Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)-24x^2\)
\(=\left(x^2-13x+30\right)\left(x^2-11x+30\right)-24x^2\)
Đặt \(t=x^2-11x+30\)
\(\Rightarrow\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)-24x^2\)
\(=t.\left(t-2x\right)-24x^2\)
\(=t^2-2xt-24x^2\)
\(=\left(t^2-2xt+x^2\right)-25x^2\)
\(=\left(t-x\right)-\left(5x\right)^2\)
\(=\left(t-6x\right)\left(t+4x\right)\)
\(=\left(x^2-17x+30\right)\left(x^2-7x+30\right)\)
Tham khảo nhé~
e) Ta có: \(x^4-2x^3+2x-1\)
\(=\left(x^4-1\right)-2x\left(x^2-1\right)\)
\(=\left(x^2+1\right)\left(x-1\right)\left(x+1\right)-2x\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x+1\right)\cdot\left(x^2-2x+1\right)\)
\(=\left(x+1\right)\cdot\left(x-1\right)^3\)
h) Ta có: \(3x^2-3y^2-2\left(x-y\right)^2\)
\(=3\left(x^2-y^2\right)-2\left(x-y\right)^2\)
\(=3\left(x-y\right)\left(x+y\right)-2\left(x-y\right)^2\)
\(=\left(x-y\right)\left(3x+3y-2x+2y\right)\)
\(=\left(x-y\right)\left(x+5y\right)\)
a) Ta có: \(x^2-y^2-2x-2y\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-2\right)\)
b) Ta có: \(x^2\left(x+2y\right)-x-2y\)
\(=\left(x+2y\right)\left(x^2-1\right)\)
\(=\left(x+2y\right)\left(x-1\right)\left(x+1\right)\)
a: \(=\left(x+1\right)\left(x^2-x+1\right)+5x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+4x+1\right)\)
\(a,=\left(x-1\right)^4-2\left(x-1\right)^2+1\\ =\left[\left(x-1\right)^2-1\right]^2\\ =\left(x^2-2x-2\right)^2\\ b,=\left[\left(x+1\right)\left(x+5\right)\right]\left[\left(x+2\right)\left(x+4\right)\right]-4\\ =\left(x^2+6x+5\right)\left(x^2+6x+8\right)-4\\ =\left(x^2+6x\right)^2+13\left(x^2+6x\right)+36\\ =\left(x^2+6x+4\right)\left(x^2+6x+9\right)\\ =\left(x+3\right)^2\left(x^2+6x+4\right)\)
a) \(=\left(2x-1\right)^2\)
b) \(=x\left(y^2-x^2+2x-1\right)=x\left[y^2-\left(x-1\right)^2\right]=x\left(y-x+1\right)\left(y+x-1\right)\)
a) \(4\left(x+1\right)^3-x-1=4\left(x+1\right)^3-\left(x+1\right)=\left(x+1\right)\left[4\left(x+1\right)^2-1\right]=\left(x+1\right)\left[2\left(x+1\right)-1\right]\left[2\left(x+1\right)+1\right]=\left(x+1\right)\left(2x+1\right)\left(2x+3\right)\)
b) \(5x\left(x-3\right)+\left(3-x\right)^2-\left(x-3\right)=5x\left(x-3\right)+\left(x-3\right)^2-\left(x-3\right)=\left(x-3\right)\left(5x+x-3-1\right)=\left(x-3\right)\left(6x-4\right)=2\left(x-3\right)\left(3x-2\right)\)
c) \(9x^2y^3-3x^4y^2-6x^3y^2+16xy^4=xy^2\left(9xy-3x^3-6x^2+16y^2\right)\)
c: \(x^4+x^3-4x^2+x+1\)
\(=x^4-x^3+2x^3-2x^2-2x^2+2x-x+1\)
\(=\left(x-1\right)\left(x^3+2x^2-2x-1\right)\)
\(=\left(x-1\right)\left[\left(x-1\right)\left(x^2+x+1\right)+2x\left(x-1\right)\right]\)
\(=\left(x-1\right)^2\cdot\left(x^2+3x+1\right)\)
a, x^4+6x^3+11x^2+6x+1
= x^4 + 6x^3 + 9x² + 2x² + 6x + 1
= x^4 + 9x² + 1 + 6x^3 + 2x² + 6x
= x^4 + 9x² + 1² + 2.x².3x + 2.x².1 + 2.3x.1
= (x² + 3x + 1)²
Mình làm được ý a nên tk 1 tk