Thực hiên các phép tính :
A=\(\left(\frac{1}{\sqrt{x-1}}+\frac{1}{\sqrt{x+1}}\right):\left(\frac{1}{\sqrt{x-1}}-\frac{1}{\sqrt{x+1}}\right)\)
với x =\(\frac{a^2+b^2}{2ab}\)
và b>a>0
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A
20 tháng 6 2021
\(A=\left(\frac{1}{\sqrt{x-1}}+\frac{1}{\sqrt{x+1}}\right):\left(\frac{1}{\sqrt{x-1}}+\frac{1}{\sqrt{x+1}}\right)\)
\(A=\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x^2-1}}\times\frac{\sqrt{x^2-1}}{\sqrt{x+1}-\sqrt{x-1}}\)
\(A=\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}\)
Thay \(x=\frac{a^2+b^2}{2ab}\)vào A, ta được :
\(A=\frac{\sqrt{\frac{a^2+b^2}{2ab}+1}+\sqrt{\frac{a^2+b^2}{2ab}-1}}{\sqrt{\frac{a^2+b^2}{2ab}+1}-\sqrt{\frac{a^2+b^2}{2ab}-1}}\)
\(A=\frac{\sqrt{\frac{\left(a+b\right)^2}{2ab}}+\sqrt{\frac{\left(b-a\right)^2}{2ab}}}{\sqrt{\frac{\left(a+b\right)^2}{2ab}}-\sqrt{\frac{\left(b-a\right)^2}{2ab}}}\)
\(A=\frac{a+b\sqrt{\frac{1}{2ab}}+\left(b-a\right)\sqrt{\frac{1}{2ab}}}{a+b\sqrt{\frac{1}{2ab}}-\left(b-a\right)\sqrt{\frac{1}{2ab}}}\)
\(A=\frac{a+b+b-a}{a+b-b+a}\)
\(A=\frac{2b}{2a}\)
\(A=\frac{b}{a}\)
Ps : Nhớ k cho tui nhó, tui đã rất cố gắng rồi đấy. :)) K để lần sau có j tui giải giúp cho :)))
# Aeri #
1 tháng 9 2020
\(A=\left(\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{\left(x-1\right)\left(x+1\right)}}\right).\left(\frac{\sqrt{\left(x-1\right)\left(x+1\right)}}{\sqrt{x+1}-\sqrt{x-1}}\right)=\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}\)
\(=\frac{\left(\sqrt{x+1}+\sqrt{x-1}\right)^2}{2}=\frac{2\left(x+\sqrt{x^2-1}\right)}{2}=x+\sqrt{x^2-1}\)
Thế vào rồi tính nhé
\(\)
1 tháng 9 2020
Ta có: \(A=\left(\frac{1}{\sqrt{x+1}}+\frac{1}{\sqrt{x-1}}\right):\left(\frac{1}{\sqrt{x+1}}-\frac{1}{\sqrt{x-1}}\right)\) \(\left(ĐK:x\ge1\right)\)
\(\Leftrightarrow A=\left(\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}.\sqrt{x-1}}\right).\left(\frac{\sqrt{x+1}.\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}\right)\)
\(\Leftrightarrow A=\frac{\left(\sqrt{x+1}+\sqrt{x-1}\right).\left(\sqrt{x+1}-\sqrt{x-1}\right)}{\left(\sqrt{x+1}-\sqrt{x-1}\right)^2}\)
\(\Leftrightarrow A=\frac{x+1-x+1}{x+1+x-1+2\sqrt{\left(x+1\right)\left(x-1\right)}}\)
\(\Leftrightarrow A=\frac{2}{2x+2\sqrt{x^2-1}}\)
Thay \(x=\frac{a^2+b^2}{2ab}\)vào phương trình \(A,\)ta có:
\(A=\frac{1}{\frac{a^2+b^2}{2ab}+\sqrt{\left(\frac{a^2+b^2}{2ab}+1\right)\left(\frac{a^2+b^2}{2ab}-1\right)}}\)
\(\Leftrightarrow A=\frac{1}{\frac{a^2+b^2}{2ab}+\sqrt{\left(\frac{a^2+2ab+b^2}{2ab}\right)\left(\frac{a^2-2ab+b^2}{2ab}\right)}}\)
\(\Leftrightarrow A=\frac{1}{\frac{a^2+b^2}{2ab}+\sqrt{\frac{\left(a+b\right)^2\left(a-b\right)^2}{\left(2ab\right)^2}}}\)
\(\Leftrightarrow A=\frac{1}{\frac{a^2+b^2}{2ab}+\frac{\left(a+b\right)\left(a-b\right)}{2ab}}\)
\(\Leftrightarrow A=\frac{1}{\frac{a^2+b^2+a^2-b^2}{2ab}}\)
\(\Leftrightarrow A=\frac{2ab}{2a^2}\)
\(\Leftrightarrow A=\frac{b}{a}\)
Chúc bn hok tốt
13 tháng 9 2018
\(B=\frac{-2a\sqrt{a}+2a^2}{\left(\sqrt{a}-\right)\left(a-1\right)}\)
\(C=-x\sqrt{x}+x+\sqrt{x}-1\)
\(D=x-\sqrt{x}+1\)
\(A=\left(\frac{1}{\sqrt{x-1}}+\frac{1}{\sqrt{x+1}}\right):\left(\frac{1}{\sqrt{x-1}}-\frac{1}{\sqrt{x+1}}\right)\)
\(=\left(\frac{\sqrt{x+1}}{x-1}+\frac{\sqrt{x-1}}{x-1}\right):\left(\frac{\sqrt{x+1}}{x-1}-\frac{\sqrt{x-1}}{x-1}\right)\)
\(=\frac{\sqrt{x+1}+\sqrt{x-1}}{x-1}:\frac{\sqrt{x+1}-\sqrt{x-1}}{x-1}\)
\(=\frac{\sqrt{x+1}+\sqrt{x-1}}{x-1}.\frac{x-1}{\sqrt{x+1}-\sqrt{x-1}}\)
=\(\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}=\frac{\left(\sqrt{x+1}-\sqrt{x-1}\right)\left(\sqrt{x+1}+\sqrt{x-1}\right)}{\left(x+1\right)-\left(x-1\right)}\)
\(=\frac{\left(\sqrt{x+1}\right)^2-\left(\sqrt{x-1}\right)^2}{x+1-x+1}=\frac{\left(x+1\right)-\left(x-1\right)}{2}\)
\(=\frac{x+1-x+1}{2}=\frac{2}{2}=1\)