uk, bn sinh năm 2006 phải không

a) 563 x 23 + 23 x 28 - 23

= 23 x (563 + 28 - 1)

= 23 x 600

= 19200

b) \(\frac{7}{11}+\frac{3}{4}+\frac{4}{11}+\frac{1}{4}-1\frac{2}{5}=\left(\frac{7}{11}+\frac{4}{11}\right)+\left(\frac{3}{4}+\frac{1}{4}\right)-\frac{7}{5}=2-\frac{7}{5}=\frac{3}{5}\)

c) \(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times...\times\left(1-\frac{1}{10}\right)=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times....\times\frac{9}{10}\)

\(=\frac{1\times2\times3\times...\times9}{2\times3\times4\times...\times10}=\frac{1}{10}\)

d) \(\frac{2}{7}\times\frac{3}{9}+\frac{2}{7}\times\frac{2}{3}=\frac{2}{7}\times\left(\frac{3}{9}+\frac{2}{3}\right)=\frac{2}{7}\times1=\frac{2}{7}\)

a ) \(563\cdot23+23\cdot28-23\)

\(=23\cdot\left(563+28-1\right)\)

\(=23\cdot600\)

\(=19200\)

b ) \(\frac{7}{11}+\frac{3}{4}+\frac{4}{11}+\frac{1}{4}-1\frac{2}{5}=\left(\frac{7}{11}+\frac{4}{11}\right)+\left(\frac{3}{4}+\frac{1}{4}\right)-\frac{7}{5}=2-\frac{7}{5}=\frac{3}{5}\)

c ) \(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{10}\right)=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{9}{10}\)

\(=\frac{1\cdot2\cdot3\cdot...\cdot9}{2\cdot3\cdot4\cdot...\cdot10}=\frac{1}{10}\)

d ) \(\frac{2}{7}\cdot\frac{3}{9}+\frac{2}{7}\cdot\frac{2}{3}=\frac{2}{7}\cdot\left(\frac{3}{9}+\frac{2}{3}\right)=\frac{2}{7}\cdot1=\frac{2}{7}\)

Bài 3: 

\(\Leftrightarrow x^3+64-x^3+25x=264\)

hay x=8

\(1,C=6x^2+23x-55-6x^2-23x-21=-76\\ 2,=\left(2x^4-x^2+2x^3-x-6x^2+6-3\right):\left(2x^2-1\right)\\ =\left[\left(2x^2-1\right)\left(x^2+x-6\right)-3\right]:\left(2x^2-1\right)\\ =x^2+x-6\left(dư.-3\right)\\ 3,\Leftrightarrow x^3+64-x^3+25x=264\\ \Leftrightarrow25x=200\Leftrightarrow x=8\)

tính theo công thức bạn nhé

bằng 13/15 bạn ạ 

a=-11/13

các bạn cho mình xin lời giải cụ thể với ạ

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

A=0-0+0-0+0-11/13

A=-11/13

\(A=\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}+\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{13}+\dfrac{13}{15}-\dfrac{9}{11}-\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{1}{3}\left(+\dfrac{7}{9}\rightarrow-\dfrac{7}{9}\right)\)

\(\Rightarrow A=\dfrac{1}{3}-\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{5}{7}+\dfrac{7}{9}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{9}{11}-\dfrac{11}{13}+\dfrac{13}{15}\)

\(\Rightarrow A=-\dfrac{11}{13}+\dfrac{13}{15}\)

\(\Rightarrow A=\dfrac{-11.15+13.13}{13.15}\)

\(\Rightarrow A=\dfrac{-165+169}{195}=\dfrac{4}{195}\)