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|3x-7|=12

=>\(\left[{}\begin{matrix}3x-7=12\\3x-7=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=19\\3x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{19}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)

23 tháng 7 2024

`|3x - 7| = 12`

`=> 3x - 7 = 12` hoặc `3x - 7 = -12`

`=> 3x = 19` hoặc `3x = -5`

`=> x =` \(\dfrac{19}{3}\) hoặc `x =` \(-\dfrac{5}{3}\)

7 tháng 2 2020

a) \(7+2x=22-3x\)

\(\Leftrightarrow5x=15\Leftrightarrow x=3\)

b) \(8x-3=5x+12\)

\(\Leftrightarrow3x=15\Leftrightarrow x=5\)

c) \(x-12+4x=25+2x-1\)

\(\Leftrightarrow3x=36\Leftrightarrow x=12\)

d) \(x+2x+3x-19=3x+5\)

\(\Leftrightarrow3x=24\Leftrightarrow x=8\)

e) \(7-\left(2x+4\right)=-\left(x+4\right)\)

\(\Leftrightarrow x=7\)

f) \(\left(x-1\right)-\left(2x-1\right)=9-x\)

\(\Leftrightarrow0x=9\) ( vô lí )

1) \(\frac{3x-1}{4}+\frac{2x-3}{3}=\frac{x-1}{2}\) Mc : 12 \(\Leftrightarrow\) \(\frac{3.\left(3x-1\right)}{12}+\frac{4.\left(2x-3\right)}{12}=\frac{6.\left(x-1\right)}{12}\) \(\Leftrightarrow\) 9x - 3 + 8x - 12 = 6x - 6 \(\Leftrightarrow\) 9x + 8x - 6x = 3 + 12 - 6 \(\Leftrightarrow\) 11x = 9 \(\Leftrightarrow\) x = 0,8 Vậy S = {0,8} 2) \(\frac{x+1}{2}-\frac{x+3}{12}=3-\frac{5-3x}{3}\) Mc : 12 \(\Leftrightarrow\)...
Đọc tiếp

1) \(\frac{3x-1}{4}+\frac{2x-3}{3}=\frac{x-1}{2}\) Mc : 12

\(\Leftrightarrow\) \(\frac{3.\left(3x-1\right)}{12}+\frac{4.\left(2x-3\right)}{12}=\frac{6.\left(x-1\right)}{12}\)

\(\Leftrightarrow\) 9x - 3 + 8x - 12 = 6x - 6

\(\Leftrightarrow\) 9x + 8x - 6x = 3 + 12 - 6

\(\Leftrightarrow\) 11x = 9

\(\Leftrightarrow\) x = 0,8

Vậy S = {0,8}

2) \(\frac{x+1}{2}-\frac{x+3}{12}=3-\frac{5-3x}{3}\) Mc : 12

\(\Leftrightarrow\) \(\frac{6.\left(x+1\right)}{12}-\frac{x+3}{12}=\frac{12.3}{12}-\frac{4.\left(5-3x\right)}{12}\)

\(\Leftrightarrow\) 6x + 6 - x + 3 = 36 - 20 - 12x

\(\Leftrightarrow\) 6x - x + 12x = -6 - 3 + 36 - 20

\(\Leftrightarrow\) 17x = 7

\(\Leftrightarrow\) x = \(\frac{7}{17}\)

Vậy S = {\(\frac{7}{17}\)}

3) x - \(\frac{x+1}{3}\) = \(\frac{2x-1}{5}\) Mc : 15

\(\Leftrightarrow\) \(\frac{15.x}{15}-\frac{5.\left(x+1\right)}{15}=\frac{3.\left(2x-1\right)}{15}\)

\(\Leftrightarrow\) 15x - 5x - 5 = 6x - 3

\(\Leftrightarrow\) 15x - 5x - 6x = 5 - 3

\(\Leftrightarrow\) 4x = 2

\(\Leftrightarrow\) x = \(\frac{2}{4}=\frac{1}{2}\)

Vậy S = {\(\frac{1}{2}\)}

4) \(\frac{2x+7}{3}-\frac{x-2}{4}=-2\) Mc : 12

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

\(\Leftrightarrow\) 8x + 28 -3x + 6 = -24

\(\Leftrightarrow\) 8x - 3x = -28 - 6 -24

\(\Leftrightarrow\) 5x = -58

\(\Leftrightarrow\) x = -11,6

Vậy S = {-11,6}

5) \(\frac{2x-3}{4}-\frac{4x-5}{3}=\frac{5-x}{6}\) Mc : 12

\(\Leftrightarrow\) \(\frac{3.\left(2x-3\right)}{12}-\frac{4.\left(4x-5\right)}{12}=\frac{2.\left(5-x\right)}{12}\)

\(\Leftrightarrow\) 6x - 9 - 16x + 20 = 10 - 2x

\(\Leftrightarrow\) 6x - 16x + 2x = 9 - 20 + 10

\(\Leftrightarrow\) -8x = -1

\(\Leftrightarrow\) x = \(\frac{1}{8}\)

Vậy S = {\(\frac{1}{8}\)}

6) \(\frac{12x+1}{4}=\frac{9x+1}{3}-\frac{3-5x}{12}\) Mc : 12

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

\(\Leftrightarrow\) 36x + 3 = 36x + 4 - 3 + 5x

\(\Leftrightarrow\) 36x - 36x - 5x = -3 + 4 - 3

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

\(\Leftrightarrow x=\frac{2}{5}\)

7) \(\frac{x+6}{4}\) - \(\frac{x-2}{6}-\frac{x+1}{3}=0\) Mc : 12

\(\Leftrightarrow\) \(\frac{3.\left(x+6\right)}{12}-\frac{2.\left(x-2\right)}{12}-\frac{4.\left(x+1\right)}{12}=0\)

\(\Leftrightarrow\) 3x + 18 - 2x + 4 - 4x - 4 = 0

\(\Leftrightarrow\) 3x - 2x - 4x = -18 - 4 + 4

\(\Leftrightarrow\) -3x = -18

\(\Leftrightarrow\) x = 6

Vậy S = {6}

8) x\(^2\) - x - 6 = 0

\(\Leftrightarrow\) x\(^2\) + 2x - 3x - 6 = 0

\(\Leftrightarrow\) x.(x + 2) - 3.(x + 2) = 0

\(\Leftrightarrow\) (x - 3).(x + 2) = 0

\(\Leftrightarrow\) x - 3 = 0 hoặc x + 2 = 0

\(\Leftrightarrow\) x = 3 hoặc x = -2

Vậy S = {3; -2}

0
22 tháng 4 2017

a)X= 3 b)X= 5 c)X= 12

d)X= 8 e)X= 7 f)X= ?

22 tháng 4 2017

Giải bài 17 trang 14 SGK Toán 8 Tập 2 | Giải toán lớp 8

Giải bài 17 trang 14 SGK Toán 8 Tập 2 | Giải toán lớp 8

a: =>-12x>12

hay x<-1

b: =>7(3x-1)-252>=21x+3(6x+1)

=>21x-7-252>=21x+18x+3

=>18x+3<=-259

=>18x<=-262

hay x<=-131/9

c: =>3(3x+5)-24x<=48+4(x+8)

=>9x+15-24x<=48+4x+32=4x+80

=>-15x+24<=4x+80

=>-19x<=56

hay x>=-56/19

19 tháng 9 2018

\(a,\left(-2\right)^5:\left(-2\right)^3=\left(-2\right)^{5-3}=\left(-2\right)^2=4\)

\(b,\left(-y\right)^7:\left(-y\right)^3=\left(-y\right)^{7-3}=\left(-y\right)^4=y^4\)

\(c,x^{12}:\left(-x\right)^{10}=x^{12}:x^{10}=x^{12-10}=x^2\)

\(d,\left(2x\right)^6:\left(2x\right)^3=\left(2x\right)^{6-3}=\left(2x\right)^3=8x^3\)

\(e,\left(-3x\right)^5:\left(-3x\right)^2=\left(-3x\right)^{5-2}=\left(-3x\right)^3=-27x^3\)

\(f,\left(xy^2\right)^4:\left(xy^2\right)^2=\left(xy^2\right)^2=x^2y^4\)

a: \(\Leftrightarrow4\left(-5x+6\right)\left(3x-7\right)=30x-240-6x-84\)

\(\Leftrightarrow4\left(-15x^2+35x+18x-42\right)=24x-324\)

\(\Leftrightarrow-60x^2+212x-168-24x+324=0\)

\(\Leftrightarrow-60x^2+188x+156=0\)

\(\Leftrightarrow15x^2-47x-39=0\)

\(\text{Δ​}=\left(-47\right)^2-4\cdot15\cdot\left(-39\right)=4549>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{47-\sqrt{4549}}{30}\\x_2=\dfrac{47+\sqrt{4549}}{30}\end{matrix}\right.\)

b: \(\Leftrightarrow6x^2+27x+4x+18-6x^2-x-12x-2=x+1-x+6\)

\(\Leftrightarrow17x+16=7\)

hay x=-9/17

c: \(\Leftrightarrow4x^2+8x+4+4x^2-4x+1-8x^2+8=11\)

=>4x+13=11

hay x=-1/2

Bài 1:

a) Ta có: \(2,3x-2\left(0,7+2x\right)=3,6-1,7x\)

\(\Leftrightarrow2,3x-1,4-4x-3,6+1,7x=0\)

\(\Leftrightarrow-5=0\)(vl)

Vậy: \(x\in\varnothing\)

b) Ta có: \(\frac{4}{3}x-\frac{5}{6}=\frac{1}{2}\)

\(\Leftrightarrow\frac{4}{3}x=\frac{1}{2}+\frac{5}{6}=\frac{8}{6}=\frac{4}{3}\)

hay x=1

Vậy: x=1

c) Ta có: \(\frac{x}{10}-\left(\frac{x}{30}+\frac{2x}{45}\right)=\frac{4}{5}\)

\(\Leftrightarrow\frac{9x}{90}-\frac{3x}{90}-\frac{4x}{90}-\frac{72}{90}=0\)

\(\Leftrightarrow2x-72=0\)

\(\Leftrightarrow2\left(x-36\right)=0\)

mà 2>0

nên x-36=0

hay x=36

Vậy: x=36

d) Ta có: \(\frac{10x+3}{8}=\frac{7-8x}{12}\)

\(\Leftrightarrow12\left(10x+3\right)=8\left(7-8x\right)\)

\(\Leftrightarrow120x+36=56-64x\)

\(\Leftrightarrow120x+36-56+64x=0\)

\(\Leftrightarrow184x-20=0\)

\(\Leftrightarrow184x=20\)

hay \(x=\frac{5}{46}\)

Vậy: \(x=\frac{5}{46}\)

e) Ta có: \(\frac{10x-5}{18}+\frac{x+3}{12}=\frac{7x+3}{6}-\frac{12-x}{9}\)

\(\Leftrightarrow\frac{2\left(10x-5\right)}{36}+\frac{3\left(x+3\right)}{36}-\frac{6\left(7x+3\right)}{36}+\frac{4\left(12-x\right)}{36}=0\)

\(\Leftrightarrow2\left(10x-5\right)+3\left(x+3\right)-6\left(7x+3\right)+4\left(12-x\right)=0\)

\(\Leftrightarrow20x-10+3x+9-42x-18+48-4x=0\)

\(\Leftrightarrow-23x+29=0\)

\(\Leftrightarrow-23x=-29\)

hay \(x=\frac{29}{23}\)

Vậy: \(x=\frac{29}{23}\)

f) Ta có: \(\frac{x+4}{5}-x-5=\frac{x+3}{2}-\frac{x-2}{2}\)

\(\Leftrightarrow\frac{2\left(x+4\right)}{10}-\frac{10x}{10}-\frac{50}{10}=\frac{25}{10}\)

\(\Leftrightarrow2x+8-10x-50-25=0\)

\(\Leftrightarrow-8x-67=0\)

\(\Leftrightarrow-8x=67\)

hay \(x=\frac{-67}{8}\)

Vậy: \(x=\frac{-67}{8}\)

g) Ta có: \(\frac{2-x}{4}=\frac{2\left(x+1\right)}{5}-\frac{3\left(2x-5\right)}{10}\)

\(\Leftrightarrow5\left(2-x\right)-8\left(x+1\right)+6\left(2x-5\right)=0\)

\(\Leftrightarrow10-5x-8x-8+12x-30=0\)

\(\Leftrightarrow-x-28=0\)

\(\Leftrightarrow-x=28\)

hay x=-28

Vậy: x=-28

h) Ta có: \(\frac{x+2}{3}+\frac{3\left(2x-1\right)}{4}-\frac{5x-3}{6}=x+\frac{5}{12}\)

\(\Leftrightarrow\frac{4\left(x+2\right)}{12}+\frac{9\left(2x-1\right)}{12}-\frac{2\left(5x-3\right)}{12}-\frac{12x}{12}-\frac{5}{12}=0\)

\(\Leftrightarrow4x+8+18x-9-10x+6-12x-5=0\)

\(\Leftrightarrow0x=0\)

Vậy: \(x\in R\)

Bài 2:

a) Ta có: \(5\left(x-1\right)\left(2x-1\right)=3\left(x+8\right)\left(x-1\right)\)

\(\Leftrightarrow5\left(x-1\right)\left(2x-1\right)-3\left(x-1\right)\left(x+8\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[5\left(2x-1\right)-3\left(x+8\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(10x-5-3x-24\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(7x-29\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\7x-29=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\7x=29\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{29}{7}\end{matrix}\right.\)

Vậy: Tập nghiệm \(S=\left\{1;\frac{29}{7}\right\}\)

b) Ta có: \(\left(3x-2\right)\left(x+6\right)\left(x^2+5\right)=0\)(1)

Ta có: \(x^2\ge0\forall x\)

\(\Rightarrow x^2+5\ge5\ne0\forall x\)(2)

Từ (1) và (2) suy ra:

\(\left[{}\begin{matrix}3x-2=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=-6\end{matrix}\right.\)

Vậy: Tập nghiệm \(S=\left\{\frac{2}{3};-6\right\}\)

c) Ta có: \(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)

\(\Leftrightarrow27x^3-8-\left(27x^3-1\right)-x+4=0\)

\(\Leftrightarrow27x^3-8-27x^3+1-x+4=0\)

\(\Leftrightarrow-x-3=0\)

\(\Leftrightarrow-x=3\)

hay x=-3

Vậy: Tập nghiệm S={-3}

d) Ta có: \(x\left(x-1\right)-\left(x-3\right)\left(x+4\right)=5x\)

\(\Leftrightarrow x^2-x-\left(x^2+x-12\right)-5x=0\)

\(\Leftrightarrow x^2-x-x^2-x+12-5x=0\)

\(\Leftrightarrow12-7x=0\)

\(\Leftrightarrow7x=12\)

hay \(x=\frac{12}{7}\)

Vậy: Tập nghiệm \(S=\left\{\frac{12}{7}\right\}\)

e) Ta có: (2x+1)(2x-1)=4x(x-7)-3x

\(\Leftrightarrow4x^2-1-4x^2+28x+3x=0\)

\(\Leftrightarrow31x-1=0\)

\(\Leftrightarrow31x=1\)

hay \(x=\frac{1}{31}\)

Vậy: Tập nghiệm \(S=\left\{\frac{1}{31}\right\}\)

2 tháng 7 2018

a)  \(x^2+3x+2\)

\(=\left(x^2+2x\right)+\left(x+2\right)\)

\(=x\left(x+2\right)+\left(x+2\right)\)

\(=\left(x+1\right)\left(x+2\right)\)

b)  \(x^2+5x+6\)

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

\(=x\left(x+2\right)+3\left(x+2\right)\)

\(=\left(x+3\right)\left(x+2\right)\)

c)  \(x^2+5x+6\)

( giống câu b -_- )

d)  \(x^2+7x+12\)

\(=\left(x^2+4x\right)+\left(3x+12\right)\)

\(=x\left(x+4\right)+3\left(x+4\right)\)

\(=\left(x+3\right)\left(x+4\right)\)

2 tháng 7 2018

\(1,x^2+3x+2\)

\(=x^2+x+2x+2\)

\(=x\left(x+1\right)+2\left(x+1\right)\)

\(=\left(x+2\right)\left(x+1\right)\)

\(2,x^2+5x+6\)

\(=x^2+2x+3x+6\)

\(=x\left(x+2\right)+3\left(x+2\right)\)

\(=\left(x+3\right)\left(x+2\right)\)

\(4,x^2+7x+12\)

\(=x^2+3x+4x+12\)

\(=x\left(x+3\right)+4\left(x+3\right)\)

\(=\left(x+4\right)\left(x+3\right)\)

\(5,x^2-4x+3\)

\(=x^2-x-3x+3\)

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

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

\(6,x^2+3x-4\)

\(=x^2-x+4x-4\)

\(=x\left(x-1\right)+4\left(x-1\right)\)

\(=\left(x+4\right)\left(x-1\right)\)

\(8,x^2-3x-10\)

\(=x^2-5x+2x-10\)

\(=x\left(x-5\right)+2\left(x-5\right)\)

\(=\left(x+2\right)\left(x-5\right)\)

27 tháng 2 2020

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

<=> 3(3x-7)=5(2x-1)

<=> 9x-21=10x-5

<=> -21+5=10x-9x

<=> x=-16

2. \(\frac{3x-7}{2}+\frac{2x-1}{3}=-16\)

<=> \(\frac{3\left(3x-7\right)}{6}+\frac{2\left(2x-1\right)}{6}=\frac{-96}{6}\)

=> 9x-21+4x-2=-96

<=> 13x-23=-96

<=> 13x=-73

<=> x=\(\frac{-73}{13}\)

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

<=> \(\frac{15x}{15}-\frac{5\left(x+1\right)}{15}=\frac{3\left(2x+1\right)}{15}\)

=> 15x-5x-5=6x+3

<=> 15x-5x-6x=3+5

<=> 4x=8

<=> x=2

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

<=>\(\frac{7-3x}{12}+\frac{9}{12}=\frac{24\left(x-2\right)}{12}+\frac{2\left[5-\left(5-2x\right)\right]}{12}\)

=> 7-3x+9=24x-48+4x

<=> -3x-24x-4x=-48-7

<=> -31x=-55

<=> x= \(\frac{55}{31}\)

5. \(\frac{2x-1}{3}-\frac{5x+2}{7}=x+13\)

<=> \(\frac{7\left(2x-1\right)}{21}-\frac{3\left(5x+2\right)}{21}=\frac{21\left(x+13\right)}{21}\)

=> 14x-7-15x-6=21x+273

<=> 14x-15x-21x=273+7+6

<=> -22x=286

<=> x= -13

27 tháng 2 2020

a/\(\Leftrightarrow3\left(3x-7\right)=5\left(2x-1\right)\Leftrightarrow9x-21=10x-5\Leftrightarrow x=-16\)

b/\(\Leftrightarrow\frac{9x-21+4x-2}{6}=-16\)\(\Leftrightarrow13x-23=-96\Leftrightarrow x=x=-\frac{73}{13}\)

c/\(\Leftrightarrow\frac{3x-x+1}{3}-\frac{2x+1}{5}=0\Leftrightarrow\left(2x+1\right)\left(\frac{1}{3}-\frac{1}{5}\right)=0\Leftrightarrow x=-\frac{1}{2}\)