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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.
\(\left|x\right|-\sqrt{1-2x+x^2}\)
\(=\left|x\right|-\sqrt{\left(1-x\right)^2}\)
\(=\left|x\right|-\left|x-1\right|\)
\(=x-x+1=1\)
Bài 3: \(3\left(\sqrt{2x^2+1}-1\right)=x\left(1+3x+8\sqrt{2x^2+1}\right)\)
\(\Leftrightarrow\left(3-8x\right)\sqrt{2x^2+1}=3x^2+x+3\)
\(\Rightarrow\left(3-8x\right)^2\left(2x^2+1\right)=\left(3x^2+x+3\right)^2\)
\(\Leftrightarrow119x^4-102x^3+63x^2-54x=0\)
\(\Leftrightarrow x\left(7x-6\right)\left(17x^2+9\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{6}{7}\end{cases}}\)
Thử lại, ta nhận được \(x=0\)là nghiệm duy nhất của phương trình
\(a,\sqrt{\left(\sqrt{x}-\sqrt{y}\right)^2\left(\sqrt{x}+\sqrt{y}\right)^2}=\left|\sqrt{x}-\sqrt{y}\right|\left(\sqrt{x}+\sqrt{y}\right)\)
\(=\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{x}+\sqrt{y}\right)\)
\(=y-x\)
\(b,\frac{3-\sqrt{x}}{x-9}=\frac{3-\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=-\frac{1}{\sqrt{x}+3}\)
\(c,\frac{x-5\sqrt{x}+6}{\sqrt{x}-3}=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\sqrt{x}-2\)
\(d,6-2x-\sqrt{9-6x+x^2}=6-2x-\sqrt{\left(3-x\right)^2}=6-2x-3+x=3-x\)
\(a,\)\(\sqrt{\left(\sqrt{x}-\sqrt{y}\right)^2\left(\sqrt{x}+\sqrt{y}\right)^2}\)
\(=|\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)|\)
\(=|\sqrt{x}^2-\sqrt{y}^2|\)
\(=|x-y|\)
Vì \(x\le y\)\(\Rightarrow x-y\ge0\)
\(\Rightarrow|x-y|=x-y\)
1. Áp dụng BĐT Bunhiakovski
a) \(\sqrt{x-2}+\sqrt{4-x}=\sqrt{\left(\sqrt{x-2}.1+\sqrt{4-x}.1\right)^2}\le\sqrt{\left(1^2+1^2\right)\left(x-2+4-x\right)}=2\)
Đẳng thức xảy ra \(\Leftrightarrow\) \(\sqrt{x-2}=\sqrt{4-x}\) \(\Leftrightarrow\) \(x=3\)
b) \(\sqrt{6-x}+\sqrt{x+2}=\sqrt{\left(\sqrt{6-x}.1+\sqrt{x+2}.1\right)^2}\le\sqrt{\left(1^2+1^2\right)\left(6-x+x+2\right)}=4\)
Đẳng thức xảy ra \(\Leftrightarrow\) \(\sqrt{6-x}=\sqrt{x+2}\) \(\Leftrightarrow\) \(x=2\)
c) \(\sqrt{x}+\sqrt{2-x}=\sqrt{\left(\sqrt{x}.1+\sqrt{2-x}.1\right)^2}\le\sqrt{\left(1^2+1^2\right)\left(x+2-x\right)}=2\)
Đẳng thức xảy ra \(\Leftrightarrow\) \(\sqrt{x}=\sqrt{2-x}\) \(\Leftrightarrow\) \(x=1\)
1.Điều kiện xđ \(x\ge2,x\le4\)
Từ ĐKXĐ ta có
\(x\ge2\Leftrightarrow x-2\ge0\Leftrightarrow\sqrt{x-2}\ge0\left(1\right)\)
\(x\le4\Leftrightarrow4-x\ge0\Leftrightarrow\sqrt{4-x}\ge0\left(2\right)\)
Từ (1),(2) cộng vế theo vế ta có:
\(\sqrt{x-2}+\sqrt{4-x}\ge0+0=0\)