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\(a,5x^2y-10xy^2=5xy\left(x-2y\right)\\ b,x^2+2xy+y^2-5x-5y=\left(x+y\right)^2-5\left(x+y\right)=\left(x+y\right)\left(x+y-5\right)\\ c,x^2-6x+8=\left(x^2-2x\right)-\left(4x-8\right)=x\left(x-2\right)-4\left(x-2\right)=\left(x-2\right)\left(x-4\right)\\ d,5x^2-10xy+5y^2-20z^2=5\left(x^2-2xy+y^2-4z^2\right)=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]=5\left(x-y-2z\right)\left(x-y+2z\right)\)
b: \(x^2-2xy+y^2-z^2\)
\(=\left(x-y\right)^2-z^2\)
\(=\left(x-y-z\right)\left(x-y+z\right)\)
d: \(x^2+4x+3=\left(x+3\right)\left(x+1\right)\)
2, a^3-3ab^2 = 5
<=> (a^3-3ab^2)^2 = 25
<=> a^6-6a^4b^2+9a^2b^4 = 25
b^3-3a^2b=10
<=> (b^3-3a^2b)^2 = 100
<=> b^6-6a^2b^4+9a^4b^2 = 100
=> 100+25 = a^6-6a^4b^2+9a^2b^4+b^6+6a^2b^4+9a^4b^2
<=> 125 = a^6+3a^4b^2+3a^3b^4+b^6 = (a^2+b^2)^3
<=> a^2+b^2 = 5
Khi đó : S = 2016.(a^2+b^2) = 2016.5 = 10080
Tk mk nha
1) \(x^2+6xy+5y^2-5y-x=\left(x^2+xy-x\right)+\left(5xy+5y^2-5y\right)\)
\(=x\left(x+y-1\right)+5y\left(x+y-1\right)\)
\(=\left(x+5y\right)\left(x+y-1\right)\)
2) Ta có : \(a^3-3ab^2-5\Rightarrow\left(a^3-3ab^2\right)^2=25\Rightarrow a^6-6a^4b^2+9a^2b^4=25\)
và \(b^3-3a^2b=10\Rightarrow\left(b^3-3a^2b\right)^2=100\Rightarrow b^6-6b^4a^2+9a^4b^2=100\)
\(\Rightarrow\)\(125=a^6+b^6+3a^2b^4+3a^4b^2\)
Hay \(125=\left(a^2+b^2\right)^2\Rightarrow a^2+b^2=5\)
Nên \(S=2016\left(a^2+b^2\right)=2016.5=10080\)
1.\(=5\left(x^2-2xy+y^2-4z^2\right)=5\left[\left(x+y\right)^2-\left(2z\right)^2\right]=5\left(x+y-2z\right)\left(x+y+2z\right)\)
2. \(=\left(-5x^2+15x\right)+\left(x-3\right)=-5x\left(x-3\right)+\left(x-3\right)=\left(1-5x\right)\left(x-3\right)\)
3. \(=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\)
4.\(=3\left(x^2-2xy+y^2-4z^2\right)=3\left[\left(x-y\right)^2-\left(2z\right)^2\right]=3\left(x-y-2z\right)\left(x-y+2z\right)\)
5. \(=\left(x^2+x\right)+\left(3x+3\right)=x\left(x+1\right)+3\left(x+1\right)=\left(x+1\right)\left(x+3\right)\)
6. \(=\left(x^2-2x+1\right)\left(x^2+2x+1\right)=\left(x-1\right)^2\left(x+1\right)^2\)
7. \(=\left(x^2+x\right)-\left(5x+5\right)=x\left(x+1\right)-5\left(x+1\right)=\left(x-5\right)\left(x+1\right)\)
\(1,=5\left[\left(x-y\right)^2-4z^2\right]=5\left(x-y-2z\right)\left(x-y+2z\right)\\ 2,=-5x^2+15x+x-3=\left(x-3\right)\left(1-5x\right)\\ 3,=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\\ 4,=3\left[\left(x-y\right)^2-4z^2\right]=3\left(x-y-2z\right)\left(x-y+2z\right)\\ 5,=x^2+x+3x+3=\left(x+3\right)\left(x+1\right)\\ 6,=\left(x^2+2x+1\right)\left(x^2-2x+1\right)=\left(x-1\right)^2\left(x+1\right)^2\\ 7,=x^2+x-5x-5=\left(x+1\right)\left(x-5\right)\)
a) \(x^4+2x^3-4x-4=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)
\(=\left(x^2+x\right)^2-\left(x+2\right)^2=\left(x^2+x-x-2\right)\left(x^2+x+x+2\right)\)
\(=\left(x^2-2\right)\left(x^2+2x+2\right)\)
a) Ta có: \(x^4+2x^3-4x-4\)
\(=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)
\(=\left(x^2+x\right)^2-\left(x+2\right)^2\)
\(=\left(x^2+x-x-2\right)\left(x^2+x+x+2\right)\)
\(=\left(x^2-2\right)\cdot\left(x^2+2x+2\right)\)
1) \(x^2+6xy+5y^2-5y-x\)
\(=\left(x^2-xy+x\right)+\left(5xy+5y^2-5y\right)\)
\(=x\left(x+y-1\right)+5y\left(x+y-1\right)\)
\(\left(x+5y\right)\left(x+y-1\right)\)
2) Ta có : \(a^3-3ab^2=5\)
\(\Rightarrow\)\(\left(a^3-3ab^2\right)^2-100=25\Rightarrow a^6-6a^4b^2+9a^2b^4=25\)
Và \(b^3-3a^2b=10\)
\(\Rightarrow\)\(\left(b^3-3a^2b\right)^2=100\Rightarrow b^6-6b^4a^2-9a^4b^2=100\)
\(\Rightarrow\)\(125=a^6+b^6+3a^2b^4+3a^4b^2\)
Hoặc \(125=\left(a^2+b^2\right)^3\Rightarrow a^2+b^2=5\)
Do đó : \(S=2016\left(a^2+b^2\right)=2016.5=10080\)
(a-b)^2=(a-b)(a-b)=a^2-ab-ab+b^2=a^2-2ba+b^2
(a-b)(a+b)=a^2+ab-ab-b^2=a^2-b^2
(a+3)^3=(a+b)^2*(a+b)
=(a^2+2ab+b^2)(a+b)
=a^3+a^2b+2a^2b+2ab^2+b^2a+b^3
=a^3+3a^2b+3ab^2+b^3
Câu hỏi của Chien - Toán lớp 8 - Học toán với OnlineMath
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