tính : B= 4x (2x+y)+ 2y(2x+y-y.y+2x)
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Ta có:
\(B=4x\left(2x+y\right)+2y\left(2x+y\right)-y\left(y+2x\right)\)
\(\Leftrightarrow B=\left(4x+2y-y\right)\left(2x+y\right)=\left(4x+y\right)\left(2x+y\right)=\left(4.\dfrac{1}{2}+\dfrac{-3}{5}\right)\left(2.\dfrac{1}{2}+\dfrac{-3}{5}\right)=\dfrac{14}{25}\)
\(log_{\sqrt{3}}\left(2x+y\right)-log_{\sqrt{3}}\left(4x^2+y^2+2xy+2\right)=\left(4x^2+y^2+2xy+2\right)-3\left(2x+y\right)-2\)
\(\Leftrightarrow log_{\sqrt{3}}\left(2x+y\right)+2+3\left(2x+y\right)=log_{\sqrt{3}}\left(4x^2+y^2+2xy+2\right)+\left(4x^2+y^2+2xy+2\right)\)
\(\Leftrightarrow log_{\sqrt{3}}\left(6x+3y\right)+\left(6x+3y\right)=log_{\sqrt{3}}\left(4x^2+y^2+2xy+2\right)+\left(4x^2+y^2+2xy+2\right)\)
Xét hàm \(f\left(t\right)=log_{\sqrt{3}}t+t\) với \(t>0\)
\(f'\left(t\right)=\dfrac{1}{t.ln\sqrt{3}}+1>0\Rightarrow f\left(t\right)\) đồng biến
\(\Rightarrow6x+3y=4x^2+y^2+2xy+2\)
\(\Leftrightarrow4x+y=\left(x+y-1\right)^2+1+3\left(x^2+1\right)-3\ge2\left(x+y-1\right)+6x-3\)
\(\Leftrightarrow4x+y\ge2\left(4x+y\right)-5\)
\(\Leftrightarrow4x+y\le5\)
\(\Rightarrow P=\dfrac{2x+y+6+\left(4x+y-5\right)}{2x+y+6}=1+\dfrac{4x+y-5}{2x+y+6}\le1\)
\(P_{max}=1\) khi \(x=y=1\)
`a)A=x(x+y)-x(y-x)`
`=x^2+xy-xy+x^2`
`=2x^2`
Thay `x=-3`
`=>A=2.9=18`
`b)B=4x(2x+y)+2y(2x+y)-y(y+2x)`
`=8x^2+4xy+4xy+2y^2-y^2-2xy`
`=8x^2+y^2+6xy`
Thay `x=1/2,y=-3/4`
`=>B=8*1/4+9/16-9/4`
`=2+9/16-9/4`
`=9/16-1/4=5/16`
\(A=16x^2-y^2-16x^2+8x=8x-y^2\\ A=8\cdot3-\left(-1\right)^2=24-1=23\\ B=64x^3-80x-64x^3-1=-80x-1\\ B=-80\cdot\dfrac{1}{5}-1=-16-1=-17\)
\(B=4x\left(2x+y\right)+2y\left(2x+y-y^2+2x\right)\)
\(=4x\left(2x+y\right)+2y\left(4x-y^2+y\right)=8x^2+4xy+8xy-2y^3+2y^2\)