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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
ĐKXĐ: \(x\ge4\)
a/ \(A=\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\left[\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\right]\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\left(\frac{x-4-x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(-3\right)}\)
\(=\frac{\sqrt{x}-2}{-3\sqrt{x}}\)
b/ A = 0 \(\Rightarrow\frac{\sqrt{x}-2}{-3\sqrt{x}}=0\Rightarrow\sqrt{x}-2=0\Rightarrow\sqrt{x}=2\Rightarrow x=4\)
Bài 1 :
a )\(A=\frac{3-\sqrt{3}}{\sqrt{3}-1}+\frac{\sqrt{35}-\sqrt{15}}{\sqrt{5}}-\sqrt{28}\)
\(A=\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\frac{\sqrt{5}\left(\sqrt{7}-\sqrt{3}\right)}{\sqrt{5}}-\sqrt{28}\)
\(A=\sqrt{3}+\sqrt{7}-\sqrt{3}-\sqrt{28}\)
\(A=\sqrt{7}-\sqrt{28}\)
\(A=\sqrt{7}-2\sqrt{7}=-\sqrt{7}\)
Vậy \(A=-\sqrt{7}\)
b)\(B=\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\frac{\sqrt{a}+\sqrt{b}}{a-b}\left(a,b>0;a\ne b\right)\)
\(B=\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}:\frac{\sqrt{a}+\sqrt{b}}{a-b}\)
\(B=\left(\sqrt{a}+\sqrt{b}\right).\frac{a-b}{\sqrt{a}+\sqrt{b}}\)
\(B=a-b\)
Vậy \(B=a-b\left(a,b>0;a\ne b\right)\)
_Minh ngụy_
Bài 2 :
a )\(B=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{x+\sqrt{x}}\left(x>0\right)\)
\(B=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\frac{x-1+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\frac{x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(B=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
Vậy \(B=\frac{\sqrt{x}-1}{\sqrt{x}+1}\left(x>0\right)\)
b) \(B=\frac{\sqrt{x}-1}{\sqrt{x}+1}\left(x>0\right)\)
Ta có : \(B>0\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}>0\)
Vì : \(\sqrt{x}\ge0\forall x\Rightarrow\)để \(B>O\)cần \(\sqrt{x}-1>0\Leftrightarrow\sqrt{x}>1\Leftrightarrow x>1\)( thỏa mãn \(x>0\))
Vậy \(x>1\)thì \(B>0\)
_Minh ngụy_
1)))))))
\(\frac{2}{\sqrt{ab}}:\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\right)^2-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\frac{2}{\sqrt{ab}}:\frac{\left(\sqrt{b}-\sqrt{a}\right)^2}{\left(\sqrt{ab}\right)^2}-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\frac{2}{\sqrt{ab}}.\frac{\left(\sqrt{ab}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\frac{2\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)^2}-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\frac{2\sqrt{ab}-a-b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\frac{-\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}=-1\)
\(\text{VT}=\left(1+\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1-\frac{x-\sqrt{x}}{\sqrt{x}-1}\right)=\left(1+\frac{\sqrt{x}.\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)\left(1-\frac{\sqrt{x}.\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right)\)
\(=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x=\text{VP(điều phải chứng minh)}\)
