a: =(căn x+4)(căn x-1)

b: =(căn x-1)(x+căn x+1)

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

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

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

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

`a, 4x^3 - 16x = 4x(x^2-4) = 4x(x-2)(x+2)`

`b, x^4 - y^4 = (x^2-y^2)(x^2+y^2) = (x-y)(x+y)(x^2+y^2)`

`c, xy^2 + x^2y + 1/4y^3`

`= y(xy + x^2 + 1/4y^2)`

`d, x^2 + 2x - y^2 + 1 = (x+1)^2 - y^2`

`= (x+1+y)(x+1-y)`

Câu 1:

Phần a đề sai nên mk sửa lại:

a, x² + 5x - 14 = x² - 2x + 7x - 14 = x(x - 2) + 7(x - 2) = (x - 2)(x + 7)

b, xz + yz - 5(x + y) = z(x + y) - 5(x + y) = (x + y)(z - 5)

Câu 2:

x² - 4x = -4

\(\Leftrightarrow\) x² - 4x + 4 = 0

\(\Leftrightarrow\) (x - 2)² = 0

\(\Leftrightarrow\) x - 2 = 0

\(\Leftrightarrow\) x = 2

Vậy x = 2

Chúc bn học tốt!

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: \(x^4+4x^2-5\)

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

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

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

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

Bài 1: 

a: \(4a^2-6b=2\left(2a^2-3b\right)\)

b: \(m^3n-2m^2n^2-mn\)

\(=mn\left(m^2-2mn-1\right)\)

Bài 1:

a) \(4a^2-6b=2\left(a^2-3b\right)\)

b) \(=mn\left(m^2-2mn-1\right)\)

Bài 2:

a) \(=4\left(u-2\right)^2+v\left(u-2\right)=\left(u-2\right)\left(4u-8+v\right)\)

b) \(=a\left(a-b\right)^3-b\left(a-b\right)^2-b^2\left(a-b\right)=\left(a-b\right)\left[a\left(a-b\right)^2-b\left(a-b\right)-b^2\right]=\left(a-b\right)\left(a^3-2a^2b+ab^2-ab+b^2-b^2\right)=\left(a-b\right)\left(a^3-2a^2b+ab^2-ab\right)\)

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

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

c) \(\left(a+b\right)^2-\left(a-2b\right)^2\)

\(=\left(a+b-a+2b\right)\left(a+b+a-2b\right)\)

\(=3b\left(2a-b\right)\)

`a, 4x^2-1 = (2x+1)(2x-1)`

`b, (x+2)^2-9 = (x+2-3)(x+2+3) = (x-1)(x+5)`

`c, (a+b)^2-(a-2b)^2 = (a+b+a-2b)(a+b-a+2b) = (2a-b)(3b)`