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\(A=\left(\dfrac{x+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{2\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\right)\cdot\dfrac{\sqrt{x}+2-\sqrt{x}+3}{\sqrt{x}+2}\)
\(=\dfrac{x+2-2x+4\sqrt{x}+x-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{5}{\sqrt{x}+2}\)
\(=\dfrac{5\left(4\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}\)
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)
\(\dfrac{\left(\sqrt{x^2+4}-2\right)\left(\sqrt{x^2+4}-2\right)\left(x+\sqrt{x}+1\right)\sqrt{x-2\sqrt{x}+1}}{x\left(x\sqrt{x}-1\right)}\\=\dfrac{\left(\left(\sqrt{x^2+4}\right)^2-4\right)\left(\left(x+\sqrt{x}+1\right)\sqrt{\left(x-1\right)^2}\right)}{x\left(x\sqrt{x}-1\right)}\\ =\dfrac{\left(x^2+4-4\right)\left(\left(x+\sqrt{x}+1\right)\left(x-1\right)\right)}{x\left(x\sqrt{x}-1\right)}\\ =\dfrac{x^2\left(x^3-1\right)}{x\left(x\sqrt{x}-1\right)}=x^2\sqrt{x}\)
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) \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}\)
\(=\dfrac{4a^2b^3}{8\sqrt{2}a^3b^3}\)
\(=\dfrac{1}{2\sqrt{2}a}\)
\(=\dfrac{\sqrt{2}}{4a}\)
b) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\)
chịu đấy :v
c) \(\sqrt{\dfrac{\left(x-2\right)^2}{\left(3-x\right)^2}}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{x-2}{3-x}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{x-2}{-\left(x-3\right)}+\dfrac{x^2-1}{x-3}\)
\(=-\dfrac{x-2}{x-3}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{-\left(x-2\right)+x^2-1}{x-3}\)
\(=\dfrac{-x+1+x^2}{x-3}\)
d) \(\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(y-2\sqrt{y}+1^2\right)}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{y-2\sqrt{y}+1}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y-2\sqrt{y}+1}}{\left(x-1\right)^2}\)
\(=\dfrac{1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y-2\sqrt{y}+1}}{x-1}\)
\(=\dfrac{\sqrt{y-2\sqrt{y}+1}}{\left(\sqrt{y}-1\right)\left(x-1\right)}\)
\(=\dfrac{\sqrt{y-2\sqrt{y}+1}}{x\sqrt{y}-\sqrt{y}-x+1}\)
e) \(4x-\sqrt{8}+\dfrac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\)
\(=4x-2\sqrt{2}+\dfrac{\sqrt{x^2\cdot\left(x+2\right)}}{\sqrt{x+2}}\)
\(=4x-2\sqrt{2}+\sqrt{x^2}\)
\(=4x-2\sqrt{x}+x\)
\(=5x-2\sqrt{2}\)
Bài 1 : Rút gọn biểu thức :
\(\left(2-\sqrt{2}\right)\left(-5\sqrt{2}\right)-\left(3\sqrt{2}-5\right)^2\)
\(=\left(-10\sqrt{2}+10\right)-\left(18-30\sqrt{2}+25\right)\)
\(=\left(-10\sqrt{2}+10\right)-\left(7-30\sqrt{2}\right)\)
\(=-10\sqrt{2}+10-7+30\sqrt{2}\)
\(=20\sqrt{2}+3\)
Bài 2:
a) ĐKXĐ : x # 4 ; x # - 4
P = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)
P =\(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2+5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{x+2\sqrt{x}+\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
P = \(\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
b ) Để P = 2 \(\Leftrightarrow\dfrac{3\sqrt{x}}{\sqrt{x}+2}\) = 2
\(\Leftrightarrow3\sqrt{x}=2\sqrt{x}+4\)
\(\Leftrightarrow\sqrt{x}=4\)
\(\Leftrightarrow x=16\)
Vậy, để P = 2 thì x = 16.
Bài 2:
a: \(P=\dfrac{a-1}{2\sqrt{a}}\cdot\left(\dfrac{\sqrt{a}\left(a-2\sqrt{a}+1\right)-\sqrt{a}\left(a+2\sqrt{a}+1\right)}{a-1}\right)\)
\(=\dfrac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{2}=-2\sqrt{a}\)
b: Để P>=-2 thì P+2>=0
\(\Leftrightarrow-2\sqrt{a}+2>=0\)
=>0<=a<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 \(x=\dfrac{4}{29+12\sqrt{5}}=\dfrac{4\left(29-12\sqrt{5}\right)}{121}\)
a, Với \(x>0;x\ne4;x\ne9\)
\(A=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)
\(=\left(\frac{4\sqrt{x}\left(2-\sqrt{x}\right)+8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(=\left(\frac{8\sqrt{x}-4x+8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{8\sqrt{x}+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}:\frac{-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{4\sqrt{x}\left(2+\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\frac{3-\sqrt{x}}{\sqrt{x}\left(2-\sqrt{x}\right)}=\frac{4\sqrt{x}}{2-\sqrt{x}}.\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{3-\sqrt{x}}=\frac{4x}{3-\sqrt{x}}\)
b, Ta có : A = -2 hay
\(\frac{4x}{3-\sqrt{x}}=-2\Rightarrow4x=-6+2\sqrt{x}\)
\(\Leftrightarrow4x+6-2\sqrt{x}=0\Leftrightarrow2\left(2x+3-\sqrt{x}\right)=0\)
\(\Leftrightarrow2x+3-\sqrt{x}=0\Leftrightarrow\sqrt{x}=2x+3\)
bình phương 2 vế ta có :
\(x=\left(2x+3\right)^2=4x^2+12x+9\)
\(\Leftrightarrow-4x^2-11x-9=0\)giải delta ta thu được : \(x=-\frac{11\pm\sqrt{23}i}{8}\)
\(a,A=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)
\(=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{2}{\sqrt{x}}\right)\)
\(=\frac{4\sqrt{x}.\left(2-\sqrt{x}\right)+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\frac{\sqrt{x}-1-2.\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{8\sqrt{x}-4x+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-1-2\sqrt{x}+4}\)
\(=\frac{\left(4x+8\sqrt{x}\right)\left(\sqrt{x}\right)\left(\sqrt{x}-2\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)\left(-\sqrt{x}+3\right)}\)
\(=\frac{-4\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}\right)\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)\left(-\sqrt{x}+3\right)}\)
\(=\frac{4x}{\sqrt{x}-3}\)
\(A=\left(\dfrac{x+2}{x-\sqrt{x}-2}-\dfrac{2\sqrt{x}}{\sqrt{x}+1}-\dfrac{1-\sqrt{x}}{\sqrt{x}-2}\right)\left(1-\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\right)\\ =\left(\dfrac{x+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\dfrac{2\sqrt{x}}{\sqrt{x}+1}-\dfrac{1-\sqrt{x}}{\sqrt{x}-2}\right)\left(\dfrac{\sqrt{x}-2}{\sqrt{x}-2}-\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\right)\\ =\left(\dfrac{x+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\dfrac{\left(1-\sqrt{x}\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-2}\\ =\dfrac{x+2-\left(2x-4\sqrt{x}\right)-\left(\sqrt{x}+1-x-\sqrt{x}\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{1}{\sqrt{x}-2}\)
\(=\dfrac{x+2-2x+4\sqrt{x}-\sqrt{x}-1+x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{1}{\sqrt{x}-2}\\ =\dfrac{4\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{1}{\sqrt{x}-2}\\ =\dfrac{4\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)^2}\)
\(A=\left(\dfrac{x+2}{x-\sqrt{x}-2}-\dfrac{2\sqrt{x}}{\sqrt{x}+1}-\dfrac{1-\sqrt{x}}{\sqrt{x}-2}\right)\left(1-\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\right)\)
\(A=\left[\dfrac{x+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\dfrac{\left(1-\sqrt{x}\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\right]\left(\dfrac{\sqrt{x}-2}{\sqrt{x}-2}-\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\right)\)
\(A=\dfrac{x+2-2x+4\sqrt{x}+x-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-2}\)
\(A=\dfrac{4\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{1}{\sqrt{x}-2}\)
\(A=\dfrac{4\sqrt{x}+1}{\left(\sqrt{x}-2\right)^2\left(\sqrt{x}+1\right)}\)