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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 \(x=\dfrac{4}{29+12\sqrt{5}}=\dfrac{4\left(29-12\sqrt{5}\right)}{121}\)
Bài 6:
a: \(\Leftrightarrow\sqrt{x^2+4}=\sqrt{12}\)
=>x^2+4=12
=>x^2=8
=>\(x=\pm2\sqrt{2}\)
b: \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>x+1=1
=>x=0
c: \(\Leftrightarrow3\sqrt{2x}+10\sqrt{2x}-3\sqrt{2x}-20=0\)
=>\(\sqrt{2x}=2\)
=>2x=4
=>x=2
d: \(\Leftrightarrow2\left|x+2\right|=8\)
=>x+2=4 hoặcx+2=-4
=>x=-6 hoặc x=2
a: \(P=\dfrac{\sqrt{x}-4+3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}:\dfrac{x-4-x}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{4\left(\sqrt{x}-1\right)}{-4}=-\sqrt{x}+1\)
b: \(x=\dfrac{8}{3+\sqrt{5}}=6-2\sqrt{5}\)
Thay \(x=6-2\sqrt{5}\) vào P, ta được:
\(P=-\left(\sqrt{5}-1\right)+1=-\sqrt{5}+2\)
bài 2 : chữa đề câu a chút nha
a) ta có : \(\sqrt{\dfrac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\dfrac{4}{\left(2+\sqrt{5}\right)^2}}\)
\(\sqrt{\left(\dfrac{2}{\sqrt{5}-2}\right)^2}-\sqrt{\left(\dfrac{2}{\sqrt{5}+2}\right)^2}=\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{\sqrt{5}+2}\)
\(=\dfrac{2\sqrt{5}+4-2\sqrt{5}+4}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}=\dfrac{8}{5-4}=8\left(đpcm\right)\)
b) ta có : \(\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}\)
\(=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\sqrt{6-2\sqrt{5}}=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)^2=\left(3+\sqrt{5}\right)\left(6-2\sqrt{5}\right)=2\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)\) \(=2\left(9-5\right)=2.4=8\left(đpcm\right)\)
Bài 1 : Mình gợi ý thôi nhé :v
\(C=\dfrac{\sqrt{x}+3}{\sqrt{x}-2}=\dfrac{\sqrt{x}-2+5}{\sqrt{x}-2}=1+\dfrac{5}{\sqrt{x}-2}\)
\(D=\dfrac{2\sqrt{x}-1}{\sqrt{x}+3}=\dfrac{2\left(\sqrt{x}+3\right)-7}{\sqrt{x}+3}=2-\dfrac{7}{\sqrt{x}+3}\)
a)
\(\dfrac{\left(\sqrt{x^2+4}-2\right)\left(\sqrt{x^2+4}-2\right)\left(x+\sqrt{x}+1\right)\sqrt{x-2\sqrt{x}+1}}{x\left(x\sqrt{x}-1\right)}\\=\dfrac{\left(\left(\sqrt{x^2+4}\right)^2-4\right)\left(\left(x+\sqrt{x}+1\right)\sqrt{\left(x-1\right)^2}\right)}{x\left(x\sqrt{x}-1\right)}\\ =\dfrac{\left(x^2+4-4\right)\left(\left(x+\sqrt{x}+1\right)\left(x-1\right)\right)}{x\left(x\sqrt{x}-1\right)}\\ =\dfrac{x^2\left(x^3-1\right)}{x\left(x\sqrt{x}-1\right)}=x^2\sqrt{x}\)
b)
\(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right)\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\\ =\left(\dfrac{\left(\sqrt{a}-2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}-\dfrac{\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\right)\left(\dfrac{a}{\sqrt{a}}-\dfrac{4}{\sqrt{a}}\right)\\ =\left(\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{a-4}\right)\left(\dfrac{a-4}{\sqrt{a}}\right)\\ =\dfrac{-8\sqrt{a}}{a-4}\cdot\dfrac{a-4}{\sqrt{a}}=-8\)
c)
\(\left(\dfrac{\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)}+\dfrac{\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)}\right)\left(1-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}+\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\sqrt{a}}{\sqrt{a}}-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{a-2\sqrt{a}+1+a+2\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}\right)\\ =\dfrac{2a+2}{a-1}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ =\dfrac{-2\left(a+1\right)}{a+1}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ =\dfrac{-2\left(\sqrt{a}-1\right)}{\sqrt{a}}\)
d)
\(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+x+1\\ =\dfrac{\sqrt{x}\left(\sqrt{x}^3-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}^3+1\right)}{x-\sqrt{x}+1}+x+1\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\\ =\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)+x+1\\ =x-\sqrt{x}-x-\sqrt{x}+x+1\\ =x-2\sqrt{x}+1\\ =\left(x-1\right)^2\)
a) \(\sqrt{\sqrt{2\sqrt{6}+6+2\sqrt{2}+2\sqrt{3}-\sqrt{5+2\sqrt{6}}}}\)
\(=\sqrt{1+\sqrt{2}+\sqrt{3}-\left(\sqrt{3}+\sqrt{2}\right)}=1\)
b) \(A=\sqrt{x^2-6x+9}-\dfrac{x^2-9}{\sqrt{9-6x+x^2}}\)
\(=\left|x-3\right|-\dfrac{\left(x-3\right)\left(x+3\right)}{\left|x-3\right|}\)
Th1: x-3 < 0
\(A=\left(3-x\right)-\dfrac{\left(x-3\right)\left(x+3\right)}{3-x}=3-x+x-3=0\)
Th2: x-3 > 0
\(A=x-3-\dfrac{\left(x-3\right)\left(x+3\right)}{x-3}=x-3-\left(x+3\right)=-6\)
c)
Đk: x >/ 1 \(B=\dfrac{\sqrt{x+\sqrt{4\left(x-1\right)}}-\sqrt{x-\sqrt{4\left(x-1\right)}}}{\sqrt{x^2-4\left(x-1\right)}}\cdot\left(\sqrt{x-1}-\dfrac{1}{\sqrt{x-1}}\right)\)
\(=\dfrac{\sqrt{x+2\sqrt{x-1}}-\sqrt{x-2\sqrt{x-1}}}{\sqrt{x^2-4\left(x-1\right)}}\cdot\dfrac{x-2}{\sqrt{x-1}}\)
\(=\dfrac{\sqrt{x-1}+1-\left|\sqrt{x-1}-1\right|}{\left|x-2\right|}\cdot\dfrac{x-2}{\sqrt{x-1}}\)
Th1: \(x-2\ge0\Leftrightarrow x\ge2\)
\(B=\dfrac{\sqrt{x-1}+1-\sqrt{x-1}+1}{x-2}\cdot\dfrac{x-2}{\sqrt{x-1}}=\dfrac{2}{\sqrt{x-1}}\)
Th2: \(x-2\le0\Leftrightarrow x\le2\)
kết hợp với đk, ta được: 1 \< x \< 2
\(=\dfrac{\sqrt{x-1}+1-\sqrt{x-1}-1}{2-x}\cdot\dfrac{x-2}{\sqrt{x-1}}=0\)
d) \(A=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}=\sqrt{x-2}+\sqrt{2}+\left|\sqrt{x-2}-\sqrt{2}\right|=\sqrt{x-2}+\sqrt{2}-\sqrt{x-2}+\sqrt{2}=2\sqrt{2}\)
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a: \(P=\left(\dfrac{x\sqrt{x}+4x+4\sqrt{x}}{x+2\sqrt{x}+4}+\dfrac{4}{\sqrt{x}-2}-\dfrac{8\sqrt{x}+32}{x\sqrt{x}-8}\right)\cdot\dfrac{\sqrt{x}+2-2}{\sqrt{x}+2}\)
\(=\dfrac{x\sqrt{x}+4x+4\sqrt{x}+4x+8\sqrt{x}+16-8\sqrt{x}-32}{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}:\dfrac{\sqrt{x}+2}{\sqrt{x}}\)
\(=\dfrac{x\sqrt{x}+8x+4\sqrt{x}-16}{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+2}\)
\(=\dfrac{x\sqrt{x}+2x+6x+12\sqrt{x}-8\sqrt{x}-16}{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+2}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(x+6\sqrt{x}-8\right)}{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+2}\)
\(=\dfrac{\sqrt{x}\left(x+6\sqrt{x}-8\right)}{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}\)
b: Khi x=9-4 căn 5 thi \(P=\dfrac{\left(\sqrt{5}-2\right)\left(9-4\sqrt{5}+6\sqrt{5}-12-8\right)}{\left(\sqrt{5}-2-2\right)\left(9-4\sqrt{5}+2\sqrt{5}-4+4\right)}\)
\(=\dfrac{\left(\sqrt{5}-2\right)\left(-11+2\sqrt{5}\right)}{\left(\sqrt{5}-4\right)\left(9-2\sqrt{5}\right)}\)