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\(a,x^2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2-2=a^2-2\)
\(x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)=a^3-3a\)
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a) \(\left(\frac{1}{x}+2\right)=\left(\frac{1}{x}+2\right)\left(x^2+1\right)\)
\(\Leftrightarrow\left(\frac{1}{x}+2\right)\left(x^2+1\right)-\left(\frac{1}{x}+2\right)=0\)
\(\Leftrightarrow\left(\frac{1}{x}+2\right)x^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{x}+2=0\\x^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=0\left(L\right)\end{cases}}\)
Vậy \(x=-\frac{1}{2}\)
s e thấy == câu này mọi ngừi ko tl vậy :v ( bài này cs cần đk ko -.- e chưa hc nên ko nắm chắc , kệ đi , cứ lm )
\(a,\left(\frac{1}{x}+2\right)=\left(\frac{1}{x}+2\right)\left(x^2+1\right)\)
\(\frac{1}{x}+2=\left(\frac{1}{x}+2\right)\left(x^2+1\right)\)
\(1+2x=x\left(\frac{1}{x}+2\right)\left(x^2+1\right)\)
\(1+2x=x^2+1+2x^3+2x\)
\(2x=x^2+2x^3+2x\)
\(0=x^2+2x^3\)
\(0=x^2\left(1+2x\right)\)
\(x=0;-\frac{1}{2}\)
a) \(\left(2x+1\right)\left(3x-2\right)=\left(2x+1\right)\left(5x-8\right)\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(3x-2\right)-\left(2x+1\right)\left(5x-8\right)=0\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(3x-2-5x+8\right)=0\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(6-2x\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}2x+1=0\\6-2x=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-0,5\\x=3\end{cases}}\)
Vậy...
b) \(ĐKXĐ:\) \(x\ne-2;\) \(x\ne4\)
\(\frac{3}{x+2}+\frac{2}{x-4}=0\)
\(\Leftrightarrow\)\(\frac{3\left(x-4\right)}{\left(x+2\right)\left(x-4\right)}+\frac{2\left(x+2\right)}{\left(x+2\right)\left(x-4\right)}=0\)
\(\Leftrightarrow\)\(\frac{3x-12+2x+4}{\left(x+2\right)\left(x-4\right)}=0\)
\(\Leftrightarrow\)\(\frac{5x-8}{\left(x+2\right)\left(x-4\right)}=0\)
\(\Rightarrow\)\(5x-8=0\)
\(\Leftrightarrow\)\(x=\frac{8}{5}\) (T/m đkxđ)
Vậy...
c) \(x^3+4x^2+4x+3=0\)
\(\Leftrightarrow\)\(x^3+3x^2+x^2+3x+x+3=0\)
\(\Leftrightarrow\)\(x^2\left(x+3\right)+x\left(x+3\right)+\left(x+3\right)=0\)
\(\Leftrightarrow\)\(\left(x+3\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\)\(x+3=0\) (do \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\) \(\forall x\))
\(\Leftrightarrow\)\(x=-3\)
Vậy...
\(=2x^2y-\frac{1}{2}x^2y^2-xy\left(2x-xy\right)\)
\(=xy\left(2x-\frac{1}{2}xy\right)-xy\left(2x-xy\right)\)
\(=xy\left(2x-\frac{1}{2}xy-2x+xy\right)\)
\(=xy.\frac{1}{2}xy\)
\(=\frac{1}{2}x^2y^2\)
Ta có: \(\left(x-\frac{1}{x}\right):\left(x+\frac{1}{x}\right)=n\Rightarrow\frac{x^2-1}{x}:\frac{x^2+1}{x}=n\Rightarrow\frac{x^2-1}{x^2+1}=n\)
\(\Rightarrow x^2-1=n\left(x^2+1\right)=nx^2+n\Rightarrow x^2-nx^2=n+1\Rightarrow x^2\left(1-n\right)=n+1\Rightarrow x^2=\frac{n+1}{1-n}\left(n\ne1\right)\)
THay vào V ta được: \(V=\left(\frac{n+1}{1-n}-\frac{1}{\frac{n+1}{1-n}}\right):\left(\frac{n+1}{1-n}+\frac{1}{\frac{n+1}{1-n}}\right)=\left(\frac{n+1}{1-n}-\frac{1-n}{n+1}\right):\left(\frac{n+1}{1-n}+\frac{1-n}{n+1}\right)\)
\(=\frac{\left(n+1\right)^2-\left(1-n\right)^2}{\left(n+1\right)\left(1-n\right)}:\frac{\left(n+1\right)^2+\left(1-n\right)^2}{\left(n+1\right)\left(1-n\right)}=\frac{n^2+2n+1-1+2n-n^2}{n^2+2n+1+1-2n+n^2}\)
\(=\frac{4n}{2n^2+2}=\frac{4n}{2\left(n^2+1\right)}=\frac{2n}{n^2+1}\)