2) \(\dfrac{\left(1+\sqrt{a}\right)^2-\left(2-\sqrt{a}\right)^2}{1-2\sqrt{a}}:\dfrac{\sqrt{a}}{3}\left(a>0,a\ne\dfrac{1}{4}\right)\)

\(=\dfrac{\left(1+\sqrt{a}-2+\sqrt{a}\right)\left(1+\sqrt{a}+2-\sqrt{a}\right)}{1-2\sqrt{a}}.\dfrac{3}{\sqrt{a}}\)

\(=\dfrac{3.\left(2\sqrt{a}-1\right)}{1-2\sqrt{a}}.\dfrac{3}{\sqrt{a}}=-\dfrac{9}{\sqrt{a}}\)

5) \(\left(5-\dfrac{a+3\sqrt{a}}{\sqrt{a}+3}\right)\left(2-\dfrac{3a+\sqrt{a}}{3\sqrt{a}+1}\right)\left(a\ge0\right)\)

\(=\left(5-\dfrac{\sqrt{a}\left(\sqrt{a}+3\right)}{\sqrt{a}+3}\right)\left(2-\dfrac{\sqrt{a}\left(3\sqrt{a}+1\right)}{3\sqrt{a}+1}\right)\)

\(=\left(5-\sqrt{a}\right)\left(2-\sqrt{a}\right)=10-7\sqrt{a}+a\)

6) \(\left(2-\dfrac{a-3\sqrt{a}}{\sqrt{a}-3}\right)\left(2-\dfrac{5\sqrt{a}-\sqrt{ab}}{\sqrt{b}-5}\right)\left(a,b\ge0,a\ne9,b\ne25\right)\)

\(=\left(2-\dfrac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}-3}\right)\left(2+\dfrac{\sqrt{a}\left(\sqrt{b}-5\right)}{\sqrt{b}-5}\right)\)

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

Ta có: \(BC=\sqrt{AB^2+AC^2}=\sqrt{6^2+8^2}=10\)

Vì CD là phân giác trong góc C 

\(\Rightarrow\dfrac{AD}{BD}=\dfrac{AC}{BC}\Rightarrow\dfrac{AD}{BD}=\dfrac{8}{10}=\dfrac{4}{5}\Rightarrow BD=\dfrac{5}{4}AD\)

Ta có: \(AD+BD=AB\Rightarrow AD+\dfrac{5}{4}AD=6\Rightarrow\dfrac{9}{4}AD=6\Rightarrow AD=\dfrac{8}{3}\)

Vì CD,CE lần lượt là phân giác trong và ngoài góc C

\(\Rightarrow CD\bot CE\Rightarrow\Delta DCE\) vuông tại C có \(AC\bot DE\)

\(\Rightarrow AD.AE=AC^2\Rightarrow AE=\dfrac{AC^2}{AD}=\dfrac{8^2}{\dfrac{8}{3}}=24\)

a) Dễ dàng thấy số mà học sinh thứ 2 mang lớn hơn số mà học sinh thứ 1 mang 4 đơn vị và tương tự cho các học sinh sau

\(\Rightarrow\) số học sinh \(=\dfrac{4001-1}{4}+1=1001\) ((số cuối -số đầu)/khoảng cách+1)

b) \(\Rightarrow\) có tổng cổng \(1001+30=1031\) (người tham gia)

\(\Rightarrow\) cần \(\left[\dfrac{1031}{45}\right]+1=23\) (xe) (\(\left[x\right]\) là phần nguyên của x)

Bài 1:1) \(10+2\sqrt{10}=\sqrt{10}\left(2+\sqrt{10}\right)\)

2) \(7+3\sqrt{7}=\sqrt{7}\left(3+\sqrt{7}\right)\)

các câu 3,4,5 bạn làm tương tự như 2 câu trên

6) \(3a\sqrt{b}+3b\sqrt{a}=3\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)\)

7) \(a^2-2a\sqrt{2}+2=a^2-2a\sqrt{2}+\left(\sqrt{2}\right)^2=\left(a-\sqrt{2}\right)^2\)

8) \(b-4=\left(\sqrt{b}\right)^2-2^2=\left(\sqrt{b}-2\right)\left(\sqrt{b}+2\right)\)

Bài 2: 1) \(\dfrac{\left(2-\sqrt{a}\right)^2-\left(\sqrt{a}+3\right)^2}{2a+\sqrt{a}}\left(a>0\right)\)

\(=\dfrac{\left(2-\sqrt{a}-\sqrt{a}-3\right)\left(2-\sqrt{a}+\sqrt{a}+3\right)}{\sqrt{a}\left(2\sqrt{a}+1\right)}\)

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

3) \(\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{\sqrt{a}-2}\left(a\ge0,a\ne4\right)\)

\(=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}+\dfrac{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}{\sqrt{a}-2}=\sqrt{a}+2-\left(2+\sqrt{a}\right)=0\)

4) \(\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)\left(1-\dfrac{a+\sqrt{a}}{1+\sqrt{a}}\right)\left(a\ge0,a\ne1\right)\)

\(=\left(1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\left(1-\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right)\)

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

mấy câu còn lại bạn làm tương tự

D.worrying

\(\sqrt{3+\sqrt{13+\sqrt{48}}}=\sqrt{3+\sqrt{13+4\sqrt{3}}}=\sqrt{3+\sqrt{\left(2\sqrt{3}\right)^2+2.2\sqrt{3}.1+1^2}}\)

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

\(=\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1^2}=\sqrt{\left(\sqrt{3}+1\right)^2}=\left|\sqrt{3}+1\right|=\sqrt{3}+1\)

\(U=\left(\dfrac{1}{2-\sqrt{5}}+\dfrac{2}{\sqrt{5}+\sqrt{3}}\right):\dfrac{1}{\sqrt{21-12\sqrt{3}}}\)

\(=\left(\dfrac{2+\sqrt{5}}{\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)}+\dfrac{2\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}\right):\dfrac{1}{\sqrt{\left(2\sqrt{3}\right)^2-2.2\sqrt{3}.3+3^2}}\)

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

\(=\left(-2-\sqrt{5}+\sqrt{5}-\sqrt{3}\right).\left|2\sqrt{3}-3\right|=\left(-2-\sqrt{3}\right)\left(2\sqrt{3}-3\right)\)

\(=-\sqrt{3}\)

\(M=\sqrt{\dfrac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\dfrac{4}{\left(2+\sqrt{5}\right)^2}}=\dfrac{2}{\left|2-\sqrt{5}\right|}-\dfrac{2}{\left|2+\sqrt{5}\right|}\)

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

\(=\dfrac{8}{1}=8\)

a) \(\dfrac{2}{\sqrt{10}-3}+\dfrac{2}{\sqrt{10}+3}=\dfrac{2\left(\sqrt{10}+3\right)+2\left(\sqrt{10}-3\right)}{\left(\sqrt{10}-3\right)\left(\sqrt{10}+3\right)}\)

\(=\dfrac{4\sqrt{10}}{1}=4\sqrt{10}\)

b) \(\dfrac{3-2\sqrt{3}}{\sqrt{3}}-\dfrac{2}{\sqrt{3}-1}=\dfrac{\sqrt{3}-2}{1}-\dfrac{2}{\sqrt{3}-1}\)

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

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

c) \(\dfrac{3}{\sqrt{5}-2}+\dfrac{3}{\sqrt{5}+2}-\dfrac{5-\sqrt{5}}{\sqrt{5}-1}\)

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

\(=\dfrac{6\sqrt{5}}{1}-\sqrt{5}=5\sqrt{5}\)

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

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

mấy câu dưới bạn làm tương tự