Giải phương trình sau: \(\sqrt[3]{\frac{1}{2}+x}+\sqrt{\frac{1}{2}-x}=1\) .
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ĐKXĐ: \(-1\le x\le1\)
Xét \(\sqrt{\left(1+x\right)^3}-\sqrt{\left(1-x\right)^3}=\left(\sqrt{1+x}-\sqrt{1-x}\right)\left[\left(1+x\right)+\left(1-x\right)+\sqrt{\left(1+x\right)\left(1-x\right)}\right]\)
\(=\left(\sqrt{1+x}-\sqrt{1-x}\right)\left(2+\sqrt{1-x^2}\right)\)
Khi đó phương trình đề trở thành:
\(\sqrt{1+\sqrt{1-x}}\left(\sqrt{1+x}-\sqrt{1-x}\right)\left(2+\sqrt{1-x^2}\right)=\frac{2+\sqrt{1-x^2}}{3}\)
Vì \(2+\sqrt{1-x^2}>0\)nên ta có thể chia 2 vế cho \(2+\sqrt{1-x^2}\):
\(\Rightarrow\sqrt{1+\sqrt{1-x^2}}\left(\sqrt{1+x}-\sqrt{1-x}\right)=\frac{1}{\sqrt{3}}\),Bình phương 2 vế:
\(\Rightarrow\left(1+\sqrt{1-x^2}\right)\left[\left(1+x\right)+\left(1-x\right)-2\sqrt{\left(1+x\right)\left(1-x\right)}\right]=\frac{1}{3}\)
\(\Leftrightarrow\left(1+\sqrt{1-x^2}\right)\left(2-2\sqrt{1-x^2}\right)=\frac{1}{3}\Leftrightarrow2\left(1+\sqrt{1-x^2}\right)\left(1-\sqrt{1-x^2}\right)=\frac{1}{3}\)\(\Leftrightarrow1-\left(1-x^2\right)=\frac{1}{3}\Leftrightarrow x^2=\frac{1}{6}\Leftrightarrow x=\pm\frac{1}{\sqrt{6}}\)
Ta xét phương trình đề: vế phải luôn không âm vì vậy vế trái phải không âm
Khi đó \(\sqrt{\left(1+x\right)^3}-\sqrt{\left(1-x\right)^3}\ge0\Leftrightarrow1+x\ge1-x\Leftrightarrow x\ge0\)
Vậy ta chỉ nhận nghiệm duy nhất là \(x=\frac{1}{\sqrt{6}}\)
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\(\frac{1}{\sqrt{x+1}+\sqrt{x+2}}+\frac{1}{\sqrt{x+2}+\sqrt{x+3}}+...+\frac{1}{\sqrt{x+2019}+\sqrt{x+2020}}=11\)
\(\Leftrightarrow\)\(\frac{\sqrt{x+2}-\sqrt{x+1}}{\left(\sqrt{x+1}+\sqrt{x+2}\right)\left(\sqrt{x+2}-\sqrt{x+1}\right)}+\frac{\sqrt{x+3}-\sqrt{x+2}}{\left(\sqrt{x+2}+\sqrt{x+3}\right)\left(\sqrt{x+3}-\sqrt{x+2}\right)}\)
\(+...+\frac{\sqrt{x+2020}-\sqrt{x+2019}}{\left(\sqrt{x+2019}+\sqrt{x+2020}\right)\left(\sqrt{x+2020}-\sqrt{x+2019}\right)}=11\)
\(\Leftrightarrow\)\(\frac{\sqrt{x+2}-\sqrt{x+1}}{x+2-x-1}+\frac{\sqrt{x+3}-\sqrt{x+2}}{x+3-x-2}+...+\frac{\sqrt{x+2020}-\sqrt{x+2019}}{x+2020-x-2019}=11\)
\(\Leftrightarrow\)\(\sqrt{x+2}-\sqrt{x+1}+\sqrt{x+3}-\sqrt{x+2}+...+\sqrt{x+2020}-\sqrt{x+2019}=11\)
\(\Leftrightarrow\)\(\sqrt{x+2020}-\sqrt{x+1}=11\)
\(\Leftrightarrow\)\(\sqrt{x+2020}=11+\sqrt{x+1}\)
\(\Leftrightarrow\)\(x+2020=121+22\sqrt{x+1}+x+1\)
\(\Leftrightarrow\)\(22\sqrt{x+1}=1898\)
\(\Leftrightarrow\)\(\sqrt{x+1}=\frac{949}{11}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+1=\frac{900601}{121}\\x+1=\frac{-900601}{121}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{900480}{121}\\x=\frac{-900722}{121}\end{cases}}\)
Chúc bạn học tốt ~
PS : sai thì thui nhá
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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)
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ĐKXĐ: \(x\ge1\)
Ta có:
\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\dfrac{x+3}{2}\\ \Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=\dfrac{x+3}{2}\\ \Leftrightarrow\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=\dfrac{x+3}{2}\\ \Leftrightarrow\sqrt{x-1}+\left|\sqrt{x-1}-1\right|=\dfrac{x+1}{2}\left(1\right)\)
Ta xét 2 trường hợp sau:
TH1: \(x\ge2\)
Khi đó:
\(\left(1\right)\Leftrightarrow2\sqrt{x-1}-1=\dfrac{x+1}{2}\\ \Leftrightarrow2\sqrt{x-1}=\dfrac{x+3}{2}\\ \Leftrightarrow16\left(x-1\right)=x^2+6x+9\\ \Leftrightarrow x^2-10x+25=0\\ \Leftrightarrow\left(x-5\right)^2=0\\ \Leftrightarrow x=5\left(TMĐK\right)\)
TH2: \(1\le x< 2\)
Khi đó:
\(\left(1\right)\Leftrightarrow1=\dfrac{x+1}{2}\Leftrightarrow x=1\left(TMĐK\right)\)
Vậy x=1 hoặc x=5