phân tích đt thành nhân tử
\(x^2-x-12\)
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1/ Đặt x2+x=t
=>(x2+x)2+4(x2+x)-12=t2+4t-12=t2+6t-2t-12=t(t+6)-2(t+6)=(t-2)(t+6)=(x2+x-2)(x2+x+6)=(x2-x+2x-2)(x2+x+6)
=[x(x-1)+2(x-1)](x2+x+6)=(x-1)(x+2)(x2+x+6)
2/ Đặt x2+x=t
=>(x2+x)2+9x2+9x+14=(x2+x)2+9(x2+x)+14=t2+9t+14=t2+2t+7t+14=t(t+2)+7(t+2)=(t+2)(t+7)=(x2+x+2)(x2+x+7)
3/ Đặt x2+5x=t
=>(x2+5x)2+10x2+50x+24=(x2+5x)2+10(x2+5x)+24=t2+10t+24=t2+4t+6t+24=t(t+4)+6(t+4)=(t+4)(t+6)=(x2+5x+4)(x2+5x+6)
=(x2+x+4x+4)(x2+2x+3x+6)=[x(x+1)+4(x+1][x(x+2)+3(x+2)]=(x+1)(x+4)(x+2)(x+3)=(x+1)(x+2)(x+3)(x+4)
\(x^2-6x+5\)
\(=x^2-x-5x+5\)
\(=x\left(x-1\right)-5\left(x-1\right)\)
\(=\left(x-1\right)\left(x-5\right)\)
(1+x2)2−4x(1−x2)
= \(-\left(1-x^2\right)^2-4x\left(1-x^2\right)\)
đặt \(\left(1-x^2\right)\)= a
ta có :
- a . a - 4x .a
= a ( - a - 4x )
thay a = \(\left(1+x^2\right)\) ta có
\(\left(1+x^2\right)\left(1-x^2-4x\right)\)
phân tích tiếp nhé !
\(\left(1+x^2\right)^2-4x\left(1-x^2\right)=1+2x^{ }+x^4-4x+4x^3\)\(=\left(x^4+2x^3-x^2\right)+\left(2x^3+4x^2-2x\right)-x^2-2x+1=x^2\left(x^2+2x-1\right)+2x\left(x^2+x-1\right)-\left(x^2+2x-1\right)\)\(\left(x^2+2x-1\right)\left(x^2+2x-1\right)=\left(x^2+2x-1\right)^2\)
\(x^2-x-12\\ =x^2-4x+3x-12\\ =x\left(x-4\right)+3\left(x-4\right)\\ =\left(x-4\right)\left(x+3\right)\)
Ta có: \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
\(=\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12\)
\(=\left(x^2+x\right)^2+3\left(x^2+x\right)-10\)
\(=\left(x^2+x+5\right)\left(x^2+x-2\right)\)
\(=\left(x^2+x+5\right)\left(x+2\right)\left(x-1\right)\)
\(x^2-x-12=x^2-x+\frac{1}{4}-\frac{1}{4}-12=\left(x-\frac{1}{2}\right)^2-\frac{49}{4}=\left(x+3\right)\left(x-4\right)\)
\(x^2-x-12=x^2-4x+3x-12=x\left(x-4\right)+3\left(x-4\right)=\left(x-4\right)\left(x+3\right)\)