Tìm x biết :
\(\left(4-2x\right)^x=4\)
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DT
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18 tháng 11 2016
\(\left|\left(x+\frac{1}{2}\right).\left|2x-\frac{3}{4}\right|\right|=2x-\frac{3}{4}\)
\(\Rightarrow\left|x+\frac{1}{2}\right|.\left|2x-\frac{3}{4}\right|=2x-\frac{3}{4}\)
\(\Rightarrow2x-\frac{3}{4}\ge0\) (1)
Lúc này ta có: \(\left|x+\frac{1}{2}\right|.\left(2x-\frac{3}{4}\right)=2x-\frac{3}{4}\)
\(\Rightarrow\left|x+\frac{1}{2}\right|.\left(2x-\frac{3}{4}\right)-\left(2x-\frac{3}{4}\right)=0\)
\(\Rightarrow\left(2x-\frac{3}{4}\right).\left(\left|x+\frac{1}{2}\right|-1\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}2x-\frac{3}{4}=0\\\left|x+\frac{1}{2}\right|-1=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}2x=\frac{3}{4}\\\left|x+\frac{1}{2}\right|=1\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{3}{8}\\x+\frac{1}{2}=1\\x+\frac{1}{2}=-1\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{3}{8}\\x=\frac{1}{2}\\x=\frac{-3}{2}\end{array}\right.\)
Mà \(x\ge\frac{3}{8}\) do \(2x-\frac{3}{4}\ge0\)
Vậy \(x\in\left\{\frac{3}{8};\frac{1}{2}\right\}\)
KG
15 tháng 5 2016
(2x+3)(x-4)+(x-5)(x-2)=(3x-5)(x-4)
<=.2x2-8x+3x-12+x2 -2x-5x+10=3x2-12x-5x+20
<=>5x=22
<=>x=22/5
vay
5 tháng 9 2018
\(\left(2x+3\right)\left(x-4\right)-\left(x-5\right)\left(x-2\right)=\left(3x+5\right)\left(x-4\right)\)
=> \(\left(2x+3\right)\left(x-4\right)-\left(3x+5\right)\left(x-4\right)-\left(x-5\right)\left(x-2\right)=0\)
=> \(\left(x-4\right)\left(-x-2\right)-\left(x-5\right)\left(x-2\right)=0\)
=> \(-x^2-2x+4x+8-x^2+2x+5x-10=0\)
=> \(-2x^2+9x-2=0\)
=> \(-\left(x^2-2x+1\right)-\left(x^2-2x+1\right)+5x=0\)
=> \(-\left(x-1\right)^2-\left(x-1\right)^2+5x=0\)
=> \(-2\left(x-1\right)^2+5x=0\)
=> \(2\left(x-1\right)^2-5x=0\)
...................
để mik giải rồi ghi lại sau nha
12 tháng 11 2017
b) \(\left(2x+3\right)^6-\left(2x+3\right)^4=0\)
\(\left(2x+3\right)^4.\left[\left(2x+3\right)^2-1\right]=0\)
\(\left(2x+3\right)^4.\left(2x+3-1\right)\left(2x+3+1\right)=0\)
\(\left(2x+3\right)^4.\left(2x+2\right)\left(2x+4\right)=0\)
\(\Rightarrow\left(2x+3\right)^4=0\) hoac \(\orbr{\begin{cases}2x+2=0\\2x+4=0\end{cases}}\)
\(\Rightarrow2x+3=0\) hoac \(\orbr{\begin{cases}2x=-2\\2x=-4\end{cases}}\)
\(\Rightarrow x=\frac{-3}{2}\) hoac \(\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)
vay \(x=\frac{-3}{2}\) hoac \(\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)
23 tháng 6 2017
\(x\left(x^2-5^2\right)-\left(x^3+2^3\right)=3\)
\(x^3-25x-x^3-8=3\)
\(-25x-8=3\)
\(-25x=3+8=11\)
\(x=\frac{-11}{25}\)
8 tháng 10 2017
\(\left(x-1\right)^3+\left(x-3\right)^3=\left(2x-4\right)^3\)
\(\Leftrightarrow\left(x-1\right)^3+\left(x-3\right)^3-\left(2x-4\right)^3=0\)
\(\Leftrightarrow\left(x-1\right)^3+\left(x-3\right)^3+\left(4-2x\right)^3=0\)
Đặt \(\left(x-1\right)=a;\left(x-3\right)=b;\left(4-2x\right)=c\)ta có:
\(a^3+b^3+c^3=0\)
\(\Leftrightarrow\left(a+b+c\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)^3=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Thay lại, ta được:
\(\left(a+b+c\right)^3=\left(x-1+x-3+4-2x\right)^3=0\)
\(\Rightarrow3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(\Leftrightarrow\left(x-1+x-3\right)\left(x-3+4-2x\right)\left(4-2x+x-1\right)=0\)
\(\Leftrightarrow\left(2x-4\right)\left(1-x\right)\left(3-x\right)=0\)
\(\Leftrightarrow2x-4=0\Leftrightarrow2x=4\Rightarrow x=2\)
hoặc \(1-x=0\Leftrightarrow x=1\)
hay \(3-x=0\Leftrightarrow x=3\)
Vậy \(x\in\left\{1;2;3\right\}\). Xong :))