\(\sqrt{2x^2+4x+6}-\sqrt{x^4-2x^2+2}=2-2x-x^2\)

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30 tháng 7 2019

Đề câu c ptrinh = 4 là phải riêng ra chứ

\(a,\frac{3x+2}{\sqrt{x+2}}=2\sqrt{x+2}\)

\(\Rightarrow3x+2=2\sqrt{x+2}.\sqrt{x+2}\)

\(\Rightarrow3x+2=2\left(x+2\right)\)

\(\Rightarrow3x+2=2x+4\)

\(\Rightarrow3x-2x=4-2\)

\(\Rightarrow x=2\)

\(b,\sqrt{4x^2-1}-2\sqrt{2x+1}=0\)

\(\Rightarrow\sqrt{\left(2x+1\right)\left(2x-1\right)}-2\sqrt{2x+1}=0\)

\(\Rightarrow\sqrt{2x+1}\left(\sqrt{2x-1}-2\right)=0\)

\(\Rightarrow\hept{\begin{cases}\sqrt{2x+1}=0\\\sqrt{2x-1}-2=0\end{cases}\Rightarrow\orbr{\begin{cases}2x+1=0\\\sqrt{2x-1}=2\end{cases}\Rightarrow}\orbr{\begin{cases}2x=-1\\2x-1=4\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{2}\\2x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{5}{2}\end{cases}}}\)

\(c,\sqrt{x-2}+\sqrt{4x-8}-\frac{2}{5}\sqrt{\frac{25x-50}{4}}=4\)

\(\Rightarrow\sqrt{x-2}+\sqrt{4\left(x-2\right)}-\frac{2}{5}\sqrt{\frac{25\left(x-2\right)}{4}}=4\)

\(\Rightarrow\sqrt{x-2}+2\sqrt{x-2}-\frac{2}{5}.\frac{5\sqrt{x-2}}{2}=4\)

\(\Rightarrow\sqrt{x-2}+2\sqrt{x-2}-\sqrt{x-2}=4\)

\(\Rightarrow2\sqrt{x-2}=4\)

\(\Rightarrow\sqrt{x-2}=2\)

\(\Rightarrow x-2=4\)

\(\Rightarrow x=6\)

\(d,\sqrt{x+4}-\sqrt{1-x}=\sqrt{1-2x}\)

\(\Rightarrow\sqrt{x+4}=\sqrt{1-2x}+\sqrt{1-x}\)

\(\Rightarrow x+4=1-2x+2\sqrt{\left(1-2x\right)\left(1-x\right)}+1-x\)

\(\Rightarrow x+4=2-3x+2\sqrt{1-3x+2x^2}\)

\(\Rightarrow x+4-2+3x=2\sqrt{1-3x+2x^2}\)

\(\Rightarrow4x+2=2\sqrt{1-3x+2x^2}\)

\(\Rightarrow2x+1=\sqrt{1-3x+2x^2}\)

\(\Rightarrow4x^2+4x+1=1-3x+2x^2\)

\(\Rightarrow4x^2-2x^2+4x+3x+1-1=0\)

\(\Rightarrow2x^2+7x=0\)

\(\Rightarrow x\left(2x+7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\2x+7=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{-7}{2}\end{cases}}}\)

\(e,\frac{2x}{\sqrt{5}-\sqrt{3}}-\frac{2x}{\sqrt{3}+1}=\sqrt{5}+1\)

\(\frac{2x\left(\sqrt{5}+\sqrt{3}\right)}{5-3}-\frac{2x\left(\sqrt{3}-1\right)}{3-1}=\sqrt{5}+1\)

\(\Rightarrow x\left(\sqrt{5}+\sqrt{3}\right)-x\left(\sqrt{3}-1\right)=\sqrt{5}+1\)

\(\Rightarrow\sqrt{5}x+\sqrt{3}x-\sqrt{3x}+x=\sqrt{5}+1\)

\(\Rightarrow\sqrt{5}x+x=\sqrt{5}+1\)

\(\Rightarrow x\left(\sqrt{5}+1\right)=\sqrt{5}+1\)

\(\Rightarrow x=1\)

20 tháng 7 2018

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20 tháng 7 2018

cho đúng nha