Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
x\(^2\) - 9 + ( x - 3)\(^2\)
=(x^2 - 3^2 ) + (x-3)^2
=(x - 3) (x+3) +(x-3)^2
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
a) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)
\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-15\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)(1)
Đặt \(x^2+5x+4=t\)
\(\Rightarrow\left(1\right)=t\left(t+2\right)-15=t^2+2t+1-16\)
\(=\left(t+1\right)^2-4^2=\left(t+5\right)\left(t-3\right)\)
\(=\left(x^2+5x+9\right)\left(x^2+5x+1\right)\)
b) \(\left(2x+5\right)^2-\left(x-9\right)^2\)
\(=\left(2x+5+x-9\right)\left(2x+5-x+9\right)\)
\(=\left(3x-4\right)\left(x+14\right)\)
\(b,x^3-2x^2-4xy^2+x\)
\(=x\left(x^2-2x-4y^2+1\right)\)
\(=x\left[\left(x^2-2x+1\right)-4y^2\right]\)
\(=x\left[\left(x-1\right)^2-\left(2y\right)^2\right]\)
\(=x\left(x-1-2y\right)\left(x-1+2y\right)\)
\(=x\left(x-2y-1\right)\left(x+2y-1\right)\)
\(---\)
\(c,\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8\)
\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-8\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\) (1)
Đặt \(y=x^2+7x+10\), thay vào (1) ta được:
\(y\left(y+2\right)-8\)
\(=y^2+2y+1-9\)
\(=\left(y+1\right)^2-3^2\)
\(=\left(y+1-3\right)\left(y+1+3\right)\)
\(=\left(y-2\right)\left(y+4\right)\)
\(=\left(x^2+7x+10-2\right)\left(x^2+7x+10+4\right)\)
\(=\left(x^2+7x+8\right)\left(x^2+7x+14\right)\)
#Ayumu
a) Ta có: \(x^4-16x^2=0\)
\(\Leftrightarrow x^2\left(x^2-16\right)=0\)
\(\Leftrightarrow x^2\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
b) Ta có: \(x^8+36x^4=0\)
\(\Leftrightarrow x^4\left(x^4+36\right)=0\)
\(\Leftrightarrow x^4=0\)
hay x=0
c) Ta có: \(\left(x-5\right)^3-x+5=0\)
\(\Leftrightarrow\left(x-5\right)\cdot\left[\left(x-5\right)^2-1\right]=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-4\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)
d) Ta có: \(5\left(x-2\right)-x^2+4=0\)
\(\Leftrightarrow5\left(x-2\right)-\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(5-x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(3-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
