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a)Đặt \(A=\dfrac{1}{8}x^3-\dfrac{3}{4}x^2+\dfrac{3}{2}x-1\)
\(A=\dfrac{1}{8}\left(x^3-6x^2+12x-8\right)\)
\(A=\dfrac{1}{8}\left(x-2\right)^3\)
b,\(x^4+2015x^2+2014x+2015=x^4+2015x^2+2015x-x+2015=x\left(x^3-1\right)+2015\left(X^2+x+1\right)=x\left(x-1\right)\left(x^2+x+1\right)+2015\left(x^2+x+1\right)=\left(x^2+x+1\right)\left(x^2-x+2015\right)\)
\(=x^4-x+2016x^2+2016x+2016.\)
\(=x\left(x^3-1\right)+2016\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2016\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2016\right)\)
2015x4 + 2016x2 + x + 2016
= (2015x4 + 2015x3 + 2015x2) + (- 2015x3 - 2015x2 - 2015x) + (2016x2 + 2016x + 2016)
= (x2 + x + 1)(2015x2 - 2015x + 2016)
Ta có: x^4 + 2016x^2 + 2015x + 2016
= x^4 + x^3 + x^2 - x^3 - x^2 - x + 2016x^2 + 2016x + 2016
= x^2(x^2 + x + 1) - x(x^2 + x + 1) + 2016(x^2 + x + 1)
= (x^2 + x + 1)(x^2 - x + 2016)
\(x^4+2016x^2+2015x+2016\)
\(=x^4-x+2016x^2+2016x+2016\)
\(=x\left(x^3-1\right)+2016\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2016\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2016\right)\)
b: \(x^4+x^2+1\)
\(=x^4+2x^2+1-x^2\)
\(=\left(x^2+1\right)^2-x^2\)
\(=\left(x^2-x+1\right)\left(x^2+x+1\right)\)
c: \(x^7+x^5+1\)
\(=x^7+x^6+x^5-x^6-x^5-x^4+x^5+x^4+x^3-x^3+1\)
\(=x^5\left(x^2+x+1\right)-x^4\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^5-x^4+x^3-x+1\right)\)
a, x2+5x = 6
=> x2+5x - 6 =0
=> x2+6x -x- 6 =0
=> x(x+6)-(x+6)=0
=>(x-1)(x+6)=0
=> x=1 hoặc x=-6
b, x2-2015x +2014=0
=> x2-2014x-x +2014=0
=>x(x-2014)-(x-2014)=0
=> (x-1)(x-2014)=0
=> x=1 hoặc x=2014
k viết lại đề!
\(a.\\ \Leftrightarrow x^2+5x-6=0\\ \Leftrightarrow x^2+6x-x-6=0\\ \Leftrightarrow x\left(x+6\right)-\left(x+6\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-6\end{matrix}\right.\\ \Rightarrow S=\left\{1;-6\right\}\)
\(b.\\ \Leftrightarrow x^2-2014x-x+2014=0\\ \Leftrightarrow x\left(x-2014\right)-\left(x-2014\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2014\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2014=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2014\end{matrix}\right.\\ \Rightarrow S=\left\{1;2014\right\}\)
x4+2014x2-2014x-x+2014
=x(x3-1)+2014(x2-x-1)
=x(x-1)(x2-x-1)+2014(x2-x-1)
=(x2-x-1)(x2-x+2014)
\(\left(x^2+2015x\right)\left(\dfrac{1}{2016}+\dfrac{1}{1008}+\dfrac{1}{672}+1\right)=2022\)
\(\Leftrightarrow\left(x^2+2015x\right).\dfrac{2022}{2016}=2022\)
\(\Leftrightarrow x^2+2015x=2016\)
\(\Leftrightarrow x^2+2015x-2016=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2016\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2016\end{matrix}\right.\)