A = \(\dfrac{4}{8\times9}\)+\(\dfrac{4}{9\times10}\)+\(\dfrac{4}{10\times11}\)+...+\(\dfrac{4}{66\times67}\)

A = 4 \(\times\)( \(\dfrac{1}{8\times9}\)+\(\dfrac{1}{9\times10}\)+\(\dfrac{1}{10\times11}\)+...+\(\dfrac{1}{66\times67}\))

A = 4\(\times\)(\(\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\)...+\(\dfrac{1}{66}-\dfrac{1}{67}\))

A = 4\(\times\)(\(\dfrac{1}{8}\) - \(\dfrac{1}{67}\))

A = 4 \(\times\) \(\dfrac{59}{536}\)

A = \(\dfrac{59}{134}\)

A = 

    \(\dfrac{3}{5}+\dfrac{5}{7}\) x \(\dfrac{3}{5}-\dfrac{9}{7}\) x \(\dfrac{3}{5}\)

\(=\dfrac{3}{5}+\left(\dfrac{5}{7}-\dfrac{9}{7}\right)\) x \(\dfrac{3}{5}\)

\(=\dfrac{3}{5}-\dfrac{4}{7}\) x \(\dfrac{3}{5}\)

\(=\dfrac{3}{5}-\dfrac{12}{35}\)

\(=\dfrac{21}{35}-\dfrac{12}{35}=\dfrac{21-12}{35}=\dfrac{9}{35}\)

\(\frac{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}{\frac{6}{5}+\frac{6}{7}-\frac{6}{11}}=\frac{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}{2.\left(\frac{3}{5}+\frac{3}{7}-\frac{3}{11}\right)}=\frac{1}{2}\)

=5^3+3^9

=125+19683

=19808

5 x 6 : 5 x 3 + 3 x 2 x 3 x 7 

= 30 : 15 + 126 

= 2 + 126 

= 128

5x6:5x3+3x2x3x7

=5x(6:3)+3x2x3x7

=5x2+126

=10+126

=136

Mình chỉ biết kết quả là 1980 thôi . sorry!

31886311680

không sao nhé bạn

Wow 1 câu olm một câu hoc24 Câu hỏi của Aino Minako

\(\left\{x^2-\left[6^2-\left(8^2-9.7\right)^3-7.5\right]^3\right\}=89-102\)

\(\rightarrow\left\{x^2-\left[6^2-\left(8^2-9.7\right)^3-7.5\right]^3\right\}=-13\)

\(\rightarrow\left\{x^2-\left[6^2-1^3-7.5\right]^3\right\}=-13\)

\(\rightarrow\left\{x^2-\left[36-1-35\right]^3\right\}=-13\)

\(\rightarrow\left\{x^2-0^3\right\}=-13\)

\(\rightarrow x^2-0=-13\)

\(\rightarrow x^2=-13\) (vô lí)

Vậy x không có giá trị thỏa mãn biểu thức.

. dùng máy tính là ra mà bạn =))

\(x^2-\left[6^2-\left(8^2-9.7\right)^3-7.5\right]^3=89-102\)

\(x^2\left[36-\left(64-63\right)^3-35\right]^3=-13\)

\(x^2-0^3=-13\)

\(x^2=-13\)( Vô lý)

=> x không xác định

=\(=-\frac{17}{35}\)nha bạn

\(=\frac{-8}{35}-\frac{9}{35}\)

\(=\frac{-17}{35}\)

............

10/7+39/70=139/70

\(9\times7+8\times9-15\times9.\\ =9\times\left(7+8-15\right)=9\times0=0.\)

\(\Rightarrow\left(9\times7+8\times9-15\times9\right):\left(1+3+5+7+...+17+19\right).\\ =0:\left(1+3+5+7+...+17+19\right).\\ =0.\)

\(=\left(15\cdot9-15\cdot9\right):\left(1+3+...+17+19\right)=0\)

=9×(7+8-15)=9×0=0

=0:(1+3+5+7+...+17+19)

=0