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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: \(B=\left(\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\right):\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+3\sqrt{x}+6-2\sqrt{x}-6}=1\)
Bài 1:
a: \(B=\dfrac{\sqrt{x}+x+\sqrt{x}-x}{1-x}\cdot\dfrac{x-1}{3-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}-3}\)
b: Để B=-1 thì \(2\sqrt{x}=-\sqrt{x}+3\)
=>3 căn x=3
=>căn x=1
hay x=1(loại)
\(1.a.A=\left(1-\dfrac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{1}{\sqrt{x}+1}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\left(x\ge0;x\ne4;x\ne9\right)\)
\(b.A< 0\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\)
\(\Leftrightarrow\sqrt{x}-2< 0\)
\(\Leftrightarrow x< 4\)
Kết hợp với ĐKXĐ , ta có : \(0\le x< 4\)
KL............
\(2.\) Tương tự bài 1.
\(3a.A=\dfrac{1}{x-\sqrt{x}+1}=\dfrac{1}{x-2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{4}{3}\)
\(\Rightarrow A_{Max}=\dfrac{4}{3}."="\Leftrightarrow x=\dfrac{1}{4}\)
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
a) A=\(\dfrac{\sqrt{x}[\left(\sqrt{x}\right)^3-1]}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
A=\(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\) A=\(\sqrt{x}\left(\sqrt{x}-1\right)-2\sqrt{x}-1+2\sqrt{x}+2\)
A=\(x-\sqrt{x}+1\)
b) A=\(\dfrac{3}{4}\)
=> \(x-\sqrt{x}+1=\dfrac{3}{4}\)
\(x-\sqrt{x}+\dfrac{1}{4}=0\)
\(\left(\sqrt{x}-\dfrac{1}{2}\right)^2=0\)
=> \(\sqrt{x}=\dfrac{1}{2}\)
=> \(x=\dfrac{1}{4}\)
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 : \(x=-\frac{11\pm\sqrt{23}i}{8}\)
\(a,A=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)
\(=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{2}{\sqrt{x}}\right)\)
\(=\frac{4\sqrt{x}.\left(2-\sqrt{x}\right)+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\frac{\sqrt{x}-1-2.\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{8\sqrt{x}-4x+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-1-2\sqrt{x}+4}\)
\(=\frac{\left(4x+8\sqrt{x}\right)\left(\sqrt{x}\right)\left(\sqrt{x}-2\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)\left(-\sqrt{x}+3\right)}\)
\(=\frac{-4\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}\right)\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)\left(-\sqrt{x}+3\right)}\)
\(=\frac{4x}{\sqrt{x}-3}\)
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.
a: \(T=1:\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right)\)
\(=1:\dfrac{x+2+x-1-x-\sqrt{x}-1}{x\sqrt{x}-1}\)
\(=\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
b: \(T-3=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\)
=>T>3
\(a,P=\dfrac{\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{2-\sqrt{x}}{\sqrt{x}}=\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{-2}{\sqrt{x}+2}\\ P=-\dfrac{3}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\\ \Leftrightarrow3\sqrt{x}+6=10\Leftrightarrow\sqrt{x}=\dfrac{4}{3}\Leftrightarrow x=\dfrac{16}{9}\left(tm\right)\)
\(P=-\dfrac{3}{5}\) sao suy ra đc \(\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\) thế