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a) \(\dfrac{x^2-5}{x+\sqrt{5}}\)(với x khác -\(\sqrt{5}\)) =\(\dfrac{\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)}{x+\sqrt{5}}\) = x-\(\sqrt{5}\) vậy \(\dfrac{x^2-5}{x+\sqrt{5}}\) = x-\(\sqrt{5}\) với x khác -\(\sqrt{5}\) b) \(\dfrac{x^2+2\sqrt{2}x+2}{x^2-2}\) ( với x khác +-\(\sqrt{2}\) ) = \(\dfrac{\left(x+\sqrt{2}\right)^2}{\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)}\) =\(\dfrac{x+\sqrt{2}}{x-\sqrt{2}}\) vậy \(\dfrac{x^2+2\sqrt{2}x+2}{x^2-2}\) =\(\dfrac{x+\sqrt{2}}{x-\sqrt{2}}\) với x khác +-\(\sqrt{2}\)
trả lời :
a)
\(M=\dfrac{x^2-2x\sqrt{2}+2}{x^2-2}=\dfrac{\left(x-\sqrt{2}\right)^2}{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}\)
\(M=\dfrac{x-\sqrt{2}}{x+\sqrt{2}}\)
b)\(N=\dfrac{x+\sqrt{5}}{x^2+2x\sqrt{5}+5}\)
\(N=\dfrac{x+\sqrt{5}}{\left(x+\sqrt{5}\right)^2}=\dfrac{1}{x+\sqrt{5}}\)
^HT^
a, Ta có :
\(M=\frac{\left(x-\sqrt{2}\right)^2}{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}\)
\(=\frac{x-\sqrt{2}}{x+\sqrt{2}}\)( với x khác cộng trừ căn 2)
b, Ta có:
\(N=\frac{x+\sqrt{5}}{\left(x+\sqrt{5}\right)^2}=\frac{1}{x+\sqrt{5}}\)
( với x khác trừ căn 5)
Chúc học tốt + k mình nha
a: \(=-4+2\sqrt{5}-\sqrt{5}+2+\sqrt{5}=2\sqrt{5}-2\)
b: \(B=\dfrac{2\sqrt{x}+4+6\sqrt{x}-3-2\sqrt{x}}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}}{6\sqrt{x}+4}\)
\(=\dfrac{\left(6\sqrt{x}+1\right)\cdot\sqrt{x}}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)\left(6\sqrt{x}+4\right)}\)
e/ \(\left(x-4\right)\sqrt{16-8x+x^2}=\left(x-4\right)\sqrt{\left(x-4\right)^2}=\left(x-4\right)\left(x-4\right)=\left(x-4\right)^2\)
f/ \(\left(2x-5\right)\sqrt{\dfrac{2}{\left(2x-5\right)^2}}=\left(2x-5\right)\cdot\dfrac{1}{\left|2x-5\right|}\cdot\sqrt{2}\)
+) với \(x>\dfrac{5}{2}\) có: \(\left(2x-5\right)\cdot\dfrac{1}{\left|2x-5\right|}\cdot\sqrt{2}=\dfrac{2x-5}{2x-5}\cdot\sqrt{2}=\sqrt{2}\)
+) với \(x< \dfrac{5}{2}\) có:
\(\left(2x-5\right)\cdot\dfrac{1}{\left|2x-5\right|}\cdot\sqrt{2}=\dfrac{2x-5}{-\left(2x-5\right)}\cdot\sqrt{2}=-1\cdot\sqrt{2}=-\sqrt{2}\)
g/ \(\sqrt{x-4\sqrt{x-4}}=\sqrt{x-4-2\cdot2\cdot\sqrt{2-4}+4}=\sqrt{\left(\sqrt{x-4}+2\right)^2}=\sqrt{x-4}+2\)
Bài 2:
a: \(A=\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{4}-\dfrac{3\sqrt{5}\left(3-\sqrt{5}\right)}{4}\)
\(=-5+3\sqrt{5}+\dfrac{5+\sqrt{5}-9\sqrt{5}+15}{4}\)
\(=-5+3\sqrt{5}+5-2\sqrt{5}=\sqrt{5}\)
b: \(B=\left(\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\right):\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+3\sqrt{x}+6-2\sqrt{x}-6}=1\)
a: \(M=\dfrac{x+6\sqrt{x}-3\sqrt{x}+18-x}{x-36}\)
\(=\dfrac{3\left(\sqrt{x}+6\right)}{x-36}=\dfrac{3}{\sqrt{x}-6}\)
b: \(N=\dfrac{x^2}{y}\cdot\sqrt{xy\cdot\dfrac{y}{x}}-x^2\)
\(=\dfrac{x^2}{y}\cdot y-x^2=0\)
a) \(C=\dfrac{2}{\sqrt{5}+1}+\sqrt{\dfrac{2}{3-\sqrt{5}}}=\dfrac{2\left(\sqrt{5}-1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}+\sqrt{\dfrac{2\left(3+\sqrt{5}\right)}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}}=\dfrac{2\left(\sqrt{5}-1\right)}{5-1}+\sqrt{\dfrac{2\left(3+\sqrt{5}\right)}{9-5}}=\dfrac{\sqrt{5}-1}{2}+\sqrt{\dfrac{6+2\sqrt{5}}{4}}=\dfrac{\sqrt{5}-1}{2}+\dfrac{\sqrt{5+2\sqrt{5}+1}}{2}=\dfrac{\sqrt{5}-1+\sqrt{\left(\sqrt{5}+1\right)^2}}{2}=\dfrac{\sqrt{5}-1+\sqrt{5}+1}{2}=\dfrac{2\sqrt{5}}{2}=\sqrt{5}\)
b) \(D=\dfrac{1}{x-\sqrt{x}}-\dfrac{2\sqrt{x}}{x-1}+\dfrac{1}{x+\sqrt{x}}=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{2x}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+1-2x+\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{2\sqrt{x}-2x}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{2\sqrt{x}\left(1-\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{-2}{\sqrt{x}+1}\)
Câu 1:
a: \(P=\dfrac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b: Để \(2P=2\sqrt{5}+5\) thì \(P=\dfrac{2\sqrt{5}+5}{2}\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+5\right)=2\left(\sqrt{x}+1\right)\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+3\right)=2\)
hay \(x=\dfrac{4}{29+12\sqrt{5}}=\dfrac{4\left(29-12\sqrt{5}\right)}{121}\)
Câu 1:
a: \(P=\dfrac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b: Để \(2P=2\sqrt{5}+5\) thì \(P=\dfrac{2\sqrt{5}+5}{2}\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+5\right)=2\left(\sqrt{x}+1\right)\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+3\right)=2\)
hay \(x=\dfrac{4}{29+12\sqrt{5}}=\dfrac{4\left(29-12\sqrt{5}\right)}{121}\)
\(M=\dfrac{x^2-2x\sqrt{2}+2}{x^2-2}=\dfrac{\left(x-\sqrt{2}\right)^2}{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}\)
\(M=\dfrac{x-\sqrt{2}}{x+\sqrt{2}}\)
hi vọng bạn hiểu
b, \(N=\dfrac{x+\sqrt{5}}{x^2+2x\sqrt{5}+5}\)
chú ý dưới mẫu nhé! khá hay đẫy, nếu ghép lại là thành dạng bình phương đấy, mời bạn xem nhé!
\(N=\dfrac{x+\sqrt{5}}{\left(x+\sqrt{5}\right)^2}=\dfrac{1}{x+\sqrt{5}}\)
thấy chưa, đơn giản quá phải k