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\(A=\left(\frac{1}{\sqrt{x-1}}+\frac{1}{\sqrt{x+1}}\right):\left(\frac{1}{\sqrt{x-1}}+\frac{1}{\sqrt{x+1}}\right)\)
\(A=\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x^2-1}}\times\frac{\sqrt{x^2-1}}{\sqrt{x+1}-\sqrt{x-1}}\)
\(A=\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}\)
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 #
\(A=\left(\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{\left(x-1\right)\left(x+1\right)}}\right).\left(\frac{\sqrt{\left(x-1\right)\left(x+1\right)}}{\sqrt{x+1}-\sqrt{x-1}}\right)=\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}\)
\(=\frac{\left(\sqrt{x+1}+\sqrt{x-1}\right)^2}{2}=\frac{2\left(x+\sqrt{x^2-1}\right)}{2}=x+\sqrt{x^2-1}\)
Thế vào rồi tính nhé
\(\)
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
a: \(P=\dfrac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{x-1-x+4}\)
\(=\dfrac{1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}-2}{3}=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
b: P=1/4
=>\(\dfrac{\sqrt{x}-2}{3\sqrt{x}}=\dfrac{1}{4}\)
=>\(4\left(\sqrt{x}-2\right)=3\sqrt{x}\)
=>\(4\sqrt{x}-8-3\sqrt{x}=0\)
=>\(\sqrt{x}=8\)
=>x=64
c: Khi \(x=4+2\sqrt{3}\) thì \(P=\dfrac{\sqrt{4+2\sqrt{3}}-2}{3\cdot\sqrt{4+2\sqrt{3}}}\)
\(=\dfrac{\sqrt{3}+1-2}{3\left(\sqrt{3}+1\right)}=\dfrac{\sqrt{3}-1}{3\sqrt{3}+3}=\dfrac{2-\sqrt{3}}{3}\)
\(a,A=\sqrt{27}+\frac{2}{\sqrt{3}-2}-\sqrt{\left(1-\sqrt{3}\right)^2}\)
\(=3\sqrt{3}+\frac{2\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\left(\sqrt{3}-1\right)\)
\(=3\sqrt{3}+\frac{2\sqrt{3}+4}{3-4}-\sqrt{3}+1\)
\(=3\sqrt{3}-2\sqrt{3}-4-\sqrt{3}+1\)
\(=-3\)
\(B=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
\(=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}}\)
b, Ta có \(B< A\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}< -3\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}+3< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-1+3\sqrt{x}}{\sqrt{x}}< 0\)
\(\Leftrightarrow\frac{4\sqrt{x}-1}{\sqrt{x}}< 0\)
\(\Leftrightarrow4\sqrt{x}-1< 0\left(Do\sqrt{x}>0\right)\)
\(\Leftrightarrow\sqrt{x}< \frac{1}{4}\)
\(\Leftrightarrow0< x< \frac{1}{2}\)(Kết hợp ĐKXĐ)
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