Theo đề ra, ta có: \(\frac{x-1}{y+2}=\frac{3}{5}\)

\(\Rightarrow\frac{x-1}{3}=\frac{y+2}{5}=\frac{x-1+y+2}{8}=\frac{23-1+2}{8}=\frac{24}{8}=3\)

\(\frac{x-1}{3}=3\Rightarrow x=3.3+1=10\)

\(\frac{y+2}{5}=3\Rightarrow y=5.3-2=13\)

a: =>x=3/7+3/5=15/35+21/35=36/35

b: =>x/35=4/5-5/7=28/35-25/35=3/35

=>x=3

c: =>x<3/4+8/4=11/4

=>\(x\in\left\{0;1;2;3\right\}\)

d: =>5/3<x<5/6+24/6=29/6

=>\(x\in\left\{2;3;4\right\}\)

e: =>x<10/12-9/12=1/12

=>x=0

f: =>2/3<x<12/6-5/6=7/6

=>x=1

a: x-4/5=3/7

=>x=3/7+4/5=43/35

b: x+3/7=4/5

=>x=4/5-3/7=13/35

c: 19/20-x=8/5-3/4

=>19/20-x=32/20-15/20=17/20

=>x=2/20=1/10

d: =>4/5*x=2/21

=>x=2/21:4/5=5/42

e: =>x:7/9=6/8

=>x=6/8*7/9=3/4*7/9=21/36=7/12

f: =>x/6=2/3-1/3=1/3

=>x=2

a: x-4/5=3/7

=>x=3/7+4/5=43/35

b: x+3/7=4/5

=>x=4/5-3/7=13/35

c: 19/20-x=8/5-3/4

=>19/20-x=32/20-15/20=17/20

=>x=2/20=1/10

d: =>4/5*x=11/21

=>x=55/84

e: =>x=6/8*7/9=42/72=7/12

f: =>x/6=2/3-1/3=1/3

=>x=2

=>x/6*9=4*9*6

=>x/6=4*6

=>x=4*36=144

X + 5/9 = 4/3   

X = 4/3 - 5/9

X = 7/9 

X - 4/9 = 1/2       

X = 1/2 + 4/9

X =  17/18 

6/13 + X = 7/6 

X = 7/6 - 6/13

X = 55/78

13/5 - X = 5/6   
X = 13/5 - 5/6

X= 53/30

X + 5/9 = 4/3 

x           = 4/3 - 5/9

x           =   7/9

  X - 4/9 = 1/2   

  x          = 1/2 + 4/9 

  x          =   17/18        

6/13 + X = 7/6     

            x = 7/6 - 6/13 

            x -  55/78

13/5 - X = 5/6 

          x  = 13/5 - 5/6

          x  =   53/30   

hc ngiu thế eim anh lười giải lắm =))))

Bài 1: 

a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)

\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)

\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)

\(\Leftrightarrow-12x^2+14x+13=0\)

\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)

b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)

\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)

hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)

ai giúp mik vs

Bài 3: 

a) Ta có: \(2x-3=x+\dfrac{1}{2}\)

\(\Leftrightarrow2x-x=\dfrac{1}{2}+3\)

\(\Leftrightarrow x=\dfrac{7}{2}\)

b) Ta có: \(4x-\left(x+\dfrac{1}{2}\right)=2x-\left(\dfrac{1}{2}-5\right)\)

\(\Leftrightarrow3x-\dfrac{1}{2}-2x+\dfrac{1}{2}-5=0\)

\(\Leftrightarrow x=5\)

`#3107.101107`

a)

\(x\times\dfrac{4}{5}\div2=\dfrac{6}{7}\\x\times\dfrac{2}{5}=\dfrac{6}{7}\\ x=\dfrac{6}{7}\div\dfrac{2}{5}\\ x=\dfrac{15}{7} \)

Vậy, \(x=\dfrac{15}{7}\)

b)

\(\dfrac{6}{5}\div x\div\dfrac{5}{4}=\dfrac{10}{15}\\ \dfrac{24}{25}\div x=\dfrac{10}{15}\\ x=\dfrac{24}{15}\div\dfrac{10}{15}\\ x=\dfrac{36}{25}\)

Vậy, \(x=\dfrac{36}{25}.\)

1) \(-4< x< 3\)

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

Tổng:

\(\left(-3\right)+\left(-2\right)+\left(-1\right)+0+1+2\)

\(=\left(-2+2\right)+\left(-1+1\right)+0-3\)

\(=-3\)

2) \(-5< x< 5\)

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

Tổng:

\(\left(-4\right)+\left(-3\right)+\left(-2\right)+\left(-1\right)+0+1+2+3+3\)

\(=\left(-4+4\right)+\left(-3+3\right)+\left(-2+2\right)+\left(-1+1\right)+0\)

\(=0\)

3) \(-10< x< 6\)

\(\Rightarrow x\in\left\{-9;-8;-7;-6;-5;-4;-3;-2;-1;0;1;2;3;4;5\right\}\)

Tổng:

\(\left(-9\right)+\left(-8\right)+\left(-7\right)++\left(-6\right)+\left(-5\right)+\left(-4\right)+\left(-3\right)+\left(-2\right)+\left(-1\right)+0+1+2+3+4+5\)

\(=-24\)

4) \(-6< x< 5\)

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

Tổng:

\(\left(-5\right)+\left(-4\right)+\left(-3\right)+\left(-2\right)+\left(-1\right)+0+1+2+3+4\)

\(=\left(-4+4\right)+\left(-3+3\right)+\left(-2+2\right)+\left(-1+1\right)+0-5\)

\(=-5\)

5) \(-5< x< 2\)

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

Tổng:

\(\left(-4\right)+\left(-3\right)+\left(-2\right)+\left(-1\right)+0+1\)

\(=\left(-1+1\right)+0+\left(-4-3-2\right)\)

\(=-6\)

Sửa đề: 3^5+3^5+3^5; 2^x

=>\(2^x=\dfrac{4^5\cdot4}{3^5\cdot3}\cdot\dfrac{6^5\cdot6}{2^5\cdot2}\)

=>\(2^x=\left(\dfrac{4}{3}\right)^6\cdot\left(\dfrac{6}{2}\right)^6=4^6=2^{12}\)

=>x=12