\(\left(\dfrac{3}{7}-2x\right)^2=\dfrac{4}{9}\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{7}-2x=\dfrac{2}{3}\\\dfrac{3}{7}-2x=-\dfrac{2}{3}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}2x=-\dfrac{5}{21}\\2x=\dfrac{23}{21}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{42}\\x=\dfrac{23}{42}\end{matrix}\right.\)

?!

:))

Spam ko zui đâu cj:)

Số dư r = 5

9-3=6

6

\(\left(-3\right)+9=6\)

Máy tính Bạn không có thì đi mượn nhé

KQ = 2007!

Làm bên dưới rồi.

Ta có; 4 + 4² + 4³ + ..... + 4¹⁹

    = (4 + 4²) + (4³ + 4⁴) + .... + (4¹⁸ + 4¹⁹)

    = 4.(1 + 4) + 4³.(1+4) + ..... + 4¹⁸.(1 + 4)

   = 4. 5 + 4³ . 5 + ......+ 4¹⁸.5

    = (4 + 4³ + .... + 4¹⁸) . 5 chia hết cho 5

= ( 4 + 4^2) + .......( 4^18 + 4^18 _)

= 4 . ( 1+4 ) + 4^3 ( 1+4 ) ......+ 4^18  ( 1 + 4 )

= 5 ( 4 + 4^3 + ..../ + 4^18 ) chia hêt cho 5

T nha

\(\left(\frac{x}{2}-\frac{1}{3}\right):\frac{1}{2}=\left(\frac{1}{4}-\frac{3}{2}\right):\left(1-\frac{5}{4}\right)\)

\(\left(\frac{3x-2}{6}\right):\frac{1}{2}=\left(-\frac{5}{4}\right):\left(-\frac{1}{4}\right)\)

\(\left(\frac{3x-2}{6}\right):\frac{1}{2}=5\)

\(\left(\frac{3x-2}{6}\right)=\frac{5}{2}\)

           Áp dụng công thức \(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=bc\) ta đc:

                    \(\Rightarrow2\left(3x-2\right)=30\)

                    \(\Rightarrow\left(3x-2\right)=15\)

                    \(\Rightarrow3x=17\)

                         \(\Rightarrow x=\frac{17}{3}\)

\(\left(\frac{x}{2}-\frac{1}{3}\right):\frac{1}{2}=\left(\frac{1}{4}-\frac{3}{2}\right):\left(\frac{1-5}{4}\right)\)

\(\left(\frac{x}{2}-\frac{1}{3}\right):\frac{1}{2}=\left(\frac{1}{4}-\frac{6}{4}\right):1\)

\(\left(\frac{x}{2}-\frac{1}{3}\right):\frac{1}{2}=-\frac{5}{4}:1\)

\(\left(\frac{x}{2}-\frac{1}{3}\right):\frac{1}{2}=-\frac{5}{4}\)

\(\frac{x}{2}-\frac{1}{3}=-\frac{5}{4}\times\frac{1}{2}\)

\(\frac{x}{2}-\frac{1}{3}=-\frac{5}{8}\)

\(\frac{x}{2}=-\frac{5}{8}+\frac{1}{3}\)

\(\frac{x}{2}=-\frac{7}{24}\)

\(x\times24=-14\)

\(x=-\frac{7}{12}\)

Hình như có gì đó sai sai Chị ah

M = \(\frac{1}{5}+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+...+\left(\frac{1}{5}\right)^{^{^{ }}50}\)

=> 5M = 1 + \(\frac{1}{5}+\left(\frac{1}{5}\right)^2+...+\left(\frac{1}{5}\right)^{49}\)

=> 5M - M = ( 1 + \(\frac{1}{5}+\left(\frac{1}{5}\right)^2+...+\left(\frac{1}{5}\right)^{49}\)) - ( \(\frac{1}{5}+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+...+\left(\frac{1}{5}\right)^{^{^{ }}50}\))

4M = 1 - \(\left(\frac{1}{5}\right)^{50}\)

=> M = \(\frac{1-\left(\frac{1}{5}\right)^{50}}{4}\)< \(\frac{1}{4}\)

Ta có\(\frac{1}{2^2}< \frac{1}{1.2}\)

\(\frac{1}{3^2}< \frac{1}{2.3}\)

.....

\(\frac{1}{100^2}< \frac{1}{99.100}\)

cộng các vế trái và vế phải với nhau ta được

\(\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

Ta có tổng vế phải là

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}< 1\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< 1\left(dpcm\right)\)