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\(S=\frac{-1+\sqrt{2}}{2-1}+\frac{-\sqrt{2}+\sqrt{3}}{3-2}+...+\frac{-\sqrt{99}+\sqrt{100}}{100-99}\)
\(=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-....-\sqrt{99}+\sqrt{100}\)
\(=-1+\sqrt{100}\)
\(\hept{\begin{cases}a=\left(x^2-x+1\right)^2\\b=x^2\end{cases}}\)
\(a^2-\left(b+1\right)a+b=0\Leftrightarrow\left(a-1\right)\left(a-b\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=1\\a=b\end{cases}\Leftrightarrow}\orbr{\begin{cases}\left(x^2-x+1\right)^2=1\\\left(x^2-x+1\right)^2=x^2\end{cases}}\)(easy)
a) \(x+1=\sqrt{2\left(x+1\right)+2\sqrt{2\left(x+1\right)+2\sqrt{4\left(x+1\right)}}}\)
<=> \(\left(x+1\right)^2=\left[\sqrt{2\left(x+1\right)+2\sqrt{2\left(x+1\right)+2\sqrt{4\left(x+1\right)}}}\right]^2\)
<=> \(x^2+2x+1=2x+2+2\sqrt{2x+2+4\sqrt{x+1}}\)
<=> \(x^2+1=2x+2+2\sqrt{2x+2+4\sqrt{x+1}}-2x\)
<=> \(x^2+1=2\sqrt{2x+2+4\sqrt{x+1}}+2\)
<=> \(x^2+1-2=2\sqrt{2x+2+4\sqrt{x+1}}\)
<=> \(x^2-1=2\sqrt{2x+2+4\sqrt{x+1}}\)
<=> \(\left(x^2-1\right)^2=\left(2\sqrt{2x+2+4\sqrt{x+1}}\right)^2\)
<=> \(x^4-2x^2+1=8x+8+16\sqrt{x+1}\)
<=> \(x^4-2x^2+1-8x=16\sqrt{x+1}+8\)
<=> \(x^4-2x^2-8x-7=16\sqrt{x+1}\)
<=> \(\left(x^4-2x^2-8x-7\right)^2=\left(16\sqrt{x+1}\right)^2\)
<=> \(x^8-4x^6-16x^5-10x^4+32x^3+92x^2+112x+49=256x+256\)
<=> \(x^8-4x^6-16x^5-10x^4+32x^3+92x^2+112x-144x-207=0\)
<=> \(\left(x+1\right)\left(x-2\right)\left(x^6+2x^5+3x^4-4x^3-9x^2+2x+69\right)=0\)
<=> \(\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)
Vì: \(x^6+2x^5+3x^4-4x^3-9x^2+2x+69\ne0\)
=> \(\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)
\(\Rightarrow\sqrt{x^2-\frac{1}{4}+\sqrt{\left(x+\frac{1}{2}\right)^2}}=\frac{2x^3}{2}+\frac{x^2}{2}+\frac{2x}{2}+\frac{1}{2}\)
\(\Rightarrow\sqrt{x^2+x+\frac{1}{2}-\frac{1}{4}}=\sqrt{x^2+x+\frac{1}{4}}=x^3+\frac{x^2}{2}+x+\frac{1}{2}\)
\(\Rightarrow\sqrt{\left(x+\frac{1}{2}\right)^2}=x+\frac{1}{2}=x^3+\frac{x^2}{2}+x+\frac{1}{2}\)
\(\Rightarrow x^3+\frac{x^2}{2}+x+\frac{1}{2}-x-\frac{1}{2}=x^3+\frac{x^2}{2}=0\Rightarrow\frac{2x^3+x^2}{2}=0\)
\(\Rightarrow2x^3+x^2=0\Rightarrow x^2\left(2x+1\right)=0\Rightarrow\hept{\begin{cases}x^2=0\Rightarrow x=0\\2x+1=0\Rightarrow2x=-1\Rightarrow x=-\frac{1}{2}\end{cases}}\)
vậy x=0 và x=-1/2
Đặt \(x\sqrt{1-x^2}=a\)
\(PT\Leftrightarrow\frac{1-a^2}{a^2}+\frac{5}{2a}+2=0\)
\(\Leftrightarrow2x^2+5x+2=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=-2\\a=-\frac{1}{2}\end{cases}}\)
Tới đây thì đơn giản rồi bạn làm tiếp nhé
\(pt\Leftrightarrow\left(\frac{4}{x^2}+\frac{x^2}{4-x^2}\right)+\frac{5}{2}\left(\frac{\sqrt{4-x^2}}{x}+\frac{x}{\sqrt{4-x^2}}\right)+2=0\)
\(\Leftrightarrow\left(\frac{\sqrt{4-x^2}}{x}+\frac{x}{\sqrt{4-x^2}}\right)^2-1+\frac{5}{2}\left(\frac{\sqrt{4-x^2}}{x}+\frac{x}{\sqrt{4-x^2}}\right)+2=0\)
Đặt \(\frac{\sqrt{4-x^2}}{x}+\frac{x}{\sqrt{4-x^2}}=t\)pt thành
\(t^2-1+\frac{5}{2}t+2=0\)\(\Rightarrow\orbr{\begin{cases}t=-2\\t=-\frac{1}{2}\end{cases}}\)(loại)
-->PT vô nghiệm
\(\frac{3}{x^2}\)hay\(\frac{4}{x^2}\)