\(\Rightarrow\left[{}\begin{matrix}-2x+12=10\\2x-12=10\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}-2x=-2\\2x=22\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=11\end{matrix}\right.\)

\(a,=5\left(x^2+2xy+y^2\right)-10y^2+5=5\left(x+y\right)^2-10y^2+5\\ =5\left(1+2\right)^2-10\cdot4+5=45-40+5=10\\ b,=7\left(x-y\right)-\left(x-y\right)^2=\left(x-y\right)\left(7-x+y\right)\\ =\left(2-2\right)\left(7-2+2\right)=0\)

\(a,f\left(-1\right)=3-\left(-1\right)=4\\ f\left(2\right)=3-2=1\\ b,y=5\Rightarrow3-x=5\Rightarrow x=-2\\ y=2\Rightarrow3-x=2\Rightarrow x=1\\ y=-1\Rightarrow3-x=-1\Rightarrow x=4\)

\(a,=\left(x+3\right)^2\\ b_1=\left(x-5\right)^2\\ b_2,=x^2-9\\ d,=\left(x+3\right)^3\\ e,=\left(x-4\right)^3\\ f,=x^3+64\\ g,=x^3-8\\ h,=\left(3x^2-\sqrt{7}\right)\left(3x^2+\sqrt{7}\right)\)

\(A=\dfrac{z-x+x-y+y-z}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=\dfrac{0}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=0\)

Tỉ số % cr và cd là \(\dfrac{60}{60\times\dfrac{5}{3}}\times100\%=60\%\)

Đàn gà có \(28:25\%=112\left(con\right)\)

Gọi số hs lớp 6C là \(a(a\in \mathbb{N^*})\)

Ta có \(2=2;3=3;4=2^2;8=2^3\)

\(\Rightarrow BCNN\left(2,3,4,8\right)=2^3\cdot3=24\\ \Rightarrow x\in BC\left(2,3,4,8\right)=B\left(24\right)=\left\{0;24;48;72;96;...\right\}\)

Mà \(38< x< 60\Rightarrow x=48\)

Vậy có 48 hs

\(PT\Leftrightarrow\left(x-1\right)\sqrt{x^2+4}-\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(\sqrt{x^2+4}-x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\\sqrt{x^2+4}=x+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x^2+4=x^2+2x+1\left(x\ge-1\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{2}\left(tm\right)\end{matrix}\right.\)

\(\dfrac{x^2+2}{x+2}=\dfrac{x^2+2x-2x-4+6}{x+2}=\dfrac{x\left(x+2\right)-2\left(x+2\right)+6}{x+2}\\ =x-2+\dfrac{6}{x+2}\in Z\\ \Leftrightarrow x+2\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\\ \Leftrightarrow x\in\left\{-8;-5;-4;-3;-1;0;1;4\right\}\left(tm\right)\)