những ai thích xem minecraft và blockman go thì hãy xem kênh youtube của mik kênh mik là M.ichibi các bn nhớ sud và chia sẻ cho nhiều người khác nhé

a) \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\)

\(=\left[-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right]\cdot\left(\sqrt{2}-\sqrt{5}\right)\)

\(=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)

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

\(=-\left(2-5\right)\)

\(=-\left(-3\right)\)

\(=3\)

b) Ta có:

\(x^2-x\sqrt{3}+1\) 

\(=x^2-2\cdot\dfrac{\sqrt{3}}{2}\cdot x+\left(\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)

\(=\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)

Mà: \(\left(x-\dfrac{\sqrt{3}}{2}\right)^2\ge0\forall x\) nên

\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\forall x\)

Dấu "=" xảy ra:

\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}=\dfrac{1}{4}\)

\(\Leftrightarrow x=\dfrac{\sqrt{3}}{2}\)

Vậy: GTNN của biểu thức là \(\dfrac{1}{4}\) tại \(x=\dfrac{\sqrt{3}}{2}\)

a)

\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\\ =\left(-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(-\sqrt{2}-\sqrt{5}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}^2-\sqrt{5}^2\right)\\ =-\left(2-5\right)\\ =-\left(-3\right)\\ =3\)

\(2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}=9\)

Bài 1:

a) \(5\sqrt{\frac{1}{5}}+\frac{1}{3}\sqrt{45}+\frac{5-\sqrt{5}}{\sqrt{5}}\)

\(=\sqrt{5}+\sqrt{5}+\sqrt{5}-1\)

\(=3\sqrt{5}-1\)

b) \(\sqrt{48}-6\sqrt{\frac{1}{3}}+\frac{\sqrt{3}-3}{\sqrt{3}}\)

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

\(=\sqrt{3}+1\)

c) \(\left(\frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{5}{\sqrt{5}}\right)\div\left(\frac{1}{\sqrt{5}-\sqrt{2}}\right)\)

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

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

\(=-3\)

Bài 2:

đk: \(x\ge1\)

Ta có: \(\sqrt{4x+4}-\sqrt{9x-9}-8\sqrt{\frac{x+1}{16}}=5\)

\(\Leftrightarrow2\sqrt{x+1}-3\sqrt{x-1}-2\sqrt{x+1}=5\)

\(\Leftrightarrow-3\sqrt{x-1}=5\)

\(\Rightarrow\sqrt{x-1}=-\frac{5}{3}\) (vô lý)

=> PT vô nghiệm

\(A=\left(\sqrt{5}+3\right)\left(5-\sqrt{15}\right)=5\sqrt{5}-5\sqrt{3}+15-3\sqrt{15}\)

Bạn ghi nhầm đề thì phải, ngoặc đầu là \(\sqrt{5}+\sqrt{3}\) mới rút gọn được theo HĐT số 3

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

\(=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=27-2=25\)

\(C=1-\left(3\sqrt{5}-2\sqrt{5}-\sqrt{3}\right)\left(2\sqrt{5}-3\sqrt{5}-\sqrt{3}\right)\)

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

\(D=\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{2}{3}}\right).\sqrt{6}=\frac{\left(3-2\right)}{\sqrt{6}}.\sqrt{6}=1\)

a) Ta có: \(A=\sqrt{8-2\sqrt{15}}\cdot\left(\sqrt{3}+\sqrt{5}\right)-\left(\sqrt{45}-\sqrt{20}\right)\)

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

\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)

\(=\left|\sqrt{5}-\sqrt{3}\right|\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)

\(=\left(\sqrt{5}-\sqrt{3}\right)\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)(Vì \(\sqrt{5}>\sqrt{3}\))

\(=5-3-\sqrt{5}\)

\(=2-\sqrt{5}\)

b) Ta có: \(B=\left(\frac{\sqrt{21}-\sqrt{3}}{\sqrt{7}-1}-\frac{\sqrt{15}-\sqrt{3}}{1-\sqrt{5}}\right)\left(\frac{1}{2}\sqrt{6}-\sqrt{\frac{3}{2}}+3\sqrt{\frac{2}{3}}\right)\)

\(=\left(\frac{\sqrt{3}\left(\sqrt{7}-1\right)}{\sqrt{7}-1}+\frac{\sqrt{3}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{3}{2}}+\sqrt{6}\right)\)

\(=\sqrt{3}+\sqrt{3}+\sqrt{6}\)

\(=2\sqrt{3}+\sqrt{6}\)

c) Ta có: \(C=2\sqrt{3}+\sqrt{7-4\sqrt{3}}+\left(\sqrt{\frac{1}{3}}-\sqrt{\frac{4}{3}}+\sqrt{3}\right):\sqrt{3}\)

\(=2\sqrt{3}+\sqrt{4-2\cdot2\cdot\sqrt{3}+3}+\sqrt{\frac{1}{3}:3}-\sqrt{\frac{4}{3}:3}+\sqrt{3:3}\)

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

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

\(=2\sqrt{3}+2-\sqrt{3}+\frac{2}{3}\)(Vì \(2>\sqrt{3}\))

\(=\sqrt{3}+\frac{8}{3}\)

d) Ta có: \(D=\left(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\right):\frac{1}{\sqrt{7-4\sqrt{3}}}\)

\(=\left(\frac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\right)\cdot\sqrt{4-2\cdot2\cdot\sqrt{3}+3}\)

\(=\frac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\cdot\sqrt{\left(2-\sqrt{3}\right)^2}\)

\(=\frac{60}{20}\cdot\left|2-\sqrt{3}\right|\)

\(=3\cdot\left(2-\sqrt{3}\right)\)(Vì \(2>\sqrt{3}\))

\(=6-3\sqrt{3}\)

mik làm bài này

linh tinh

bn ơi

cho mik

xin 1 L-I-K-E

b,  
 
 
 
 

d, 
 
 
 
 
 
 
 
 
 

\(\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}\)

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

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

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

\(=8\)