a ) A = 1/15 . 75/ x + 2 + 3/8 . 64/ 3x + 6

     A = 1.75 / 15.( x + 2 ) + 3.64/8.( 3x + 6 )

     A = 1.5/1.( x + 2 ) + 1.8/1.( x + 2 )

b ) A = 1.5/1.( x + 2 ) + 1.8/1.( x + 2 )

     A = 5/ x+ 2 + 8/ x + 2

     A = 5 + 8 / x + 2

     A = 13/ x + 2 

Thay x = 37

     A = 13 / 37 + 2

     A = 13 / 39

     A = 1/3 

Câu 2 : 

B = 1 + 3 + 3^2 + 3^3 + ... + 3^199 

3B = 3 + 3^2 + 3^3 + 3^4 + ... + 3^200

3B - B = ( 3 + 3^2 + 3^3 + 3^4 + ... 3^200 ) - ( 1 + 3 + 3^2 + 3^3 + ... + 3^199 )

2B = 3^200 - 1

B = 3^200 - 1 / 2 

Câu 1 Tính 

\(S=\frac{1}{2}+\frac{1}{6}+\frac{1}{20}+...+\frac{1}{2352}+\frac{1}{2450}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{4.5}+...+\frac{1}{48.49}+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{48}-\frac{1}{49}+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}=\frac{49}{50}\)

Câu 2 Tính 

\(P=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{99}\right)\left(1-\frac{1}{100}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{98}{99}.\frac{99}{100}\)

\(=\frac{1.2.3...98.99}{2.3.4...99.100}=\frac{1}{100}\)

Câu 3 

a) Ta có : M = 1 + 3 + 3² + 3³ + ... + 3¹¹⁸ + 3¹¹⁹ (1)

=> 3M = 3 + 3² + 3³ + 3⁴ + ... + 3¹¹⁹ + 3¹²⁰  (2)

Lấy (2) trừ (1) theo vế ta có : 

3M - M = (3 + 3² + 3³ + 3⁴ + ... + 3¹¹⁹ + 3¹²⁰) - ( M = 1 + 3 + 3² + 3³ + ... + 3¹¹⁸ + 3¹¹⁹)

=>  2M = 3¹²⁰ - 1

=>    M = \(\frac{3^{120}-1}{2}\)

b) M = 1 + 3 + 3² + 3³ + ... + 3¹¹⁸ + 3¹¹⁹

        = (1 + 3 + 3²) + (3³ + 3⁴ + 3⁵) + ... + (3¹¹⁷ + 3¹¹⁸ + 3¹¹⁹)

        = (1 + 3 + 3²) + 3³(1 + 3 + 3²) + ... + 3¹¹⁷(1 + 3 + 3²)

        = 13 + 3³.13 + ... + 3¹¹⁷.13

        = 13(1 + 3³ + ... + 3¹¹⁷) \(⋮\)13

=> M \(⋮\)13

M = 1 + 3 + 3² + 3³ + ... + 3¹¹⁸ + 3¹¹⁹

= (1 + 3 + 3² + 3³) + (3⁴ + 3⁵ + 3⁶ + 3⁷) + ... + (3¹¹⁶ + 3¹¹⁷ + 3¹¹⁸ + 3¹¹⁹)

= (1 + 3 + 3² + 3³) + 3⁴(1 + 3 + 3² + 3³) + ... + 3¹¹⁶(1 + 3 + 3² + 3³)

= 40 + 3⁴.40 + ... + 3¹¹⁶.40

= 40(1 + 3⁴ + ... + 3¹¹⁶) 

= 5.8.(1 + 3⁴ + ... + 3¹¹⁶)  \(⋮\)5

4) Tính 

A = 2¹⁰⁰ - 2⁹⁹ - 2⁹⁸ - ... - 2² - 2 - 1

=> 2A =  2¹⁰¹ - 2¹⁰⁰ - 2⁹⁹ - 2⁹⁸ - 2² - 2 - 1

Lấy 2A trừ A theo vế ta có : 

2A - A = (2¹⁰¹ - 2¹⁰⁰ - 2⁹⁹ - 2⁹⁸ - 2² - 2 - 1) - (2¹⁰⁰ - 2⁹⁹ - 2⁹⁸ - ... - 2² - 2 - 1)

=>   A = 2¹⁰¹ - 2¹⁰⁰ - 2¹⁰⁰ + 1

=>   A = 2¹⁰¹ - (2¹⁰⁰ + 2¹⁰⁰) + 1

=>   A  = 2¹⁰¹ - 2¹⁰⁰ . 2 + 1

=>   A = 1

Câu 5 a) C = 1.2 + 2.3 + 3.4 + ... + 99.100

=> 3C = 1.2.3 + 2.3.3 + 3.4.3 + .... + 99.100.3

          = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 99.100.(101 - 98)

          = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100

          = 99.100.101 

=> C = 99.100.101 : 3 =  333300

b) Ta có : D = 2² + 4² + 6² + ... + 98²

                    = 2²(1² + 2²  + 3² + ... + 49²

                    =  2² .(1² + 2²  + 3² + ... + 49²)

                    = 2².(1.1 + 2.2 + 3.3 + ... + 49.49)

                    = 2².[1.(2 - 1) + 2..(3 - 1) + 3(4 - 1) + ... + 49(50 - 1)]

                    = 2².[(1.2 + 2.3 + 3.4 + ... + 49.50) - (1 + 2 + 3 + 4 + ... + 49)]

Đặt E = 1.2 + 2.3 + 3.4 + ... + 49.50

=> 3E = 1.2.3 + 2.3.3 + 3.4.3 + .... + 49.50.3

          = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 49.50.(51 - 48)

          = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 49.50.51 - 48.49.50

          = 49.50.51 

=> E = 49.50.51/3 = 41650

Khi đó D = 2².[41650 - (1 + 2 + 3 + 4 + ... + 49)]

               = 2².[41650 - 49(49 + 1)/2]

               = 2².[41650 - 1225 

               = 2².40425

               = 161700

=> D = 161700

\(\begin{array}{l}a)A = (2 - \frac{1}{2} - \frac{1}{8}):(1 - \frac{3}{2} - \frac{3}{4})\\ = (\frac{{16}}{8} - \frac{4}{8} - \frac{1}{8}):(\frac{4}{4} - \frac{6}{4} - \frac{3}{4})\\ = \frac{{11}}{8}:\frac{{ - 5}}{4}\\ = \frac{{11}}{8}.\frac{4}{{ - 5}}\\ = \frac{{ - 11}}{{10}}\\b)B = 5 - \frac{{1 + \frac{1}{3}}}{{1 - \frac{1}{3}}}\\ = 5 - \frac{{\frac{3}{3} + \frac{1}{3}}}{{\frac{3}{3} - \frac{1}{3}}}\\ = 5 - \frac{{\frac{4}{3}}}{{\frac{2}{3}}}\\ = 5 - \frac{4}{3}:\frac{2}{3}\\ = 5 - \frac{4}{3}.\frac{3}{2}\\ = 5 - 2\\ = 3\end{array}\)

Chú ý:

Khi thực hiện phép cộng hai phân số, nếu phân số thu được chưa tối giản thì ta rút gọn thành phân số tối giản.

a) ĐKXĐ: \(x\ne-2;x\ne2\), rút gọn:

\(A=\left[\frac{3\left(x-2\right)-2x\left(x+2\right)+2\left(2x^2+3\right)}{2\left(x-2\right)\left(x+2\right)}\right]\div\frac{2x-1}{4\left(x-2\right)}\)

\(A=\frac{3x-6-2x^2-4x+4x^2+6}{2\left(x-2\right)\left(x+2\right)}\cdot\frac{4\left(x-2\right)}{2x-1}=\frac{4\left(2x^2-x\right)}{x\left(x+2\right)\left(2x-1\right)}=\frac{4x\left(2x-1\right)}{x\left(x+2\right)\left(2x-1\right)}=\frac{4}{x+2}\)

b) Ta có: \(\left|x-1\right|=3\Leftrightarrow\hept{\begin{cases}x-1=3\\x-1=-3\end{cases}\Leftrightarrow\hept{\begin{cases}x=4\left(n\right)\\x=-2\left(l\right)\end{cases}}}\)

=> Khi \(x=4\)thì \(A=\frac{4}{4+2}=\frac{4}{6}=\frac{2}{3}\)

c) \(A< 2\Leftrightarrow\frac{4}{x+2}< 2\Leftrightarrow4< 2x+4\Leftrightarrow0< 2x\Leftrightarrow x>0\)Vậy \(A< 2,\forall x>0\)

d) \(\left|A\right|=1\Leftrightarrow\left|\frac{4}{x+2}\right|=1\Leftrightarrow\hept{\begin{cases}\frac{4}{x+2}=1\\\frac{4}{x+2}=-1\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\left(l\right)\\x=-6\left(n\right)\end{cases}}}\)Vậy \(\left|A\right|=1\)khi và chỉ khi x = -6

a) Cách 1:

\(\begin{array}{l}(8 + 2\frac{1}{3} - \frac{3}{5}) - (5 + 0,4) - (3\frac{1}{3} - 2)\\ = (8 + \frac{7}{3} - \frac{3}{5}) - (5 + \frac{4}{{10}}) - (\frac{{10}}{3} - 2)\\ = 8 + \frac{7}{3} - \frac{3}{5} - 5 - \frac{2}{5} - \frac{{10}}{3} + 2\\ = (8 - 5 + 2) + (\frac{7}{3} - \frac{{10}}{3}) - (\frac{3}{5} + \frac{2}{5})\\ = 5 + \frac{{ - 3}}{3} - \frac{5}{5}\\ = 5 + ( - 1) - 1\\ = 3\end{array}\)

Cách 2:

\(\begin{array}{l}(8 + 2\frac{1}{3} - \frac{3}{5}) - (5 + 0,4) - (3\frac{1}{3} - 2)\\ = (8 + \frac{7}{3} - \frac{3}{5}) - (5 + \frac{4}{{10}}) - (\frac{{10}}{3} - 2)\\ = (\frac{{120}}{{15}} + \frac{{35}}{{15}} - \frac{9}{{15}}) - (\frac{{25}}{5} + \frac{2}{5}) - (\frac{{10}}{3} - \frac{6}{3})\\ = \frac{{146}}{{15}} - \frac{{27}}{5} - \frac{4}{3}\\ = \frac{{146}}{{15}} - \frac{{81}}{{15}} - \frac{{20}}{{15}}\\ = \frac{{45}}{{15}}\\ = 3\end{array}\)

b)

\(\begin{array}{l}(7 - \frac{1}{2} - \frac{3}{4}):(5 - \frac{1}{4} - \frac{5}{8})\\ = (\frac{{28}}{4} - \frac{2}{4} - \frac{3}{4}):(\frac{{40}}{8} - \frac{2}{8} - \frac{5}{8})\\ = \frac{{23}}{4}:\frac{{33}}{8}\\ = \frac{{23}}{4}.\frac{8}{{33}}\\ = \frac{{46}}{{33}}\end{array}\)

a/ \(\frac{-9}{10}.\frac{5}{14}+\frac{1}{10}.\left(\frac{-9}{2}\right)+\frac{1}{7}.\left(-\frac{9}{10}\right)\)

= \(-\frac{9}{10}.\left(\frac{5}{14}+\frac{1}{7}\right)+\frac{1}{10}.\left(-\frac{9}{2}\right)\)

= \(-\frac{9}{10}.\frac{1}{2}+\frac{1}{10}.\left(-\frac{9}{2}\right)\)

= \(\frac{-9}{20}+\left(-\frac{9}{20}\right)=\frac{-18}{20}=\frac{-9}{10}\)

b/ \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{11}\right).132\)

\(=\left(\frac{1}{2}.132\right)+\left(\frac{1}{3}.132\right)+\left(\frac{1}{4}.132\right)+\left(\frac{1}{6}.132\right)\)\(+\left(\frac{1}{11}.132\right)\)

\(=66+44+33+22+12=177\)

c/ \(-\frac{2}{3}.\left(\frac{8}{9}.\frac{8}{13}-\frac{8}{27}.\frac{8}{13}+\frac{4}{3}.\frac{22}{39}\right)\)

= \(-\frac{2}{3}.\left[\frac{8}{13}\left(\frac{8}{9}-\frac{8}{27}\right)+\frac{88}{117}\right]\)

= \(-\frac{2}{3}.\left(\frac{8}{13}.\frac{16}{27}+\frac{88}{117}\right)\)

= còn lại làm nốt nha! bận ròy

gidkjbibvvfrxdrfdfsddf

\(A=-1-4=-5\)

\(B=\frac{4}{3}.\frac{15}{7}-16\)

\(B=\frac{20}{7}-16\)

\(B=\frac{-92}{7}\)

\(C=\frac{28}{15}.0,25.3+\left(\frac{8}{15}-\frac{1}{4}\right)\div1\frac{23}{24}\)

\(C=1,4+\frac{17}{60}\div\frac{47}{24}\)

\(C=1,4+\frac{34}{235}\)

\(C=\frac{363}{235}\)

\(A=\frac{-15}{8}+\frac{7}{8}-4\)

\(=-1-4=-5\)

\(B=\left(4-2\frac{2}{3}\right).2\frac{1}{7}-1\frac{3}{5}:\frac{1}{10}\)

\(=\frac{4}{3}.\frac{15}{7}-\frac{8}{5}:\frac{1}{10}\)

\(=\frac{20}{7}-16=\frac{-92}{7}\)

\(C=1\frac{13}{15}.\left(0,5\right)^2.3+\left(\frac{8}{15}-25\%\right):1\frac{23}{24}\)

\(=\frac{28}{15}.\frac{1}{4}.3+\frac{17}{60}:\frac{47}{24}\)

\(=\frac{7}{15}.3+\frac{17}{60}:\frac{47}{24}\)

\(=\frac{7}{5}+\frac{34}{235}=\frac{363}{235}\)

\(A=-1-4=-5\)

\(B=\frac{4}{3}.\frac{15}{7}-16\)

\(B=\frac{20}{7}-16\)

\(B=\frac{-92}{7}\)

\(C=\frac{28}{15}.0,25.3+\left(\frac{8}{15}-\frac{1}{4}\right):1\frac{23}{24}\)

\(C=1,4+\frac{17}{60}:\frac{47}{24}\)

\(C=1,4+\frac{34}{235}\)

\(C=\frac{363}{235}\)