\(\dfrac{5\left(x-1\right)+2}{6}-\dfrac{7x-1}{4}=\dfrac{2\left(2x+1\right)}{7}\)

⇔ \(\dfrac{5x-3}{6}-\dfrac{7x-1}{4}=\dfrac{4x+2}{7}\)

⇔ \(\dfrac{5x-3}{6}-\dfrac{7x-1}{4}=\dfrac{4x+2}{7}\)

⇔ \(\dfrac{140x-84}{168}-\dfrac{294x-42}{168}=\dfrac{96x+48}{168}\)

⇔ 140x-84-294x+42=96x+48

⇔ -154x-42=96x+48

⇔ -250x=90

⇔ x=\(\dfrac{-9}{26}\)

Vậy phương trình đã cho có tập nghiệm S={\(\dfrac{-9}{26}\)}

Copy có khác, ko đọc đc j!!! ʌl

Câu 3:

1)

a) Ta có:

hay x=-1

Vậy: x=-1

b) Ta có:

hay y=0

Vậy: y=0

c) Ta có:

hay x=15

Vậy: x=15

d) Ta có:

Vì 3≠0

nên x-5=0

hay x=5

Vậy: x=5

a) 3x - 2 = 2x - 3

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

\(\Leftrightarrow\) x + 1 = 0

\(\Rightarrow\) x = -1

b) 3 - 4y + 24 + 6y = y + 27 + 3y

\(\Leftrightarrow\) 3 - 4y + 24 + 6y - y - 27 - 3y = 0

\(\Leftrightarrow\) -2y = 0

\(\Rightarrow\) y = 0

c)7 - 2x = 22 - 3x

\(\Leftrightarrow\) 7 - 2x - 22 + 3x = 0

\(\Leftrightarrow\) -15 + x = 0

\(\Rightarrow\) x = 15

d) 8x - 3 = 5x + 12

\(\Leftrightarrow\) 8x - 3 - 5x - 12 = 0

\(\Leftrightarrow\)3x -15 = 0

\(\Leftrightarrow\) 3x = 15

\(\Rightarrow\) x = 5

e) x - 12 + 4x = 25 + 2x - 1

\(\Leftrightarrow\) x - 12 + 4x - 25 - 2x + 1 = 0

\(\Leftrightarrow\) 3x - 36 = 0

\(\Leftrightarrow\) 3x = 36

\(\Rightarrow\) x = 12

f ) x + 2x + 3x - 19 = 3x + 5

\(\Leftrightarrow\) x + 2x + 3x - 19 - 3x - 5 = 0

\(\Leftrightarrow\)3x - 24 = 0

\(\Leftrightarrow\) 3x = 24

\(\Rightarrow\) x = 8

g) 11+ 8x - 3 = 5x - 3 +x

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

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

\(\Leftrightarrow\)2x = -11

\(\Rightarrow\)x = \(-\frac{11}{2}\)

h) 4 - 2x +15 = 9x + 4 -2

\(\Leftrightarrow\)19 - 2x = 7x + 4

\(\Leftrightarrow\)-2x - 7x = 4 - 19

\(\Leftrightarrow\)-9x = -15

\(\Rightarrow\)x = \(\frac{15}{9}\) = \(\frac{5}{3}\)

2)

a) \(5-\left(x-6\right)=4\cdot\left(3-2\right)\)

\(\Leftrightarrow5-x+6=12-8\)

\(\Leftrightarrow11-x=4\)

\(\Rightarrow x=7\)

b) \(2x\cdot\left(x+2\right)^2-8x^2=2\cdot\left(x-2\right)\cdot\left(x^2+2x+4\right)\)

\(\Leftrightarrow2x\cdot\left(x^2+4x+4\right)-8x^2=2\cdot\left(x^3-8\right)\)

\(\Leftrightarrow2x^3+8x^2+8x-8x^2-2x^3+16=0\)

\(\Leftrightarrow8x+16=0\)

\(\Rightarrow x=-2\)

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

\(\Leftrightarrow7-2x-4=-x-4\)

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

\(\Leftrightarrow-x=-7\)

\(\Rightarrow x=7\)

d) \(\left(x-2\right)^3+\left(3x-1\right)\cdot\left(3x+1\right)=\left(x+1\right)^3\)

\(\Leftrightarrow x^3-6x^2+12x-8+9x^2-1-x^3-3x^2-3x-1=0\)

\(\Leftrightarrow9x-10=0\)

\(\Rightarrow x=\frac{10}{9}\)

e)\(\left(x+1\right)\cdot\left(2x-3\right)=\left(2x-1\right)\cdot\left(x+5\right)\)

\(\Leftrightarrow2x^3-3x+2x-3-2x^2-10x+x+5=0\)

\(\Leftrightarrow2-10x=0\)

\(\Rightarrow x=\frac{2}{10}=\frac{1}{5}\)

f)\(\left(x-1\right)^3-x\cdot\left(x+1\right)^2=5x\cdot\left(2-x\right)-11\cdot\left(x+2\right)\)

\(\Leftrightarrow x^3-3x^2+3x-1-x^3-2x^2-x-10x+5x^2+11x+22=0\)

\(\Leftrightarrow3x+21=0\)

\(\Rightarrow x=-7\)

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

\(\Leftrightarrow x-1-2x+1-9+x=0\)

\(\Leftrightarrow-9=0\)

\(\Rightarrow\) Phương trình vô nghiệm

h)\(\left(x-3\right)\cdot\left(x+4\right)-2\cdot\left(3x-2\right)=\left(x-4\right)^2\)

\(\Leftrightarrow x^2+4x-3x-12-6x+4=x^2-8x+16\)

\(\Leftrightarrow x^2-5x-8=x^2-8x+16\)

\(\Leftrightarrow x^2-5x-8-x^2+8x-16=0\)

\(\Leftrightarrow3x-24=0\)

\(\Rightarrow x=8\)

i)\(x\cdot\left(x+3\right)^2-3x=\left(x+2\right)^3+1\)

\(\Leftrightarrow x^3+6x^2+9x-3x=x^3+6x^2+12x+8+1\)

\(\Leftrightarrow x^3+6x^2+6x=x^3+6x^2+12x+9\)

\(\Leftrightarrow x^3+6x^2+6x-x^3-6x^2-12x-9=0\)

\(\Leftrightarrow-6x-9=0\)

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

j)\(\left(x+1\right)\cdot\left(x^2-x+1\right)-2x=x\cdot\left(x+1\right)\cdot\left(x-1\right)\)

\(\Leftrightarrow\left(x^3+1\right)-2x=x\left(x^2-1\right)\)

\(\Leftrightarrow x^3+1-2x-x^3+x=0\)

\(\Leftrightarrow1-x=0\)

\(\Rightarrow x=1\)

a) 2x-(3x-5x)=4(x+3) 

2x - 3x + 5x = 4x +12

4x = 4x + 12

0x= 12 => ko có giá trị nào của x thỏa mãn( cái kết luận này mik ko bik đúng hay sai)

b) 5(x-3)-4=2(x-1)+7

5x-15 - 4 = 2x-2 + 7

5x-19 = 2x+5

5x-2x = 5+19

3x = 24

x= 8

c) 4(x+3)=-7X+17

4x +12 = -7x + 17

4x+7x = 17-12

11x = 5

x = 5/11

  1)      2x - (3x -5x) = 4(x+3)

\(\Leftrightarrow\)2x +2x = 4x +12

\(\Leftrightarrow\)4x = 4x +12

\(\Leftrightarrow\)0x = 12

Vậy phương trình đã cho vô nghiệm
2)        5(x-3) - 4 = 2(x-1) +7

\(\Leftrightarrow\)5x - 15 - 4 = 2x - 2 +7

\(\Leftrightarrow\)    5x - 1   = 2x +5

\(\Leftrightarrow\)    5x - 2x = 5 +1

\(\Leftrightarrow\)        3x   =   6

\(\Leftrightarrow\)         x    =   2

Vậy tập nghiệm của phương trình là S= {2}

 3)      4(x + 3) = -7x + 17

\(\Leftrightarrow\)4x + 12 = -7x +17

\(\Leftrightarrow\)4x + 7x = 17 - 12

\(\Leftrightarrow\)   11x    =     5

\(\Leftrightarrow\)     x     =    \(\frac{5}{11}\)

Vậy tập nghiệm của phương trình là S={   \(\frac{5}{11}\)}

pt 1 <=> 2x -3x +5x =4x +12

       <=>0x =12

     <=>vo lý vậy pt vô nhiệm

pt <=>5x -19 -2x +2 -7 =0

  <=> 3x = -24

 <=> x= -8 

pt3 4x +12 +7x -17 =0

<=> 11x =5

<=> x =5/11

câu a bài 1:(2x+1)(3x-2)=(5x-8)(2x+1)

<=>(2x+1)(3x-2)-(5x-8)(2x+1)=0

<=>(2x+1)(3x-2-5x+8)=0

<=>(2x+1)(6-2x)=0

bước sau tự làm nốt nha !

câu b:gợi ý: tách 4x^2-1thành (2x-1)(2x+1) rồi làm như câu a

Bài 2: 

a: \(\dfrac{1}{2x-3}-\dfrac{3}{x\left(2x-3\right)}=\dfrac{5}{x}\)

\(\Leftrightarrow x-3=5\left(2x-3\right)=10x-15\)

=>-9x=-12

hay x=4/3

b: \(\Leftrightarrow x\left(x+2\right)-x+2=2\)

=>x²+2x-x+2=2

=>x²+x=0

=>x=0(loại) hoặc x=-1(nhận)

c: \(\dfrac{x+1}{x-2}+\dfrac{x-1}{x+2}=\dfrac{2\left(x^2+2\right)}{\left(x-2\right)\left(x+2\right)}\)

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

=>4=4(luôn đúng)

Vậy: S={x|x<>2; x<>-2}

Đây là những bài cơ bản mà bạn!

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

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

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

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

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

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

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

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

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

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

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

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

Bài 5 :

a, Ta có : \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)

=> \(\frac{3\left(2x+1\right)^2}{15}-\frac{5\left(x-1\right)^2}{15}=\frac{7x^2-14x-5}{15}\)

=> \(3\left(2x+1\right)^2-5\left(x-1\right)^2=7x^2-14x-5\)

=> \(12x^2+12x+3-5x^2+10x-5-7x^2+14x+5=0\)

=> \(36x+3=0\)

=> \(x=-\frac{1}{12}\)

Vậy phương trình trên có nghiệm là \(S=\left\{-\frac{1}{12}\right\}\)

b, Ta có : \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\)

=> \(\frac{5\left(7x-1\right)}{30}+\frac{60x}{30}=\frac{6\left(16-x\right)}{30}\)

=> \(5\left(7x-1\right)+60x=6\left(16-x\right)\)

=> \(35x-5+60x-96+6x=0\)

=> \(101x-101=0\)

=> \(x=1\)

Vậy phương trình trên có tạp nghiệm là \(S=\left\{1\right\}\)

c, Ta có : \(\frac{\left(x-2\right)^2}{3}-\frac{\left(2x-3\right)\left(2x+3\right)}{8}+\frac{\left(x-4\right)^2}{6}=0\)

=> \(\frac{8\left(x-2\right)^2}{24}-\frac{3\left(2x-3\right)\left(2x+3\right)}{24}+\frac{4\left(x-4\right)^2}{24}=0\)

=> \(8\left(x-2\right)^2-3\left(2x-3\right)\left(2x+3\right)+4\left(x-4\right)^2=0\)

=> \(8\left(x^2-4x+4\right)-3\left(4x^2-9\right)+4\left(x^2-8x+16\right)=0\)

=> \(8x^2-32x+32-12x^2+27+4x^2-32x+64=0\)

=> \(-64x+123=0\)

=> \(x=\frac{123}{64}\)

Vậy phương trình có nghiệm là \(S=\left\{\frac{123}{64}\right\}\)

a, <=> x = -4 

b, <=> 6x + 2 = -2x + 5 <=> 8x = 3 <=> x = 3/8 

c, <=> 5x + 2x - 2 = 4x + 7 <=> 2x = 9 <=> x = 9 /2 

d, <=> 10x^2 - 10x^2 - 15x = 15 <=> x = -1 

a, <=> x = -4 

b, <=> 6x + 2 = -2x + 5 <=> 8x = 3 <=> x = 3/8 

c, <=> 5x + 2x - 2 = 4x + 7 <=> 2x = 9 <=> x = 9 /2 

d <=> 10x^2 - 10x^2 - 15x = 15 <=> x = -1 

a) x=-4

b)4x=3

x=3/4

c)3x=9

x=3

d) 15x=15

x=1