5 ( x - 1 ) - 7 ( x - 2 ) = 2x - 39

<=> 5x - 5 - 7x + 14 = 2x - 39

<=> 5x - 7x - 2x = -39 + 5 - 14

<=> -4x = -48

<=> x = 12

x - 3 - 14.( x-2 )= -3x -3\(\Rightarrow\chi-3-28-14\chi-28=-3\chi-3\)

\(\Rightarrow\chi-3-28+3=-3\chi-3\)

\(\Rightarrow\chi-28=11\chi\)

\(\Rightarrow\chi-11\chi=28\)

\(\Rightarrow10\chi=28\Rightarrow\chi=2,8\left(kot.m\chi\inℤ\right)\) 

HTDT-sai-rùi

x-3-14(x-2)=-3x-3

x-3+28-14x=-3x-3

-13x+35=-3x-3

-13x+3x=-3-35

-10x=-38

x=38/10=19/5

vt rõ đề câu a đi lm 1 thể

\(\dfrac{2}{7}\)\(x\) - \(\dfrac{1}{3}\) = \(\dfrac{3}{5}\)\(x\) - 1

\(\dfrac{2}{7}\)\(x\) - \(\dfrac{3}{5}\)\(x\) =  - 1 + \(\dfrac{1}{3}\)

 (\(\dfrac{2}{7}\) - \(\dfrac{3}{5}\))\(x\) =  - \(\dfrac{2}{3}\)

  - \(\dfrac{11}{35}\)\(x\)     = - \(\dfrac{2}{3}\)

         \(x\)     = - \(\dfrac{2}{3}\) : (- \(\dfrac{11}{35}\))

         \(x\)     = \(\dfrac{70}{33}\)

Vậy \(x=\dfrac{70}{33}\) 

|3\(x\) + 7| - 3\(x\) = 7

|3\(x\) + 7| = 7  +3\(x\) Vì 7 + 3\(x\) > 0  nên 3\(x\) >  -7 ⇒ \(x\) >  -\(\dfrac{7}{3}\)

Với \(x\) >  - \(\dfrac{7}{3}\) ta có:

3\(x\) + 7 = 7 + 3\(x\) 

        7  =7 (luôn đúng)

Vậy \(x\) \(\ge\) - \(\dfrac{7}{3}\) 

\(1,\\ \left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\\ \Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\end{matrix}\right.\)

\(2,\\ a,\left|2x-3\right|>5\Leftrightarrow\left[{}\begin{matrix}2x-3< -5\\2x-3>5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< -1\\x>4\end{matrix}\right.\\ b,\left|3x-1\right|\le7\Leftrightarrow\left[{}\begin{matrix}3x-1\le7\\1-3x\le7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\le\dfrac{8}{3}\\x\ge-2\end{matrix}\right.\\ c,\cdot x< -\dfrac{3}{2}\\ \Leftrightarrow5-3x+\left(-2x-3\right)=7\Leftrightarrow2-5x=7\Leftrightarrow x=-1\left(ktm\right)\\ \cdot-\dfrac{3}{2}\le x\le\dfrac{5}{3}\\ \Leftrightarrow\left(5-3x\right)+\left(2x+3\right)=7\Leftrightarrow8-x=7\Leftrightarrow x=1\left(tm\right)\\ \cdot x>\dfrac{5}{3}\\ \Leftrightarrow\left(3x-5\right)+\left(2x+3\right)=7\Leftrightarrow5x-2=7\Leftrightarrow x=\dfrac{9}{5}\left(tm\right)\\ \Leftrightarrow S=\left\{1;\dfrac{9}{5}\right\}\)

Mai lam

1) x (x-2016) + 2015 (2016-x) = 0

 x (x-2016) - 2015 (x- 2016) = 0

(x-2015)(x-2016) =0

\(\Rightarrow\orbr{\begin{cases}x-2015=0\\x-2016=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2015\\x=2016\end{cases}}}\)

Vậy x= 2015; 2016

2) -5x (x-15) + (15-x) = 0

-5x (x-15) - (x-15) =0

(-5x -1) (x-15) =0

\(\Rightarrow\orbr{\begin{cases}-5x-1=0\\x-15=0\end{cases}\Rightarrow\orbr{\begin{cases}-5x=1\\x=15\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{5}\\x=15\end{cases}}}\)

Vậy x= -1/5; 15

3) 3x (3x-7) - (7-3x) =0

3x(3x-7) + (3x -7) =0

(3x+1) (3x-7) =0

\(\Rightarrow\orbr{\begin{cases}3x+1=0\\3x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=-1\\3x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{3}\\x=\frac{7}{3}\end{cases}}}\)

Vậy x= -1/3 ; 7/3

219 - 7(x + 1) = 10²

7(x + 1) = 219 - 10²

7(x + 1) = 219 - 100

7(x + 1) = 119

x + 1 = 119 : 7

x + 1 = 17

x = 17 - 1

x = 16

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

(3x - 2⁴) . 7³ = 2.7⁴

(3x - 16).7³ = 2.7⁴

3x - 16 = 2.7⁴:7³

3x - 16 = 2.7

3x - 16 = 14

3x = 14 + 16

3x = 30

x = 30 : 3

x = 10

Lời giải:

$10+2x=4^5:4^3=4^2=16$

$2x=16-10=6$

$x=6:2=3$

---------

$2x-6^2:18=3.2^2$
$2x-2=12$

$2x=14$

$x=14:2=7$

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

$70-5(x-3)=3.2^5=96$

$5(x-3)=70-3.2^5=-26$

$x-3=\frac{-26}{5}$

$x=3+\frac{-26}{5}=\frac{-11}{5}$

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

$19-7(x+1)=10^2=100$

$7(x+1)=19-100=-81$

$x+1=\frac{-81}{7}$

$x=\frac{-81}{7}-1=\frac{-88}{7}$

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

$(3x-2^4).7^3=2.7^4$

$3x-16=2.7^4:7^3=2.7=14$

$3x=14+16=30$

$x=30:3=10$

tích mình đi

ai tích mình 

mình tích lại 

thanks

\(3x^2-\left(x+2\right)\left(3x-1\right)=-7\)

\(\Rightarrow3x^2-\left(3x^2+6x-x-2\right)=-7\)

\(\Rightarrow3x^2-3x^2-5x+2=-7\)

\(\Rightarrow-5x+2=-7\)

\(\Rightarrow-5x=-9\)

\(\Rightarrow x=\frac{9}{5}\)

Vậy \(x=\frac{9}{5}\)

\(3x^2-\left(x+2\right)\left(3x-1\right)=-7\) \(7\)

<=> \(3x^2-\left(3x^2-x+6x-2\right)=-7\)

<=> \(3x^2-3x^2+x-6x+2=-7\)

<=> \(-5x+2=-7\)

<=> \(-5x=-9\)

<=> \(x=\frac{9}{5}\)