\(5\sqrt{34}+\left|6-\sqrt{34}\right|\)

\(6>\sqrt{34}\)

\(5\sqrt{34}+6-\sqrt{34}\)

\(4\sqrt{34}+6\)

=) ko bt

a) Ta có:

\(\sqrt{\frac{289}{225}}=\sqrt{\frac{\sqrt{289}}{\sqrt{225}}}=\sqrt{\frac{17^2}{15^2}}=\frac{17}{15}\)

b) Ta có:

\(\sqrt{2\frac{14}{25}}=\sqrt{\frac{64}{25}}=\sqrt{\frac{\sqrt{64}}{\sqrt{25}}}=\sqrt{\frac{8^2}{5^2}}=\frac{8}{5}\)

c) Ta có:

\(\sqrt{\frac{0,25}{9}}=\sqrt{\frac{\sqrt{0,25}}{\sqrt{9}}}=\sqrt{\frac{0,5^2}{3^2}}=\frac{0,5}{3}=\frac{1}{6}\)

d) Ta có:

\(\sqrt{\frac{8,1}{1,6}}=\sqrt{\frac{81.0,1}{16.0,1}}=\sqrt{\frac{81}{16}}=\sqrt{\frac{\sqrt{81}}{\sqrt{16}}}=\sqrt{\frac{9^2}{4^2}}=\frac{9}{4}\)

a)Ta có: \(\sqrt{\frac{289}{225}}=\frac{\sqrt{289}}{\sqrt{225}}=\frac{17}{15}\)

b) Ta có: \(\sqrt{2\frac{14}{25}}=\sqrt{\frac{64}{25}}=\frac{\sqrt{64}}{\sqrt{25}}=\frac{8}{5}\)

c) Ta có: \(\sqrt{\frac{0,25}{9}}=\frac{\sqrt{0,25}}{\sqrt{9}}=\frac{0,5}{3}=\frac{1}{6}\)

d)Ta có : \(\sqrt{\frac{8,1}{1,6}}=\frac{\sqrt{8,1}}{\sqrt{1,6}}=\frac{\sqrt{8,1}.100}{\sqrt{1,6}.100}=\frac{\sqrt{81}}{\sqrt{16}}=\frac{9}{4}\)

\(\sqrt{x}=0\)  đúng 

Vì 

\(\sqrt{x}\ge0\forall x\)

\(\sqrt{x}=0\)  là đúng

=> Vì căn của 0 = 0

a) đk: \(\hept{\begin{cases}x\ge1\\x\ne3\end{cases}}\)

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

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

b) \(x=4\left(2-\sqrt{3}\right)\Rightarrow x-1=7-4\sqrt{3}=\left(2-\sqrt{3}\right)^2\)

\(\Rightarrow P=\sqrt{x-1}+\sqrt{2}=2-\sqrt{3}+\sqrt{2}\)

a) \(B=\left(\frac{\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}\right)\div\frac{\sqrt{x}+2}{x-4}\)

\(=\frac{\sqrt{x}+\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\frac{x-4}{\sqrt{x}+2}\)

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

b) \(C=A\left(B-2\right)=\frac{\sqrt{x}+2}{\sqrt{x}-2}\cdot\left(\frac{2\sqrt{x}+2}{\sqrt{x}+2}-2\right)\)

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

Để C nguyên => \(2-\sqrt{x}\inƯ\left(2\right)\Rightarrow2-\sqrt{x}\in\left\{\pm1;\pm2\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{0;1;3;4\right\}\Leftrightarrow x\in\left\{0;1;9;16\right\}\)

Đề 1:

Câu 1: Chọn \(C.\)\(9cm\).

Câu 2: Chọn \(D.\)\(25cm\).

Câu 3: Chọn \(B.\)\(BA^2=BC.BH\).

Câu 4: Chọn \(C.\)\(\sqrt{HB.HC}\).

Đề 2:

Câu 1: Chọn \(B.\)\(8\).

Câu 2: Chọn \(B.\)\(6\).

Câu 3: Chọn \(C.\)\(8\).

Câu 4: Chọn \(A.\)\(\frac{1}{AI^2}=\frac{1}{AD^2}+\frac{1}{4HC^2}\).

chịu rồi , nhiều quá

Câu 1 

\(AC^2=CH\cdot CB\)   

\(6^2=4\cdot BC\)   

\(36=4\cdot BC\)   

\(BC=9\)   ( Chọn C )