Cho A=\(\frac{\sqrt{1-\sqrt{1-x^2}}[\sqrt{\left(1+x\right)^3}+\sqrt{\left(1-x\right)^3}]}{2-\sqrt{1-x^2}}\)
a) Rút gọn A
b) Tìm x biết \(A\ge\frac{1}{2}\)
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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: \(\left\{{}\begin{matrix}0\le x\le1\\\sqrt{2}x\ge\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow\dfrac{1}{2\sqrt{x}}\le x\le1\)
5 tháng 2 2022
a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 3 2022
Đặt \(\left\{{}\begin{matrix}\sqrt{1+x}=a\\\sqrt{1-x}=b\end{matrix}\right.\) \(\Rightarrow2=a^2+b^2\)
\(A=\dfrac{\sqrt{1-ab}\left(a^3+b^3\right)}{a^2+b^2-ab}=\dfrac{\sqrt{\dfrac{2}{2}-ab}\left(a+b\right)\left(a^2+b^2-ab\right)}{a^2+b^2-ab}\)
\(=\sqrt{\dfrac{a^2+b^2}{2}-ab}\left(a+b\right)=\left(a+b\right)\sqrt{\dfrac{\left(a-b\right)^2}{2}}=\dfrac{\left|a-b\right|\left(a+b\right)}{\sqrt{2}}\)
\(=\pm\dfrac{a^2-b^2}{\sqrt{2}}=\pm\dfrac{2x}{\sqrt{2}}=\pm\sqrt{2}x\)
b.
\(A\ge\dfrac{1}{2}\Rightarrow\left[{}\begin{matrix}\sqrt{2}x\ge\dfrac{1}{2}\left(x\ge0\right)\\-\sqrt{2}x\ge\dfrac{1}{2}\left(x\le0\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x\ge\dfrac{\sqrt{2}}{4}\\x\le-\dfrac{\sqrt{2}}{4}\end{matrix}\right.\)
Kết hợp ĐKXĐ \(\Rightarrow\left[{}\begin{matrix}\dfrac{\sqrt{2}}{4}\le x\le1\\-1\le x\le-\dfrac{\sqrt{2}}{4}\end{matrix}\right.\)
KT
11 tháng 7 2018
Bài 1:
a) \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{2}-\frac{2\left(\sqrt{3}-1\right)}{2}\)
\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)
b) \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)
\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)
c) ĐK: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)
\(=1-a+a=1\)
a, A xác định khi : \(-1\le x\le1\)
\(=\frac{\sqrt{\frac{\left(\sqrt{1-x}-\sqrt{1+x}\right)^2}{2}}.\left[\left(\sqrt{1+x}+\sqrt{1-x}\right)\left(2-\sqrt{1-x^2}\right)\right]}{2-\sqrt{1-x^2}}\)
\(=\frac{\left|\sqrt{1-x}-\sqrt{1+x}\right|.\left(\sqrt{1+x}+\sqrt{1-x}\right)}{\sqrt{2}}=\hept{\begin{cases}\sqrt{2x}khi0\le x\le1\\-\sqrt{2x}khi-1\le x\le0\end{cases}}\)
b, \(A\ge\frac{1}{2}\)
Khi \(0\le x\le1\)thì \(\sqrt{2x}\ge\frac{1}{2}\Leftrightarrow x\ge\frac{1}{2\sqrt{2}}\)
Khi \(-1\le x\le0\)thì \(-\sqrt{2x}\ge\frac{1}{2}\Leftrightarrow x\le-\frac{1}{2\sqrt{2}}\)
Vậy \(A\ge\frac{1}{2}\)\(\Leftrightarrow-1\le x\le-\frac{1}{2\sqrt{2}}\)hoặc \(\frac{1}{2\sqrt{2}}\le x\le1\)