\left(-0,2\right)^x=\frac{1}{25}\)
\(\left(-0,2\right)^x=\left(-\frac{1}{5}\right)^2\)
\(\left(-0,2\right)^x=\left(-0,2\right)^2\)
\(x=2\)
c,/x-1/=1/2
Nếu
\(x-1\ge0\)
\(x\ge1\)
suy ra x-1=1/2
x=3/2(thỏa mãn điều kiện )
nếu \(x-1\le0\)
\(x\le1\)
suy ra x-1=-1/2
x=1/2 (thỏa mãn điều kiện )
Vậy ...
nha !!!
\(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x-15=0\\2x-15=1\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x=15\\2x=16\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{15}{2}\\x=8\end{array}\right.\)
bai nay thieu = -1