Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Tham khảo:https://hoc247.net/hoi-dap/toan-8/phan-tich-da-thuc-x-7-x-2-1-thanh-nhan-tu-faq417522.html
\(=x^7+x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2+x^2-x^2+x-x+1\\ =\left(x^7+x^6+x^5\right)-\left(x^6+x^5+x^4\right)+\left(x^4+x^3+x^2\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)\)
a) \(x^3+3x^2+3x=0\Rightarrow x\left(x^2+3x+3\right)=0\Rightarrow x\left[\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\right]=0\Rightarrow x=0\)
(do \(\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\))
b) \(x^3+6x^2+12x=0\Rightarrow x\left(x^2+6x+12\right)=0\Rightarrow x\left[\left(x+3\right)^2+4\right]=0\Rightarrow x=0\)
(do (x+3)2+4≥4>0)
a: Ta có: \(x^3+3x^2+3x=0\)
\(\Leftrightarrow x\left(x^2+3x+3\right)=0\)
hay x=0
b: Ta có: \(x^3+6x^2+12x=0\)
\(\Leftrightarrow x\left(x^2+6x+12\right)=0\)
hay x=0
a) \(2xy-y+6x-3=\left(2xy+6x\right)-\left(y+3\right)=2x\left(y+3\right)-\left(y+3\right)=\left(2x-1\right)\left(y+3\right)\)
b) \(x^2-2xy-x+2y=\left(x^2-2xy\right)-\left(x-2y\right)=x\left(x-2y\right)-\left(x-2y\right)=\left(x-1\right)\left(x-2y\right)\)
\(\left(x^2+2x\right)^2-2x^2-4x-3=0\Leftrightarrow x^4+4x^3+4x^2-2x^2-4x-3=0\Leftrightarrow x^4+4x^3+2x^2-4x-3=0\Leftrightarrow\left(x-1\right)\left(x+1\right)^2\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=3\end{matrix}\right.\)
Ta có: \(\left(x^2+2x\right)^2-2x^2-4x-3=0\)
\(\Leftrightarrow\left(x^2+2x\right)^2-2\left(x^2+2x\right)-3=0\)
\(\Leftrightarrow\left(x^2+2x-3\right)\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2\cdot\left(x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-3\\x=1\end{matrix}\right.\)
Tôi làm tạm theo cách này nhé.
\(x^4-2x^2-114x-1295\)
\(=\frac{d}{dx}\left(x^4-2x^2-114x-1295\right)\)
\(=4x^3-4x-114-0\)
\(=4x^3-4x-114\)
Bạn Phương Lê Nhật ơi!!!!
Đây là Toán 8 bạn ạ
Bạn giải mk ko hiểu j cả
Giải cụ thể đc ko bạn ạ
x^3-2x-4
=x^3-4x+2x-4
=x(x^2-4)+2(x-2)
=x(x-2)(x+2)+2(x-2)
=(x^2+2x+2)(x-2)
Nhớ cho mình nha!!!
Dễ thôi
\(x^3-x^2-4\)
\(=\left(x^3-8\right)-\left(x^2-4\right)\)
\(=\left(x-2\right)\left(x^2+2x+4\right)-\left(x-2\right)\left(x+2\right)\)
\(=\left(x-2\right)\left(x^2+x+2\right)\)
x4y4 + x2y2 + 2xy
= ( x3y3 + xy + 2 ) xy
= ( x3y3 + x2y2 - x2y2 - xy + 2xy + 2 ) xy
= [ x2y2 ( xy + 1 ) - xy ( xy + 1 ) + 2 ( xy + 1)] xy
= ( x2y2 - xy + 2 ) ( xy + 1 ) xy
\(x^2+4x=2\)
\(\Rightarrow x^2+4x-2=0\)
\(\Rightarrow\left(x^2+4x+4\right)-6=0\)
\(\Rightarrow\left(x+2\right)^2-\left(\sqrt{6}\right)^2=0\)
\(\Rightarrow\left(x+2-\sqrt{6}\right).\left(x+2+\sqrt{6}\right)\)
\(\Rightarrow\orbr{\begin{cases}x+2-\sqrt{6}=0\\x+2+\sqrt{6}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\sqrt{6}-2\\x=-\sqrt{6}-2\end{cases}}\)
\(9x^2-18x=7\)
\(\Rightarrow9x^2-18x-7=0\)
\(\Rightarrow\left(9x^2-18x+9\right)-16=0\)
\(\Rightarrow[\left(3x\right)^2-2.3x.3+3^2]-4^2=0\)
\(\Rightarrow\left(3x-3\right)^2-4^2=0\)
\(\Rightarrow\left(3x-3-4\right).\left(3x-3+4\right)=0\)
\(\Rightarrow\left(3x-7\right).\left(3x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-7=0\\3x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}3x=7\\3x=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=\frac{-1}{3}\end{cases}}\)