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a) \(A=x^2-4y^2+x-2y\)
\(=\left(x-2y\right)\left(x+2y\right)+\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+2y+1\right)\)
Thay vào
b) tương tự
Tại x=1 ; y=2 thay vào BT ta có
A= \(1-4.2^2+1-2.2=\)-18
ý b) cũng thay v thoy
Ta có A=x(x+2)+y(y-2)-2xy+37
= x^2 +2x + y^2 - 2y - 2xy +37
=(x^2 +y^2 -2xy +1 +2x - 2y) +36
=(x -y +1)^2 +36
= (7+1)^2 +36 = 64 +36 =100
nhầm xíu nhá mk lm lại :
\(A=\frac{xz}{z\left(xy+x+1\right)}+\frac{xyz}{xz\left(yz+y+1\right)}+\frac{z}{xz+z+1}\)\(=\frac{xz}{xyz+xz+z}+\frac{1}{xyz^2+xyz+xz}+\frac{z}{xz+z+1}\)
\(=\frac{xz}{1+xz+z}+\frac{1}{z+1+xz}+\frac{z}{xz+z+1}\)
\(=\frac{xz+z+1}{xz+z+1}=1\)
\(A=\frac{x}{xy+x+1}+\frac{y}{yz+y+1}+\frac{z}{xz+z+1}=\frac{xz}{z\left(xy+x+1\right)}+\frac{xyz}{xz\left(yz+y+1\right)}+\frac{z}{xz+z+1}\)
\(=\frac{xy}{xyz+xz+z}+\frac{1}{xyz^2+xyz+xz}+\frac{z}{xz+z+1}=\frac{xy}{xz+z+1}+\frac{1}{xz+z+1}+\frac{z}{xz+z+1}\)
\(=\frac{xy+1+z}{xz+z+1}=1\)
vậy A=1
Ta có:\(10=2xyz\)
=> \(P=\frac{1}{2x+2xz+1}+\frac{2xy}{y+2xy+10}+\frac{10z}{10z+yz+10}\)
\(=\frac{1}{2x+2xz+1}+\frac{2xy}{y+2xy+2xyz}+\frac{2xyz^2}{2xyz^2+yz+2xyz}\)
\(=\frac{1}{2x+2xz+1}+\frac{2x}{1+2x+2xz}+\frac{2xz}{2xz+1+2x}\)
\(=1\)
Vậy P=1
\(a.\)
\(x\left(x+z\right)+y\left(y-z\right)-2xy+37\)
\(=x^2+xz+y^2-yz-2xy+37\)
\(=\left(x^2-2xy+y^2\right)+z\left(x-y\right)+37\)
\(=\left(x-y\right)^2+z\left(x-y\right)+37\)
\(=7^2+x.7^2+37\)
\(=86+49x\)
\(b.\)
\(x^2+4y^2-2x+10+4xy-4y\)
\(=\left(x^2+4xy+4y^2\right)-2\left(x+2y\right)+10\)
\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
\(=5^2-2.5+10\)
\(=25\)