\(0,1.\sqrt{400}+0,2.\sqrt{1600}\\ =0,1.20+0,2.40\\ =2+8\\ =10\)

=10

Thấy em giỏi chx lớp 6 làm đc bài lớp 7

\(0,1.\sqrt{400}+0,2.\sqrt{1600}=0,1.20+0,2.40\)\(=2+8=10\)
 

\(\begin{array}{l}a)\sqrt {1600}  = 40;\\b)\sqrt {0,16}  = 0,4;\\c)\sqrt {2\frac{1}{4}}  = \sqrt {\frac{9}{4}}  = \frac{3}{2}\end{array}\)

\(\text{a)}\sqrt{1600}=40\)

\(\text{b)}\sqrt{0,16}=0,4\)

\(\text{c)}\sqrt{2\dfrac{1}{4}}=\sqrt{\dfrac{9}{4}}=\dfrac{3}{2}\)

\(\begin{array}{l}a)\sqrt {0,49}  + \sqrt {0,64}  = 0,7 + 0,8 = 1,5;\\b)\sqrt {0,36}  - \sqrt {0,81}  = 0,6 - 0,9 =  - 0,3;\\c)8.\sqrt 9  - \sqrt {64}  = 8.3 - 8 = 24 - 8 = 16;\\d)0,1.\sqrt {400}  + 0,2.\sqrt {1600}  = 0,1.20 + 0,2.40 = 2 + 8 = 10\end{array}\)

AB=AC \(\Rightarrow\Delta ABC\) cân tại A

\(\Rightarrow AH\) đồng thời là phân giác và trung tuyến

\(\Rightarrow\left\{{}\begin{matrix}\widehat{BAH}=\dfrac{1}{2}\widehat{A}=60^0\\BH=\dfrac{1}{2}BC=6\end{matrix}\right.\)

Trong tam giác vuông ABH:

\(tan\widehat{BAH}=\dfrac{BH}{AH}\Rightarrow AH=\dfrac{BH}{tan\widehat{BAH}}=\dfrac{6}{tan60^0}=2\sqrt{3}\)

\(a,=2\sqrt{3}+9\sqrt{3}-10\sqrt{3}=\sqrt{3}\\ b,=\left|1-\sqrt{2}\right|+\sqrt{5}=\sqrt{2}-1+\sqrt{5}\)

a)\(\sqrt{12}+3\sqrt{27}-\sqrt{300}=2\sqrt{3}+9\sqrt{3}-10\sqrt{3}=\sqrt{3}\)

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

Xét phân thức phụ sau:

Ta có: \(\frac{1}{n\sqrt{n+1}+\left(n+1\right)\sqrt{n}}=\frac{1}{\sqrt{n\left(n+1\right)}}\cdot\left(\frac{1}{\sqrt{n}+\sqrt{n+1}}\right)\)

\(=\frac{1}{\sqrt{n\left(n+1\right)}}\cdot\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\frac{1}{\sqrt{n\left(n+1\right)}}\cdot\left(\sqrt{n+1}-\sqrt{n}\right)\)

\(=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

Thay vào ta được:

\(BT=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{399}}-\frac{1}{\sqrt{400}}\)

\(BT=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{400}}=1-\frac{1}{20}=\frac{19}{20}\)

Đặt biểu thức đã cho là A

Tổng quát ta có: Với \(a\inℕ^∗\)ta có:

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

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

\(=\frac{\sqrt{a+1}}{\sqrt{a}.\sqrt{a+1}}-\frac{\sqrt{a}}{\sqrt{a}.\sqrt{a+1}}=\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{a+1}}\)

Áp dụng kết quả trên ta có:

Với \(n=1\)\(\Rightarrow\frac{1}{2\sqrt{1}+1\sqrt{2}}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}\)

Với \(n=2\)\(\Rightarrow\frac{1}{3\sqrt{2}+2\sqrt{3}}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\)

Với \(n=3\)\(\Rightarrow\frac{1}{4\sqrt{3}+3\sqrt{4}}=\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}\)

.....................

Với \(n=399\)\(\Rightarrow\frac{1}{400\sqrt{399}+399\sqrt{400}}=\frac{1}{\sqrt{399}}-\frac{1}{\sqrt{400}}\)

\(\Rightarrow A=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+......+\frac{1}{\sqrt{399}}-\frac{1}{\sqrt{400}}\)

\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{400}}=1-\frac{1}{20}=\frac{19}{20}\)

\(E=2\sqrt{3}+3\sqrt{3^3}-\sqrt{100.3}\\ =2\sqrt{3}+9\sqrt{3}-10\sqrt{3}\\ =\left(2+9-10\right)\sqrt{3}=\sqrt{3}\)

\(F=\sqrt{3^2.2}+4\sqrt{18}=\sqrt{18}+4\sqrt{18}=\left(1+4\right)\sqrt{18}=5\sqrt{18}\)

\(G=2\sqrt{3}-4\sqrt{3^3}+5\sqrt{4^2.3}=2\sqrt{3}-12\sqrt{3}+20\sqrt{3}=\left(2-12+20\right)\sqrt{3}=10\sqrt{3}\)

\(H=\left(3\sqrt{25.2}-5\sqrt{9.2}+3\sqrt{2^3}\right)\sqrt{2}\\ =\left(15\sqrt{2}-15\sqrt{2}+6\sqrt{2}\right)\sqrt{2}\\ =6\sqrt{2}.\sqrt{2}=6\)

a, \(\dfrac{2}{5}\sqrt{75}-0,5\sqrt{48}+\sqrt{300}-\dfrac{2}{3}\sqrt{12}=2\sqrt{3}-2\sqrt{3}+10\sqrt{3}-\dfrac{4}{3}\sqrt{3}=\dfrac{26}{3}\sqrt{3}\)

b, \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\dfrac{3}{3+\sqrt{6}}=\dfrac{\sqrt{3}\left(3\sqrt{3}-2\right)}{\sqrt{2}\left(3\sqrt{3}-2\right)}+\dfrac{3}{\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)}\)

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

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

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

c, \(3\sqrt{2}-2\sqrt{3}+2\sqrt{3}+3\sqrt{2}=6\sqrt{2}\)

d, \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\sqrt{\left(\sqrt{6}-3\right)^2}+\sqrt{\left(2\sqrt{6}+3\right)^2}\)

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

e, Ghi đúng đề.

\(\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}=\dfrac{a+b-2\sqrt{ab}+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)

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

\(=\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}=2\sqrt{b}\)