1)           P = 2/3.5 + 2/5.7 + 2/7.9 + 2/9.11 + 2/11.13 + 2/13.15

              P= (1/3-1/5) + (1/5-1/7) + (1/7-1/9) + (1/9-1/11) + (1/11-1/13) + (1/13-1/15)

              P=1/3-1/15= 4/15

2) a/ 0,2:1+3/5+80%

= 2/10:8/5+8/10

= 2/10.5/8+8/10

= 1/8 + 4/5 = 5/40 + 32/40 = 37/40

    b/ 0,5:5/4-2+1/5

= 5/10:5/4-11/5

= 5/10.4/5-11/5

=2/5-11/5 = -9/5

a)Ta có:

\(A=4\frac{25}{16}+25\left(\frac{9}{16}:\frac{125}{64}\right):\frac{-27}{8}\)

\(\Rightarrow A=\frac{89}{16}+25.\frac{36}{125}:\frac{-27}{8}\)

\(\Rightarrow A=\frac{89}{16}+\frac{36}{5}:\frac{-27}{8}\)

\(\Rightarrow A=\frac{89}{16}+\frac{-32}{15}\)

\(\Rightarrow A=\frac{823}{240}\)

Vậy A=.....

b)Ta có:

\(C=\frac{2^3}{3.5}+\frac{2^3}{5.7}+\frac{2^3}{7.9}+...+\frac{2^3}{101.103}\)

\(\Rightarrow C=\frac{2^2.2}{3.5}+\frac{2^2.2}{5.7}+\frac{2^2.2}{7.9}+...+\frac{2^2.2}{101.103}\)

\(\Rightarrow C=2^2\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{101.103}\right)\)

\(\Rightarrow C=4\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{101}-\frac{1}{103}\right)\)

\(\Rightarrow C=4\left(\frac{1}{3}-\frac{1}{103}\right)\)

\(\Rightarrow C=4.\frac{100}{309}\)

\(\Rightarrow C=\frac{400}{309}\)

Vậy C=.....

B, C=2^3/3.5 + 2^3/5.7+......+2^3/101.103

C= 2^3(1/3-1/5+1/5-1/7+....+1/101-1/103)

C=8(1/3-1/103)

C=8.100/309

C=800/309

VẬY C= 800/309

a) 3.5² - 162:3⁴ + 3.3² = 3.5² - 2 + 3³ = 73 + 27 = 100        

a) \(3.5^2-16:2^2\)

\(=3.25-16.4\)

\(=75-64\)

\(=11\)

b) \(17.85+15.17-120\)

\(=17.\left(85+15\right)-120\)

\(=17.100-120\)

\(=1700-120\)

\(=1580\)

c) \(\left(3^{15}.4+5.3^{15}\right):3^{16}\)

\(=\left[3^{15}.\left(4+5\right)\right]:3^{16}\)

\(=\left[3^{15}.9\right]:3^{16}\)

\(=\left[3^{15}.3^2\right]:3^{16}\)

\(=3^{17}:3^{16}\)

\(=3\)

a. 3.5² - 2⁴ : 2² = 3.5²- 2² = 3.25-4= 75-4 =71

b. 17.(85+15) - 120 = 17.100-120=1700-120=1580

c. |3¹⁵.(4+5)| :3¹⁶= (3¹⁵.9) : 3¹⁶= 3¹⁵.3² : 3¹⁶= 3¹⁷:3¹⁶=3

dấu | là ngoặc vuông bn nhé

a, 3.5^2-16:2^2

=3.25-16:4

=75-4

=71

b, 17.85+15.17-120

=17.(85+15)-120

=17.100-120

=1700-120

=1580

c, (3^15.4+5.3^15):3^16

=[3^15.(4+5)]:3^16

=(3^15.9):3^16

=(3^15.3^2):3^16

=3^17:3^16

=3^1

=3

a.  

\(M=1.\left[\frac{1}{3}-\frac{1}{5}+.....\frac{1}{97}-\frac{1}{99}\right]\)

\(M=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)

b.

\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{197}-\frac{1}{199}\right]\)

\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{199}\right]=\frac{291}{995}\)

mk đầu tiên nha bạn

\(\dfrac{2^3}{3\cdot5}+\dfrac{2^3}{5\cdot7}+...+\dfrac{2^3}{101\cdot103}\)

\(=2^2\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{101\cdot103}\right)\)

\(=4\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{101}-\dfrac{1}{103}\right)\)

\(=4\cdot\left(\dfrac{1}{3}-\dfrac{1}{103}\right)\)

\(=4\cdot\dfrac{100}{309}=\dfrac{400}{309}\)

Giải:

M=\(\dfrac{3}{3.5}+\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{95.97}+\dfrac{3}{97.99}\) 

M=\(\dfrac{3}{2}.\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{95.97}+\dfrac{2}{97.99}\right)\) 

M=\(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{99}\right)\) 

M=\(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\) 

M=\(\dfrac{3}{2}.\dfrac{32}{99}\) 

M=\(\dfrac{16}{33}\) 

Chúc bạn học tốt!

 M= 3/3.5 + 3/5.7 + 3/7.9 +.......+ 3/95.97 + 3/97.99

=3.1/2 ( 2/3.5+...+2/97.99)

=3.1/2(1/3- 1/5+...+1/97+1/99)

=3.1/2(1/3- 1/99)

=(3/2).(32/99)

=96/891

n/xét

3/3.5=(3/3-3/5).1/2

3/5.7=(3/5-3/7).1/2

...

3/97.99=(3/97-3/99).1/2

vậy M=(3/3-3/5).1/2+(3/5-3/7).1/2+...+(3/97-3/99).1/2

⇒M=1/2.(3/3-3/7+3/5-3/7+...+3/97-3/99)

      =1/2.(3/3-3/99)

     =1/2.32/33

   M =16/33

VẬY M=16/33

A) 195 x 63 + 37 x195

195 x ( 63 + 37 ) = 195 x 100

= 19500

B) 7^12 : 7^10 - 36 : 3^2

= 7^12-10 - 36 : 9

= 7^2 - 4 = 49 - 4 = 45

C) 180 - [130-(12-10)^3] + 2023^0

= 180 -(130-2^3)+1

= 180-130+8+1

= 50 + 8+1 = 59

\(a,195.63+37.195\)

\(=195\left(63+37\right)\)

\(=195.100\)

\(=19500\)

\(b,7^{12}:7^{10}-36:3^2\)

\(=7^2-2^2.3^2:3^2\)

\(=7^2-2^2\)

\(=49-4=44\)

\(c,180-\left[130-\left(12-10\right)^3\right]+2023^0\)

\(=180-\left[130-2^3\right]+1\)

\(=180-\left(130-8\right)+1\)

\(=180-122+1\)

\(=59\)

Có gì k hiểu thì ib nhá