ta có:   L = \(\frac{7}{3}+\frac{11}{3^2}+\frac{15}{3^3}+...+\frac{403}{3^{100}}\)

<=> \(3L=7+\frac{11}{3}+\frac{15}{3^2} +..+\frac{403}{3^{99}}\)

=> \(3L-L=\left(7+\frac{11}{3}+\frac{15}{3^2}+...+\frac{403}{3^{99}}\right)-\left(\frac{7}{3}+\frac{11}{3^2}+...+\frac{403}{3^{100}}\right)\)

<=> \(2L=7+\left(\frac{11}{3}-\frac{7}{3}\right)+\left(\frac{15}{3^2}-\frac{11}{3^2}\right)+...+\left(\frac{403}{3 ^{99}}-\frac{399}{3^{99}}\right)-\frac{403}{3^{100}}\)

<=> \(2L=7+4\cdot\frac{1}{3}+4\cdot\frac{1}{3^2}+..+4\cdot\frac{1}{3^{99}}-\frac{403}{3^{100}}\)

<=> \(2L=7+4\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\frac{403}{3^{100}}\)

<=>\(2L=7+4\left[\frac{1}{2}\cdot\left(1-\frac{1}{3^{99}}\right)\right]-\frac{403}{3^{100}}\)

<=> \(2L=7+2-\frac{2}{3^{99}}-\frac{403}{3^{100}}\)

<=> \(L=3,5+1-\frac{1}{3^{99}}-\frac{403}{2\cdot3^{100}}\)

<=> \(L=4,5-\frac{1}{3^{99}}-\frac{403}{2\cdot3^{100}}

\(A=\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}\)

\(=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}\)

\(=\frac{2}{3}\)

sai đề bạn ơi

để mk sửa lại

a, 1 - 7x = 3x - 4

=> -7x - 3x = - 4 - 1

=> - 10x = - 5

=> x = 1/2

vậy_

b, đặt  \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)

\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)

\(3A-A=1-\frac{1}{3^{99}}\)

\(A=\frac{1-\frac{1}{3^{99}}}{2}\)

mk chỉ bt lm mấy phần hui à!

d)\(\frac{5}{17}+\frac{-4}{7}-\frac{20}{31}+\frac{12}{17}-\frac{11}{31}\)\(=\left(\frac{5}{17}+\frac{12}{17}\right)+\left(\frac{-20}{31}-\frac{11}{31}\right)+\frac{-4}{7}\)

\(=\frac{17}{17}+\frac{-31}{31}+\frac{-4}{7}\)\(=1+\left(-1\right)+\frac{-4}{7}\)\(=0+\frac{-4}{7}\)\(=-\frac{4}{7}\)

e)\(\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{20}{7}-\frac{13}{3}+\frac{13}{23}}\)

mk xl,mk ko bt lm các phần còn lại!

1.

a) \(\frac{16}{24}-\frac{1}{3}=\frac{16}{24}-\frac{8}{24}=\)\(\frac{8}{24}=\frac{1}{3}\)

b) \(\frac{4}{5}-\frac{12}{60}=\frac{48}{60}-\frac{12}{60}=\frac{36}{60}=\frac{9}{15}\)

3.

a)\(\frac{17}{6}-\frac{2}{6}=\frac{17-2}{6}=\frac{15}{6}\)

b) \(\frac{16}{15}-\frac{11}{15}=\frac{16-11}{15}=\frac{5}{15}=\frac{1}{3}\)

c) \(\frac{19}{12}-\frac{13}{12}=\frac{19-13}{12}=\frac{6}{12}=\frac{1}{2}\)

a) \(A=\frac{4}{3}+\frac{7}{3^2}+\frac{10}{3^3}+...+\frac{301}{3^{100}}\)

\(\Rightarrow3A=4+\frac{7}{3}+\frac{10}{3^2}+...+\frac{301}{3^{100}}\)

\(\Rightarrow3A-A=\left(4+\frac{7}{3}+\frac{10}{3^2}+...+\frac{301}{3^{99}}\right)-\left(\frac{4}{3}+\frac{7}{3^2}+...+\frac{301}{3^{100}}\right)\)

\(\Rightarrow2A=4+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{301}{3^{100}}\)

Đặt \(F=1+\frac{1}{3}+...+\frac{1}{3^{98}}\)

\(\Rightarrow3F=3+1+...+\frac{1}{3^{97}}\)

\(\Rightarrow3F-F=\left(3+...+\frac{1}{3^{97}}\right)-\left(1+...+\frac{1}{3^{98}}\right)\)

\(\Rightarrow2F=3-\frac{1}{3^{98}}< 3\)

\(\Rightarrow F< \frac{3}{2}\)

\(\Rightarrow2A< 4+\frac{3}{2}\)

\(\Rightarrow2A< \frac{11}{2}\)

\(\Rightarrow A< \frac{11}{4}\left(đpcm\right)\)

2. \(B=\frac{11}{3}+\frac{17}{3^2}+\frac{23}{3^3}+...+\frac{605}{3^{100}}\)

\(\Rightarrow3B=11+\frac{17}{3}+\frac{23}{3^2}+...+\frac{605}{3^{99}}\)

\(\Rightarrow3B-B=\left(11+...+\frac{605}{3^{99}}\right)-\left(\frac{11}{3}+...+\frac{605}{3^{100}}\right)\)

\(\Rightarrow2B=11+2+\frac{2}{3}+...+\frac{2}{3^{98}}-\frac{605}{3^{100}}\)

Đặt \(D=2+\frac{2}{3}+...+\frac{2}{3^{98}}\)

\(\Rightarrow3D=6+2+...+\frac{2}{3^{97}}\)

\(\Rightarrow2D=6-\frac{2}{3^{98}}< 6\)( làm tắt )

\(\Rightarrow2D< 6\)

\(\Rightarrow D< 3\)

\(\Rightarrow2B< 11+3\)

\(\Rightarrow2B< 14\)

\(\Rightarrow B< 7\left(đpcm\right)\)

Phần cuối cũng tương tự 2 phần mình vừa làm nhé

Bạn tự làm nốt nhé đánh mệt lắm

2: \(=\dfrac{0.8}{\dfrac{16}{25}-\dfrac{1}{25}}+\dfrac{\dfrac{71}{75}\cdot\dfrac{7}{4}}{\dfrac{119}{36}\cdot\dfrac{36}{17}}\)

\(=\dfrac{4}{5}\cdot\dfrac{5}{3}+\dfrac{71}{300}=\dfrac{471}{300}=\dfrac{157}{100}\)

3: \(=\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{2}{6}-\dfrac{2}{8}+\dfrac{2}{10}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\)

=2/7-2/7=0

\(a)\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)

\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{-2}{5}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{-1}{4}+\frac{2}{7}+\frac{5}{7}+\frac{3}{5}\)

\(\Rightarrow\frac{2}{6}+\frac{1}{6}+\frac{-3}{5}\le x< -1+1+\frac{3}{5}\)

\(\Rightarrow\frac{1}{2}+\frac{-3}{5}\le x< \frac{3}{5}\)

\(\Rightarrow\frac{-1}{10}\le x< \frac{6}{10}\)

\(\Rightarrow-1\le x< 6\)

\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5\right\}\)

Bài b tương tự

bạn ơi bạn giải câu b được ko. mk ko biết làm câu b