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câu b đk x>= -1/4
\(x+\sqrt{x+\dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}}}=2\)
\(x+\sqrt{\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2}=2\)
\(\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2=2\)
\(x+\dfrac{1}{4}=\left(\sqrt{2}-\dfrac{1}{2}\right)^2\)
\(x=\left(\sqrt{2}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)
\(x=\left(\sqrt{2}-\dfrac{1}{2}-\dfrac{1}{2}\right)\left(\sqrt{2}-\dfrac{1}{2}+\dfrac{1}{2}\right)\)
\(x=\sqrt{2}\left(\sqrt{2}-1\right)=2-\sqrt{2}\)
\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}=\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=|\sqrt{x-1}+1|+|\sqrt{x-1}-1|=\left[{}\begin{matrix}\sqrt{x-1}+1+\sqrt{x-1}-1\left(x\ge2\right)\\\sqrt{x-1}+1+1-\sqrt{x-1}\left(1\le x< 2\right)\end{matrix}\right.=\left[{}\begin{matrix}2\sqrt{x-1}\left(x\ge2\right)\\2\left(1\le x< 2\right)\end{matrix}\right.\)
\(\hept{\begin{cases}a^2=x^2y^2+\left(1+x^2\right)\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\\b^2=y^2\left(1+x^2\right)+x^2\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\end{cases}}\)
\(\Rightarrow a^2-b^2=1\)
\(\Rightarrow a^2=1+b^2\)
a: \(=\dfrac{1}{x-y}\cdot x^2\cdot\left(x-y\right)=x^2\)
b: \(=\sqrt{27\cdot48}\cdot\left|a-2\right|=36\left(a-2\right)\)
c: \(=\left(\sqrt{2012}+\sqrt{2011}\right)^2\)
d: \(=\dfrac{8}{7}\cdot\dfrac{-x}{y+1}\)
e: \(=\dfrac{11}{12}\cdot\dfrac{x}{-y-2}=\dfrac{-11x}{12\left(y+2\right)}\)
Đặt \(N=\sqrt{x-1+2\sqrt{x-2}}+\sqrt{x-1-2\sqrt{x-2}}\)
\(\Rightarrow N^2=x-1+2\sqrt{x-2}+x-1-2\sqrt{x-2}+2\sqrt{\left(x-1+2\sqrt{x-2}\right)\left(x-1-2\sqrt{x-2}\right)}\)
\(\Leftrightarrow N^2=2x-2+2\sqrt{\left(x-1\right)^2-\left(2\sqrt{x-2}\right)^2}\)
\(\Leftrightarrow N^2=2x-2+2\sqrt{x^2-2x+1-4\left(x-2\right)}\)
\(\Leftrightarrow N^2=2x-2+2\sqrt{x^2-2x+1-4x+8}\)
\(\Leftrightarrow N^2=2x-2+2\sqrt{x^2-6x+9}\)
\(\Leftrightarrow N^2=2x-2+2\sqrt{\left(x-3\right)^2}\)
\(\Leftrightarrow N^2=2x-2+2\left|x-3\right|\)
* Với \(x\ge3\)thì \(N^2=2x-2+2\left(x-3\right)=4x-8=4\left(x-2\right)\Rightarrow N=2\sqrt{x-2}\)
* Với \(2\le x\le4\)thì \(N^2=2x-2-2\left(x-3\right)=4\Rightarrow N=\sqrt{4}=2\)
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