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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:
\(\sqrt{\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{5^2}}=\frac{1}{2}+\frac{1}{3}-\frac{1}{5}\) ; ... ; \(\sqrt{\frac{1}{2^2}+\frac{1}{799^2}+\frac{1}{801^2}}=\frac{1}{2}+\frac{1}{799}-\frac{1}{801}\)
\(\Rightarrow P=\frac{1}{2}+1-\frac{1}{3}+\frac{1}{2}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2}+\frac{1}{799}-\frac{1}{801}\)
\(=\frac{1}{2}\cdot400+\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{799}-\frac{1}{801}\right)\)
\(=200+\frac{800}{801}=\frac{161000}{801}=\frac{a}{b}\Rightarrow\hept{\begin{cases}a=161000\\b=801\end{cases}}\)
\(\Rightarrow Q=161000-801\cdot200=800\)
Ta có :
\(a^2=72+\sqrt{72+\sqrt{72+\sqrt{72+.......}}}\)
\(\Leftrightarrow a^2=72+a\Leftrightarrow a^2-a-72=0\Leftrightarrow\left(a-9\right)\left(a+8\right)=0\)
\(\Rightarrow\orbr{\begin{cases}a=9\\a=-8\end{cases}}\)
Mà a > 0 nên a = 9 \(\Rightarrow\left[a\right]=9\)
#)Giải :
a) A = √(3+√5)-√(3-√5)-√2
<=>A√2=√(6+2√5)-√(6-2√5)-2
<=>A√2=√(√5+1)^2-√(√5-1)-2
<=>A√2=√5+1-√5+1-2
<=>A√2=0
<=>A=0
=>√(3+√5)-√(3-√5)-√2 =0
b) B=√(4-√7)-√ (4+√7)+√7
<=>B√2=√(8-2√7)-√(8+2√7)+2√7
<=>B√2=√(√7-1)^2-√(√7+1)^2+2√7
<=>B√2=√7-1-√7-1+2√7
<=>B√2=2√7-2
<=>B=(2√7-2)/√2
=√14-√2
#~Will~be~Pens~3
Câu a) hình như sai đề đúng không bạn ?
b) \(B=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{7}\)
Xét \(\left(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\right)^2\)
\(=4-\sqrt{7}-2\sqrt{\left(4-\sqrt{7}\right)\left(4+\sqrt{7}\right)}+4+\sqrt{7}\)
\(=8-2\sqrt{16-7}\)
\(=8-2\cdot3\)
\(=2\)
\(\Rightarrow\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}=-\sqrt{2}\)( vì \(\sqrt{4-\sqrt{7}}< \sqrt{4+\sqrt{7}}\))
Khi đó : \(B=-\sqrt{2}+\sqrt{7}\)
Góp ý nhẹ với bạn ๖²⁴ʱŤ.Ƥεɳɠʉїɳş༉ ( Team TST 14 ) là không biết thì đừng làm nhé
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
\(\sqrt{53-20\sqrt{7}}=a+b\sqrt{7}\)
\(\Leftrightarrow a+b\sqrt{7}=-5+2\sqrt{7}\)
=> a=-5; b=2