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a: \(=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}=\sqrt{ab}-\sqrt{ab}=0\)
b: \(=\dfrac{\left(\sqrt{x}-2\sqrt{y}\right)^2}{\sqrt{x}-2\sqrt{y}}+\dfrac{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)
\(=\sqrt{x}-2\sqrt{y}+\sqrt{y}=\sqrt{x}-\sqrt{y}\)
c: \(=\sqrt{x}+2-\dfrac{x-4}{\sqrt{x}-2}\)
\(=\sqrt{x}+2-\sqrt{x}-2=0\)
d) \(\dfrac{4}{\sqrt{7}-\sqrt{3}}+\dfrac{6}{3+\sqrt{3}}+\dfrac{\sqrt{7}-7}{\sqrt{7}-1}\)
\(=\dfrac{4\left(\sqrt{7}+\sqrt{3}\right)}{\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}+\sqrt{3}\right)}+\dfrac{6\left(3-\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}-\dfrac{7-\sqrt{7}}{\sqrt{7}-1}\)
\(=\dfrac{4\left(\sqrt{7}+\sqrt{3}\right)}{4}+\dfrac{6\left(3-\sqrt{3}\right)}{6}-\dfrac{\sqrt{7}\left(\sqrt{7}-1\right)}{\sqrt{7}-1}\)
\(=\sqrt{7}+\sqrt{3}+3-\sqrt{3}-\sqrt{7}=3\)
b) \(\dfrac{\sqrt{5}-\sqrt{15}}{1-\sqrt{3}}-\sqrt{21+4\sqrt{5}}=\dfrac{\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}-\sqrt{20+2\sqrt{20}+1}\)
\(=\sqrt{5}-\sqrt{\left(\sqrt{20}+1\right)^2}=\sqrt{5}-\left(\sqrt{20}+1\right)=\sqrt{5}-2\sqrt{5}-1=-1-\sqrt{5}\)
A = \(x^2+3x-7=x^2+2x\frac{3}{2}+\frac{9}{4}-\frac{37}{4}\)
\(=\left(x+\frac{3}{2}\right)^2-\frac{37}{4}\ge-\frac{37}{4}\)
\(\Rightarrow\)min A = \(-\frac{37}{4}\Leftrightarrow x=-\frac{3}{2}\)
B = \(x-5\sqrt{x}-1\) ĐKXĐ: \(x\ge0\)
\(=x-2\sqrt{x}\frac{5}{2}+\frac{25}{4}-\frac{29}{4}=\left(\sqrt{x}-\frac{5}{2}\right)^2-\frac{29}{4}\ge-\frac{29}{4}\)
\(\Rightarrow\)min B = \(-\frac{29}{4}\Leftrightarrow x=\frac{25}{4}\)( thỏa mãn)
C = \(\frac{-4}{\sqrt{x}+7}\) ĐKXĐ:\(x\ge0\)
Ta có: \(\sqrt{x}+7\ge7\Rightarrow\frac{4}{\sqrt{x}+7}\le\frac{4}{7}\)\(\Leftrightarrow\frac{-4}{\sqrt{x}+7}\ge-\frac{4}{7}\)
\(\Rightarrow\)min C = \(-\frac{4}{7}\Leftrightarrow x=0\)
D = \(\frac{\sqrt{x}+1}{\sqrt{x}+3}\) ĐKXĐ:\(x\ge0\)
\(=1-\frac{2}{\sqrt{x}+3}\ge1-\frac{2}{3}=\frac{1}{3}\)
\(\Rightarrow\)min D = \(\frac{1}{3}\Leftrightarrow x=0\)
E = \(\frac{x+7}{\sqrt{x}+3}\) ĐKXĐ:\(x\ge0\)
\(=\frac{x-9+16}{\sqrt{x}+3}=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)+16}{\sqrt{x}+3}=\sqrt{x}-3+\frac{16}{\sqrt{x}+3}=\sqrt{x}+3+\frac{16}{\sqrt{x}+3}-6\ge2\sqrt{16}-6=2\)
\(\Rightarrow\)min E = \(2\Leftrightarrow x=1\)(thỏa mãn)
F = \(\frac{x^2+3x+5}{x^2}\) ĐKXĐ: \(x\ne0\)
\(\Leftrightarrow\)\(x^2\left(F-1\right)-3x-5=0\)
△ = \(3^2+20\left(F-1\right)\ge0\)\(\Leftrightarrow F\ge\frac{11}{20}\)
\(\Rightarrow\)min F = \(\frac{11}{20}\Leftrightarrow x=-\frac{10}{3}\)( thỏa mãn)
\(\sqrt{x^2-x-6}=\sqrt{x-3}\)
Tự xét điều kiện nha
\(\Leftrightarrow x^2-x-6=x-3\)
\(\Leftrightarrow x^2-2x-3=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
\(\sqrt{x^2-x}=\sqrt{3x-5}\)
\(\Leftrightarrow x^2-x=3x-5\)
\(\Leftrightarrow x^2-4x+5=0\)
vô nghiệm
6/ Đặt \(\hept{\begin{cases}\sqrt[4]{x}=a\\\sqrt[4]{2-x}=b\end{cases}}\)
\(\Rightarrow b^4+a^4=2\)
Từ đó ta có: a + b = 2
Ta có: \(a^4+b^2\ge\frac{\left(a^2+b^2\right)^2}{2}\ge\frac{\left(a+b\right)^4}{8}=\frac{16}{8}=2\)
Dấu = xảy ra khi a = b = 1
=> x = 1
Bài 1: Tính
a) Ta có: \(\frac{\sqrt{6+\sqrt{11}}-\sqrt{7-\sqrt{33}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{\sqrt{12+2\sqrt{11}}-\sqrt{14-2\sqrt{33}}}{\sqrt{12}+2}\)
\(=\frac{\sqrt{11+2\cdot\sqrt{11}\cdot1+1}-\sqrt{11-2\cdot\sqrt{11}\cdot\sqrt{3}+3}}{2\sqrt{3}+2}\)
\(=\frac{\sqrt{\left(\sqrt{11}+1\right)^2}-\sqrt{\left(\sqrt{11}-\sqrt{3}\right)^2}}{2\sqrt{3}+2}\)
\(=\frac{\left|\sqrt{11}+1\right|-\left|\sqrt{11}-\sqrt{3}\right|}{2\left(\sqrt{3}+1\right)}\)
\(=\frac{\sqrt{11}+1-\left(\sqrt{11}-\sqrt{3}\right)}{2\left(1+\sqrt{3}\right)}\)(Vì \(\left\{{}\begin{matrix}\sqrt{11}>1>0\\\sqrt{11}>\sqrt{3}\end{matrix}\right.\))
\(=\frac{\sqrt{11}+1-\sqrt{11}+\sqrt{3}}{2\left(1+\sqrt{3}\right)}\)
\(=\frac{1+\sqrt{3}}{2\left(1+\sqrt{3}\right)}=\frac{1}{2}\)
b) Ta có: \(\frac{5\sqrt{3}-3\sqrt{5}}{\sqrt{5}-\sqrt{3}}+\frac{2}{4+\sqrt{15}}-\frac{5\sqrt{5}+3\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
\(=\frac{\sqrt{15}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}+\frac{2}{4+\sqrt{15}}-\frac{\left(\sqrt{5}+\sqrt{3}\right)\left(8-\sqrt{15}\right)}{\sqrt{5}+\sqrt{3}}\)
\(=\sqrt{15}+\frac{2}{4+\sqrt{15}}-\left(8-\sqrt{15}\right)\)
\(=\sqrt{15}+\frac{2}{4+\sqrt{15}}-8+\sqrt{15}\)
\(=2\sqrt{15}-8+\frac{2}{4+\sqrt{15}}\)
\(=\frac{2\sqrt{15}\left(4+\sqrt{15}\right)}{4+\sqrt{15}}-\frac{8\left(4+\sqrt{15}\right)}{4+\sqrt{15}}+\frac{2}{4+\sqrt{15}}\)
\(=\frac{8\sqrt{15}+30-32-8\sqrt{15}+2}{4+\sqrt{15}}\)
\(=\frac{0}{4+\sqrt{15}}=0\)
Bài 2: Rút gọn
Ta có: \(B=\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\left(\frac{1+\sqrt{a}}{a-1}\right)^2\)
\(=\left(\frac{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}+a\right)}{1+\sqrt{a}}-\sqrt{a}\right)\cdot\left(\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)^2\)
\(=\left(1-\sqrt{a}+a-\sqrt{a}\right)\cdot\left(\frac{1}{\sqrt{a}-1}\right)^2\)
\(=\left(a-2\sqrt{a}+1\right)\cdot\frac{1}{\left(\sqrt{a}-1\right)^2}\)
\(=\frac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)^2}=1\)
Bài 3:
a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\notin\left\{9;4\right\}\end{matrix}\right.\)
b) Ta có: \(A=\frac{\sqrt{x}+2}{\sqrt{x}-3}-\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{3-3\sqrt{x}}{x-5\sqrt{x}+6}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{3-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-4-\left(x-2\sqrt{x}-3\right)+3-3\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x-3\sqrt{x}-1-x+2\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{-\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{1}{3-\sqrt{x}}\)
c) Để A<-1 thì A+1<0
\(\Leftrightarrow\frac{1}{3-\sqrt{x}}+1< 0\)
\(\Leftrightarrow\frac{-1}{\sqrt{x}-3}+\frac{\sqrt{x}-3}{\sqrt{x}-3}< 0\)
\(\Leftrightarrow\frac{-1+\sqrt{x}-3}{\sqrt{x}-3}< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-4}{\sqrt{x}-3}< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}-4>0\\\sqrt{x}-3< 0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}-4< 0\\\sqrt{x}-3>0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}>4\\\sqrt{x}< 3\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}< 4\\\sqrt{x}>3\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< 16\\x>9\end{matrix}\right.\Leftrightarrow9< x< 16\)