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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: Khi x=64 thì \(A=\dfrac{3\cdot8+1}{8+2}=\dfrac{25}{10}=\dfrac{5}{2}\)
b: \(B=\dfrac{2\sqrt{x}-4-\sqrt{x}+5}{x-4}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}+2}\)
a) Thay x = 81 vào A ta có:
\(A=\dfrac{4\sqrt{81}}{\sqrt{81}-5}=\dfrac{4\cdot9}{9-5}=\dfrac{4\cdot9}{4}=9\)
b) \(B=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+2}+\dfrac{5-2\sqrt{x}}{x+\sqrt{x}-2}\left(x\ne1;x\ge0\right)\)
\(B-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+2}+\dfrac{5-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(B=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\dfrac{5-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(B=\dfrac{x-4+\sqrt{x}-1+5-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(B=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(B=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(B=\dfrac{\sqrt{x}}{\sqrt{x}+2}\)
c) \(\dfrac{A}{B}< 4\) khi
\(\dfrac{4\sqrt{x}}{\sqrt{x}-5}:\dfrac{\sqrt{x}}{\sqrt{x}+2}< 4\)
\(\Leftrightarrow\dfrac{4\left(\sqrt{x}+2\right)}{\sqrt{x}-5}< 4\)
\(\Leftrightarrow\dfrac{4\sqrt{x}+8-4\left(\sqrt{x}-4\right)}{\sqrt{x}-5}< 0\)
\(\Leftrightarrow\dfrac{24}{\sqrt{x}-5}< 0\)
\(\Leftrightarrow\sqrt{x}-5< 0\)
\(\Leftrightarrow x< 25\)
Kết hợp với đk:
\(0\le x< 5\)
a: Thay x=16 vào A, ta được:
\(A=\dfrac{2\cdot4}{4+3}=\dfrac{8}{7}\)
a: Khi x=16 thì \(A=\dfrac{2\cdot\sqrt{16}}{\sqrt{16}+3}=\dfrac{2\cdot4}{4+3}=\dfrac{8}{7}\)
b: P=A+B
\(=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{7\sqrt{x}+3}{9-x}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{7\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)+7\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3+7\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3x+5\sqrt{x}+6}{x-9}\)
Bài 1:
a) \(B=\sqrt{1-4x+4x^2}\)
\(=\sqrt{\left(1-2x\right)^2}\)
\(=\left|1-2x\right|\)
Nếu \(x\le\frac{1}{2}\)thì: \(B=1-2x\)
Nếu \(x>\frac{1}{2}\)thì: \(B=2x-1\)
b) Tại \(x=-7\)thì: \(B=1-2.\left(-7\right)=15\)
