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17 tháng 8 2020

Bài làm:

a) \(A=\left(\sqrt{3}+1\right)^2+\frac{5}{4}\sqrt{48}-\frac{2}{\sqrt{3+1}}\)

\(A=3+2\sqrt{3}+1+\sqrt{\frac{25.48}{16}}-\frac{2}{\sqrt{4}}\)

\(A=4+2\sqrt{3}+\sqrt{25.3}-\frac{2}{2}\)

\(A=4+2\sqrt{3}+5\sqrt{3}-1\)

\(A=3+7\sqrt{3}\)

b) \(\frac{4}{3-\sqrt{5}}-\frac{3}{\sqrt{5}+\sqrt{2}}-\frac{1}{\sqrt{2}-1}\)

\(=\frac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}-\frac{3\left(\sqrt{5}-\sqrt{2}\right)}{\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)}-\frac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)

\(A=\frac{4\left(3+\sqrt{5}\right)}{9-5}-\frac{3\left(\sqrt{5}-\sqrt{2}\right)}{5-2}-\frac{\sqrt{2}+1}{2-1}\)

\(A=3+\sqrt{5}-\sqrt{5}+\sqrt{2}-\sqrt{2}-1\)

\(A=2\)

17 tháng 8 2020

Phần b mình viết nhầm tên thành A, bn sửa thành B nhé

c) \(C=\sqrt{4-2\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)

\(C=\sqrt{3-2\sqrt{3}+1}-\sqrt{4+4\sqrt{3}+3}\)

\(C=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)

\(C=\sqrt{3}-1-2-\sqrt{3}\)

\(C=-3\)

6 tháng 8 2020

Bài 1

a, Với \(x=9\)thì \(A=\frac{3+\sqrt{x}}{\sqrt{x}}=\frac{3}{\sqrt{x}}+1=\frac{3}{3}+1=2\)

b, Để \(A=\frac{5}{2}\)thì \(\frac{3+\sqrt{x}}{\sqrt{x}}=\frac{3}{\sqrt{x}}+1=\frac{5}{2}< =>\frac{3}{\sqrt{x}}=\frac{3}{2}< =>x=4\)

Bài 2

a, \(B=\frac{\sqrt{x}-2}{\sqrt{x}}+\frac{4\sqrt{x}+2}{x+\sqrt{x}}\left(đk:x>0\right)\)

\(=1-\frac{2}{\sqrt{x}}+\frac{4\sqrt{x}+2}{x+\sqrt{x}}=\frac{x+5\sqrt{x}+2}{x+\sqrt{x}}-\frac{2}{\sqrt{x}}\)

\(=\frac{x\sqrt{x}+5x+2\sqrt{x}-2x-2\sqrt{x}}{x\sqrt{x}+x}=\frac{x\sqrt{x}+3x}{x\sqrt{x}+x}\)

\(=1+\frac{2x}{x\left(\sqrt{x}+1\right)}=1+\frac{2}{\sqrt{x}+1}=\frac{\sqrt{x}+3}{\sqrt{x}+1}\)

6 tháng 8 2020

\(A=\frac{3+\sqrt{x}}{\sqrt{x}}\)Thay x = 9 ta có : 

\(VT=\frac{3+\sqrt{9}}{\sqrt{9}}=\frac{3+3}{3}=2\)

Bài ra ta có : \(A=\frac{3+\sqrt{x}}{\sqrt{x}}=\frac{5}{2}\)

\(\Leftrightarrow\frac{3}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}}=\frac{5}{2}\Leftrightarrow\frac{3}{\sqrt{x}}+1=\frac{5}{2}\)

\(\Leftrightarrow\frac{3}{\sqrt{x}}=\frac{3}{2}\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)

28 tháng 3 2018

\(A=\left(\frac{2+\sqrt{x}}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{2-\sqrt{x}}-\frac{\sqrt{x}+2}{\sqrt{x}-3}\right)\) \(:\left(2-\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)

\(A=\left[\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+2}{\sqrt{x}-3}\right]\) 

 \(:\left[\frac{2\left(\sqrt{x}+1\right)-\sqrt{x}}{\sqrt{x}+1}\right]\)

\(A=\left[\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+2}{\sqrt{x}-3}\right]\)

\(:\left[\frac{2\sqrt{x}+2-\sqrt{x}}{\sqrt{x}+1}\right]\)

\(A=\left[\frac{\sqrt{x}+2+x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\right]\)  \(:\left[\frac{\sqrt{x}+2}{\sqrt{x}+1}\right]\)

\(A=\left[\frac{\sqrt{x}+x-7-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]:\frac{\sqrt{x}+2}{\sqrt{x}+1}\)

\(A=\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\frac{\sqrt{x}+2}{\sqrt{x}+1}\)

\(A=\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)

\(a,ĐKXĐ:a\ge0;a\ne4\)

\(P=\frac{\sqrt{a}+1}{\sqrt{a}-2}+\frac{2\sqrt{a}}{\sqrt{a}+2}-\frac{5\sqrt{a}+2}{a-4}\)

\(=\frac{\sqrt{a}+1}{\sqrt{a}-2}+\frac{2\sqrt{a}}{\sqrt{a}+2}-\frac{5\sqrt{a}+2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)

\(=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)+2\sqrt{a}\left(\sqrt{a}-2\right)-5\sqrt{a}-2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)

\(=\frac{a+3\sqrt{a}+2+2a-4\sqrt{a}-5\sqrt{a}-2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)

\(=\frac{3a-6\sqrt{a}}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)

\(=\frac{3\sqrt{a}\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=\frac{3\sqrt{a}}{\sqrt{a}+2}\)

\(b,P=2\Rightarrow\frac{3\sqrt{a}}{\sqrt{a}+2}=2\)

\(\Rightarrow3\sqrt{a}=2\left(\sqrt{a}+2\right)\)

\(\Rightarrow3\sqrt{a}=2\sqrt{a}+4\)

\(\Rightarrow3\sqrt{a}-2\sqrt{a}=4\)

\(\Rightarrow\sqrt{a}=4\)

\(\Rightarrow a=16\)