a: \(=\dfrac{2x+x-2-x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{2}{x+2}\)

b: x^2-x-6=0

=>(x-3)(x+2)=0

=>x=3(nhận) hoặc x=-2(loại)

Khi x=3 thì \(E=\dfrac{2}{3+2}=\dfrac{2}{5}\)

c: Để E nguyên thì \(x+2\in\left\{1;-1;2;-2\right\}\)

=>\(x\in\left\{-1;-3;0;-4\right\}\)

\(A=\left(2x-1\right)^2+9\ge9\\ A_{min}=9\Leftrightarrow x=\dfrac{1}{2}\\ B=2\left(x^2-2\cdot\dfrac{3}{4}x+\dfrac{9}{16}\right)+\dfrac{1}{8}=2\left(x-\dfrac{3}{4}\right)^2+\dfrac{1}{8}\ge\dfrac{1}{8}\\ B_{min}=\dfrac{1}{8}\Leftrightarrow x=\dfrac{3}{4}\\ C=\left(4x^2+4xy+y^2\right)+2\left(2x+y\right)+1+\left(y^2+4y+4\right)-4\\ C=\left[\left(2x+y\right)^2+2\left(2x+y\right)+1\right]+\left(y+2\right)^2-4\\ C=\left(2x+y+1\right)^2+\left(y+2\right)^2-4\ge-4\\ C_{min}=-4\Leftrightarrow\left\{{}\begin{matrix}2x=-1-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-2\end{matrix}\right.\)

\(D=\left(3x-1-2x\right)^2=\left(x-1\right)^2\ge0\\ D_{min}=0\Leftrightarrow x=1\\ G=\left(9x^2+6xy+y^2\right)+\left(y^2+4y+4\right)+1\\ G=\left(3x+y\right)^2+\left(y+2\right)^2+1\ge1\\ G_{min}=1\Leftrightarrow\left\{{}\begin{matrix}3x=-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-2\end{matrix}\right.\)

\(H=\left(x^2-2xy+y^2\right)+\left(x^2+2x+1\right)+\left(2y^2+4y+2\right)+2\\ H=\left(x-y\right)^2+\left(x+1\right)^2+2\left(y+1\right)^2+2\ge2\\ H_{min}=2\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=-1\\y=-1\end{matrix}\right.\Leftrightarrow x=y=-1\)

Ta luôn có \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz\ge0\\ \Leftrightarrow x^2+y^2+z^2\ge xy+yz+xz\\ \Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz\ge3xy+3yz+3xz\\ \Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\\ \Leftrightarrow\dfrac{3^2}{3}\ge xy+yz+xz\\ \Leftrightarrow K\le3\\ K_{max}=3\Leftrightarrow x=y=z=1\)

a) \(=\left(x-y\right)\left(3+5x\right)\)

b) \(=\left(x^2-6x+9\right)-y^2=\left(x-3\right)^2-y^2=\left(x-y-3\right)\left(x+y-3\right)\)

c) \(=5\left(x^2-xy-2x+2y\right)=5\left[x\left(x-y\right)-2\left(x-y\right)\right]=5\left(x-y\right)\left(x-2\right)\)

Trong SGK có chỉ đó b

Ko hiểu thì kb vs mik

mik chỉ thêm cho

Chỉ mình với nha

k: \(=\left(2-3x\right)\left(4+6x+9x^2\right)\)

i: \(=3\left(x^2-2xy+y^2\right)=3\left(x-y\right)^2\)

\(g,27+27x+9x^2+x^3=\left(3+x\right)^3\\ i,2x^2+2y^2-x^2z+z-y^2z-2=\left(2x^2-x^2z\right)+\left(2y^2-y^2z\right)-\left(2-z\right)=x^2\left(2-z\right)+y^2\left(2-z\right)-\left(2-z\right)=\left(x^2+y^2-1\right)\left(2-z\right)\)

\(k,8-27x^2=2^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)

\(l,3x^2-6xy+3y^2=3\left(x^2-2xy+y^2\right)=3\left(x-y\right)^2\)

Câu 1 : \(\frac{x}{2}=\frac{2y}{5}=\frac{4z}{7}\)\(\Rightarrow\)\(\frac{1}{4}.\frac{x}{2}=\frac{1}{4}.\frac{2y}{5}=\frac{1}{4}.\frac{4z}{7}\)\(\Leftrightarrow\)\(\frac{x}{8}=\frac{y}{10}=\frac{z}{7}\)                                                             \(\Rightarrow\)\(\frac{3x}{24}=\frac{5y}{50}=\frac{7z}{49}=\frac{3x+5y+7z}{24+50+49}=\frac{123}{123}=1\)

\(\frac{3x}{24}=1\Rightarrow3x=24\Rightarrow x=8\)

\(\frac{5y}{50}=1\Rightarrow5y=50\Rightarrow y=10\)

\(\frac{7z}{49}=1\Rightarrow7z=49\Rightarrow z=7\)

Vậy x,y,z lần lượt là 8,10,7

Áp dụng tính chất dãy tỉ số bằng nhau

\(\frac{x}{5}=\frac{y}{7}=\frac{z}{9}=\frac{x-y+z}{5-7+9}=\frac{315}{7}=45\)

  suy ra:   x/5 = 45   =>  x  =  225

               y/7 = 45  =>  y  =  315

               z/9 = 45  =>  z  =  405

được không hãy giúp mình đi nha hứa sẽ cho like