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A=x^4+6x^3+7x^2–6x+1=x^4+(6x^3–2x^2)+(9x^2–6x+1)
= x^4+2x^2(3x–1)+(3x–1)^2 =(x^2+3x–1)^2
chỉnh lại tí
Đặt P(x)=x4+6x3+7x2- 6x+1
Đặt y=x2-1
=>y2=x4-2x2+1
P(x)=x4-2x2+1+6x3-6x+9x2
=(x2-1)2+6x(x2-1)+9x2
Q(y)=y2+6xy+9x2
=(y+3x)2
P(x)=(x2-1+3x)2
\(a,4x^2-y^2-1-4x\)
\(\Rightarrow\left(4x^2+4x+1\right)-y^2\)
\(=\left(2x+1\right)^2-y^2=\left(2x+1-y\right)\left(2x+1+y\right)\)
\(b,6x^2-7x-20=6x^2-15x+8x-20\)
\(=\left(6x^2-15x\right)+\left(8x-20\right)\)
\(=3x\left(2x-5\right)+4\left(2x-5\right)\)
\(=\left(3x+4\right)\left(2x-5\right)\)
\(7x-6x^2-2=-\left(6x^2-7x+2\right)\)
\(=-\left(6x^2-3x-4x+2\right)\)
\(=-\left[3x\left(2x-1\right)-2\left(2x-1\right)\right]\)
\(=-\left(3x-2\right)\left(2x-1\right)\)
\(7x-6x^2-2\)
\(=-6x^2+7x-2\)
\(=\left(x-\frac{2}{3}\right)\left(x-\frac{1}{2}\right)\)
\(6x^2+xy-7x-2y^2+7y-5=-\left(y-2x-1\right)\left(2y+3x-5\right)\)
Trả lời:
\(6x^2+xy-7x-2y^2+7y-5=-\left(y-2x-1\right)\left(2y+3x-5\right)\)
Hok_tốt nha
a) Ta có : 6x2 - 13x + 6 = 6x2 - 9x - 4x + 6 = 3x(2x - 3) - 2(2x - 3) = (3x - 2)(2x - 3)
b) Ta có: 6x2 + 7x - 3 = 6x2 + 9x - 2x - 3 = 3x(2x + 3) - (2x + 3) = (3x - 1)(2x + 3)
\(a,6x^2-13x+6\)
\(=6x^2-9x-4x+6\)
\(=3x\cdot\left(2x-3\right)-x\cdot\left(2x-3\right)\)
\(=\left(2x-3\right)\cdot\left(3x-x\right)\)
\(=\left(2x-3\right)\cdot2x\)
\(b,6x^2+7x-3\)
\(=6x^2-2x+9x-3\)
\(=2x\cdot\left(3x-1\right)+3\cdot\left(3x-1\right)\)
\(=\left(3x-1\right)\cdot\left(2x+3\right)\)
Ta có: \(6x^2-7x+2=6x^2-3x-4x+2\)
\(=\left(6x^2-3x\right)-\left(4x-2\right)\)
\(=3x\left(2x-1\right)-2\left(2x-1\right)\)
\(=\left(2x-1\right)\left(3x-2\right)\)
\(=\left(x+1\right)\left(x-2\right)\left(2x^2-x-4\right)\)
\(6x^2-7x-20\)
\(=6x^2+8x-15x-20\)
\(=2x\cdot\left(3x+4\right)-5\left(3x+4\right)\)
\(=\left(3x+4\right)\left(2x-5\right)\)
\(6x^2-7x+1=6x^2-6x+1\)
\(\left(x-1\right)-\left(x-1\right)=\left(6x-1\right)\left(x-1\right)\)