a: =>10x-4=15-9x

=>19x=19

hay x=1

b: \(\Leftrightarrow3\left(10x+3\right)=36+4\left(8x+6\right)\)

=>30x+9=36+32x+24

=>30x-32x=60-9

=>-2x=51

hay x=-51/2

c: \(\Leftrightarrow2x+\dfrac{6}{5}=5-\dfrac{13}{5}-x\)

=>3x=6/5

hay x=2/5

d: \(\Leftrightarrow\dfrac{7x}{8}-\dfrac{5\left(x-9\right)}{1}=\dfrac{20x+1.5}{6}\)

\(\Leftrightarrow21x-120\left(x-9\right)=4\left(20x+1.5\right)\)

=>21x-120x+1080=80x+60

=>-179x=-1020

hay x=1020/179

e: \(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)

=>35x-5+60x=96-6x

=>95x+6x=96+5

=>x=1

f: \(\Leftrightarrow6\left(x+4\right)+30\left(-x+4\right)=10x-15\left(x-2\right)\)

=>6x+24-30x+120=10x-15x+30

=>-24x+96=-5x+30

=>-19x=-66

hay x=66/19

a: =>10x-14=15-9x

=>19x=29

hay x=29/19

b: \(\Leftrightarrow3\left(10x+3\right)=36+4\left(8x+6\right)\)

=>30x+9=36+32x+24

=>30x+9=32x+60

=>-2x=51

hay x=-51/2

c: \(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)

=>35x-5+60x=96-6x

=>101x=101

hay x=1

d: \(\Leftrightarrow12\left(\dfrac{1}{2}-\dfrac{3}{2}x\right)=-5x+6\)

\(\Leftrightarrow6-18x+5x-6=0\)

=>-13x=0

hay x=0

\(a,\dfrac{5x-7}{3}=\dfrac{5-3x}{2}\\ \Leftrightarrow2\left(5x-7\right)=3\left(5-3x\right)\\ \Leftrightarrow10x-14=15-9x\\ \Leftrightarrow10x-14-15+9x=0\\ \Leftrightarrow19x-19=0\\ \Leftrightarrow x=1\)

\(b,\dfrac{10x+3}{12}=1+\dfrac{6+8x}{9}\\ \Leftrightarrow\dfrac{3\left(10x+3\right)}{36}=\dfrac{36}{36}+\dfrac{4\left(6+8x\right)}{36}\\ \Leftrightarrow30x+9=36+24+32x\\ \Leftrightarrow36+24+32x-30x-9=0\\ \Leftrightarrow2x+51=0\\ \Leftrightarrow x=-\dfrac{51}{2}\)

\(c,\dfrac{7x-1}{6}+2x=\dfrac{16-x}{5}\\ \Leftrightarrow\dfrac{7x-1+12x}{6}=\dfrac{16-x}{5}\\ \Leftrightarrow5\left(19x-1\right)=6\left(16-x\right)\\ \Leftrightarrow95x-5=96-6x\\ \Leftrightarrow95x-5-96+6x=0\\ \Leftrightarrow101x-101=0\\ \Leftrightarrow x=1\)

\(d,4\left(0,5-1,5x\right)=-\dfrac{5x-6}{3}\\ \Leftrightarrow12\left(0,5-1,5x\right)=6-5x\\ \Leftrightarrow6-18x=6-5x\\ \Leftrightarrow6-5x-6+18x=0\\ \Leftrightarrow13x=0\\ \Leftrightarrow x=0\)

a) Ta có: \(\dfrac{x-3}{5}=6-\dfrac{1-2x}{3}\)

\(\Leftrightarrow\dfrac{3\left(x-3\right)}{15}=\dfrac{90}{15}-\dfrac{5\left(1-2x\right)}{15}\)

\(\Leftrightarrow3x-9=90-5+10x\)

\(\Leftrightarrow3x-9=10x+85\)

\(\Leftrightarrow3x-10x=85+9\)

\(\Leftrightarrow-7x=94\)

hay \(x=-\dfrac{94}{7}\)

Vậy: \(S=\left\{-\dfrac{94}{7}\right\}\)

b) Ta có: \(\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}\)

\(\Leftrightarrow\dfrac{2\left(3x-2\right)}{12}-\dfrac{60}{12}=\dfrac{3\left(3-2x-14\right)}{12}\)

\(\Leftrightarrow6x-4-60=9-6x-42\)

\(\Leftrightarrow6x-64=-6x-33\)

\(\Leftrightarrow6x+6x=-33+64\)

\(\Leftrightarrow12x=31\)

hay \(x=\dfrac{31}{12}\)

Vậy: \(S=\left\{\dfrac{31}{12}\right\}\)

c) Ta có: \(3\left(x-1\right)+3=5x\)

\(\Leftrightarrow3x-3+3=5x\)

\(\Leftrightarrow3x-5x=0\)

\(\Leftrightarrow-2x=0\)

hay x=0

Vậy: S={0}

d) Ta có: \(\dfrac{x+1}{100}+\dfrac{x+2}{99}=\dfrac{x+3}{98}+\dfrac{x+4}{97}\)

\(\Leftrightarrow\dfrac{x+1}{100}+1+\dfrac{x+2}{99}+1=\dfrac{x+3}{98}+1+\dfrac{x+4}{97}+1\)

\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}=\dfrac{x+101}{98}+\dfrac{x+101}{97}\)

\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}-\dfrac{x+101}{98}-\dfrac{x+101}{97}=0\)

\(\Leftrightarrow\left(x+101\right)\left(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\right)=0\)

mà \(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\ne0\)

nên x+101=0

hay x=-101

Vậy: S={-101}

a) \(\dfrac{x-3}{5}=6-\dfrac{1-2x}{3}\\ \Leftrightarrow\dfrac{3\left(x-3\right)}{15}=\dfrac{90-5\left(1-2x\right)}{15}\\ \Leftrightarrow3x-9=90-5+10x\\ \Leftrightarrow3x-10x=90-5+9\\ \Leftrightarrow-7x=94\\ \Leftrightarrow x=\dfrac{-94}{7}\)

Vậy \(x=\dfrac{-94}{7}\) là nghiệm của pt

b) \(\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}\\ \Leftrightarrow\dfrac{2\left(3x-2\right)-60}{12}=\dfrac{9-6\left(x+7\right)}{12}\\ \Leftrightarrow6x-4-60=9-6x-42\\ \Leftrightarrow6x+6x=9-42+4+60\\ \Leftrightarrow12x=31\\ \Leftrightarrow x=\dfrac{31}{12}\)

Vậy \(x=\dfrac{31}{12}\) là nghiệm của pt

c) \(3\left(x-1\right)+3=5x\\ \Leftrightarrow3x+3+3=5x\\ \Leftrightarrow5x-3x=3+3\\ \Leftrightarrow2x=6\\ \Leftrightarrow x=3\)

Vậy x = 3 là nghiệm của pt

d) \(\dfrac{x+1}{100}+\dfrac{x+2}{99}=\dfrac{x+3}{98}+\dfrac{x+4}{97}\\ \Leftrightarrow\left(\dfrac{x+1}{100}+1\right)+\left(\dfrac{x+2}{99}+1\right)=\left(\dfrac{x+3}{98}+1\right)+\left(\dfrac{x+4}{97}+1\right)\\ \Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}-\dfrac{x+101}{98}-\dfrac{x+101}{97}=0\\ \Leftrightarrow\left(x+101\right)\left(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\right)=0\\ \Leftrightarrow x+101=0\\ \Leftrightarrow x=-101\)

Vậy x = -101 là nghiệm của pt

e) \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}=-4\\ \Leftrightarrow\left(\dfrac{59-x}{41}+1\right)+\left(\dfrac{57-x}{43}+1\right)+\left(\dfrac{53-x}{45}+1\right)+\left(\dfrac{53-x}{47}+1\right)=0\\ \Leftrightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}=0\\ \Leftrightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}\right)=0\\ \Leftrightarrow100-x=0\\ \Leftrightarrow x=100\)

Vậy x = 100 là nghiệm của pt

f) \(\dfrac{x-90}{10}+\dfrac{x-76}{12}+\dfrac{x-58}{14}+\dfrac{x-36}{16}+\dfrac{x-15}{17}=15\\ \Leftrightarrow\left(\dfrac{x-90}{10}-1\right)+\left(\dfrac{x-76}{12}-2\right)+\left(\dfrac{x-58}{14}-3\right)+\left(\dfrac{x-36}{16}-4\right)+\left(\dfrac{x-15}{17}-5\right)=0\\ \Leftrightarrow\dfrac{x-100}{10}+\dfrac{x-100}{12}+\dfrac{x-100}{14}+\dfrac{x-100}{16}+\dfrac{x-100}{17}=0\\ \Leftrightarrow\left(x-100\right)\left(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\right)=0\\ \Leftrightarrow x-100=0\\ \Leftrightarrow x=100\)

Vậy x = 100 là nghiệm của pt

e) Ta có: \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}=-4\)

\(\Leftrightarrow\dfrac{59-x}{41}+1+\dfrac{57-x}{43}+1+\dfrac{55-x}{45}+1+\dfrac{53-x}{47}+1=0\)

\(\Leftrightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}=0\)

\(\Leftrightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}\right)=0\)

mà \(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}>0\)

nên 100-x=0

hay x=100

Vậy: S={100}

f) Ta có: \(\dfrac{x-90}{10}+\dfrac{x-76}{12}+\dfrac{x-58}{14}+\dfrac{x-36}{16}+\dfrac{x-15}{17}=15\)

\(\Leftrightarrow\dfrac{x-90}{10}-1+\dfrac{x-76}{12}-2+\dfrac{x-58}{14}-3+\dfrac{x-36}{16}-4+\dfrac{x-15}{17}-5=0\)

\(\Leftrightarrow\dfrac{x-100}{10}+\dfrac{x-100}{12}+\dfrac{x-100}{14}+\dfrac{x-100}{16}+\dfrac{x-100}{17}=0\)

\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\right)=0\)

mà \(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}>0\)

nên x-100=0

hay x=100

Vậy: S={100}

k,\(\dfrac{x}{3}-\dfrac{2x+1}{2}=\dfrac{x}{6}-x\)

giúp mk câu k nhé đề bài như trên

b: \(\Leftrightarrow4x+8-9=4x-4\)

=>-1=-4(loại)

d: \(\Leftrightarrow3\left(x-2\right)+2\left(x+1\right)=8x\)

=>8x=3x-6+2x+2=5x-4

=>3x=-4

=>x=-4/3

f: \(\Leftrightarrow3\left(x+2\right)+4\left(2x-3\right)=2\left(x-12\right)\)

=>3x+6+8x-12=2x-24

=>11x-6=2x-24

=>9x=-18

=>x=-2

a,\(x-\frac{5x+2}{6}=\frac{7-3x}{4}\)

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

=>\(\frac{12x-10x-4}{12}=\frac{21-9x}{12}\)

=>(khử mẫu)

=>\(12x-10x-4=21-9x\)

=>11x=25

=>x=25/11

b: \(\Leftrightarrow3\left(10x+3\right)=36+4\left(8x+6\right)\)

=>30x+9=36+32x+24

=>32x+60=30x+9

=>2x=-51

=>x=-51/2

c: \(\Leftrightarrow2x-3\left(2x+1\right)=x+6x\)

=>7x=2x-6x-3

=>7x=-4x-3

=>11x=-3

=>x=-3/11

d: \(\Leftrightarrow4\left(x+2\right)-6x=3\left(1-2x+1\right)\)

=>4x+8-6x=3(-2x+2)

=>-2x+8+6x-6=0

=>4x+2=0

=>x=-1/2

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

\(\Leftrightarrow\dfrac{6\left(x+4\right)}{30}-\dfrac{30\left(x-4\right)}{30}=\dfrac{10x}{30}-\dfrac{15\left(x-2\right)}{30}\)

\(\Leftrightarrow6x+24-30x+120=10x-15x+30\)

\(\Leftrightarrow-24x+144=-5x+30\)

\(\Leftrightarrow-24x+144+5x-30=0\)

\(\Leftrightarrow-19x+114=0\)

\(\Leftrightarrow-19x=-114\)

hay x=6

Vậy: x=6

3) Ta có: \(\dfrac{10x+3}{12}=1+\dfrac{6+8x}{9}\)

\(\Leftrightarrow\dfrac{3\left(10x+3\right)}{36}=\dfrac{36}{36}+\dfrac{4\left(6+8x\right)}{36}\)

\(\Leftrightarrow30x+9=36+24+32x\)

\(\Leftrightarrow30x+9-60-32x=0\)

\(\Leftrightarrow-2x-51=0\)

\(\Leftrightarrow-2x=51\)

hay \(x=-\dfrac{51}{2}\)

Vậy: \(x=-\dfrac{51}{2}\)

4) Ta có: \(\dfrac{x+1}{3}-\dfrac{x-2}{6}=\dfrac{2x-1}{2}\)

\(\Leftrightarrow\dfrac{2\left(x+1\right)}{6}-\dfrac{x-2}{6}=\dfrac{3\left(2x-1\right)}{6}\)

\(\Leftrightarrow2x+2-x+2=6x-3\)

\(\Leftrightarrow x+4-6x+3=0\)

\(\Leftrightarrow-5x+7=0\)

\(\Leftrightarrow-5x=-7\)

hay \(x=\dfrac{7}{5}\)

Vậy: \(x=\dfrac{7}{5}\)

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

\(2\left(5x-2\right)=3\left(5-3x\right)\)

\(10x-4=15-9x\)

\(10x+9x=15+4\)

\(19x=19\)

\(x=1\)

Vậy \(x=1\)

2) Ta có: ⇔6(x+4)30−30(x−4)30=10x30−15(x−2)30⇔6(x+4)30−30(x−4)30=10x30−15(x−2)30

hay x=6

Vậy: x=6

3) Ta có: ⇔3(10x+3)36=3636+4(6+8x)36⇔3(10x+3)36=3636+4(6+8x)36

hay x=−512x=−512

4) Ta có: ⇔2(x+1)6−x−26=3(2x−1)6⇔2(x+1)6−x−26=3(2x−1)6

hay x=75

`a,` \(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)

`<=> (5(5x+2))/30 - (10(8x-1))/30 = (6(4x+2))/30 - (5.30)/30`

`<=> 5(5x+2) - 10(8x-1) =6(4x+2) - 5.30`

`<=> 25x + 10 - 80x + 10 = 24x+12 - 150`

`<=> -55x +20 = 24x-138`

`<=> -55x -24x=-138-20`

`<=>-79x=-158`

`<=> x=2`

Vậy pt có nghiệm `x=2`

`b,` \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\)

ĐKXĐ : \(\left\{{}\begin{matrix}x-2\ne0\\x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne2\\x\ne0\end{matrix}\right.\)

Ta có : `(x+2)/(x-2) -1/x = 2/(x(x-2))`

`<=> (x(x+2))/(x(x-2)) - (x-2)/(x(x-2))  = 2/(x(x-2))`

`=> x^2 +2x - x +2 = 2`

`<=> x^2 + x =0`

`<=>x(x+1)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\)

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

Vậy pt có nghiệm `x=-1`

`c,2x^3 + 6x^2 =x^2 +3x`

`<=> 2x^3 + 6x^2 -x^2 -3x=0`

`<=> 2x^3 + 5x^2 -3x=0`

`->` Đề có sai ko ạ ?

`d,` \(\left|x-4\right|+3x=5\) `(1)`

Thường hợp `1` : `x-4 >= 0<=> x >=0` thì phương trình `(1)` thở thành :

`x-4 = 5-3x`

`<=> x+3x=5+4`

`<=> 4x=9`

`<=> x= 9/4 (t//m)`

Trường hợp `2` : `x-4< 0<=> x<0` thì phương trình `(1)` trở thành :

`-(x-4) =5-3x`

`<=> -x +4=5-3x`

`<=> -x+3x=5-4`

`<=> 2x =1`

`<=>x=1/2 ( kt//m)`

Vậy phương trình có nghiệm `x=9/4`

đây là phương trình mà đâu phải bất phương trình đâu

thực hiện các phép biến đổi để đưa các phương trình đã cho về các phương trình tương đương có dạng ax+b=0 hoặc ax=-b,ta được:

a)5x-2/3=5-3x/2⇔2(5x-2)=3(5-3x)⇔10x-4=15-9x⇔10x+9x=15+4⇔19x=19⇔x=1

phương trình có 1 nghiệm x=1

`b,2(x+1)=5x-7`

`=>2x+2=5x-7`

`=>3x=9`

`=>x=3`

`c,3-4x(25-2x)=8x^2+x-300`

`<=>3-100x+8x^2=8x^2+x-300`

`<=>101x=303`

`<=>x=3`

`d,(10x+3)/12=1+(6+8x)/9`

`<=>(10x+3)/12=(8x+15)/9`

`<=>30x+9=32x+60`

`<=>2x=-51`

`<=>x=-51/2`