Tìm số nguyên x, biết:
1) -16 + 23 + x = - 16

7+x=-16

    x=-16-7

    x=-23
2) 2x – 35 = 15

2x=15+35

2x=50

  x=50:2

  x=25
3) 3x + 17 = 12

3x=12-17

3x=-5

  x=-5/3
4) (2x – 5) + 17 = 6

2x-5=6-17

2x-5=-11

2x=-11+5

2x=-6

  x=-6:2

  x=-3
5) 10 – 2(4 – 3x) = -4

2(4-3x)=10-(-4)

2(4-3x)=14

4-3x=14:2

4-3x=7

3x=4-7

3x=-3

  x=-3:3

  x=-1
6) - 12 + 3(-x + 7) = -18

3(-x+7)=-18-(-12)

3(x+7)=-6

x+7=-6:3

x+7=-2

    x=-2-7

    x=-9

tự đi mà làm

ủng hộ mình lên 330 điểm nha các bạn

=) vào ngay quả bảng phá dấu GTTĐ, cay thế :< 

a, \(3x+\frac{2x}{3}-3=\frac{5}{2}x-2\Leftrightarrow\frac{18x+4x-18}{6}=\frac{15x-12}{6}\)

\(\Rightarrow22x-18=15x-12\Leftrightarrow7x=6\Leftrightarrow x=\frac{6}{7}\)

Vậy pt có nghiệm x = 6/7 

b, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}+\frac{x+1}{3}=\frac{x+7}{12}\)

\(\Leftrightarrow\frac{9\left(2x+1\right)-2\left(5x+3\right)+4\left(x+1\right)}{12}=\frac{x+7}{12}\)

\(\Rightarrow18x+9-10x-6+4x+4=x+7\)

\(\Leftrightarrow12x+7=x+7\Leftrightarrow11x=0\Leftrightarrow x=0\)

Vậy pt có nghiệm là x = 0 

c, \(\frac{3x}{x-3}-\frac{x-3}{x+3}=2\)ĐK : \(x\ne\pm3\)

\(\Leftrightarrow\frac{3x\left(x+3\right)-\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\frac{2\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)

\(\Rightarrow3x^2+9x-x^2+6x-9=2\left(x^2-9\right)\)

\(\Leftrightarrow2x^2+15x-9=2x^2-18\Leftrightarrow15x+9=0\Leftrightarrow x=-\frac{9}{15}=-\frac{3}{5}\)

Vậy pt có nghiệm là x = -3/5 

d, Sửa đề :  \(\frac{x+10}{2003}+\frac{x+6}{2007}+\frac{x+2}{2011}+3=0\)

\(\Leftrightarrow\frac{x+10}{2003}+1+\frac{x+6}{2007}+1+\frac{x+2}{2011}+1=0\)

\(\Leftrightarrow\frac{x+2013}{2003}+\frac{x+2013}{2007}+\frac{x+2013}{2011}=0\)

\(\Leftrightarrow\left(x+2013\right)\left(\frac{1}{2003}+\frac{1}{2007}+\frac{1}{2011}\ne0\right)=0\Leftrightarrow x=-2013\)

Vậy pt có nghiệm là x = -2013 

e, \(4\left(x+5\right)-3\left|2x-1\right|=10\)

\(\Leftrightarrow4x+20-3\left|2x-1\right|=10\Leftrightarrow-3\left|2x-1\right|=-10-4x\)

\(\Leftrightarrow\left|2x-1\right|=\frac{10+4x}{3}\)

ĐK : \(\frac{10+4x}{3}\ge0\Leftrightarrow10+4x\ge0\Leftrightarrow x\ge-\frac{10}{4}=-\frac{5}{2}\)

TH1 : \(2x-1=\frac{10+4x}{3}\Rightarrow6x-3=10+4x\Leftrightarrow2x=13\Leftrightarrow x=\frac{13}{2}\)( tm )

TH2 : \(2x-1=\frac{-10-4x}{3}\Rightarrow6x-3=-10-4x\Leftrightarrow10x=-7\Leftrightarrow x=-\frac{7}{10}\)( tm )

f, để mình xem lại đã, quên cách phá GTTĐ rồi :v :> 

tách đi bạn

a) (2x - 3)(6 - 2x) = 0

=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)

b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)

c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)

d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)

e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)

f)\(\dfrac{x-12}{4}=\dfrac{1}{2}=\dfrac{x-12}{4}=\dfrac{2}{4}\)

⇒\(x-12=2\)

   \(x=2+12\)

  x = 14

g)2\(\dfrac{1}{4}.\left(x-7\dfrac{1}{3}\right)=1,5\)

\(\dfrac{9}{4}.\left(x-\dfrac{22}{3}\right)=1,5\) 

      \(\left(x-\dfrac{22}{3}\right)=\dfrac{3}{2}:\dfrac{9}{4}\)

       \(x-\dfrac{22}{3}=\dfrac{2}{3}\)

       \(x=\dfrac{2}{3}+\dfrac{22}{3}\) 

      \(x=8\)

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

\(\Leftrightarrow-5(5-2x)=3\left(4x-1\right)\)

\(\Leftrightarrow10x-25=12x-3\)

\(\Leftrightarrow10x-12x=25-3\)

\(\Leftrightarrow-2x=22\)

\(\Leftrightarrow x=-11\)

b,  \(\frac{\left(12-3x\right)}{32}=\frac{6}{\left(4-x\right)}\)

\(\Leftrightarrow\frac{3\left(4-x\right)}{32}=\frac{6}{\left(4-x\right)}\)

\(\Leftrightarrow3(4-x)\left(4-x\right)=32.6\)

\(\Leftrightarrow(4-x)\left(4-x\right)=32.2\)

\(\Leftrightarrow(4-x)^2=64\)

\(\Leftrightarrow\orbr{\begin{cases}4-x=8\\4-x=-8\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-4\\x=12\end{cases}}\)

c,  \(\frac{\left(10-2x\right)}{6}=\frac{27}{\left(5-x\right)}\)

\(\Leftrightarrow\frac{2\left(5-x\right)}{6}=\frac{27}{\left(5-x\right)}\)

\(\Leftrightarrow2(5-x)\left(5-x\right)=27.6\)

\(\Leftrightarrow(5-x)\left(5-x\right)=27.3\)

\(\Leftrightarrow(5-x)^2=81\)

\(\Leftrightarrow\orbr{\begin{cases}5-x=9\\5-x=-9\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-4\\x=14\end{cases}}\)

a, \(\frac{5-2x}{3}=\frac{4x-1}{-5}\Leftrightarrow-25+10x=12x-3\Leftrightarrow-22-2x=0\Leftrightarrow x=-11\)

b, \(\frac{12-3x}{32}=\frac{6}{4-x}\Leftrightarrow\frac{12-3x}{32}=\frac{18}{12-3x}\)

\(\Leftrightarrow\left(12-3x\right)^2=576\Leftrightarrow12-3x=\pm2\)\(\Leftrightarrow\orbr{\begin{cases}x=\frac{10}{3}\\x=\frac{14}{3}\end{cases}}\)

c, \(\frac{10-2x}{6}=\frac{27}{5-x}\Leftrightarrow\frac{10-2x}{6}=\frac{54}{10-2x}\)

\(\Leftrightarrow\left(10-2x\right)^2=324\Leftrightarrow10-2x=\pm18\)\(\Leftrightarrow\orbr{\begin{cases}x=14\\x=-4\end{cases}}\)

\(\frac{5-2x}{3}=\frac{4x-1}{-5}\)

=> -5(5 - 2x) = 3(4x - 1)

=> -25 + 10x = 12x - 3

=> -25 + 10x - 12x +3  = 0

=> (-25 + 3) + (10x - 12x) = 0

=> -22 - 2x = 0

=> 2x = -22

=> x = -11

\(\frac{12-3x}{32}=\frac{6}{4-x}\)

=> (12 - 3x)(4 - x) = 32.6

=> 12(4 - x) - 3x(4 - x) = 192

=> 48 - 12x - 12x + 3x² = 192

=> 48 - 24x + 3x² = 192

=> 3(x² - 8x + 16) = 192

=> x² - 8x + 16 = 64

=>  x² - 4x - 4x + 16 = 64

=> x(x - 4) - 4(x - 4) = 64

=> (x - 4)² = 64

=> (x - 4)² = (\(\pm\)8)²

=> \(\orbr{\begin{cases}x-4=8\\x-4=-8\end{cases}}\Rightarrow\orbr{\begin{cases}x=12\\x=-4\end{cases}}\)

\(\frac{10-2x}{6}=\frac{27}{5-x}\)

=> (10 - 2x)(5 - x) = 27.6

=> 10(5 - x) - 2x(5 - x) = 162

=> 50 - 10x - 10x + 2x² = 162

=> 50 - 20x + 2x² = 162

=> 2(x² - 10x + 25) = 162

=> x² - 10x + 25 = 81

=> x² - 5x - 5x + 25 = 81

=> x(x - 5) - 5(x - 5) = 81

=> (x - 5)² = 81

=> (x - 5)² = (\(\pm\)9)²

=> \(\orbr{\begin{cases}x-5=9\\x-5=-9\end{cases}}\Rightarrow\orbr{\begin{cases}x=14\\x=-4\end{cases}}\)

a) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}\\ \dfrac{x}{5}+\dfrac{1}{2}=\dfrac{3}{5}\\ \dfrac{x}{5}=\dfrac{3}{5}-\dfrac{1}{2}\\ \dfrac{x}{5}=\dfrac{6}{10}-\dfrac{5}{10}\\ \dfrac{x}{5}=\dfrac{1}{10}\\ \dfrac{2x}{10}=\dfrac{1}{10}\\ \Rightarrow2x=1\\ x=1:2\\ x=0,5=\dfrac{1}{2}\)

b) \(x+\dfrac{3}{15}=\dfrac{1}{3}\\ x=\dfrac{1}{3}-\dfrac{3}{15}\\ x=\dfrac{5}{15}-\dfrac{3}{15}\\ x=\dfrac{2}{15}\)

c) \(x-\dfrac{12}{4}=\dfrac{1}{2}\\ x-3=\dfrac{1}{2}\\ x=\dfrac{1}{2}+3\\ x=\dfrac{1}{2}+\dfrac{6}{2}\\ x=\dfrac{7}{2}\)

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

a. \(\dfrac{2x+5}{10}=\dfrac{6}{10}\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\)

b. \(\dfrac{15x+3}{15}=\dfrac{5}{15}\Leftrightarrow15x=2\Leftrightarrow x=\dfrac{2}{15}\)

c. \(\dfrac{4x-12}{4}=\dfrac{2}{4}\Leftrightarrow4x=14\Leftrightarrow x=\dfrac{7}{2}\)

d. \(\dfrac{1+x}{2x}=\dfrac{5x}{2x}\Leftrightarrow-4x=-1\Leftrightarrow x=\dfrac{1}{4}\)

e. \(\dfrac{-4\left(2x-5\right)}{6\left(2x-5\right)}-\dfrac{2}{6\left(2x-5\right)}=\dfrac{9\left(2x-5\right)}{6\left(2x-5\right)}\)

\(\Leftrightarrow-8x+20-2=18x-45\)

\(\Leftrightarrow-26x=-63\Leftrightarrow x=\dfrac{63}{26}\)

f. \(\left(3x-1\right)\left(\dfrac{-1}{2x+5}\right)=0\)

\(\left[{}\begin{matrix}3x-1=0\Rightarrow3x=1\Rightarrow x=\dfrac{1}{3}\\\dfrac{-1}{2x+5}=0\Rightarrow2x+5=0\Rightarrow2x=-5\Rightarrow x=\dfrac{-5}{2}\end{matrix}\right.\)

g. \(\dfrac{6x}{12x}-\dfrac{8}{12x}=\dfrac{7x}{12x}\Leftrightarrow-x=8\Leftrightarrow x=-8\)

h. \(\dfrac{45}{60x}+\dfrac{12x}{6x}=\dfrac{10x}{60x}\Leftrightarrow2x=-45\Leftrightarrow x=\dfrac{-45}{2}\)

3) \(-12+2x-9+x=0\\ -21+3x=0\\ 3x=21\\ x=7\)

4)
\(11+\left(15-x\right)=1\)
\(15-x=1-11\)
\(15-x=-10\)
\(x=15-\left(-10\right)\)
\(x=25\)

5) 
\(4-\left(27-3\right)=x-\left(13-4\right)\)
\(4-24=x-9\)
\(x-9=-20\)
\(x=-20+9\)
\(x=-11\)
 

\(3.-12+\left(2x-9\right)+x=0.\)

\(\Leftrightarrow-12+2x-9+x=0.\Leftrightarrow3x=21.\Leftrightarrow x=7.\)

Vậy \(x=7.\)

\(4.11+\left(15-x\right)=1.\Leftrightarrow11+15-x=1.\Leftrightarrow26-x=1.\Leftrightarrow x=25.\)

Vậy \(x=25.\)

\(5.4-\left(27-3\right)=x-\left(13-4\right).\Leftrightarrow4-24=x-9.\Leftrightarrow-20=x-9.\Leftrightarrow x=-11.\)

Vậy \(x=-11.\)

\(6.8-\left(x-10\right)=23-\left(-4+12\right).\Leftrightarrow8-x+10=23-8.\Leftrightarrow18-x=15.\Leftrightarrow x=3.\)

Vậy \(x=3.\)

\(7.105-5\left(10-5x\right)=-20.\Leftrightarrow105-50+25x=-20.\Leftrightarrow25x=-75.\Leftrightarrow x=-3.\)

Vậy \(x=-3.\)

\(8.\left(x-1\right)\left(8-2x\right)\left(3x+123\right)=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0.\\8-2x=0.\\3x+123=0.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=4.\\x=-41.\end{matrix}\right.\)

Vậy \(x\in\left\{1;4;-41\right\}.\)

\(9.\left(x^2-25\right)\left(x+10\right)=0.\)

\(\Leftrightarrow\left(x-5\right)\left(x+5\right)\left(x+10\right)=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0.\\x+5=0.\\x+10=0.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5.\\x=-5.\\x=-10.\end{matrix}\right.\)

Vậy \(x\in\left\{5;-5;-10\right\}.\)

\(10.x\left(x^2+5\right)=0.\Leftrightarrow x=0.\)