a) \(5^{3n+2}=25^{n+3}\)

\(\Leftrightarrow5^{3n+2}=5^{2\left(n+3\right)}\)

\(\Leftrightarrow3n+2=2\left(n+3\right)\)

\(\Leftrightarrow3n+2=2n+6\)

\(\Leftrightarrow n=4\)

b) \(6.5^{n-2}+10^2=2.5^3\left(n>1\right)\)

\(\Leftrightarrow6.5^{n-2}=2.5^3-10^2\)

\(\Leftrightarrow6.5^{n-2}=2.5^3-2^2.5^2\)

\(\Leftrightarrow6.5^{n-2}=2.5^2\left(5-2\right)\)

\(\Leftrightarrow6.5^{n-2}=2.5^2.3\)

\(\Leftrightarrow5^{n-2}=5^2\)

\(\Leftrightarrow n-2=2\)

\(\Leftrightarrow n=4\)

b: Ta có: \(\left(\sqrt{7-3\sqrt{5}}\right)\cdot\left(7+3\sqrt{5}\right)\cdot\left(3\sqrt{2}+\sqrt{10}\right)\)

\(=\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\left(7+3\sqrt{5}\right)\)

\(=4\left(7+3\sqrt{5}\right)\)

\(=28+12\sqrt{5}\)

Lời giải:

a. 

$A=\sqrt{8+\sqrt{55}}-\sqrt{8-\sqrt{55}}-\sqrt{125}$
$\sqrt{2}A=\sqrt{16+2\sqrt{55}}-\sqrt{16-2\sqrt{55}}-\sqrt{250}$

$=\sqrt{(\sqrt{11}+\sqrt{5})^2}-\sqrt{(\sqrt{11}-\sqrt{5})^2}-5\sqrt{10}$

$=|\sqrt{11}+\sqrt{5}|-|\sqrt{11}-\sqrt{5}|-5\sqrt{10}$

$=2\sqrt{5}-5\sqrt{10}$

$\Rightarrow A=\sqrt{10}-5\sqrt{5}$

b.

$B=\sqrt{7-3\sqrt{5}}.(7+3\sqrt{5})(3\sqrt{2}+\sqrt{10})$

$B\sqrt{2}=\sqrt{14-6\sqrt{5}}(7+3\sqrt{5})(3\sqrt{2}+\sqrt{10})$

$=\sqrt{(3-\sqrt{5})^2}(7+3\sqrt{5}).\sqrt{2}(3+\sqrt{5})$

$=(3-\sqrt{5})(7\sqrt{2}+3\sqrt{10})(3+\sqrt{5})$

$=(3^2-5)(7\sqrt{2}+3\sqrt{10})$

$=4(7\sqrt{2}+3\sqrt{10})=28\sqrt{2}+12\sqrt{10}$

$\Rightarrow B=28+12\sqrt{5}$

c.

$C=\sqrt{2}(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{6+\sqrt{35}}$

$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{12+2\sqrt{35}}$

$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{(\sqrt{7}+\sqrt{5})^2}

$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})(\sqrt{7}+\sqrt{5})$

$=(7-5)(6-\sqrt{35})$

$=2(6-\sqrt{35})=12-2\sqrt{35}$

1. \(=\sqrt{\left(\sqrt{\frac{7}{2}}+\sqrt{\frac{3}{2}}\right)^2}+\sqrt{\left(\sqrt{\frac{7}{2}}-\sqrt{\frac{3}{2}}\right)^2}-2\sqrt{4\sqrt{7}}=\frac{7}{2}+\frac{3}{2}+\frac{7}{2}-\frac{3}{2}-2\sqrt{4\sqrt{7}}\)

\(=7-2\sqrt{4\sqrt{7}}\)

cho hỏi tại sao có số \(\frac{7}{2};\frac{3}{2}\)zậy chỉ với

tinh den moi tay a

Ngu vl sử dụng công thức để tính trên mtct gà

1000010000803

chúc bạn học tốt~

Sai rồi họk toán vs đamê à

\(3A=9+99+....+99....99=10-1+10^2-1+...+10^{50}-1\)

\(=\left(10+10^2+...+10^{50}\right)-50\)

Đến đây dễ hơn rồi.

\(\frac{9}{5}B=9+99+...+99.99\)tương tự A

C tương tự A.

tự làm

50) \(\sqrt{98-16\sqrt{3}}=4\sqrt{6}-\sqrt{2}\)

51) \(\sqrt{2-\sqrt{3}}=\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}=\dfrac{\sqrt{3}-1}{\sqrt{2}}=\dfrac{\sqrt{6}-\sqrt{2}}{2}\)

52) \(\sqrt{4+\sqrt{15}}=\dfrac{\sqrt{8+2\sqrt{15}}}{\sqrt{2}}=\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{10}+\sqrt{6}}{2}\)

53) \(\sqrt{5-\sqrt{21}}=\dfrac{\sqrt{10-2\sqrt{21}}}{\sqrt{2}}=\dfrac{\sqrt{14}-\sqrt{6}}{2}\)

54) \(\sqrt{6-\sqrt{35}}=\dfrac{\sqrt{12-2\sqrt{35}}}{\sqrt{2}}=\dfrac{\sqrt{14}-\sqrt{10}}{2}\)

55) \(\sqrt{2+\sqrt{3}}=\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}=\dfrac{\sqrt{6}+\sqrt{2}}{2}\)

56) \(\sqrt{4-\sqrt{15}}=\dfrac{\sqrt{8-2\sqrt{15}}}{\sqrt{2}}=\dfrac{\sqrt{10}-\sqrt{6}}{2}\)

Can bac 8

  2x1=2     3x1=3     4x1=4      5x1=5     2:2= 1     3:3= 1    4:4=1    5:5=1

   2x2= 4    3x2=6     4x2= 8     5x2=10     4:2=   2   6:3= 2     8:4=2    10:5=2

   2x3=6     3x3= 9    4x3=12      5x3=15     6:2=3      9:3= 3     12:4=3  15:5=3

   2x4=8    3x4=12     4x4=16      5x4=20     8:2=4      12:3=4    16:4=4  20:5=4

   2x5=10     3x5= 15    4x5= 20     5x5= 25    10:2=5   15:3=  5  20:4=5  25:5=5

   2x6=12     3x6=18     4x6= 24     5x6=30     12:2=6    18:3= 6   24:4=6  30:5=6

   2x7= 14    3x7=21     4x7= 28     5x7=35     14:2=7    21:3=7    28:4=7  35:5=7

   2x8=16     3x8=24     4x8= 32     5x8=50     16:2=8    24:3=8    32:4=8  40:5=8

   2x9= 18    3x9= 27    4x9= 36     5x9=45     18:2=9    27:3= 9   36:9=9  45:5=9

   2x10=20   3x10=30   4x10= 40   5x10=50   20:2=10    30:3=10    40:4= 10 50:5=10