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\(A=\left(\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{\left(x-1\right)\left(x+1\right)}}\right).\left(\frac{\sqrt{\left(x-1\right)\left(x+1\right)}}{\sqrt{x+1}-\sqrt{x-1}}\right)=\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}\)
\(=\frac{\left(\sqrt{x+1}+\sqrt{x-1}\right)^2}{2}=\frac{2\left(x+\sqrt{x^2-1}\right)}{2}=x+\sqrt{x^2-1}\)
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
\(A=\left(\frac{1}{\sqrt{x-1}}+\frac{1}{\sqrt{x+1}}\right):\left(\frac{1}{\sqrt{x-1}}+\frac{1}{\sqrt{x+1}}\right)\)
\(A=\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x^2-1}}\times\frac{\sqrt{x^2-1}}{\sqrt{x+1}-\sqrt{x-1}}\)
\(A=\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}\)
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, \(B=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{2\sqrt{x}+1}{x+\sqrt{x}}\) (ĐKXĐ: \(x>0\))
\(=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{x-1+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
b, \(\frac{A}{B}=\frac{2+\sqrt{x}}{\sqrt{x}}:\frac{\sqrt{x}+2}{\sqrt{x}+1}=\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(\frac{A}{B}>\frac{3}{2}\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}}-\frac{3}{2}>0\)
\(\Leftrightarrow\frac{2\sqrt{x}+2-3\sqrt{x}}{2\sqrt{x}}>0\)
\(\Leftrightarrow2-\sqrt{x}>0\)
\(\Leftrightarrow\sqrt{x}< 2\Leftrightarrow x< 4\)
Kết hợp với điều kiện \(x>0\)ta có: \(0< x< 4\)
Vậy với \(0< x< 4\)thì \(\frac{A}{B}>\frac{3}{2}\)
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_
Bài 1 :
a) \(P=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}}{x-2\sqrt{x}+1}\)
\(P=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right).\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)
\(P=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}-1}{\sqrt{x}}\)
\(P=\frac{\sqrt{x}+1}{x}\)
b) \(P>\frac{1}{2}\)
\(\Leftrightarrow\frac{\sqrt{x}+1}{x}>\frac{1}{2}\)
\(\Leftrightarrow\frac{\sqrt{x}+1}{x}-\frac{1}{2}>0\)
\(\Leftrightarrow\frac{\sqrt{x}+1-2x}{x}>0\)
\(\Leftrightarrow\sqrt{x}-2x+1>0\left(x>0\right)\)
\(\Leftrightarrow\sqrt{x}+x^2-2x+1-x^2>0\)
\(\Leftrightarrow\sqrt{x}+x^2+\left(x-1\right)^2>0\left(\forall x>0\right)\)
Vậy P > 1/2 với mọi x> 0 ; x khác 1
Bài 2 :
a) \(K=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}+a}+\frac{2}{a-1}\right)\)
\(K=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{1}{\sqrt{a}\left(\sqrt{a}+1\right)}+\frac{2}{a-1}\right)\)
\(K=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1+2\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}\left(a-1\right)\left(\sqrt{a}+1\right)}\)
\(K=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\sqrt{a}\left(a-1\right)\left(\sqrt{a}-1\right)}{a-1+2a+2\sqrt{a}}\)
\(K=\frac{\left(a-1\right)^2}{3a+2\sqrt{a}-1}\)
b) \(a=3+2\sqrt{2}=2+2\sqrt{2}+1=\left(\sqrt{2}+1\right)^2\)( thỏa mãn ĐKXĐ )
Thay a vào biểu thức K , ta có :
\(K=\frac{\left(3+2\sqrt{2}-1\right)^2}{3\left(3+2\sqrt{2}\right)+2\sqrt{\left(\sqrt{2}+1\right)^2}-1}\)
\(K=\frac{\left(2+2\sqrt{2}\right)^2}{9+6\sqrt{2}+2\left|\sqrt{2}+1\right|-1}\)
\(K=\frac{\left(2+2\sqrt{2}\right)^2}{8+6\sqrt{2}+2\sqrt{2}+2}\)
\(K=\frac{\left(2+2\sqrt{2}\right)^2}{10+8\sqrt{2}}\)