tính A nếu \(x\ge\sqrt{2}\)
A= \(\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x}-1}\)
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10 tháng 8 2023
a: ĐKXĐ: x^2-1>=0
=>x>=1 hoặc x<=-1
\(A=\sqrt{x^2-1+2\sqrt{x^2-1}+1}-\sqrt{x^2-1-2\sqrt{x^2-1}+1}\)
\(=\left|\sqrt{x^2-1}+1\right|-\left|\sqrt{x^2-1}-1\right|\)
x>=căn 2
=>x^2>=2
=>x^2-1>=1
=>căn x^2-1>=1
=>căn(x^2-1)-1>=0
=>\(A=\sqrt{x^2-1}+1-\sqrt{x^2+1}+1=2\)
14 tháng 6 2019
\(A=\sqrt{x^2-1+2\sqrt{x^2-1}+1}-\sqrt{x^2-1-2\sqrt{x^2-1}+1}\)
\(=\sqrt{\left(\sqrt{x^2-1}+1\right)^2}-\sqrt{\left(\sqrt{x^2-1}-1\right)^2}\)
\(=\left|\sqrt{x^2-1}+1\right|-\left|\sqrt{x^2-1}-1\right|\)
a) A có nghĩa <=> \(x^2-1\ge0\Leftrightarrow x^2\ge1\Leftrightarrow\orbr{\begin{cases}x\ge1\\x\le-1\end{cases}}\)
b) Nếu \(x\ge\sqrt{2}\)khi đó \(\sqrt{x^2-1}-1\ge\sqrt{\left(\sqrt{2}\right)^2-1}-1=0\)
Ta có: \(A=\sqrt{x^2-1}+1-\left(\sqrt{x^2-1}-1\right)=2\)
13 tháng 6 2015
a) ĐK; x>1; x<-1
b)\(A=\sqrt{x^2-1+2\sqrt{x^2-1}+1}-\sqrt{x^2-1-2\sqrt{x^2-1}+1}=\sqrt{\left(\sqrt{x^2-1}+1\right)^2}-\sqrt{\left(\sqrt{x^2-1}-1\right)^2}\)\(=\sqrt{x^2-1}+1-\left|\sqrt{x^2-1}-1\right|\)
Nếu \(x\ge\sqrt{2}\Rightarrow x^2\ge2\Leftrightarrow x^2-1\ge1\Leftrightarrow\sqrt{x^2-1}\ge1\Leftrightarrow\sqrt{x^2-1}-1\ge0\Rightarrow\left|\sqrt{x^2-1}-1\right|=\sqrt{x^2-1}-1\)
\(\Leftrightarrow A=\sqrt{x^2-1}+1-\sqrt{x^2-1}+1=2\)
Đúng nha
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 9 2020
ĐKXĐ: \(x^2-1\ge0\Leftrightarrow\left[{}\begin{matrix}x\ge1\\x\le-1\end{matrix}\right.\)
\(A=\sqrt{x^2-1+2\sqrt{x^2-1}+1}+\sqrt{x^2-1-2\sqrt{x^2-1}+1}\)
\(A=\sqrt{\left(\sqrt{x^2-1}+1\right)^2}+\sqrt{\left(\sqrt{x^2-1}-1\right)^2}\)
\(A=\sqrt{x^2-1}+1+\left|\sqrt{x^2-1}-1\right|\)
Do \(x\ge\sqrt{2}\Rightarrow\sqrt{x^2-1}-1\ge0\)
\(\Rightarrow A=\sqrt{x^2-1}+1+\sqrt{x^2-1}-1=2\sqrt{x^2-1}\)
8 tháng 10 2019
c,C= \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\left(x\ge1\right)\)
=\(\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}\)
=\(\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|\) (1)
TH1: \(\sqrt{x-1}< 1\) hay \(1\le x< 2\)
Từ (1)=>C= \(\sqrt{x-1}+1+1-\sqrt{x-1}\)=2
TH2: \(\sqrt{x-1}\ge1\) hay \(x\ge2\)
Từ (1) =>C=\(\sqrt{x-1}+1+\sqrt{x-1}-1\)=\(2\sqrt{x-1}\)
d, D=\(\sqrt{13+30\sqrt{2}+\sqrt{9+4\sqrt{2}}}=\sqrt{13+30\sqrt{2}+\sqrt{8+2\sqrt{8}+1}}=\sqrt{13+30\sqrt{2}+\sqrt{\left(\sqrt{8}+1\right)^2}}\)
=\(\sqrt{13+30\sqrt{2}+\sqrt{8}+1}=\sqrt{14+30\sqrt{2}+2\sqrt{2}}\)
=\(\sqrt{14+32\sqrt{2}}\)
8 tháng 10 2019
a)\(\frac{x-y}{\sqrt{x}-\sqrt{y}}=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}=\sqrt{x}+\sqrt{y}\)
b)\(\frac{x-2\sqrt{x}+1}{\sqrt{x}-1}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}=\sqrt{x}-1\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 6 2021
ĐKXĐ: \(-1\le x\le1\)
Đặt \(\left\{{}\begin{matrix}\sqrt{1-x}=a\\\sqrt{1+x}=b\end{matrix}\right.\) \(\Rightarrow a^2+b^2=2\) ta được:
\(A=\dfrac{\sqrt{1-ab}\left(a^3+b^3\right)}{2-ab}=\dfrac{\sqrt{\dfrac{a^2+b^2}{2}-ab}\left(a+b\right)\left(a^2+b^2-ab\right)}{a^2+b^2-ab}\)
\(=\sqrt{\dfrac{a^2+b^2-2ab}{2}}\left(a+b\right)=\dfrac{\left|a-b\right|\left(a+b\right)}{\sqrt{2}}\)
\(=\dfrac{\left|\sqrt{1-x}-\sqrt{1+x}\right|\left(\sqrt{1-x}+\sqrt{1+x}\right)}{\sqrt{2}}\)
- Với \(-1\le x\le0\Rightarrow A=\dfrac{\left(\sqrt{1-x}-\sqrt{1+x}\right)\left(\sqrt{1-x}+\sqrt{1+x}\right)}{\sqrt{2}}=-\sqrt{2}x\)
- Với \(0\le x\le1\Rightarrow A=\dfrac{\left(\sqrt{1+x}-\sqrt{1-x}\right)\left(\sqrt{1+x}+\sqrt{1-x}\right)}{\sqrt{2}}=\sqrt{2}x\)
b.
TH1: \(\left\{{}\begin{matrix}-1\le x\le0\\-\sqrt{2}x\ge\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow-1\le x\le-\dfrac{1}{2\sqrt{2}}\)
TH2: \(\left\{{}\begin{matrix}0\le x\le1\\\sqrt{2}x\ge\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow\dfrac{1}{2\sqrt{x}}\le x\le1\)
9 tháng 6 2017
a, ĐKXĐ: \(\left\{{}\begin{matrix}x\le-1\\x\ge1\end{matrix}\right.\)
\(A=\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}\)
\(A=\sqrt{\left(\sqrt{x^2-1}+1\right)^2}-\sqrt{\left(\sqrt{x^2-1}-1\right)^2}\)
\(A=\sqrt{x^2-1}+1-\left|\sqrt{x^2-1}-1\right|\)
nếu \(\left\{{}\begin{matrix}x\le-\sqrt{2}\\x\ge\sqrt{2}\end{matrix}\right.\) thì \(A=\sqrt{x^2-1}+1-\sqrt{x^2-1}+1=2\) (1)
nếu \(-\sqrt{2}< x< \sqrt{2}\) thì \(A=2\sqrt{x^2-1}\)
vì \(x\ge\sqrt{2}\) thuộc khoảng (1) nên \(A=2\)
Sửa đề:
\(A=\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}\)
\(A=\sqrt{x^2-1+2\sqrt{x^2-1}+1}-\sqrt{x^2-1-2\sqrt{x^2-1}+1}\)
\(A=\sqrt{\sqrt{x^2-1}^2+2\sqrt{x^2-1}+1}-\sqrt{\sqrt{x^2-1}^2-2\sqrt{x^2-1}+1}\)
\(A=\sqrt{\left(\sqrt{x^2-1}+1\right)^2}-\sqrt{\left(\sqrt{x^2-1}-1\right)^2}\)
\(A=\left|\sqrt{x^2-1}+1\right|-\left|\sqrt{x^2-1}-1\right|\)
\(A=\sqrt{x^2-1}+1-\sqrt{x^2-1}+1\)
\(A=2\)