a) x−5/7-13/14 =1                                                                                                         

   x                 =5/7+13/14

    x                 =23/14

=>1=14/14

    x                  =14/14+23/14

    x                  =37/14

  b) 3/5 +x+1 1/5 =11/3 

   Chuyển 1 1/5=6/5

              x=11/3-6/5-3/5

              x=28/15  

a)

\(\begin{array}{l}x + \left( { - \frac{1}{5}} \right) = \frac{{ - 4}}{{15}}\\x = \frac{{ - 4}}{{15}} + \frac{1}{5}\\x = \frac{{ - 4}}{{15}} + \frac{3}{{15}}\\x = \frac{{ - 1}}{{15}}\end{array}\)                 

Vậy \(x = \frac{{ - 1}}{{15}}\).

b)

\(\begin{array}{l}3,7 - x = \frac{7}{{10}}\\x = 3,7 - \frac{7}{{10}}\\x = \frac{{37}}{{10}} - \frac{7}{{10}}\\x=\frac{30}{10}\\x = 3\end{array}\)

Vậy \(x = 3\).

c)

\(\begin{array}{l}x.\frac{3}{2} = 2,4\\x.\frac{3}{2} = \frac{{12}}{5}\\x = \frac{{12}}{5}:\frac{3}{2}\\x = \frac{{12}}{5}.\frac{2}{3}\\x = \frac{8}{5}\end{array}\)

Vậy \(x = \frac{8}{5}\)       

d)

\(\begin{array}{l}3,2:x =  - \frac{6}{{11}}\\\frac{{16}}{5}:x =  - \frac{6}{{11}}\\x = \frac{{16}}{5}:\left( { - \frac{6}{{11}}} \right)\\x = \frac{{16}}{5}.\frac{{ - 11}}{6}\\x = \frac{{ - 88}}{{15}}\end{array}\)

Vậy \(x = \frac{{ - 88}}{{15}}\).

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

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

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

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

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

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

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

Bài b tương tự

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

a, x( x - 3) - ( x + 2)( x - 1) = 3

<=> x² - 3x - x² + x - 2x + 2 = 3

<=>  x² - 3x - x² + x - 2x = 3 - 2

<=> -4x = 1

<=> x =\(-\frac{1}{4}\)

Vậy_

b,  \(x-\frac{x-1}{5}+\frac{x+2}{6}=4+\frac{x+1}{3}\)

\(\Leftrightarrow\frac{30x}{30}-\frac{\left(x-1\right)6}{30}+\frac{\left(x+2\right)5}{30}=\frac{120}{30}+\frac{\left(x+1\right)10}{30}\)

=> 30x - 6x + 6 + 5x + 10 = 120 + 10x + 10

<=> 30x - 6x + 5x - 10x = 120 + 10 - 6 - 10

<=> 19x = 114

<=> x = 6

Vậy _

\(a,\left(\frac{1}{7}x-\frac{2}{7}\right)\left(-\frac{1}{5}x+\frac{3}{5}\right)\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)

TH1 : \(\frac{1}{7}x-\frac{2}{7}=0\Rightarrow\frac{x-2}{7}=0\Rightarrow x-2=0\Leftrightarrow x=2\)

TH2 : \(-\frac{1}{5}x+\frac{3}{5}=0\Rightarrow\frac{-x+3}{5}=0\Rightarrow-x+3=0\Leftrightarrow x=3\)

TH3 : \(\frac{1}{3}x+\frac{4}{3}=0\Rightarrow\frac{x+4}{3}=0\Rightarrow x+4=0\Leftrightarrow x=-4\)

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

\(b,\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)

\(\Rightarrow\frac{5}{30}x+\frac{3}{30}x-\frac{8}{30}x+1=0\)

\(\Rightarrow\frac{5x+3x-8x}{30}+1=0\)

\(\Rightarrow1=0\)( vô lý )\(\Rightarrow x\in\varnothing\)

a) (1/7x - 2/7)(-1/5x + 3/5)(1/3x + 4/3) = 0

3 trường hợp:

TH1: 1/7x - 2/7 = 0 <=> 1/7x = 0 + 2/7 <=> 1/7x = 2/7 <=> x = 2.7/7 = 2

=> x = 2

TH2: -1/5x + 3/5 = 0 <=> -1/5x = 0 - 3/5 <=> -1/5x = -3/5 <=> x = (-3/5).(-5) = 3

=> x = 3

TH3: 1/3x + 4/3 = 0 <=> 1/3x = 0 - 4/3 <=> 1/3x = -4/3 <=> x = x = 3.(-4/3) = -4

=> x = -4

Vậy: x = 2, 3, -4

b) 1/6x + 1/10x - 4/15x + 1 = 0

<=> 1/6x + 1/10x - 4/15x = 0 - 1

<=> 1/6x + 1/10x - 4/15x = -1

<=> 1/6x.30 + 1/10x.30 - 4/15x.30 = -1.30

<=> 5x + 3x - 8x = -30

<=> 0 = -30

=> không có x thỏa mãn

\(\frac{1}{4}\cdot\frac{2}{6}\cdot\frac{3}{8}\cdot\frac{4}{10}\cdot....\cdot\frac{30}{62}\cdot\frac{31}{64}=2^x\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot.....\cdot\frac{30}{31}\cdot\frac{31}{32}\right)=2^x\)

\(\Leftrightarrow\frac{1}{32}=2^{x+1}\)

Làm nốt.

ko làm được câu này hay câu b ib với tớ nha.khẳng định tối giải.

Câu b) tạm thời ko bít làm =.= 

Bài 1 : 

\(d)\) \(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=2x\)

\(\Leftrightarrow\)\(\frac{4^5.4}{3^5.3}.\frac{6^5.6}{2^5.2}=2x\)

\(\Leftrightarrow\)\(\frac{4^6}{3^6}.\frac{6^6}{2^6}=2x\)

\(\Leftrightarrow\)\(\frac{2^{12}}{3^6}.\frac{2^6.3^6}{2^6}=2x\)

\(\Leftrightarrow\)\(\frac{2^{12}}{3^6}.\frac{3^6}{1}=2x\)

\(\Leftrightarrow\)\(2^{12}=2x\)

\(\Leftrightarrow\)\(x=\frac{2^{12}}{2}\)

\(\Leftrightarrow\)\(x=2^{11}\)

\(\Leftrightarrow\)\(x=2048\)

Vậy \(x=2048\)

Chúc bạn học tốt ~ 

Bài 1 : 

\(a)\) Ta có : 

\(4+\frac{x}{7+y}=\frac{4}{7}\)

\(\Leftrightarrow\)\(\frac{x}{7+y}=\frac{4}{7}-4\)

\(\Leftrightarrow\)\(\frac{x}{7+y}=\frac{-24}{7}\)

\(\Leftrightarrow\)\(\frac{x}{-24}=\frac{7+y}{7}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có : 

\(\frac{x}{-24}=\frac{7+y}{7}=\frac{x+7+y}{-24+7}=\frac{22+7}{-17}=\frac{29}{-17}=\frac{-29}{17}\)

Do đó : 

\(\frac{x}{-24}=\frac{-29}{17}\)\(\Rightarrow\)\(x=\frac{-29}{17}.\left(-24\right)=\frac{696}{17}\)

\(\frac{7+y}{7}=\frac{-29}{17}\)\(\Rightarrow\)\(y=\frac{-29}{17}.7-7=\frac{-322}{17}\)

Vậy \(x=\frac{696}{17}\) và \(y=\frac{-322}{17}\)

Chúc bạn học tốt ~ 

2.

Ta có 1+2+...+n=n.(n+1):2

=>P=\(1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+...+\)\(\frac{1}{16}.\frac{16.17}{2}\)=1+\(\frac{3}{2}+\frac{4}{2}+...+\frac{17}{2}\)=1+\(\frac{1}{2}.\left(3=4+..=17\right)\)

=1+\(\frac{1}{2}.153=1+\frac{153}{2}=\frac{155}{2}\)

a) \(\frac{11}{15}+\frac{5}{7}+\frac{2}{7}+\frac{4}{15}=\left(\frac{11}{15}+\frac{4}{15}\right)+\left(\frac{5}{7}+\frac{2}{7}\right)\)

\(=2\)

b) \(\frac{5}{9}\times\frac{1}{2}\times\frac{5}{9}\times\frac{6}{4}=\frac{25}{81}\times\frac{3}{4}=\frac{25}{108}\)

c) \(\frac{7}{8}\div\frac{1}{2}+\frac{9}{8}\div\frac{1}{2}=\left(\frac{7}{8}+\frac{9}{8}\right)\div\frac{1}{2}\)

\(=2\div\frac{1}{2}=4\)

d) \(\frac{17}{10}+\frac{1}{2}-\frac{7}{10}=\left(\frac{17}{10}-\frac{7}{10}\right)+\frac{1}{2}\)

\(=1+\frac{1}{2}=\frac{3}{2}\)

a) \(\frac{11}{15}+\frac{5}{7}+\frac{2}{7}+\frac{4}{15}\)

\(=\left(\frac{11}{15}+\frac{4}{15}\right)+\left(\frac{5}{7}+\frac{2}{7}\right)\)

\(=1+1\)

\(=2\)

b) \(\frac{5}{9}.\frac{1}{2}.\frac{5}{9}.\frac{6}{4}\)

\(=\left(\frac{5}{9}\right)^2\left(\frac{1}{2}.\frac{6}{4}\right)\)

\(=\frac{25}{81}.\frac{3}{4}\)

\(=\frac{25}{108}\)

c) \(\frac{7}{8}:\frac{1}{2}+\frac{9}{8}:\frac{1}{2}\)

\(=\frac{7}{8}.2+\frac{9}{8}.2\)

\(=2\left(\frac{7}{8}+\frac{9}{8}\right)\)

\(=2.\frac{16}{8}\)

\(=2.2\)

\(=4\)

d) \(\frac{17}{10}+\frac{1}{2}-\frac{7}{10}\)

\(=\left(\frac{17}{10}-\frac{7}{10}\right)+\frac{1}{2}\)

\(=1+\frac{1}{2}\)

\(=\frac{2}{2}+\frac{1}{2}\)

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