A

\(3\times m+5\times n+2\times m+n+m\)

\(=3\times m+2\times m+m+5\times n+n\)

\(=m\times\left(3+2+1\right)+n\times\left(5+1\right)\)

\(=6\times m+6\times n\)

\(=6\times\left(m+n\right)\)

\(=6\times25\)

\(=150\)

\(2C_4H_{10}+13O_2\overset{t^o}{--->}8CO_2+10H_2O\)

a. \(S_{ABCD}=\dfrac{1}{2}.6.\left(12+12+9\right)=99\left(cm^2\right)\)

\(2x^3-12x^2+18x=2x\left(x^2-6x+9\right)=2x\left(x-3\right)^2\)

VÍ DỤ MẪU:

Ví dụ 1:

\(M=\dfrac{\left(2x^2-2\right)^2}{\left(x^2-2x+1\right)\left(x^2+2x+1\right)}\left(x\ne\pm1\right)\)

\(=\dfrac{\left(2x^2-2\right)^2}{\left(x-1\right)^2.\left(x+1\right)^2}=\dfrac{4\left(x^2-1\right)^2}{\left(x^2-1\right)^2}=4\)

Vậy .......

ví dụ 2:

\(N=\dfrac{\left(y+1\right)\left(y^2+5y+6\right)}{\left(y+3\right)\left(y^2+3y+2\right)}\left(y\ne-1;y\ne-2;y\ne-3\right)\)

\(=\dfrac{y^3+5y^2+6y+y^2+5y+6}{y^3+3y^2+2y+3y^2+9y+6}=\dfrac{y^3+6y^2+11y+6}{y^3+6y^2+11y+6}=1\)

Vậy ............

Ví dụ 3:

\(P=\dfrac{xy+2x+2y+4}{x+2}\left(x\ne-2\right)\)

\(=\dfrac{x\left(y+2\right)+2\left(y+2\right)}{x+2}=\dfrac{\left(x+2\right)\left(y+2\right)}{x+2}=y+2\)

Vậy ................

BÀI TẬP TỰ LUYỆN:

\(\dfrac{x^3+27y^3}{x^2-3xy+9y^2}=3.\dfrac{x^3+y^3}{x^2-xy+y^2}\)

\(\Leftrightarrow\dfrac{\left(x+3y\right)\left(x^2-3xy+9y^2\right)}{x^2-3xy+9y^2}=3.\dfrac{\left(x+y\right)\left(x^2-xy+y^2\right)}{x^2-xy+y^2}\)

\(\Leftrightarrow x+3y=3\left(x+y\right)\)

\(\Leftrightarrow x+3y=3x+3y\)

\(\Leftrightarrow2x=0\)

\(\Leftrightarrow x=0\)

Chọn A. x = 0

\(B=-y^2+4y\)

\(=4+4y-y^2-4\)

\(=-\left(2-y\right)^2-4\le-4\)

Dấu ''='' xảy ra khi \(2-y=0\Leftrightarrow y=2\)

Vậy B đạt GTLN là -4 khi y = 2

A