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a)\(\left(xy+1\right)^2-\left(x+y\right)^2\)
\(=\left[\left(xy+1\right)-\left(x+y\right)\right]\left(xy+1+x+y\right)\)
\(=\left(xy+1-x-y\right)\left(xy+1+x+y\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(y-1\right)\left(y+1\right)\)
b)\(4a^2b^2-\left(a^2+b^2-c^2\right)^2\)
\(=\left(2ab\right)^2-\left(a^2+b^2-c^2\right)^2\)
\(=\left(2ab-a^2-b^2+c^2\right)\left(2ab+a^2+b^2-c^2\right)\)
\(=\left(a-b+c\right)\left(-a+b+c\right)\left(a+b-c\right)\left(a+b+c\right)\)
c)\(\left(a+b+c\right)^2+\left(a+b-c\right)^2\)
\(=a^2+b^2+c^2+2ab+2bc+2ca+a^2+b^2+c^2+2ab-2bc-2ac\)
\(=2a^2+2b^2+2c^2+4ab\)
\(=2\left(a^2+b^2+c^2+2ab\right)\)
d)\(x^3-7x-6\)
\(=x^3+3x^2+2x-3x^2-9x-6\)
\(=x\left(x^2+3x+2\right)-3\left(x^2+3x+2\right)\)
\(=\left(x^2+3x+2\right)\left(x-3\right)\)
\(=\left(x^2+x+2x+2\right)\left(x-3\right)\)
\(=\left[x\left(x+1\right)+2\left(x+1\right)\right]\left(x-3\right)\)
\(=\left(x+1\right)\left(x+2\right)\left(x-3\right)\)
1/ Ta có : \(P\left(x\right)=-x^2+13x+2012=-\left(x-\frac{13}{2}\right)^2+\frac{8217}{4}\le\frac{8217}{4}\)
Dấu "=" xảy ra khi x = 13/2
Vậy Max P(x) = 8217/4 tại x = 13/2
2/ Ta có : \(x^3+3xy+y^3=x^3+3xy.1+y^3=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1\)
3/ \(a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=0\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\)
\(\Leftrightarrow ab+bc+ac=-\frac{1}{2}\) \(\Leftrightarrow\left(ab+bc+ac\right)^2=\frac{1}{4}\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=\frac{1}{4}\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=\frac{1}{4}\)(vì a+b+c=0)
Ta có : \(a^2+b^2+c^2=1\Leftrightarrow\left(a^2+b^2+c^2\right)^2=1\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=1\)
\(\Leftrightarrow a^4+b^4+c^4=1-2\left(a^2b^2+b^2c^2+c^2a^2\right)=1-\frac{2.1}{4}=\frac{1}{2}\)
bài 1:
a)\(A=x^3+y^3+xy=1^3+\left(-1\right)^3+1.\left(-1\right)=1-1-1=-1\)
b)\(B=\sqrt{x^2+y^2}=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=\left|10\right|=10\)
c)\(C=10x+10y+15=10\left(x+y\right)+15=10.1+15=25\)
d)\(D=x^2y+y^2x+5=xy\left(x+y\right)+5=xy.0+5=5\)
e)\(E=4x+7x^2y^2+3y^4+5y^2=?????\)
Bài 2:
bạn chỉ cần tìm nhân tử chung r gộp lại dưới dạng tích
VD: 10x+5xy=5x(2+y)
a)\(x+y=a\Rightarrow\left(x+y\right)^2=a^2\)
\(\Rightarrow x^2+2xy+y^2=a^2\Rightarrow x^2+y^2=a^2-2xy\Rightarrow x^2+y^2=a^2-2b\)
Bài 1:
a: \(A=x^2-2xy+y^2+x^2+2xy+y^2-2x^2-2y^2=0\)
b: \(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y+3xy^2-y^3\)
\(=6x^2y\)
c: \(=\left(a+b-c-a+c\right)\left(a+b-c+a-c\right)-2ab+2bc\)
\(=b\left(2a-2c\right)-2ab+2bc\)
=0
a: \(=x^2-4x+4+y^2+2y+1\)
\(=\left(x-2\right)^2+\left(y+1\right)^2\)
b: \(=x^2+10x+25+x^2-2xy+y^2\)
\(=\left(x+5\right)^2+\left(x-y\right)^2\)
c: \(=a^2+2ab+b^2+4b^2+4b+1\)
\(=\left(a+b\right)^2+\left(2b+1\right)^2\)
d: \(=2\left(x^2+b^2\right)\)