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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:
\(\sqrt{\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{5^2}}=\frac{1}{2}+\frac{1}{3}-\frac{1}{5}\) ; ... ; \(\sqrt{\frac{1}{2^2}+\frac{1}{799^2}+\frac{1}{801^2}}=\frac{1}{2}+\frac{1}{799}-\frac{1}{801}\)
\(\Rightarrow P=\frac{1}{2}+1-\frac{1}{3}+\frac{1}{2}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2}+\frac{1}{799}-\frac{1}{801}\)
\(=\frac{1}{2}\cdot400+\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{799}-\frac{1}{801}\right)\)
\(=200+\frac{800}{801}=\frac{161000}{801}=\frac{a}{b}\Rightarrow\hept{\begin{cases}a=161000\\b=801\end{cases}}\)
\(\Rightarrow Q=161000-801\cdot200=800\)
Xét \(\Delta=\left(2m-1\right)^2-8\left(m-1\right)=4m^2-12m+9=\left(2m-3\right)^2\ge0\)
=> PT luôn có 2 nghiệm x1,x2 với mọi m
Theo hệ thức Viet ta có \(\hept{\begin{cases}x_1+x_2=\frac{1-2m}{2}\\x_1x_2=\frac{m-1}{2}\end{cases}}\)
\(\Rightarrow A=\left(x_1+x_2\right)^2-3x_1x_2=\left(\frac{1-2m}{2}\right)^2-\frac{3\left(m-1\right)}{2}\)
\(=\frac{1-4m+4m^2-6m+6}{4}=\frac{4m^2-10m+7}{4}\)
\(=\frac{\left(2m-\frac{5}{2}\right)^2+\frac{3}{4}}{4}\ge\frac{3}{16}\)
Dấu "=" xảy ra khi \(2m=\frac{5}{2}\Rightarrow m=\frac{5}{4}\Rightarrow\frac{a}{b}=\frac{5}{4}\)
\(\Rightarrow4a=5b\Rightarrow2a=\frac{5b}{2}\)
lúc đó \(P=\frac{5b}{2}+2b=\frac{9b}{2}\)
1. \(x=\frac{1}{9}\) thỏa mãn đk: \(x\ge0;x\ne9\)
Thay \(x=\frac{1}{9}\) vào A ta có:
\(A=\frac{\sqrt{\frac{1}{9}}+1}{\sqrt{\frac{1}{9}}-3}=-\frac{1}{2}\)
2. \(B=...\)
\(B=\frac{3\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{4x+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{3x-9\sqrt{x}+x+3\sqrt{x}-4x-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{-6\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
3. \(P=A:B=\frac{\sqrt{x}+1}{\sqrt{x}-3}:\frac{-6\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{\sqrt{x}+3}{-6}\)
Vì \(\sqrt{x}+3\ge3\forall x\)\(\Rightarrow\frac{\sqrt{x}+3}{-6}\le\frac{3}{-6}=-\frac{1}{2}\)
hay \(P\le-\frac{1}{2}\)
Dấu "=" xảy ra <=> x=0
Áp dụng: \(\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\)
\(x=\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\)
=> \(x^3=\left(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\right)^3\)
\(=20+14\sqrt{2}+20-14\sqrt{2}+3\sqrt[3]{\left(20+14\sqrt{2}\right)\left(20-14\sqrt{2}\right)}\left(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\right)\)
\(=40+6x\)
=> \(x^3-6x=40\)
ta có \(x^3=\left(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\right)^3\)\(=20+14\sqrt{2}+3\sqrt[3]{\left(20+14\sqrt{2}\right)^2}.\sqrt[3]{20-14\sqrt{2}}+20-14\sqrt{2}\)\(+3\sqrt[3]{20+14\sqrt{2}}.\sqrt[3]{\left(20-14\sqrt{2}\right)^2}=\)\(40+3\sqrt[3]{\left(20+14\sqrt{2}\right)\left(20-14\sqrt{2}\right)}\left(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\right)\)
\(=40+3\sqrt[3]{20^2-14\sqrt{2}^2}.x\)x này là đề bài cho nên thay vào nha bạn
\(=40+3.2.x\)\(hay\)\(x^3=6x+40\Leftrightarrow x^3-6x=40\)(đây là kết quả cần tìm)
Bài này không cần giải phương trình dưới đâu nhé!
Liên hợp ta có:
\(\sqrt{x^2-3x+14}-\sqrt{x^2-3x+8}=2\)
<=> \(\frac{\left(x^2-3x+14\right)-\left(x^2-3x+8\right)}{\sqrt{x^2-3x+14}+\sqrt{x^2-3x+8}}=2\)
<=> \(\frac{6}{\sqrt{x^2-3x+14}+\sqrt{x^2-3x+8}}=2\)
<=> \(\sqrt{x^2-3x+14}+\sqrt{x^2-3x+8}=\frac{6}{2}=3\)
Vậy B = 3.