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a)\(đkx\ge1,x\ne-1\)
\(\sqrt{\dfrac{x-1}{x+1}}=2\)
\(\Leftrightarrow\dfrac{x-1}{x+1}=4\)
\(\Leftrightarrow x-1=4x-4\)
\(\Leftrightarrow x=1\)(nhận)
Vậy S=\(\left\{1\right\}\)
c)đk\(25x^2-10x+1=\) \(\left(5x-1\right)^2\ge0\Leftrightarrow x\ge\dfrac{1}{5}\)
\(\sqrt{25x^2-10x+1}+2x=1\)
\(\Leftrightarrow\sqrt{\left(5x-1\right)^2}+2x=1\)
\(\Leftrightarrow5x-1+2x=1\)
\(\Leftrightarrow x=\dfrac{2}{7}\)(nhận)
Vậy S=\(\left\{\dfrac{2}{7}\right\}\)
c: Ta có: \(\sqrt{25x^2-10x+1}+2x=1\)
\(\Leftrightarrow\left|5x-1\right|=1-2x\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=1-2x\left(x\ge\dfrac{1}{5}\right)\\5x-1=2x-1\left(x< \dfrac{1}{5}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{7}\left(nhận\right)\\x=0\left(nhận\right)\end{matrix}\right.\)
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b: \(BC=\sqrt{89}\left(cm\right)\)
\(\sin\widehat{B}=\dfrac{5\sqrt{89}}{89}\)
\(\Leftrightarrow\widehat{B}\simeq32^0\)
\(\widehat{C}=58^0\)
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\(n=\sqrt{2}\left(\sqrt{3}+1\right)\sqrt{2-\sqrt{3}}\\ n=\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}\\ n=\left(\sqrt{3}+1\right)\sqrt{\left(\sqrt{3}-1\right)^2}\\ n=\left(\sqrt{3}+1\right)\left|\sqrt{3}-1\right|\\ n=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\\ n=3-1=2\)
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a) Ta có: \(\sqrt{3x-2}>5\)
nên 3x-2>25
\(\Leftrightarrow3x>27\)
hay x>9
b) Ta có: \(\sqrt{2x-7}< 9\)
\(\Leftrightarrow2x-7< 81\)
\(\Leftrightarrow2x< 88\)
hay x<44
Kết hợp ĐKXĐ, ta được: \(\dfrac{7}{2}\le x< 44\)