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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.
\(\left|x\right|-\sqrt{1-2x+x^2}\)
\(=\left|x\right|-\sqrt{\left(1-x\right)^2}\)
\(=\left|x\right|-\left|x-1\right|\)
\(=x-x+1=1\)
P=\(\sqrt{\frac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(x\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1}\)
=\(\sqrt{\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1}\)
=\(\sqrt{x-\sqrt{x}-x-\sqrt{x}+x+1}\)
=\(\sqrt{x-2\sqrt{x}+1}\)
=\(\sqrt{\left(\sqrt{x}-1\right)^2}\)
=\(\sqrt{x}-1\)
\(\hept{\begin{cases}-1\le x\le1\\2-\sqrt{1-x^2}\end{cases}\Rightarrow-1\le x\le1\left(^∗\right)}\)
Đặt : \(\hept{\begin{cases}\sqrt{1+x}=a\\\sqrt{1-x}=b\end{cases}\Rightarrow\hept{\begin{cases}a^2+b^2=2\\a,b\ge0\end{cases}}}\)
A tồn tại mọi x thuộc ( * )
\(A=\frac{\sqrt{1-ab}\left(a^3+b^3\right)}{2-ab}=\frac{\sqrt{a^2-2ab+b^2}\left(a+b\right)\left(a^2+b^2-ab\right)}{2-ab}\)
\(A=\frac{\sqrt{2}\sqrt{\left(a-b\right)^2}\left(a+b\right)\left(2-ab\right)}{\left(2-ab\right)}\) . Vói đk ( \(I\)) \(A=\frac{\sqrt{2}}{2}!a-b!\left(a+b\right)\)
\(\orbr{\begin{cases}\hept{\begin{cases}a\ge b\Leftrightarrow0\le x\le1\\A=\frac{\sqrt{2}}{2}\left[\left(1+x\right)-\left(1-x\right)\right]=\frac{\sqrt{2}}{2}x\end{cases}}\\\hept{\begin{cases}a< b\Leftrightarrow-1\le x< 0\\A=\frac{-\sqrt{2}}{2}\left[\left(1+x\right)-\left(1-x\right)\right]=\frac{-\sqrt{2}}{2}x\end{cases}}\end{cases}}\)
\(\Rightarrow A=\frac{\sqrt{2}}{2}!x!\) . Với x thỏa mãn điều kiện ( * )
Với điều kiện \(0\)\(\le x\le1\)ta có
P= \(\sqrt{\frac{\sqrt{x\left(x\sqrt{x-1}\right)}}{x+\sqrt{x}+1}-\sqrt{\frac{\sqrt{x\left(x\sqrt{x+1}\right)}}{x-\sqrt{x}+1}+x+1}}\)
sử dụng hằng đẳng thức bậc 3 : \(x^3\)- \(y^3\)và \(x^3\)+ \(y^3\)
ta có P = \(\sqrt{\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)+x+1}\)= \(\sqrt{x-2\sqrt{x}+1}=\)\(\sqrt{\left(\sqrt{x}-1\right)^2}\)=\(\left|\sqrt{x}-1\right|\)
P= \(1-\sqrt{x}\)
\(\frac{A}{\sqrt{2}}\)=\(\frac{\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}}{\sqrt{2x-1+2\sqrt{2x-1}+1}-\sqrt{2x-1-2\sqrt{2x-1}+1}}\) (DK \(x\ge1\)
\(=\frac{\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|}{\sqrt{2x-1}+1-\left|\sqrt{2x-1}-1\right|}\)
vs \(x\ge2\) \(\frac{\sqrt{x-1}+1+\sqrt{x-1}-1}{\sqrt{2x-1}+1-\sqrt{2x-1}+1}=\frac{2\sqrt{x-1}}{2}=\sqrt{x-1}\) \(\Rightarrow A=\sqrt{2x-2}\)
vs \(1\le x< 2\) \(\frac{\sqrt{x-1}+1+1-\sqrt{x-1}}{\sqrt{2x-1}+1-1+\sqrt{2x-1}}=\frac{1}{\sqrt{2x-1}}\) \(\Rightarrow A=\frac{\sqrt{2}}{\sqrt{2x-1}}\)
\(\sqrt{2X-1}\ge1\Leftrightarrow X\ge1\)NEN SUY RA THEO CACH LAM CUA TO
THOI U AM BUSY SEE YOU AGAIN
Ta có: \(R=\sqrt{x^2-2x+1}+\sqrt{x^2+2x+1}\)
\(=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+1\right)^2}\)
\(=\left|x-1\right|+\left|x+1\right|\)
Ta có: \(-1\le x\le1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\le1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+1\ge0\\x-1\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left|x+1\right|=x+1\\\left|x-1\right|=1-x\end{matrix}\right.\)
\(\Leftrightarrow R=x+1+1-x=2\)