2x - 35 = 15

      2x = 15 + 35 

      2x = 50

        x = 50 : 2

        x = 25 

( x - 1 ) + 5 = 17

          x - 1 = 17 - 5 

          x - 1 = 12

               x = 12 + 1

               x = 13 

\(-(35-x)-(37-x)=33\\\Rightarrow-35+x-37+x=33\\\Rightarrow(-35-37)+(x+x)=33\\\Rightarrow-72+2x=33\\\Rightarrow2x=33+72\\\Rightarrow2x=105\\\Rightarrow x=105:2\\\Rightarrow x=52,5\\---\)

\((-14)+x-7=(-10)\\\Rightarrow x-7=(-10)+14\\\Rightarrow x-7=4\\\Rightarrow x=4+7\\\Rightarrow x=11\)

-(35  - \(x\)) - (37 - \(x\)) = 33

- 35 + \(x\)  - 37 + \(x\)  = 33

  (\(x\) + \(x\)) - (35 + 37) = 33

    2\(x\) - 72  = 33

    2\(x\)          = 33 + 72

    2\(x\)         = 105

      \(x\)        = \(\dfrac{105}{2}\)

(-14) + \(x\) - 7 = (-10)

 -14 + \(x\)  - 7 = -10

 \(x\) - (14 + 7)  = -10

  \(x\) - 21          = -10

  \(x\)                 = -10 + 21

  \(x\)                 = 11

Ta có:\(\left(4x+28\right)⋮\left(2x+1\right)\Rightarrow\left[\left(4x+2\right)+26\right]⋮\left[2\left(2x+1\right)\right]\)

           \(\Rightarrow\left[\left(4x+2\right)+26\right]⋮\left(4x+2\right)\)

            \(\Rightarrow26⋮\left(4x+2\right)\)(t/c chia hết của 1 tổng)

         Vì \(x\in N\Rightarrow4x+2\in N\)(1)

           \(\Rightarrow\left(4x+2\right)\in\left\{2;13;26\right\}\)

             \(\Rightarrow x\in\left\{0;\frac{11}{4};6\right\}\)

   Từ (1)=>\(x\in\left\{0;6\right\}\)

                 Có gì ko hiểu thì kbạn với mình nha

\(-\dfrac{4}{7}-x=\dfrac{3}{5}-2x\\ \Rightarrow-x+2x=\dfrac{3}{5}+\dfrac{4}{7}\\ \Rightarrow x=\dfrac{21}{35}+\dfrac{20}{35}\\ \Rightarrow x=\dfrac{41}{35}\)

Vậy `x=41/35`

__

\(\dfrac{3}{7}x-\dfrac{2}{3}x=\dfrac{10}{21}\\ \Rightarrow\left(\dfrac{3}{7}-\dfrac{2}{3}\right)x=\dfrac{10}{21}\\ \Rightarrow\left(\dfrac{9}{21}-\dfrac{14}{21}\right)x=\dfrac{10}{21}\\ \Rightarrow\dfrac{-5}{21}x=\dfrac{10}{21}\\ \Rightarrow x=\dfrac{10}{21}:\left(-\dfrac{5}{21}\right)\\ \Rightarrow x=-2\)

Vậy `x=-2`

a)

-4/7 - x = 3/5 - 2x

2x - x = 3/5 + 4/7

x = 41/35

Vậy x = 41/35

b)

3/7.x - 2/3.x = 10/21

x(3/7 - 2/3) = 10/21

x.(-5/21) = 10/21

x = 10/21 : (-5/21) = -2

Vậy x = -2

\(\frac{x}{7}=\frac{x+36}{35}\)

\(\Rightarrow x.35=7.\left(x+36\right)\)

\(\Rightarrow35x=7x+252\)

\(\Rightarrow35x-7x=252\)

\(\Rightarrow28x=252\)

\(\Rightarrow x=9\)

/ laf chia af

\(\frac{x}{7}=x+\frac{36}{35}\)

\(\Leftrightarrow x=7x+\frac{36}{5}\)

\(\Leftrightarrow6x=-\frac{36}{5}\)

\(\Leftrightarrow x=-\frac{6}{5}\)

a) \(\frac{x}{y}=\frac{5}{7}\)=>\(\frac{x}{5}=\frac{y}{7}=>\left(\frac{x}{5}\right)^2=\left(\frac{y}{7}\right)^2=\frac{xy}{5.7}\)

=>\(\frac{x^2}{25}=\frac{y^2}{49}=\frac{35}{35}=1\)

=> \(x^2=25;y^2=49\)

=>\(x=\pm5;y=\pm7\)

Sao phần d vẫn như thế vậy ? Bạn không sửa đề à ?

a, Ta có

\(\frac{x}{y}=\frac{5}{7}\Rightarrow\frac{x}{5}=\frac{y}{7}\)

Đặt \(\frac{x}{5}=\frac{y}{7}=k\Rightarrow\hept{\begin{cases}x=5k\\y=7k\end{cases}}\)

Ta lại có

\(xy=5k\cdot7k=35\)

\(\Rightarrow35k^2=35\)

\(\Rightarrow k^2=1\)

\(\Rightarrow k=\pm1\)

Với \(k=1\Rightarrow\hept{\begin{cases}x=5\\y=7\end{cases}}\)

Với \(k=-1\Rightarrow\hept{\begin{cases}x=-5\\y=-7\end{cases}}\)

Vậy ..............

làm xong có k ko?