\(1\dfrac{1}{2}x1\dfrac{1}{3}x1\dfrac{1}{4}x1\dfrac{1}{5}x1\dfrac{1}{6}x1\dfrac{1}{7}x1\dfrac{1}{8}x1\dfrac{1}{9}\)

\(=\dfrac{3}{2}x\dfrac{4}{3}x\dfrac{5}{4}x\dfrac{6}{5}x\dfrac{7}{6}x\dfrac{8}{7}x\dfrac{9}{8}x\dfrac{10}{9}\)

\(=x^7.\dfrac{3.4.5.6.7.8.9.10}{2.3.4.5.6.7.8.9}\)

\(=x^7.\dfrac{10}{2}\)

\(=5x^7\)

\(=\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\times...\times\dfrac{9}{8}\times\dfrac{10}{9}=\dfrac{10}{2}=5\)

\(1\dfrac{1}{3}\times1\dfrac{1}{8}\times1\dfrac{1}{15}\times1\dfrac{1}{24}\times1\dfrac{1}{35}\)

= \(\dfrac{4}{3}\times\dfrac{9}{8}\times\dfrac{16}{15}\times\dfrac{25}{24}\times\dfrac{36}{35}\)

= \(\dfrac{4\times9\times16\times25\times36}{3\times8\times15\times24\times35}\)

= \(\dfrac{1\times2\times2\times25\times36}{1\times2\times15\times24\times35}\)

= \(\dfrac{4\times25\times36}{30\times24\times35}\)

= \(\dfrac{1\times25\times36}{30\times6\times35}=\dfrac{1}{7}\)

\(1\dfrac{1}{3}\times1\dfrac{1}{8}\times1\dfrac{1}{15}\times1\dfrac{1}{24}\times1\dfrac{1}{35}\)

\(=\dfrac{4}{3}\times\dfrac{9}{8}\times\dfrac{16}{15}\times\dfrac{25}{24}\times\dfrac{36}{35}\)

\(=\dfrac{2\times2}{1\times3}\times\dfrac{3\times3}{2\times4}\times\dfrac{4\times4}{3\times5}\times\dfrac{5\times5}{4\times6}\times\dfrac{6\times6}{5\times7}\)

\(=\left(\dfrac{2}{1}\times\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\times\dfrac{6}{5}\right)\times\left(\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\times\dfrac{5}{6}\times\dfrac{6}{7}\right)\)

\(=6\times\dfrac{2}{7}=\dfrac{12}{7}\)

= \(\frac{11}{10}\cdot\frac{12}{11}\cdot\frac{13}{12}\cdot\frac{14}{13}\cdot\frac{15}{14}\cdot\frac{16}{15}\cdot\frac{17}{16}\)

=11/10 x 12/11 x 13/12 x 14/13 x 15/14 x 16/15 x 17/16

= \(\frac{17}{10}\)

=\(\frac{11}{10}\)x \(\frac{12}{11}\)x .......... x \(\frac{16}{15}\)x\(\frac{17}{16}\)

= \(\frac{11^1x12^1x......x16^1x17}{10x11^1x...x15^1x16^1}\)( những số có số nhỏ ở trên là rút gọn với số khác VD:11 rút gọn cho 11 )

=\(\frac{1x1x......x1x17}{10x1x.......x1x1}\)

=\(\frac{17}{10}\)

= 1,7

                                                                            Bài giải

      \(1\frac{1}{10}\text{ x }1\frac{1}{11}\text{ x }1\frac{1}{12}\text{ x }1\frac{1}{13}\text{ x }1\frac{1}{14}\text{ x }1\frac{1}{15}\text{ x }1\frac{1}{16}\)

\(=\frac{11}{10}\text{ x }\frac{12}{11}\text{ x }\frac{13}{12}\text{ x }\frac{14}{13}\text{ x }\frac{15}{14}\text{ x }\frac{16}{15}\text{ x }\frac{17}{16}\)

\(=\frac{11\text{ x }12\text{ x }13\text{ x }14\text{ x }15\text{ x }16\text{ x }17}{10\text{ x }11\text{ x }12\text{ x }13\text{ x }14\text{ x }15\text{ x }16}\)

\(=\frac{17}{10}\)

a) Ta có: \(x^2-11x-26=0\)

nên a=1; b=-11; c=-26

Áp dụng hệ thức Viet, ta được:

\(x_1+x_2=\dfrac{-b}{a}=\dfrac{-\left(-11\right)}{1}=11\)

và \(x_1x_2=\dfrac{c}{a}=\dfrac{-26}{1}=-26\)

giải giùm mình ạ:(

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{3}{2}\\x_1x_2=-\dfrac{1}{2}\end{matrix}\right.\)

\(A=\dfrac{1}{x_1-3}+\dfrac{1}{x_2-3}=\dfrac{x_2-3+x_1-3}{\left(x_1-3\right)\left(x_2-3\right)}=\dfrac{x_1+x_2-6}{x_1x_2-3\left(x_1+x_2\right)+9}\)

\(=\dfrac{\dfrac{3}{2}-6}{-\dfrac{1}{2}-3.\dfrac{3}{2}+9}=...\) (em tự bấm máy)

\(B=x_1^2x_2-4-x_1x_2+x_1x_2^2=x_1x_2\left(x_1+x_2\right)-4-x_1x_2\)

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

\(C=1-\left(x_1^2+x_2^2\right)=1-\left(x_1+x_2\right)^2+2x_1x_2=1-\left(\dfrac{3}{2}\right)^2+2.\left(-\dfrac{1}{2}\right)=...\)

\(D=x_1^3x_2^3+x_1^3+x_2^3=\left(x_1x_2\right)^3+\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)\)

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

a)  2 x 2   –   17 x   +   1   =   0

Có a = 2; b = -17; c = 1

Δ   =   b 2   –   4 a c   =   ( - 17 ) 2   –   4 . 2 . 1   =   281   >   0 .

Theo hệ thức Vi-et: phương trình có hai nghiệm x1; x2 thỏa mãn:

x 1 + x 2 = − b / a = 17 / 2 x 1 x 2 = c / a = 1 / 2

b)  5 x 2   –   x   –   35   =   0

Có a = 5 ; b = -1 ; c = -35 ;

Δ   =   b 2   –   4 a c   =   ( - 1 ) 2   –   4 . 5 . ( - 35 )   =   701   >   0

Theo hệ thức Vi-et, phương trình có hai nghiệm x1; x2 thỏa mãn:

x 1 + x 2 = − b / a = 1 / 5 x 1 ⋅ x 2 = c / a = − 35 / 5 = − 7

c)  8 x 2   –   x   +   1   =   0

Có a = 8 ; b = -1 ; c = 1

Δ   =   b 2   –   4 a c   =   ( - 1 ) 2   –   4 . 8 . 1   =   - 31   <   0

Phương trình vô nghiệm nên không tồn tại x1 ; x2.

d)  25 x 2   +   10 x   +   1   =   0

Có a = 25 ; b = 10 ; c = 1

Δ   =   b 2   –   4 a c   =   10 2   –   4 . 25 . 1   =   0

Khi đó theo hệ thức Vi-et có:

x 1 + x 2 = − b / a = − 10 / 25 = − 2 / 5 x 1 x 2 = c / a = 1 / 25

`1:(1 1/2 xx 1 1/3 xx 1 1/4 xx ... xx 1 1/99 )`

`= 1 : (3/2 xx 4/3 xx 5/4 xx ... xx 100/99)`

`= 1 : 100/2`

`= 1 xx 2/100`

`= 2/100`

`=1/50`