![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow x^5-1=4x^4+4x^3+4x^2+4x+4\)
\(\Leftrightarrow\left(x-1\right)\left(x^4+x^3+x^2+x+1\right)=4\left(x^4+x^3+x^2+x+1\right)\)
\(\Leftrightarrow\left(x-5\right)\left(x^4+x^3+x^2+x+1\right)=0\)
\(\Leftrightarrow x=5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3xy^3+6x^3y+xy=xy\left(3y^2+6x^2+1\right)\)
\(4x^3+8x^2+4x=4x\left(x^2+2x+1\right)=4x\left(x+1\right)^2\)
\(4x^2-4x+1-y^2=\left(2x-1\right)^2-y^2=\left(2x-1-y\right)\left(2x-1+y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(A=\left(2x+3\right)\left(4x^2-6x+9\right)-2\left(4x^3-1\right)\)
\(=8x^3+27-8x^3+2\)
=29
b: Ta có: \(B=\left(64x^3-1\right)-\left(4x-3\right)\left(16x^2+3\right)\)
\(=64x^3-1-64x^3-12x-48x^2+9\)
\(=-12x+8\)
c: Ta có: \(2\left(x^3+y^3\right)-3\left(x^2+y^2\right)\)
\(=2\left(x^2+xy+y^2\right)-3\left(-2xy\right)\)
\(=2x^2+2xy+2y^2+6xy\)
\(=2x^2+8xy+2y^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
c) Ta có: \(C=4x^2+y^2-4xy+8x-4y+4\)
\(=\left(2x-y\right)^2+2\cdot\left(2x-y\right)\cdot2+2^2\)
\(=\left(2x-y+2\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow4x^2\left(x-1\right)-\left(x-1\right)=0\Leftrightarrow\left(2x-1\right)\left(2x+1\right)\left(x-1\right)=0\Leftrightarrow x=\dfrac{1}{2};x=-\dfrac{1}{2};x=1\)
\(4x^3-4x^2-x+1=0\)
<=>\(\left(2x+1\right)\left(x-1\right)\left(2x-1\right)=0\)
<=>\(\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=1\\x=\dfrac{1}{2}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`(3x-2)(4x+5)=0`
\(< =>\left[{}\begin{matrix}3x-2=0\\4x+5=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{4}\end{matrix}\right.\)
\(\left(3x-2\right)\left(4x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\4x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\4x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{-5}{4}\end{matrix}\right.\)
\(\text{Vậy phương trình có tập nghiệm là }S=\left\{\dfrac{2}{3};\dfrac{-5}{4}\right\}\)
Ta có : \(x^4+4x^3+4x^2-4x-5=0\)
\(\Leftrightarrow\left(x^4+4x^3+5x^2\right)-\left(x^2+4x+5\right)=0\)
\(\Leftrightarrow x^2\left(x^2+4x+5\right)-\left(x^2+4x+5\right)=0\)
\(\Leftrightarrow\left(x^2+4x+5\right)\left(x^2-1\right)=0\)
Ta thấy : \(x^2+4x+5=\left(x+2\right)^2+1\ge1>0\forall x\)
\(\Rightarrow x^2-1=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\) ( thỏa mãn )
Vậy pt đã cho có tập nghiệm \(S=\left\{-1,1\right\}\)
\(x^4+4x^3+4x^2-4x-5=0\)
\(\Leftrightarrow x^4+x^3+3x^3+3x^2+x^2+x-5x-5=0\)
\(\Leftrightarrow x^3\left(x+1\right)+3x^2\left(x+1\right)+x\left(x+1\right)-5\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^3+3x^2+x-5\right)=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)