tìm x biết
\(x\left(x-1\right)=4x-4\)
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1: Ta có: \(4x^2-36=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
2: Ta có: \(\left(x-1\right)^2+x\left(4-x\right)=11\)
\(\Leftrightarrow x^2-2x+1+4x-x^2=11\)
\(\Leftrightarrow2x=10\)
hay x=5
vì \(\left(4x^2-4x+1\right)^{2022}\ge0\left(\forall x\right)\),\(\left(y^2-\dfrac{4}{5}y+\dfrac{4}{25}\right)^{2022}\ge0\left(\forall y\right)\),\(\left|x+y+z\right|\ge0\)
mà \(\left(4x^2-4x+1\right)^{2022}+\left(y^2+\dfrac{4}{5}y+\dfrac{4}{25}\right)^{2022}+\left|x+y-z\right|=0\)
=>\(\left\{{}\begin{matrix}4x^2-4x+1=0\\y^2+\dfrac{4}{5}y+\dfrac{4}{25}=0\\x+y-z=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x-1=0\\y+\dfrac{2}{5}=0\\x+y-z=0\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-2}{5}\\\dfrac{1}{2}-\dfrac{2}{5}-z=0\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-2}{5}\\z=\dfrac{1}{10}\end{matrix}\right.\)
KL: vậy \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-2}{5}\\z=\dfrac{1}{10}\end{matrix}\right.\)
\(\frac{1}{\left(x-1\right)x}+\frac{1}{\left(x-2\right)\left(x-1\right)}+\frac{1}{\left(x-3\right)\left(x-2\right)}+\frac{1}{\left(x-4\right)\left(x-3\right)}=\frac{x}{x^2-4x}\)
\(\Leftrightarrow\)\(\frac{1}{x-1}-\frac{1}{x}+\frac{1}{x-2}-\frac{1}{x-1}+\frac{1}{x-3}-\frac{1}{x-2}+\frac{1}{x-4}-\frac{1}{x-3}=\frac{x}{x\left(x-4\right)}\)
\(\Leftrightarrow\)\(-\frac{1}{x}+\frac{1}{x-4}=\frac{1}{x-4}\)
\(\Leftrightarrow\)\(\frac{-\left(x-4\right)+x}{x\left(x-4\right)}=\frac{x}{x\left(x-4\right)}\)
\(\Leftrightarrow\)\(4-x+x=x\)
\(\Leftrightarrow x=4\)
lo nói mk làm cách lâu chứ m cx hỏi người khác!!!!!!!!!!!
Em đăng bài quả môn toán nhận hỗ trợ nhanh nhất nha
a)\(\left(x-1\right)^3+3\left(x+1\right)^2=\left(x^2-2x+4\right)\left(x+2\right)\)
\(\Leftrightarrow x^3-3x^2+3x-1+3\left(x^2+2x+1\right)=x^3+8\)
\(\Leftrightarrow-3x^2+3x+3x^2+6x+3=9\)
\(\Leftrightarrow9x=6\Leftrightarrow x=\frac{2}{3}\)
b) \(x^2-4=8\left(x-2\right)\)
\(\Leftrightarrow x^2-4=8x-16\)
\(\Leftrightarrow x^2-8x+12=0\)
\(\Leftrightarrow x^2-2x-6x+12=0\)
\(\Leftrightarrow x\left(x-2\right)-6\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=2\end{cases}}\)
c) \(x^2-4x+4=9\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)^2=9\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)^2-9\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-11\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-11=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=11\end{cases}}\)
d) \(4x^2-12x+9=\left(5-x\right)^2\)
\(\Leftrightarrow\left(2x-3\right)^2=\left(5-x\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=5-x\\2x-3=x-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{8}{3}\\x=-2\end{cases}}\)
Vì \(\left(4x-1\right)^2=\left(1-4x\right)^4.\)(*)
Đặt \(\left(4x-1\right)^2=t\) ( điều kiện \(t\ge0\)) \(\Leftrightarrow1-4x=-t^2\)
nên phương trình (*) \(\Leftrightarrow t=-t^2\)
\(\Leftrightarrow t^2+t=0\)
\(\Leftrightarrow t=0\) hoặc \(t=-1\)( loại do \(t\ge0\))
Ta có \(t=0\Leftrightarrow\left(4x-1\right)^2=0\Leftrightarrow4x-1=0\)
\(\Leftrightarrow x=\frac{1}{4}\)
Vậy phương trình có 1 nghiệm \(x=\frac{1}{4}.\)
dễ thấy |x+2/7| > 0;|x+4/7|>0;|x+3 1/7| >0
=>|x+2/7|+|x+4/7|+|x+3 1/7| > 0;mà VT=VP
nên 4x>0
ta có: \(\left|x+\frac{2}{7}\right|+\left|x+\frac{4}{7}\right|+\left|x+3\frac{1}{7}\right|=4x=>x+\frac{2}{7}+x+\frac{4}{7}+x+\frac{22}{7}=4x=>3x+4=4x=>x=4\)
vậy x=4
x(x - 1) = 4x - 4
<=> x2 - x = 4x - 4
<=> x2 - x - 4x + 4 = 0
<=> x2 - 5x + 4 = 0
<=> (x - 1)(x - 4) = 0
<=> x - 1 = 0 hoặc x - 4 = 0
<=> x = 0 + 1 hoặc x = 0 + 4
=> x = 1 hoặc x = 4
\(x\left(x-1\right)=4x-4\)\(\Leftrightarrow x\left(x-1\right)=4\left(x-1\right)\)
\(\Leftrightarrow x\left(x-1\right)-4\left(x-1\right)=0\)\(\Leftrightarrow\left(x-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=1\end{cases}}\)
Vậy \(x=1\)hoặc \(x=4\)