\(\left(x-5\right)\left(x-1\right)\left(x+3\right)\left(x+7\right)+60\)

\(=\left(x^2+2x-35\right)\left(x^2+2x-3\right)+60\)

\(=\left(x^2+2x\right)^2-38\left(x^2+2x\right)+105+60\)

\(=\left(x^2+2x\right)^2-3\left(x^2+2x\right)-35\left(x^2+2x\right)+165\)

\(=\left(x^2+2x-3\right)\left(x^2+2x-35\right)\)

\(=\left(x+3\right)\left(x-1\right)\left(x+7\right)\left(x-5\right)\)

làm ơn trả lời câu hỏi của mình đi

\(\left(x+y\right)^2+3\left(x+y\right)-10=\left[\left(x+y\right)^2+2\left(x+y\right).\dfrac{3}{2}+\dfrac{9}{4}\right]-\dfrac{49}{4}\)

\(=\left(x+y+\dfrac{3}{2}\right)^2-\dfrac{49}{4}=\left(x+y+\dfrac{3}{2}-\dfrac{7}{2}\right)\left(x+y+\dfrac{3}{2}+\dfrac{7}{2}\right)=\left(x+y-2\right)\left(x+y+5\right)\)

\(\left(x+y\right)^2+3\left(x+y\right)-10\)

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

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

Đa thức này không phân tích được thành nhân tử.

Bạn coi lại đề.

Ta có: \(4\left(x+5\right)\left(x+6\right)\left(x+10\right)\left(x+12\right)+3x^2\)

\(=4\left(x^2+60+17x\right)\left(x^2+60x+16x\right)+3x^2\)

\(=4\left[\left(x^2+60\right)^2+33x\left(x^2+60\right)+272x^2\right]+3x^2\)

\(=4\left(x^2+60\right)^2+132x\left(x^2+60\right)+1091x^2\)

Cái này đã là nhân tử rồi mà bạn

\(x^4-x^3-x+1=\left(x^4-x^3\right)-\left(x-1\right)=x^3\left(x-1\right)-\left(x-1\right)=\left(x^3-1\right)\left(x-1\right)=\left(x-1\right)^2.\left(x^2+x+1\right)\)

x⁴ - x³ - x + 1

= (x⁴ - x³) - (x - 1)

= x³(x - 1) - (x - 1)

= (x³ - 1)(x - 1)

\(=4ab\left(ab+ax+bx+x^2\right)=4a^2b^2+4a^2bx+4ab^2x+4abx^2\)

4ab(a + x)(b + x) là nhân tử rồi nhé

\(A=\left(x+1\right)\left(x-4\right)\left(x+2\right)\left(x-8\right)+4x^2\)

\(A=\left[\left(x+1\right)\left(x-8\right)\right]\left[\left(x-4\right)\left(x+2\right)\right]+4x^2\)

\(A=\left(x^2-7x-8\right)\left(x^2-2x-8\right)+4x^2\)

Đặt \(p=x^2-4,5x-8\)ta có :

\(A=\left(p-2,5x\right)\left(p+2,5x\right)+4x^2\)

\(A=p^2-\left(2,5x\right)^2+4x^2\)

\(A=p^2-6,25x^2+4x^2\)

\(A=p^2-2,25x^2\)

\(A=p^2-\left(1,5x\right)^2\)

\(A=\left(p-1,5x\right)\left(p+1,5x\right)\)

Thay \(p=x^2-4,5x-8\)vào A ta có :

\(A=\left(x^2-4,5x-8-1,5x\right)\left(x^2-4,5x-8+1,5x\right)\)

\(A=\left(x^2-6x-8\right)\left(x^2-3x-8\right)\)

\(\left(x+1\right)\left(x-4\right)\left(x+2\right)\left(x-8\right)+4x^2\)

\(=\left(x+1\right)\left(x-8\right)\left(x-4\right)\left(x+2\right)+4x^2\)

\(=\left(x^2-7x-8\right)\left(x^2-2x-8\right)+4x^2\)

  Đặt \(x^2-2x-8=t\)

  Ta có : \(\left(t-5x\right)t+4x^2\)

\(=t^2-5xt+4x^2\)

\(=t^2-2.\frac{5}{2}xt+\frac{25}{4}x^2-\frac{9}{4}x^2\)

\(=\left(t-\frac{5}{2}\right)^2-\frac{9}{4}x^2\)

\(=\left(t-\frac{5}{2}-\frac{3}{2}x\right)\left(t-\frac{5}{2}+\frac{3}{2}x\right)\)

    Học tốt ~~