1/2+ y/1=3/x

1/2+2y/2=3/x

1+2y/2=3/x

1+2y.x=2.3

vì 1+2y là lẻ suy ra thuộc ước lẻ của 6 thuộc -+1,-+3

nếu x=1, y= 0

nếu x=-1 thì y=-2

nếu x=3 thì y= 1

nếu x=-3 thì y= -2

mk ko chắc lắm, tóm lại cách làm là vậy đó

\(3\left(x-2\right)+4\left(x-5\right)=23\)

\(\Rightarrow3x-6+4x-20-23=0\)

\(\Rightarrow7x-49=0\)

\(\Rightarrow x=7\)

    3(x-2)+4(x-5)=23

<=>3x-6+4x-20=23

<=>7x-26=23

<=>7x=49

<=>x=7

Vậy x=7

3 ( x -2 ) + 4(x-5)= 23

=> 3x-6+4x-20   =23

=> 7x-26            =23

=>  7x                 =23+26=49

=>         x            =\(\dfrac{49}{7}\)=7

Vậy x=7.

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

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

\(\Leftrightarrow x=5\)

         x + 1/27 = x/9

<=>  27x+1/27 = 3x/27

 <=> 27x+1 = 3x

  <=> 27x-3x= - 1

<=> 24x= -1

<=> x = -1/24

\(\dfrac{3}{7}+\dfrac{a}{b}+\dfrac{2}{3}=\dfrac{1}{2}\)

\(\dfrac{3}{7}+\dfrac{a}{b}=\dfrac{1}{2}-\dfrac{2}{3}\)

\(\dfrac{3}{7}+\dfrac{a}{b}=-\dfrac{1}{6}\)

\(\dfrac{a}{b}=-\dfrac{1}{6}-\dfrac{3}{7}\)

\(\dfrac{a}{b}=-\dfrac{25}{42}\)

_____________

\(\dfrac{a}{b}-\dfrac{4}{9}+\dfrac{1}{10}=\dfrac{1}{7}\)

\(\dfrac{a}{b}-\dfrac{4}{9}=\dfrac{1}{7}-\dfrac{1}{10}\)

\(\dfrac{a}{b}-\dfrac{4}{9}=\dfrac{3}{70}\)

\(\dfrac{a}{b}=\dfrac{3}{70}+\dfrac{4}{9}\)

\(\dfrac{a}{b}=\dfrac{307}{630}\) 

Giup mình với ah.

1- Tính :

A= 5. | x- 5 | - 3x + 1

2 - Tìm các số nguyên x,y ; sao cho :

a) 5/x - y/3 = 1/6                        b) 5/x + y/4 = 1/8

3- Tìm giá trị lớn nhất của Q = 27-2x/12-x ( x là số nguyên)

---------------------------------------------------------------------------------------------

1) \(A=5.\left|x-5\right|-3x+1\)

\(A=\left[{}\begin{matrix}5.\left(x-5\right)-3x+1\left(x-5\ge0\right)\\5.\left(5-x\right)-3x+1\left(x-5< 0\right)\end{matrix}\right.\)

\(A=\left[{}\begin{matrix}5x-25-3x+1\left(x\ge5\right)\\25-5x-3x+1\left(x< 5\right)\end{matrix}\right.\)

\(A=\left[{}\begin{matrix}2x-24\left(x\ge5\right)\\26-8x\left(x< 5\right)\end{matrix}\right.\)

3:

\(Q=\dfrac{27-2x}{12-x}=\dfrac{2x-27}{x-12}\)

\(\Leftrightarrow Q=\dfrac{2x-24-3}{x-12}=2-\dfrac{3}{x-12}\)

Để Q lớn nhất thì \(2-\dfrac{3}{x-12}\) lớn nhất

=>\(\dfrac{3}{x-12}\) nhỏ nhất

=>x-12 là số nguyên âm lớn nhất

=>x-12=-1

=>x=11

Vậy: \(Q_{min}=2-\dfrac{3}{11-12}=2+3=5\) khi x=11

Bài 2:

a: \(\dfrac{5}{x}-\dfrac{y}{3}=\dfrac{1}{6}\)

=>\(\dfrac{15-xy}{3x}=\dfrac{1}{6}\)

=>\(15-xy=\dfrac{x}{2}\)

=>\(30-2xy=x\)

=>x+2xy=30

=>x(2y+1)=30

mà x,y nguyên

nên \(\left(x;2y+1\right)\in\left\{\left(30;1\right);\left(-30;-1\right);\left(2;15\right);\left(-2;-15\right);\left(10;3\right);\left(-10;-3\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(30;0\right);\left(-30;-1\right);\left(2;7\right);\left(-2;-8\right);\left(10;1\right);\left(-10;-2\right)\right\}\)

b: \(\dfrac{5}{x}+\dfrac{y}{4}=\dfrac{1}{8}\)

=>\(\dfrac{20+xy}{4x}=\dfrac{1}{8}\)

=>\(\dfrac{40+2xy}{8x}=\dfrac{x}{8x}\)

=>40+2xy=x

=>x-2xy=40

=>x(1-2y)=40

mà x,y nguyên

nên \(\left(x;1-2y\right)\in\left\{\left(40;1\right);\left(-40;-1\right);\left(8;5\right);\left(-8;-5\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(40;0\right);\left(-40;1\right);\left(8;-2\right);\left(-8;3\right)\right\}\)

-3. /x-1/= -27

\(|x-1|=9\)

\(\Rightarrow\orbr{\begin{cases}x-1=9\\x-1=-9\end{cases}\Leftrightarrow\orbr{\begin{cases}x=10\\x=-8\end{cases}}}\)

\(\Rightarrow x\in\left\{-8;10\right\}\)

học tốt

\(-3\left|x-1\right|=-27\)

\(\left|x-1\right|=9\)

\(\Rightarrow\orbr{\begin{cases}x-1=9\\x-1=-9\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=10\\x=-8\end{cases}}\)

-3.|x-1|=-27

   |x-1|=-27:(-3)

    |x-1|=9.          Hoặc.          |x-1|=-9