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Xét bài toán phụ sau:
Nếu \(a+b+c=0\Leftrightarrow\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\) \(\left(a,b,c\ne0\right)\)
Thật vậy
Ta có: \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)}\)
\(=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-2\cdot\frac{a+b+c}{abc}}=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-2\cdot\frac{0}{abc}}\)
\(=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\)
Bài toán được chứng minh
Quay trở lại, ta sẽ áp dụng bài toán phụ vào bài chính:
Ta có: \(P=\sqrt{\frac{1}{2^2}+\frac{1}{1^2}+\frac{1}{3^2}}+\sqrt{\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{5^2}}+...+\sqrt{\frac{1}{2^2}+\frac{1}{779^2}+\frac{1}{801^2}}\)
Vì \(2+1+\left(-3\right)=0\) nên:
\(\sqrt{\frac{1}{2^2}+\frac{1}{1^2}+\frac{1}{3^2}}=\sqrt{\frac{1}{2^2}+\frac{1}{1^2}+\frac{1}{\left(-3\right)^2}}=\sqrt{\left(\frac{1}{2}+\frac{1}{1}-\frac{1}{3}\right)^2}=\frac{1}{2}+1-\frac{1}{3}\)
Tương tự ta tính được:
\(\sqrt{\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{5^2}}=\frac{1}{2}+\frac{1}{3}-\frac{1}{5}\) ; ... ; \(\sqrt{\frac{1}{2^2}+\frac{1}{799^2}+\frac{1}{801^2}}=\frac{1}{2}+\frac{1}{799}-\frac{1}{801}\)
\(\Rightarrow P=\frac{1}{2}+1-\frac{1}{3}+\frac{1}{2}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2}+\frac{1}{799}-\frac{1}{801}\)
\(=\frac{1}{2}\cdot400+\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{799}-\frac{1}{801}\right)\)
\(=200+\frac{800}{801}=\frac{161000}{801}=\frac{a}{b}\Rightarrow\hept{\begin{cases}a=161000\\b=801\end{cases}}\)
\(\Rightarrow Q=161000-801\cdot200=800\)
a) ĐKXĐ : \(0\le a\ne1\)
\(\frac{\sqrt{a}-a}{\sqrt{a}-1}=\frac{-\sqrt{a}\left(1-\sqrt{a}\right)}{1-\sqrt{a}}=-\sqrt{a}\)
b) ĐKXĐ : \(b\ne0,a\ne-\sqrt{b}\)
\(\frac{a-\sqrt{b}}{\sqrt{b}}:\frac{\sqrt{b}}{a+\sqrt{b}}=\frac{a-\sqrt{b}}{\sqrt{b}}.\frac{a+\sqrt{b}}{\sqrt{b}}=\frac{a^2-b}{b}=\frac{a^2}{b}-1\)
c) \(2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}=\sqrt{5}\left(2-5-4+11\right)\)\(=4\sqrt{5}\)
d) \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}=\left(2\sqrt{7}-2\sqrt{2}.\sqrt{7}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}\)
\(=7\left(2-2\sqrt{2}+1\right)+14\sqrt{2}=7\left(2-2\sqrt{2}+1+2\sqrt{2}\right)=7.3=21\)
e) \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\)
a, \(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)-\sqrt{2}\left(\sqrt{3}-1\right)\)
\(=3-1-\sqrt{6}+\sqrt{2}=2+\sqrt{2}-\sqrt{6}\)
b, \(=\sqrt{300.0,04}+2\left|\sqrt{3}-\sqrt{5}\right|\)
\(=2\sqrt{3}+2\left(\sqrt{5}-\sqrt{3}\right)\)
\(=2\sqrt{3}+2\sqrt{5}-2\sqrt{3}=2\sqrt{5}\)
c, \(=\sqrt{196}-2\sqrt{98}+\sqrt{49}+7\sqrt{8}\)
\(=14-14\sqrt{2}+7+14\sqrt{2}=21\)
d, \(=15\sqrt{5}+5\sqrt{20}-3\sqrt{45}\)
\(=15\sqrt{5}+10\sqrt{5}-9\sqrt{5}=16\sqrt{5}\)
Bài 1: Rút gọn
a) Ta có: \(\left(\sqrt{3}-\sqrt{2}+1\right)\cdot\left(\sqrt{3}-1\right)\)
\(=\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-1\right)-\sqrt{2}\cdot\left(\sqrt{3}-1\right)\)
\(=3-1-\sqrt{6}+\sqrt{2}\)
\(=2-\sqrt{2}-\sqrt{6}\)
b) Ta có: \(0.2\cdot\sqrt{\left(-10\right)^2\cdot3}+2\cdot\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}\)
\(=0.2\cdot\sqrt{\left(-10\right)^2}\cdot\sqrt{3}+2\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=0.2\cdot10\cdot\sqrt{3}+2\sqrt{5}-2\sqrt{3}\)
\(=2\sqrt{3}+2\sqrt{5}-2\sqrt{3}\)
\(=2\sqrt{5}\)
c) Ta có: \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}\)
\(=\sqrt{196}-2\cdot\sqrt{98}+\sqrt{49}+7\sqrt{8}\)
\(=14-\sqrt{392}+7+\sqrt{392}\)
=21
d) Ta có: \(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}\)
\(=15\sqrt{5}+5\sqrt{20}-3\sqrt{45}\)
\(=\sqrt{5}\left(15+5\cdot2-3\cdot3\right)\)
\(=16\sqrt{5}\)
1) Bạn đánh nhầm \(\sqrt{x}+3\rightarrow\sqrt{x+3}\); \(\sqrt{x}-3\rightarrow\sqrt{x-3}\)
Sửa : \(ĐKXĐ:x\ne\pm\sqrt{3}\)
a) \(M=\frac{x-\sqrt{x}}{x-9}+\frac{1}{\sqrt{x}+3}-\frac{1}{\sqrt{x}-3}\)
\(\Leftrightarrow M=\frac{x-\sqrt{x}+\sqrt{x}-3-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow M=\frac{x-\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow M=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow M=\frac{\sqrt{x}+2}{\sqrt{x}+3}\)
b) Để \(M=\frac{3}{4}\)
\(\Leftrightarrow\frac{\sqrt{x}+2}{\sqrt{x}+3}=\frac{3}{4}\)
\(\Leftrightarrow4\sqrt{x}+8=3\sqrt{x}+9\)
\(\Leftrightarrow\sqrt{x}-1=0\)
\(\Leftrightarrow\sqrt{x}=1\)
\(\Leftrightarrow x=1\)(tm)
Vậy để \(A=\frac{3}{4}\Leftrightarrow x=1\)
c) Khi x = 4
\(\Leftrightarrow M=\frac{\sqrt{4}+2}{\sqrt{4}+3}\)
\(\Leftrightarrow M=\frac{2+2}{2+3}\)
\(\Leftrightarrow M=\frac{4}{5}\)
Vậy khi \(x=4\Leftrightarrow M=\frac{4}{5}\)
Khi \(x=1,44\): \(A=\frac{1,44+7}{\sqrt{1,44}}=\frac{8,44}{1,2}=\frac{211}{30}\)
\(B=\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}-1}{\sqrt{x}-3}-\frac{2x-\sqrt{x}-3}{x-9}\)(ĐK: \(x\ge0,x\ne9\))
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{2x-\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x-3\sqrt{x}+2x+5\sqrt{x}-3-2x+\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(S=\frac{1}{B}+A=\frac{\sqrt{x}-3}{\sqrt{x}}+\frac{x+7}{\sqrt{x}}=\frac{x+\sqrt{x}+4}{\sqrt{x}}=\sqrt{x}+\frac{4}{\sqrt{x}}+1\)
\(\ge2\sqrt{\sqrt{x}.\frac{4}{\sqrt{x}}}+1=5\)
Dấu \(=\)khi \(\sqrt{x}=\frac{4}{\sqrt{x}}\Leftrightarrow x=4\)(thỏa mãn)
a: \(=\left(2\sqrt{7}+\sqrt{7}+2\sqrt{14}\right)\cdot\sqrt{7}-\left(51+14\sqrt{2}\right)\)
\(=3\sqrt{7}\cdot\sqrt{7}+2\sqrt{14}\cdot\sqrt{7}-51-14\sqrt{2}\)
\(=21-51=-30\)
b: \(=\dfrac{\sqrt{10}}{2}+\dfrac{\sqrt{10}-\sqrt{6}}{2}=\dfrac{2\sqrt{10}-\sqrt{6}}{2}\)
c: \(=\dfrac{\left(\sqrt{5}+\sqrt{3}\right)^2}{\sqrt{5}+\sqrt{3}}+\dfrac{\left(\sqrt{5}-\sqrt{2}\right)^2}{\sqrt{5}-\sqrt{2}}\)
\(=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{2}\)
\(=2\sqrt{5}+\sqrt{3}-\sqrt{2}\)
