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

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

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

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

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

Ta có: \(\dfrac{10x+3}{12}=\dfrac{15-8x}{9}\)

\(\Leftrightarrow\dfrac{3\left(10x+3\right)}{36}=\dfrac{4\left(15-8x\right)}{36}\)

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

\(\Leftrightarrow30x+32x=60-9=51\)

\(\Leftrightarrow62x=51\)

\(\Leftrightarrow x=\dfrac{51}{62}\)

Vậy: \(x=\dfrac{51}{62}\)

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

`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`

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

bài này đề bài là chứng minh hay là giải bất phương trình vậy bạn

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

5: \(\Leftrightarrow9\left(x^2-5x-4\right)=36\left(x+1\right)+8\left(x^2-10x\right)\)

\(\Leftrightarrow9x^2-45x-36-36x-36-8x^2+80x=0\)

\(\Leftrightarrow x^2-x-72=0\)

=>(x-9)(x+8)=0

=>x=9 hoặc x=-8

6: \(\Leftrightarrow x^2-9=9x-x^2-9+x\)

\(\Leftrightarrow2x^2-10x=0\)

=>2x(x-5)=0

=>x=0 hoặc x=5

5, <=> 9x^2 - 45x - 36 = 36x + 36 + 8x^2 - 80x 

<=> x^2 - x - 72 = 0 <=> x = 9 ; x = -8 

6, <=> x^2 - 9 = 9x - x^2 - 9 + x = 10x - x^2 - 9 

<=> 2x^2 - 10x = 0 <=> x = 0 ; x = 5 

7, <=> (x-1)^2 = (3x+3)^2 

<=> (x-1-3x-3)(x-1+3x+3) = 0

<=> (-2x-4)(4x+2) = 0 <=> x = -2;x=-1/2

8, = (x^2-10x-15)(x^2-10x+25)

Thay =vào biểu thức ta được :

3,5√0,7 − 1,4 + 2,5√0,7 − 2,6 / 0,7 + 3√0,7  −2,7

=

b)Thay =-√5 vào biểu thức. 

=30−30(−√5)/−10+10(−√5)=