Bn ấn vào câu hỏi của bn sẽ rs những câu tương tự có đáp án nhé!!Chúc bn lm đc bài này nha!!

Trả lời:

A=(x-1)(x+2)(x-3)(x+4)-144

A= (x²-5x-14)(x²-5x-24)-144 (1)

đặt m=x²-5x-14

=> A= m.(m-10)-144

A=m²-10m-144

A= (m-18)(m+8)

thay m vào, ta có:

A= (x²-5x-32)(x²-5x-6)

A=(x²-5x-32)(x+1)(x-6)

a) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\left(1\right)=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-15=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)

Đặt \(t=x^2+5x+4\)

(1) trở thành: \(t\left(t+2\right)-15=t^2+2t+1-16=\left(t+1\right)^2-4^2=\left(t-3\right)\left(t+5\right)\)

Thay t: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15=\left(x^2+5x+4-3\right)\left(x^2+5x+4+5\right)=\left(x^2+5x+1\right)\left(x^2+5x+9\right)\)

b) \(\left(2x+5\right)^2-\left(x-9\right)^2=\left(2x+5-x+9\right)\left(2x+5+x-9\right)=\left(x+14\right)\left(3x-4\right)\)

a: Ta có: \(\left(x+1\right)\cdot\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)

\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+24-15\)

\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+9\)

\(=\left(x^2+5x+1\right)\left(x^2+5x+9\right)\)

b: \(\left(2x+5\right)^2-\left(x-9\right)^2\)

\(=\left(2x+5-x+9\right)\left(2x+5+x-9\right)\)

\(=\left(x+15\right)\left(3x-4\right)\)

1A:

a: \(x^3+2x=x\left(x^2+2\right)\)

b: \(3x-6y=3\left(x-2y\right)\)

c: \(5\left(x+3y\right)-15x\left(x+3y\right)\)

\(=5\left(x+3y\right)\left(1-3x\right)\)

d: \(3\left(x-y\right)-5x\left(y-x\right)\)

\(=3\left(x-y\right)+5x\left(x-y\right)\)

\(=\left(x-y\right)\left(5x+3\right)\)

1A. a. x(x²+2) 

b. 3(x-2y)

c. 5(x+3y)(1-3x) 

d. (x-y) (3-5x)

1B. a. 2x(2x-3)

b.xy(x²-2xy+5)

c. 2x(x+1)(x+2)

d. 2x(y-1)+2y(y-1)=2(y-1)(x-y)

1B:

a: \(4x^2-6x=2x\left(2x-3\right)\)

b: \(x^3y-2x^2y^2+5xy\)

\(=xy\left(x^2-2xy+5\right)\)

a: =(căn x+4)(căn x-1)

b: =(căn x-1)(x+căn x+1)

a: 6x-2y=2(3x-y)

b: =(x-y)(x-2)(x+2)

Lời giải:
a. Không phân tích được nữa

b. $x^2(x-y)+4(y-x)=x^2(x-y)-4(x-y)=(x-y)(x^2-4)=(x-y)(x-2)(x+2)$
c. $x^3+2x^2y+xy^2-4x=x(x^2+2xy+y^2-4)$

$=x[(x^2+2xy+y^2)-4]=x[(x+y)^2-2^2]=x(x+y-2)(x+y+2)$

ko phân tích dc

b: =(x-y)(x-2)(x+2)

`a)(x+2)^2+2(x^2-4)+(x-2)^2`

`=(x+2)^2+2(x-2)(x+2)+(x-2)^2`

`=(x+2+x-2)^2=(2x)^2=4x^2`

`b)x^2-x+1/4`

`=x^2-2.x .1/2+1/4=(x-1/2)^2`

`c)(x+y)^3-(x-y)^3`

`=(x+y-x+y)[(x+y)^2+(x+y)(x-y)+(x-y)^2]`

`=2y(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2)`

`=2y(3x^2+y^2)`

a) \(\left(x+2\right)^2+2\left(x^2-4\right)+\left(x-2\right)^2\)

\(=\left(x+2\right)^2+2\left(x+2\right)\left(x-2\right)+\left(x-2\right)^2\)

\(=\left(x+2+x-2\right)^2=\left(2x\right)^2=4x^2\)

b) \(x^2-x+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2\)

c) \(\left(x+y\right)^3-\left(x-y\right)^3=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x^3-3x^2y+3xy^2-y^3\right)=6x^2y+2y^3=2y\left(3x^2+y^2\right)\)

a:=2(x-2)+y(x-2)

=(x-2)(y+2)

b: \(=\left(x+y\right)^2-4\)

=(x+y+2)(x+y-2)

a.

\(=\left(x+1\right)^3-\left(3z\right)^3\)

\(=\left(x+1+3z\right)\left[\left(x+1\right)^2+3z\left(x+1\right)+9z^2\right]\)

\(=\left(x+3z+1\right)\left(x^2+2x+1+3zx+3z+9z^2\right)\)

b.

\(=\left(x-y\right)^2-z\left(x-y\right)\)

\(=\left(x-y\right)\left(x-y-z\right)\)

c.

\(=x^4-1+4x^2-4\)

\(=\left(x^2-1\right)\left(x^2+1\right)+4\left(x^2-1\right)\)

\(=\left(x^2-1\right)\left(x^2+5\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)

a) Ta có: \(x^3+3x^2+3x+1-27z^3\)

\(=\left(x+1\right)^3-\left(3z\right)^3\)

\(=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)

b) Ta có: \(x^2-2xy+y^2-zx+yz\)

\(=\left(x-y\right)^2-z\left(x-y\right)\)

\(=\left(x-y\right)\left(x-y-z\right)\)

c) Ta có: \(x^4+4x^2-5\)

\(=x^4+4x^2+4-9\)

\(=\left(x^2+2\right)^2-3^2\)

\(=\left(x^2-1\right)\left(x^2+5\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)

a) \(x^2 (x+1)-2x(x+1)+x+1 \\ =(x+1)(x^2-2x+1)\\=(x+1)(x-1)^2\)

b) \(4x^2 -8x+3 \\= (2x^2)-2.2x .2 + 2^2 -1 \\=(2x-2)^2-1^2\\=(2x-2+1)(2x-2-1)\\= (2x-1)(2x-3)\)

`b)x^3+y^3+z^3-3xyz`

`=x^3+3xy(x+y)+z^3-3xy(x+y)-3xyz`

`=(x+y)^3+z^3-3xy(x+y+z)`

`=(x+y+z)[(x+y)^2-z(x+y)+z^2]-3xy(x+y)`

`=(x+y+z)(x^2+2xy+y^2-zx-yz-3xy+z^2)`

`=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)`

giúp mk vs mk cần gấp T^T

Bạn xem lại đề câu b) nhé đề đúng là `x^3+y^3+z^3-3xyz`.Còn được thì xem lại câu c) nhé do số quá xấu.

`a)x^4+3x^3-6x^2-8`

`=x(x^3+3x^2-6x-8)`

`=x[(x-2)(x^2+2x+4)+3x(x-2)]`

`=x(x-2)(x^2+5x+4)`

`=x(x-2)(x^2+x+4x+4)`

`=x(x-2)(x+1)(x+4)`

`c)(x+1)(x+3)(x+5)(x+7)-5`

`=[(x+1)(x+7)][(x+3)(x+5)]-5`

`=(x^2+8x+7)(x^2+8x+15)-5`

`=(x^2+8x+11)^2-4^2-5`

`=(x^2+8x+11)^2-21`

`=(x^2+8x+11-sqrt21)(x^2+8x+11+sqrt21)`