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\(\left|x+y\right|\text{nhỏ nhất }\Rightarrow x+y=0\Rightarrow x=-y\)
thay xy=1 và x+y=0, ta có:
\(M=2x^2+2\left(-x^2\right)+3.1-\left(x+y\right)-3=4x^2=\left(2x\right)^2\)
\(H=2x^2+9y^2-6xy-6y-12y+2004\)
\(\Rightarrow2H=4x^2+18y^2-12xy-12x-24y+4008\)
\(=\left(4x^2-12xy+9y^2\right)+9y^2-12x-24y+4008\)
\(=\left(2x-3y\right)^2-6\left(2x-3y\right)+9+9y^2-42y+49+3950\)
\(=\left(2x-3y-3\right)^2+\left(3y-7\right)^2+3950\ge3950\)
\(\Rightarrow2H\ge3950\)
\(\Rightarrow H\ge1975\)
Dấu "=" tại \(\hept{\begin{cases}x=5\\y=\frac{7}{3}\end{cases}}\)
\(J=x^2+xy+y^2-3x-3y+1999\)
\(=\left(x^2+xy+\frac{y^2}{4}\right)+\frac{3y^2}{4}-3x-3y+1999\)
\(=\left(x+\frac{y}{2}\right)^2-3\left(x+\frac{y}{2}\right)+\frac{9}{4}+3\left(\frac{y^2}{4}-\frac{y}{2}+\frac{1}{4}\right)+1996\)
\(=\left(x+\frac{y}{2}-\frac{3}{2}\right)^2+3\left(\frac{y}{2}-\frac{1}{2}\right)^2+1996\ge1996\)
Dấu "=" tại \(\hept{\begin{cases}x=1\\y=1\end{cases}}\)
\(M=x^2\left(x+y-2\right)-y\left(x+y-2\right)+y+x-2+1\)
\(=1\)
\(N=x^2\left(x-2\right)-xy^2+2xy+2\left(x+y-2\right)+2\)
Ta có : \(x+y-2=0\Rightarrow x+2=-y\)
\(\Rightarrow N=-x^2y-xy^2+2xy+2\)
\(N=-xy\left(x+y-2\right)+2=2\)
\(P=x^3\left(x+y-2\right)+x^2y\left(x+y-2\right)-x\left(x+y-2\right)+3=3\)
\(E=\left(x^3+3xy^2+3x^2y+y^3\right)+3\left(x+y\right)-3\left(x^2+2xy+y^2\right)+2016\)
\(=\left(x+y\right)^3+3\left(x+y\right)-3\left(x+y\right)^2+2016\)
\(=21^3+3.21-3.21^2+2016\)
\(=\left(21-1\right)^3+2017=8000+2017=10017\)
Mình không viết lại đề nha ~
\(E=\left(x^3+3xy^2+3x^2y+y^3\right)+\left(3y+3x\right)+\left(3x^2+6xy+3y^2\right)+2016\)
\(E=\left(x+y\right)^3+3\left(x+y\right)+3\left(x+y\right)^2+2016\)
\(E=\left(x+y\right)[\left(x+y\right)^2+3+\left(x+y\right)]+2016\)
\(E=21\left(21^2+3+21\right)+2016\)
\(E=21.465+2016\)
\(E=9765+2016=11781\)
P/s: Ko chắc lắm.
\(A=x^3+y^3+6xy-3x-3y+1\)
\(A=\left(x+y\right)\left(x^2-xy+y^2\right)-3\left(x+y\right)+6xy+1\)
\(A=\left(x+y\right)\left(x^2+2xy+y^2-2xy-xy\right)-3\left(x+y\right)+6xy+1\)
\(A=\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]-3\left(x+y\right)+6xy+1\)
\(A=\left(x+y\right)\left[\left(x+y\right)^2-3xy-3\right]+6xy+1\)
Thay x+y=2 vào biểu thức, ta có:
\(A=2\left(2^2-3xy-3\right)+6xy+1\)
\(A=2\left(1-3xy\right)+6xy+1\)
\(A=2-6xy+6xy+1\)
\(A=3\)
\(B=x^2-y^2+4y+1\)
\(B=\left(x-y\right)\left(x+y\right)+4y+1\)
\(B=2\left(x-y\right)+4y+1\)
\(B=2x-2y+4y+1\)
\(B=2x+2y+1\)
\(B=2\left(x+y\right)+1=2.2+1=5\)
Ta có: \(\left(x+y\right)^2\ge4xy=4\)
Mà (x+y)2 nhỏ nhất
\(\Rightarrow\left(x+y\right)^2=4\)
\(\Rightarrow\orbr{\begin{cases}x+y=2\\x+y=-2\end{cases}}\)
Lại có: \(M=3x^2-2x+3y^2-2y+6xy+1\)
\(=3\left(x^2+2xy+y^2\right)-2\left(x+y\right)+1\)
\(=3\left(x+y\right)^2-2\left(x+y\right)+1\)
Thay vào mà tính