a)       

\(\begin{array}{l}\,\,\,\,\,\,\,\frac{{5x - 2}}{3} = \frac{{5 - 3x}}{2}\\\frac{{2\left( {5x - 2} \right)}}{6} = \frac{{3\left( {5 - 3x} \right)}}{6}\\\,2\left( {5x - 2} \right) = 3\left( {5 - 3x} \right)\\\,\,\,\,\,\,10x - 4 = 15 - 9x\\\,\,\,10x + 9x = 15 + 4\\\,\,\,\,\,\,\,\,\,\,\,\,\,19x = 19\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 19:19\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 1.\end{array}\)

Vậy phương trình có nghiệm \(x = 1\).

b)      

\(\begin{array}{l}\,\,\,\,\,\,\frac{{10x + 3}}{{12}} = 1 + \frac{{6 + 8x}}{9}\\\frac{{3\left( {10x + 3} \right)}}{{36}} = \frac{{36}}{{36}} + \frac{{4\left( {6 + 8x} \right)}}{{36}}\\\,3\left( {10x + 3} \right) = 36 + 4\left( {6 + 8x} \right)\\\,\,\,\,\,\,\,30x + 9 = 36 + 24 + 32x\\\,\,\,\,\,\,\,30x + 9 = 60 + 32x\\\,30x - 32x = 60 - 9\\\,\,\,\,\,\,\,\,\,\,\,\,\, - 2x = 51\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x =  - \frac{{51}}{2}.\end{array}\)

Vậy phương trình có nghiệm \(x =  - \frac{{51}}{2}\).

c)       

\(\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{7x - 1}}{6} + 2x = \frac{{16 - x}}{5}\\\frac{{5\left( {7x - 1} \right)}}{{30}} + \frac{{30.2x}}{{30}} = \frac{{6\left( {16 - x} \right)}}{{30}}\\\,\,5\left( {7x - 1} \right) + 30.2x = 6\left( {16 - x} \right)\\\,\,\,\,\,\,\,\,\,\,35x - 5 + 60x = 96 - 6x\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,95x - 5 = 96 - 6x\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,95x + 6x = 96 + 5\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,101x = 101\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 101:101\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 1\end{array}\)

Vậy phương trình có nghiệm \(x = 1\).

a) 4 ( x + 5 )( x + 6 )( x + 10 )( x + 12 ) = 3x²
Do x = 0 không là nghiệm pt nên chia 2 vế pt cho \(x^2\ne0\), ta được :

\(\frac{4}{x^2}\left(x^2+60+17x\right)\left(x^2+60+16x\right)=3\)

\(\Leftrightarrow4\left(x+\frac{60}{x}+17\right)\left(x+\frac{60}{x}+16\right)=3\)

Đến đây ta đặt  \(x+\frac{60}{x}+16=t\left(1\right)\)

Ta được :

\(4t\left(t+1\right)=3\Leftrightarrow4t^2+4t-3=0\Leftrightarrow\left(2t+3\right)\left(2t-1\right)=0\)

Từ đó ta lắp vào ( 1 ) tính được x 

Đây là những bài cơ bản mà bạn!

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

\(< =>\frac{\left(5x-2\right).2}{6}=\frac{\left(5-3x\right).3}{6}\)

\(< =>\left(5x-2\right).2=\left(5-3x\right).3\)

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

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

\(< =>19x=19< =>x=1\)

\(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)

\(< =>\frac{\left(10x+3\right).3}{36}=\frac{36}{36}+\frac{\left(6+8x\right).4}{36}\)

\(< =>\left(10x+3\right).3=36+\left(6+8x\right).4\)

\(< =>30x+9=36+24+32x\)

\(< =>32x-30x=9-36-24\)

\(< =>2x=9-60=-51< =>x=-\frac{51}{2}\)

a) 7x - 35 = 0

<=> 7x = 0 + 35

<=> 7x = 35

<=> x = 5

b) 4x - x - 18 = 0

<=> 3x - 18 = 0

<=> 3x = 0 + 18

<=> 3x = 18

<=> x = 5

c) x - 6 = 8 - x

<=> x - 6 + x = 8

<=> 2x - 6 = 8

<=> 2x = 8 + 6

<=> 2x = 14

<=> x = 7

d) 48 - 5x = 39 - 2x

<=> 48 - 5x + 2x = 39

<=> 48 - 3x = 39

<=> -3x = 39 - 48

<=> -3x = -9

<=> x = 3

có bị viết nhầm thì thông cảm nha!

la`thu'hai nga`y 19 nhe

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

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

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

\(11x=25\)

\(x=\frac{24}{11}\)

\(b,\frac{10x+3}{12}=1+\frac{6+8x}{9}\)

\(\frac{10x+3}{12}=\frac{15+8x}{9}\)

\(9\left(10x+3\right)=12\left(15+8x\right)\)

\(3\left(10x+3\right)=4\left(8x+15\right)\)

\(30x+9=32x+60\)

\(-2x=51\)

\(x=-\frac{51}{2}\)

\(c,\frac{x}{3}-\frac{2x+1}{2}=\frac{x}{6}-x\)

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

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

\(x=3\)

P/s: Bn xem lại đề bài phần d nha!

=.= hk tốt!!

câu d sao bn đề đúng r ạ

tức là đề có 2 dấu = hả bn?

a: \(\Leftrightarrow4\left(x^2+60+17x\right)\left(x^2+60+16x\right)=3x^2\)

\(\Leftrightarrow4\cdot\left[\left(x^2+60\right)^2+33x\left(x^2+60\right)+272x^2\right]=3x^2\)

=>4(x^2+60)^2+132x(x^2+60)+1085x^2=0

=>4(x^2+60)^2+62x(x^2+60)+70x(x^2+60)+1085x^2=0

=>2(x^2+60)(2x^2+120+31x)+35x(2x^2+120+31x)=0

=>(2x^2+120+35x)(2x^2+31x+120)=0

=>\(x\in\left\{\dfrac{-35\pm\sqrt{265}}{4};-\dfrac{15}{2};-8\right\}\)

b: Đặt x^2-3x=a

Phương trình sẽ là \(\dfrac{1}{a+3}+\dfrac{2}{a+4}=\dfrac{6}{a+5}\)

\(\Leftrightarrow\dfrac{a+4+2a+6}{\left(a+3\right)\left(a+4\right)}=\dfrac{6}{a+5}\)

=>(3a+10)(a+5)=6(a^2+7a+12)

=>6a^2+42a+72=3a^2+15a+10a+50

=>3a^2+17a+22=0

=>x=-2 hoặc x=-11/3

\(a,4x-6< 7x-12\)

\(\Leftrightarrow6< 3x\Leftrightarrow x>2\)

\(b,\frac{3x-7}{4}\ge2-\frac{x+5}{3}\)

\(\Leftrightarrow3\left(3x-7\right)\ge24-4\left(x+5\right)\)

\(\Leftrightarrow13x\ge25\Leftrightarrow x\ge\frac{25}{13}\)

\(c,\frac{3x-8}{-7}\ge1-\frac{x+2}{-3}\)

\(\Leftrightarrow-3\left(3x-8\right)\ge21+7\left(x+2\right)\)

\(\Leftrightarrow-16x\ge11\)

\(\Leftrightarrow x\le-\frac{11}{16}\)

\(d,-12-8x>3+2x-\left(5-7x\right)\)

\(\Leftrightarrow14>17x\Leftrightarrow x< \frac{14}{17}\)

\(e,-1+\frac{x-1}{-3}\le\frac{x+2}{-9}\)

\(\Leftrightarrow-9-3\left(x-1\right)\le-\left(x+2\right)\)

\(\Leftrightarrow-2x\le4\Leftrightarrow x\ge-2\)