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bằng phương pháp nào zậy bn????
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a) \(2x\left(x-3\right)^2+5x\left(3-x\right)\)
\(=2x\left(x-3\right)^2-5x\left(x-3\right)\)
\(=\left(x-3\right)\left[2x\left(x-3\right)-5x\right]\)
\(=\left(x-3\right)\left(2x^2-6x-5x\right)\)
\(=\left(x-3\right)\left(2x^2-11x\right)\)
\(=x\left(x-3\right)\left(2x-11\right)\)
b) \(\left(x+3\right)^2-4\left(y^2-2y+1\right)\)
\(=\left(x+3\right)^2-2^2\left(y-1\right)^2\)
\(=\left(x+3\right)^2-\left[2\left(y-1\right)\right]^2\)
\(=\left[\left(x+3\right)-2\left(y-1\right)\right]\left[\left(x+3\right)+2\left(y-1\right)\right]\)
\(=\left(x+3-2y+2\right)\left(x+3+2y-2\right)\)
\(=\left(x-2y+5\right)\left(x+2y+1\right)\)
a) \(2x.\left(x-3\right)^2+5x.\left(-x+3\right)=2x.\left(x-3\right)^2-5x.\left(x-3\right)\)
\(=\left(x-3\right).\left(2x^2-11x\right)=\left(x-3\right).x.\left(2x-11\right)\)
b) \(\left(x+3\right)^2-4.\left(y^2-2y+1\right)=\left(x+3\right)^2-2^2.\left(y-1\right)^2\)
\(=\left(x+3\right)^2-\left[2.\left(y-1\right)\right]^2=\left(x-2y+1\right).\left(x+2y+5\right)\)
a)\(3x^2-8x+4\)
\(=3x^2-2x-6x+4\)
\(=x\left(3x-2\right)-2\left(3x-2\right)\)
\(=\left(x-2\right)\left(3x-2\right)\)
b)\(4x^4+81\)
\(=4x^4+36x^2+81-36x^2\)
\(=\left(2x^2+9\right)^2-36x^2\)
\(=\left(2x^2-6x+9\right)\left(2x^2+6x+9\right)\)
c)\(x^8+98x^4+1\)
\(=\left(x^8+2x^4+1\right)+96x^4\)
\(=\left(x^4+1\right)^2+16x^2\left(x^4+1\right)+64x^4-16x^2\left(x^4+1\right)+32x^4\)
\(=\left(x^4+8x^2+1\right)^2-16x^2\left(x^4-2x^2+1\right)\)
\(=\left(x^4+8x^2+1\right)^2-16x^2\left(x^4-2x^2+1\right)\)
\(=\left(x^4+8x^2+1\right)^2-\left(4x^3-4x\right)^2\)
\(=\left(x^4+4x^3+8x^2-4x+1\right)\left(x^4-4x^3+8x^2+4x+1\right)\)
d)\(x^4+6x^3+7x^2-6x+1\)
\(=x^4+3x^3-x^2+3x^3+9x^2-3x-x^2-3x+1\)
\(=x^2\left(x^2+3x-1\right)+3x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)\)
\(=\left(x^2+3x-1\right)\left(x^2+3x-1\right)\)\(=\left(x^2+3x-1\right)^2\)
a) \(4x^3y-12x^2y^3-8x^4y^3\)
\(=4x^2y\left(x-3y^2-2x^2y^2\right)\)
b) \(2x^2+4x+2-2y^2\)
\(=2\left(x^2+2x+1-y^2\right)\)
\(=2\left[\left(x+1\right)^2-y^2\right]\)
\(=2\left(x-y+1\right)\left(x+y+1\right)\)
c) \(x^3-2x^2+x-xy^2\)
\(=x\left(x^2-2x+1-y^2\right)\)
\(=x\left[\left(x-1\right)^2-y^2\right]\)
\(=x\left(x-y-1\right)\left(x+y-1\right)\)
d) \(x\left(x-2y\right)+3\left(2y-x\right)\)
\(=x\left(x-2y\right)-3\left(x-2y\right)\)
\(=\left(x-3\right)\left(x-2y\right)\)
e) \(x^2+4\)
\(=\left(x^4+4x^2+4\right)-4x^2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
f) \(5x^2-7x-6\)
\(=\left(5x^2-10x\right)+\left(3x-6\right)\)
\(=5x\left(x-2\right)+3\left(x-2\right)\)
\(=\left(5x+3\right)\left(x-2\right)\)
Bài 1:tìm x ,biết:
a) (2x - 1)(3x + 2) - 6x(x + 1) = 0
\(\Leftrightarrow6x^2+x-2-6x^2-6x=0\)
\(\Leftrightarrow-5x=2\)
\(\Leftrightarrow x=\frac{-2}{5}\)
b) \(\left(4x-1\right)^2-\left(2x+1\right)\left(8x-3\right)=0\)
\(\Leftrightarrow16x^2-8x+1-16x^2-2x+3=0\)
\(\Leftrightarrow-10x=-4\)
\(\Leftrightarrow x=\frac{2}{5}\)
c) \(4x^2-1=2\left(2x+1\right)\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-1\right)-2\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{3}{2}\end{cases}}\)
2a) \(4x^2-9y^2-6y-1=4x^2-\left(3y+1\right)^2\)
\(=\left(2x-3y-1\right)\left(2x+3y+1\right)\)
b) \(4x^2-1-2x\left(2x-1\right)=\left(2x-1\right)\left(2x+1\right)-2x\left(2x-1\right)\)
\(=1.\left(2x-1\right)\)
c) \(x^2-8x-4y^2+16=\left(x-4\right)^2-4y^2\)
\(=\left(x-4-2y\right)\left(x-4+2y\right)\)
d) \(9x^2-12x-y^2+4=\left(3x-2\right)^2-y^2\)
\(=\left(3x-2-y\right)\left(3x-2+y\right)\)
e) \(4x^2+10x-5=4x^2+2.2.\frac{5}{2}x+\frac{25}{4}-\frac{25}{4}-5\)
\(=\left(2x+\frac{5}{2}\right)^2-\frac{45}{4}\)
\(=\left(2x+\frac{5+3\sqrt{5}}{2}\right)\left(2x+\frac{5-3\sqrt{5}}{2}\right)\)
a)x4-1=(x2-1)(x2+1)=(x-1)(x+1)(x2+1)
b)x2-y2-2x+2y=(x-y)(x+y)-2(x-y)=(x-y)(x+y-2)
c)x2-6x-y2+9=(x2-6x+9)-y2=(x-3)2-y2=(x-y-3)(x+y-3)
d)5x2+3(x+y)2-5y2
=5(x2-y2)+3(x+y)2
=5(x-y)(x+y)+3(x+y)2
=(x+y)(5x-5y+3x+3y)
=(x+y)(8x-2y)
a)\(2x^2+8x+8\)
\(\Leftrightarrow2\left(x^2+4x+4\right)\)
\(\Leftrightarrow2\left(x+2\right)^2\)
P/s tham khảo nha
b)\(y^2-x^2+2y+1\)
\(\Leftrightarrow\left(y+1\right)^2-x^2\)
\(\Leftrightarrow\left(y+1-x\right)\left(y+1+x\right)\)
P/s tham khảo nha