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1/ \(\sqrt{2x-1+2\sqrt{2x-1}+1}+\sqrt{2x-1-2\sqrt{2x-1}+1}\)
\(=\sqrt{\left(\sqrt{2x-1}+1\right)^2}+\sqrt{\left(\sqrt{2x-1}-1\right)^2}\)
\(=\left|\sqrt{2x-1}+1\right|+\left|\sqrt{2x-1}-1\right|\)
\(=\sqrt{2x-1}+1+1-\sqrt{2x-1}\)
\(=2\)
2/ ĐKXĐ: \(a^2-1\ge0\Rightarrow a^2\ge1\Rightarrow\left[{}\begin{matrix}a\ge1\\a\le-1\end{matrix}\right.\)
3/ \(4\left|x\right|-\sqrt{\left(5x-1\right)^2}=4\left|x\right|-\left|5x-1\right|\)
\(=4x-\left(5x-1\right)=1-x\)
4/ \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}< \sqrt{7}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge0\\x< 7\end{matrix}\right.\) \(\Rightarrow0\le x< 7\)
5/ \(M=\sqrt{3-2\sqrt{2.3}+2}=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\left|\sqrt{3}-\sqrt{2}\right|=\sqrt{3}-\sqrt{2}\)
6/ \(\left|x\right|-\sqrt{\left(x-1\right)^2}=\left|x\right|-\left|x-1\right|=x-\left(x-1\right)=1\)
1.
\(\sqrt{2x+2\sqrt{2x-1}}+\sqrt{2x-2\sqrt{2x-1}}\)
\(=\sqrt{2x-1+2\sqrt{2x-1}+1}+\sqrt{2x-1-2\sqrt{2x-1}+1}\)
\(=\sqrt{\left(\sqrt{2x-1}+1\right)^2}+\sqrt{\left(\sqrt{2x-1}-1\right)^2}\)
\(=\left|\sqrt{2x-1}+1\right|+\left|\sqrt{2x-1}-1\right|\)
\(=\sqrt{2x-1}+1+1-\sqrt{2x-1}=2\)
2.
\(\sqrt{a^2-1}\text{ xác định }\Leftrightarrow a^2-1\ge0\)
\(\Leftrightarrow\left(a-1\right)\left(a+1\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a-1\ge0\\a+1\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}a-1\le0\\a+1\le0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a\ge1\\a\le-1\end{matrix}\right.\)
3.
\(4\left|x\right|-\sqrt{1+25x^2-10x}\)
\(=4\left|x\right|-\sqrt{\left(5x-1\right)^2}\)
\(=4\left|x\right|-\left|5x-1\right|\)
\(=4x-5x+1=1-x\)
4.
ĐKXĐ: \(x\ge0\)
\(-\sqrt{x}>-\sqrt{7}\)
\(\Leftrightarrow\sqrt{x}< \sqrt{7}\)
\(\Leftrightarrow\text{ }x< 7\)
Vậy bât phương trình có nghiệm \(0\le x< 7\)
5.
\(\sqrt{5-2\sqrt{6}}=\sqrt{2-2\sqrt{2}.\sqrt{3}+3}\)
\(=\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}\)
\(=\sqrt{3}-\sqrt{2}\)
6.
\(\left|x\right|-\sqrt{1-2x+x^2}\)
\(=\left|x\right|-\sqrt{\left(1-x\right)^2}\)
\(=\left|x\right|-\left|x-1\right|\)
\(=x-x+1=1\)
ĐKXĐ: ...
\(P=\left(\frac{\sqrt{x}}{\sqrt{x}\left(1-\sqrt{x}\right)}-\frac{1-\sqrt{x}}{\sqrt{x}\left(1-\sqrt{x}\right)}\right):\left(\frac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(1-\sqrt{x}\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right)\)
\(=\frac{\left(2\sqrt{x}-1\right)}{\sqrt{x}\left(1-\sqrt{x}\right)}:\left(\frac{2\sqrt{x}-1}{1-\sqrt{x}}+\frac{\sqrt{x}\left(2\sqrt{x}-1\right)}{x-\sqrt{x}+1}\right)\)
\(=\frac{\left(2\sqrt{x}-1\right)}{\sqrt{x}\left(1-\sqrt{x}\right)}:\left(\frac{\left(2\sqrt{x}-1\right)\left(x-\sqrt{x}+1+\sqrt{x}-x\right)}{\left(1-\sqrt{x}\right)\left(x-\sqrt{x}+1\right)}\right)\)
\(=\frac{\left(2\sqrt{x}-1\right)}{\sqrt{x}\left(1-\sqrt{x}\right)}:\left(\frac{2\sqrt{x}-1}{\left(1-\sqrt{x}\right)\left(x-\sqrt{x}+1\right)}\right)=\frac{\left(2\sqrt{x}-1\right)}{\sqrt{x}\left(1-\sqrt{x}\right)}.\frac{\left(1-\sqrt{x}\right)\left(x-\sqrt{x}+1\right)}{\left(2\sqrt{x}-1\right)}=\frac{x-\sqrt{x}+1}{\sqrt{x}}\)
\(x=7-4\sqrt{3}=\left(2-\sqrt{3}\right)^2\Rightarrow\sqrt{x}=2-\sqrt{3}\)
\(\Rightarrow P=\frac{7-4\sqrt{3}-2+\sqrt{3}+1}{2-\sqrt{3}}=\frac{6-3\sqrt{3}}{2-\sqrt{3}}=3\)
Câu c hơi nghi ngờ cái đề, cấp 2 làm sao giải được BPT bậc 3 kiểu này?
a) Ta có: \(P=\left(\frac{2x}{x\sqrt{x}+\sqrt{x}-x-1}-\frac{1}{\sqrt{x}-1}\right):\left(1+\frac{\sqrt{x}}{x+1}\right)\)
\(=\left(\frac{2x}{\left(x+1\right)\left(\sqrt{x}-1\right)}-\frac{x+1}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right):\left(\frac{x+1}{x+1}+\frac{\sqrt{x}}{x+1}\right)\)
\(=\frac{x-1}{\left(x+1\right)\left(\sqrt{x}-1\right)}\cdot\frac{x+1}{x+1+\sqrt{x}}\)
\(=\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
b) Ta có: \(x=\frac{1}{\sqrt{5}-2}-\frac{1}{\sqrt{5}+2}\)
\(=\frac{\sqrt{5}+2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}-\frac{\sqrt{5}-2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\)
\(=\frac{\sqrt{5}+2-\sqrt{5}+2}{\left(\sqrt{5}\right)^2-2^2}\)
\(=\frac{4}{5-4}=\frac{4}{1}=4\)
Thay x=4 vào biểu thức \(P=\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\), ta được:
\(P=\frac{\sqrt{4}+1}{4+\sqrt{4}+1}=\frac{2+1}{4+2+1}=\frac{3}{7}\)
Vậy: \(\frac{3}{7}\) là giá trị của biểu thức \(P=\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\) tại \(x=\frac{1}{\sqrt{5}-2}-\frac{1}{\sqrt{5}+2}\)
\(a,M=\left(\frac{\sqrt{x}+1}{\sqrt{2x}+1}+\frac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}-1\right):\left(1+\frac{\sqrt{x}+1}{\sqrt{2x}+1}-\frac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}\right)\)
\(=\left(\frac{2x-2\sqrt{2}x+2\sqrt{2x}-1}{2x-1}-1\right):\left(1+\frac{\sqrt{x}+1}{\sqrt{2x+1}}-\frac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}\right)\)
\(=\left(\frac{-2\sqrt{2}x+2\sqrt{2x}}{2x-1}\right):\left(1+\frac{x\sqrt{2}-\sqrt{x}+\sqrt{2x}-1-\left(2x+\sqrt{2x}+x\sqrt{2}+\sqrt{x}\right)}{2x-1}\right)\)
\(=\left(\frac{-2\sqrt{2}x+2\sqrt{2x}}{2x-1}\right):\left(\frac{-2\sqrt{x}-2}{2x-1}\right)\)
\(=\frac{-\sqrt{2}x+\sqrt{2x}}{\sqrt{x}-1}\)
\(=\frac{-\sqrt{2x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(=-\sqrt{2x}\)
\(b,x=\frac{1}{2}\left(3+2\sqrt{2}\right)\)
\(x=\frac{1}{2}\left(1+2\sqrt{2}+2\right)\)
\(x=\frac{1}{2}\left(1+\sqrt{2}\right)^2\)
Thay \(x=\frac{1}{2}\left(1+\sqrt{2}\right)^2\) vào \(M=-\sqrt{2x}\) ta được:
\(M=-\sqrt{2.\frac{1}{2}\left(1+\sqrt{2}\right)^2}\)
\(M=-1-\sqrt{2}\)
Vậy ..............