Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: Ta có: \(\left(x^2+x-1\right)^2-\left(x^2+2x+3\right)^2\)
\(=\left(x^2+x-1-x^2-2x-3\right)\left(x^2+x-1+x^2+2x+3\right)\)
\(=\left(-x-4\right)\left(2x^2+3x+2\right)\)
b: Ta có: \(\left(x-3\right)^2-16\)
\(=\left(x-3-4\right)\left(x-3+4\right)\)
\(=\left(x+1\right)\left(x-7\right)\)
c: \(y^2+16y+64=\left(y+8\right)^2\)
\(a,5\left(x-y\right)-3x\left(y-x\right)=5\left(x-y\right)+3x\left(x-y\right)=\left(5+3x\right)\left(x-y\right)\\ b,x^2-4xy+4y^2=\left(x-2y\right)^2\\ c,\left(x+1\right)^2+x\left(5-x\right)=0\\ \Rightarrow x^2+2x+1+5x-x^2=0\\ \Rightarrow7x+1=0\\ \Rightarrow7x=-1\\ \Rightarrow x=-\dfrac{1}{7}\)
a: =(x-y)(5+3x)
c: \(\Leftrightarrow x^2-2x+1+5x-x^2=0\)
hay x=-1/3
\(\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-5\right)+1\)(2)
\(=\left(x-2\right)\left(x-5\right)\left(x-3\right)\left(x-4\right)+1\)
\(=\left(x^2-7x+10\right)\left(x^2-7x+12\right)+1\)(1)
Đặt \(x^2-7x+10=t\)
\(\Rightarrow\left(1\right)=t\left(t+2\right)+1=t^2+2t+1=\left(t+1\right)^2\)
Mà \(x^2-7x+10=t\)nên \(\left(2\right)=\left(x^2-7x+11\right)^2\)
Vậy \(\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-5\right)+1\)\(=\left(x^2-7x+11\right)^2\)
\(\left(x+1\right)^3-x\left(x-3\right)\left(x+3\right)-6\left(x-1\right)\left(x+2\right)=13\)
\(\Leftrightarrow x^3+3x^2+3x+1-x\left(x^2-9\right)-6\left(x^2+x-2\right)=13\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+9x-6x^2-6x+12=13\)
\(\Leftrightarrow-3x^2+6x=0\)
\(\Leftrightarrow-3\left(x^2-2\right)=0\)
\(\Leftrightarrow x^2-2=0\Leftrightarrow x^2=2\)
\(\Leftrightarrow x=\pm\sqrt{2}\)