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\(a,6x^2-9x=3x\left(x-3\right)\)
\(b,x^3-2x^2-3x+6\)
\(=\left(x^3-2x^2\right)-\left(3x-6\right)\)
\(=x^2\left(x-2\right)-3\left(x-2\right)\)
\(=\left(x^2-3\right)\left(x-2\right)\)
\(e,2x\left(x-y\right)-3y\left(x-y\right)\)
\(=\left(2x-3y\right)\left(x-y\right)\)
a) 6x2 - 9x
= 3x (2x - 3)
b) x3 - 2x2 - 3x + 6
= x2(x - 2) - 3 (x - 2)
=(x - 2) (x2 - 3)
c) x2 - 4x + 4 - 9y2
= (x - 2)2 - 9y2
=(x - 2 - 3y)(x - 2 + 3y)
e) 2x(x - y) - 3y(x - y)
= (x - y)(2x - 3y)
xin lỗi mình học ngu nên không biết làm nhìu nha
a,x^4+2x^3-4x-4
=(x^3+2x^3)-(4x+4)
=x^3(x+2)-4(x+2)
=(x^3-4)(x+2)
\(X^4+2X^3-4X-4\)
\(=\left(X^2\right)^2+2X^3-4X-2^2\)
\(=\left[\left(X^2\right)^2-2^2\right]+\left[2X^3-4X\right]\)
\(=\left(X^2+2\right)\left(X^2-2\right)+2X\left(X^2-2\right)\)
\(=\left(X^2-2\right)\left(X^2+2+2X\right)\)
Bài 2
\(a,x^3+2x^2+x\)
\(=x.\left(x^2+2x+1\right)\)
\(b,xy+y^2-x-y\)
\(=y.\left(x+y\right)-\left(x+y\right)\)
\(=\left(y-1\right).\left(x+y\right)\)
bài 3
\(a,3x.\left(x^2-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x=0\\x^2=4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=2,x=-2\end{cases}}\)
vậy x=0,x=2 hay x=-2
\(b,xy+y^2-x-y=0\)
\(y.\left(x+y\right)-\left(x+y\right)=0\)
\(\left(y-1\right).\left(x+y\right)=0\)
\(\Rightarrow\orbr{\begin{cases}y-1=0\\x+y=0\end{cases}\Rightarrow\orbr{\begin{cases}y=1\\x=-1\end{cases}}}\)
vậy x=-1, y=1
a) Ta có: \(2x^4+3x^3-9x^2-3x+2\)
\(=2x^4-2x^3-2x^2+5x^3-5x^2-5x-2x^2+2x+2\)
\(=2x^2\left(x^2-x-1\right)+5x\left(x^2-x-1\right)-2\left(x^2-x-1\right)\)
\(=\left(x^2-x-1\right)\left(2x^2+5x-2\right)\)
Gợi ý:
a) Đặt \(x^2+3x+1=a\)
b) \(\left(x^2+8x+7\right)\left(x+3\right)\left(x+5\right)+15\)
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
Đặt \(x^2+8x+11=a\)
c) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(x^2+7x+11=a\)
d) \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)
Đặt \(12x^2+11x-1=a\)
Câu hỏi của Nguyễn Tấn Phát - Toán lớp 8 - Học toán với OnlineMath
Em tham khảo câu e nhé!
b, A=[(a+1)(a+7)][(a+3)(a+5)]+15
=>A=(a2+8a+7)(a2+8a+15)+15
Đặt a2+8a+11= t
=>a2+8a+7= t-4 và a2+8a+15= t+4
=>A=(t-4)(t+4)+15
=>A=t2-16+15
=t2-1=(t-1)(t+1)
Thay t = a2+8a+11
=>A=(a2+8a+11-1)(a2+8a+11+1)
=>A=(a2+8a+10)(a2+8a+12)
a) \(x^2+2xy+7x+7y+y^2+10\)
\(=\left(x+y\right)^2+7\left(x+y\right)+\frac{49}{4}-\frac{9}{4}\)
\(=\left(x+y+\frac{7}{2}\right)^2-\frac{9}{4}\)
\(=\left(x+y+\frac{7}{2}-\frac{3}{2}\right)\left(x+y+\frac{7}{2}+\frac{3}{2}\right)\)
\(=\left(x+y-2\right)\left(x+y+5\right)\)
Sửa lại đề là: \(3x^2+10x+3\)
\(=3x^2+9x+x+3\)
\(=\left(3x^2+9x\right)+\left(x+3\right)\)
\(=3x.\left(x+3\right)+\left(x+3\right)\)
\(=\left(x+3\right).\left(3x+1\right)\)
\(1.\)
\(x^2-2x+1-xy-y=\left(x-1\right)^2-y\left(x-1\right)=\left(x-1\right)\left(x-1-y\right)\)
\(2.\)
\(x^3-4x^2+4x-2x+2=x\left(x^2-4x+4\right)-2\left(x-1\right)=x\left(x-2\right)^2-2\left(x-1\right)\)
\(3.\)
\(10x-25-x^2+4y^2=4y^2-\left(x^2-10x+25\right)=4y^2-\left(x-5\right)^2=\left(2y+x-5\right)\left(2y-x+5\right)\)
\(4.\)
\(4x^2-2x+2xy-y=2x\left(2x-1\right)+y\left(2x-1\right)=\left(2x-1\right)\left(2x+y\right)\)
\(5.\)
\(4x\left(x-3\right)^2-3x^2+9x=4x\left(x-3\right)^2-3x\left(x-3\right)=\left(x-3\right)\left(4x^2-12x-3x\right)\)
Lời giải:
a.
$x^4+10x^3+26x^2+10x+1$
$=(x^4+10x^3+25x^2)+x^2+10x+1$
$=(x^2+5x)^2+2(x^2+5x)+1-x^2$
$=(x^2+5x+1)^2-x^2=(x^2+5x+1-x)(x^2+5x+1+x)$
$=(x^2+4x+1)(x^2+6x+1)$
b.
$x^4+x^3-4x^2+x+1$
$=(x^4-x^2)+(x^3-x^2)+(x-x^2)+(1-x^2)$
$=x^2(x-1)(x+1)+x^2(x-1)-x(x-1)-(x-1)(x+1)$
$=(x-1)[x^2(x+1)+x^2-x-(x+1)]$
$=(x-1)(x^3+2x^2-2x-1)$
$=(x-1)[(x^3-1)+(2x^2-2x)]=(x-1)[(x-1)(x^2+x+1)+2x(x-1)]$
$=(x-1)(x-1)(x^2+x+1+2x)=(x-1)^2(x^2+3x+1)$