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1)
a) \(2x-6=0\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\)
b) \(x\times\left(x+2\right)-3\times\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\times\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
c) \(\frac{x-6}{x+1}=\frac{x^2}{x-1}\)
nhân chéo lên, ngại chết đc
Bài giải
\(\frac{1}{2}\left(x+1\right)+\frac{1}{4}\left(x+3\right)=3\cdot\frac{1}{3}\cdot\left(x+20\right)\)
\(\frac{1}{2}\left[\left(x+1\right)+\frac{1}{2}\left(x+3\right)\right]=x+20\)
\(\frac{1}{2}\left[x+1+\frac{1}{2}x+\frac{3}{2}\right]=x+20\)
\(\frac{1}{2}\left[x\left(1+\frac{1}{2}\right)+1+\frac{3}{2}\right]=x+20\)
\(\frac{1}{2}\left[\frac{3}{2}x+\frac{5}{2}\right]=x+20\)
\(\frac{3}{4}x+\frac{5}{4}=x+20\)
\(\frac{3}{4}x-x=20-\frac{5}{4}\)
\(\frac{-1}{4}x=\frac{75}{4}\)
\(x=\frac{75}{4}\text{ : }\frac{-1}{4}\)
\(x=-75\)
\(\frac{1}{2}\left(x+1\right)+\frac{1}{4}\left(x+3\right)=3\cdot\frac{1}{3}\cdot\left(x+20\right)\)
\(\frac{1}{2}\left[\left(x+1\right)+\frac{1}{2}\left(x+3\right)\right]=x+20\)
\(\frac{1}{2}\left[x+1+\frac{1}{2}x+\frac{3}{2}\right]=x+20\)
\(\frac{1}{2}\left[x\left(1+\frac{1}{2}\right)+1+\frac{3}{2}\right]=x+20\)
\(\frac{1}{2}\left[\frac{3}{2}x+\frac{5}{2}\right]=x+20\)
\(\frac{3}{4}x+\frac{5}{4}=x+20\)
\(\frac{3}{4}x-x=20-\frac{5}{4}\)
\(\frac{-1}{4}x=\frac{75}{4}\)
\(x=\frac{75}{4}\text{ : }\frac{-1}{4}\)
\(x=-75\)
\(a,\Leftrightarrow9x^2=-36\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow3\left(x+4\right)-x\left(x+4\right)=0\\ \Leftrightarrow\left(3-x\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x^2-x-2x^2+3x+2=0\\ \Leftrightarrow2x=-2\Leftrightarrow x=-1\\ d,\Leftrightarrow\left(2x-3-2x\right)\left(2x-3+2x\right)=0\\ \Leftrightarrow-3\left(4x-3\right)=0\\ \Leftrightarrow x=\dfrac{3}{4}\\ e,\Leftrightarrow\dfrac{1}{3}x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ f,\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Ta có \(x^2+y^2+z^2\ge xy+yz+zx\)
Đẳng thức xảy ra khi x = y = z
Bạn áp dụng vào nhé.
Ngọc cứ làm tắt thì vài người hiểu chứ vài bạn không biết đâu :)
Ta có :
\(x^2+y^2+z^2=xy+xz+yz\)
\(\Rightarrow x^2+y^2+z^2-xy-xz-yz=0\)
\(\Rightarrow2\left(x^2+y^2+z^2-xy-xz-yz\right)=0\)
\(\Rightarrow x^2+y^2-2xy+y^2+z^2-2yz+x^2+z^2-2xz=0\)
\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2=0\)
Mà \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(x-z\right)^2\ge0\\\left(y-z\right)^2\ge0\end{cases}}\)
\(\Rightarrow x-y=x-z=y-z=0\)
\(\Rightarrow x=y=z\)
\(\Rightarrow x^{2016}=y^{2016}=z^{2016}\)
Mà \(x^{2016}+y^{2016}+z^{2016}=3^{2016}\)
\(\Rightarrow x^{2016}=y^{2016}=z^{2016}=\frac{3^{2016}}{3}=3^{2015}\)
\(\Rightarrow x=y=z=\sqrt[2016]{3^{2015}}=\sqrt[2016]{\frac{3^{2016}}{3}}=\frac{3}{\sqrt[2016]{3}}\)
X/2=y/6
X=y/6 . 2 = y/3=1/3 y
Thay vào ta có
1/3y+ y=6
4/3y=6
Y=18/4
X=18/4 . 1/3=18/12=3/2
Đến đây bạn tự tính x-y nha
<=>\(\frac{24-18-X}{6}\)=\(\frac{3X}{6}\)
<=>\(6-X=3X\)
<=>\(-4X=-6\)
<=>\(X=\frac{3}{2}\)