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- Toán lớp 8
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Bài 1
\(a,5x^2-10xy+5y^2\)
\(=5\cdot\left(x^2-2xy+y^2\right)\)
\(=5\cdot\left(x-y\right)^2\)
\(b,x^2-y^2+6y-9\)
\(=x^2-\left(y^2-6y+9\right)\)
\(=x^2-\left(y-3\right)^2\)
\(=\left(x-y+3\right)\cdot\left(x+y-3\right)\)
\(c,3x^4-75x^2y^2\)
\(=3x^2\cdot\left(x^2-25y^2\right)\)
\(=3x^2\cdot\left(x-5y\right)\cdot\left(x+5y\right)\)
\(d,x^4y+xy^4\)
\(=xy\left(x^3+y^3\right)\)
\(=xy\cdot\left(x+y\right)\cdot\left(x^2-xy+y^2\right)\)
\(x\left(x+2\right)\left(x+3\right)\left(x+5\right)+9\)
\(=\left(x^2+5x+6\right)\left(x^2+5x\right)+9\)
Đặt \(t=x^2+5x\)ta được;
\(t\left(t+6\right)+9=t^2+6t+9\)
\(=\left(t+3\right)^2=\left(x^2+5x+3\right)^2\)
b)\(x^2+2xy+y^2+2x+2y-15\)
\(=\left(x+y+1\right)^2-4^2\)
\(=\left(x+y+1+4\right)\left(x+y+1-4\right)\)
\(=\left(x+y-3\right)\left(x+y+5\right)\)
c)\(4x^4y^4+1=\left(2x^2y^2-2xy+1\right)\left(2x^2y^2+2xy+1\right)\)
\(x^3+4x^2+4x+3\)
\(=x^3+3x^2+x^2+3x+x+3\)
\(=x^2\left(x+3\right)+x\left(x+3\right)+\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2+x+1\right)\)
\(x^2-y^2+4y-4\)
\(=x^2-\left(y^2-4y+4\right)\)
\(=x^2-\left(y-2\right)^2\)
\(=\left(x-y+2\right)\left(x+y-2\right)\)
\(x^4+x^3y-xy^3-y^4\)
\(=x^3\left(x+y\right)-y^3\left(x+y\right)\)
\(=\left(x+y\right)\left(x^3-y^3\right)\)
\(=\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)\)
Chúc bạn học tốt.
Cách 1: \(x^2-2xy+y^2+4x-4y-5=\left(y^2-xy+y\right)+\left(-xy+x^2-x\right)+\left(-5y+5x-5\right)\)
\(=y\left(y-x+1\right)-x\left(y-x+1\right)-5\left(y-x+1\right)=\left(y-x+1\right)\left(y-x-5\right)\)
Cách 2: \(x^2-2xy+y^2+4x-4y-5=\left(x^2+y^2+2^2-2xy+4x-4y\right)-9\)
\(=\left(y-x-2\right)^2-3^2=\left(y-x-2-3\right)\left(y-x-2+3\right)=\left(y-x-5\right)\left(y-x+1\right)\)
1, \(=\left(2y\right)^2-\left(x^2-2x+1\right)=\left(2y\right)^2-\left(x-1\right)^2=\left(2y-x+1\right)\left(2y+x-1\right)\)
2, \(=2\left(x^2-y^2\right)+8\left(x+1\right)=2\left(x+1\right)\left(x-1\right)+8\left(x+1\right)=2\left(x+1\right)\left(x-1+4\right)=2\left(x+1\right)\left(x+3\right)\)
3, \(=\left(x^2+6x+9\right)-\left(2y\right)^2=\left(x+3\right)^2-\left(2y\right)^2=\left(x+3-2y\right)\left(x+3+2y\right)\)
4, \(=\left(x+y\right)^2-1=\left(x+y-1\right)\left(x+y+1\right)\)
\(4y^2-x^2+2x-1\)
\(=4y^2-\left(x^2-2x+1\right)\)
\(=\left(2y\right)^2-\left(x-1\right)^2\)
\(=\left(2y-x+1\right)\left(2y+x-1\right)\)
hk tốt
^^
\(x^2-4y^2+4y-1=x^2-\left(2y-1\right)^2=\left(x+2y-1\right)\left(x-2y+1\right)\)
\(x^4+3x^3-9x-9\)
\(=x^4-9+3x^3-9x\)
\(=\left(x^2-3\right)\left(x^2+3\right)+3x\left(x^2-3\right)\)
\(=\left(x^2-3\right)\left(x^2+3+3x\right)\)
\(a,x^2-y^2-4y-4\\ =x^2-\left(y^2+4y+4\right)\\ =x^2-\left(y+2\right)^2\\ =\left(x-y-2\right)\left(x+y+2\right)\\ b,x^2-y^2-6y-9\\ =x^2-\left(y^2+6y+9\right)\\ =x^2-\left(y+3\right)^2\\ =\left(x-y-3\right)\left(x+y+3\right)\\ c,4x^2-4y^2+12y-9\\ =\left(2x\right)^2-\left(2y-3\right)^2\\ =\left(2x-2y+3\right)\left(2x+2y-3\right)\)