a: =>(x-2)(2x+5)=0

=>x-2=0 hoặc 2x+5=0

=>x=2 hoặc x=-5/2

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

=>\(\dfrac{2x^2+2x-x^2+x}{x^2-1}=1\)

=>x^2+3x=x^2-1

=>3x=-1

=>x=-1/3

\(a,\Leftrightarrow\left(x-2\right)\left(2x+5\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x-2=0\\2x+5=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=2\\x=\dfrac{5}{2}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm S = \(\left\{2;\dfrac{5}{2}\right\}\)

\(c,\Leftrightarrow2x.\left(x+1\right)-x.\left(x-1\right)=\left(x-1\right)\left(x+1\right)\)              ( ĐKXĐ: \(x\ne-1;x\ne1\) )

\(\Leftrightarrow2x^2+2x-x^2+x=x^2-1\\ \Leftrightarrow x^2-x^2+3x=-1\\ \Leftrightarrow3x=-1\\ \Leftrightarrow x=-\dfrac{1}{3}\)  ( nhận )

Vậy phương trình có tập nghiệm S = \(\left\{-\dfrac{1}{3}\right\}\)

a. 3(x-2)-10=5(2x + 1)

<=> 3x - 6 - 10 = 10x + 5

<=> 3x - 10x = 5 + 6 + 10

<=> -7x = 21

<=> x = -3

b. 3x + 2=8 -2(x-7)

<=> 3x + 2 = 8 - 2x + 14

<=> 3x + 2x = 8 + 14 - 2

<=> 5x = 20

<=> x = 4

c. 2x-(2+5x)= 4(x + 3)

<=> 2x - 2 - 5x = 4x + 12

<=> 2x - 5x - 4x = 12 + 2

<=> -7x = 14

<=> x = -2

d. 5-(x +8)=3x + 3(x-9)

<=> 5 - x - 8 = 3x + 3x - 27

<=> -x - 3x - 3x = -27 + 8 - 5

<=> -7x = -24

<=> x = 24/7

e. 3x - 18 + x= 12-(5x + 3)

<=> 3x - 18 + x = 12 - 5x - 3

<=> 3x + x - 5x = 12 - 3 + 18

<=> -x = 27

<=> x = - 27

a. 3(x-2)-10=5(2x + 1)

<=> 3x - 6 - 10 = 10x + 5

<=> 3x - 10x = 5 + 6 + 10

<=> -7x = 21

<=> x = -3

b. 3x + 2=8 -2(x-7)

<=> 3x + 2 = 8 - 2x + 14

<=> 3x + 2x = 8 + 14 - 2

<=> 5x = 20

<=> x = 4

c. 2x-(2+5x)= 4(x + 3)

<=> 2x - 2 - 5x = 4x + 12

<=> 2x - 5x - 4x = 12 + 2

<=> -7x = 14

<=> x = -2

d. 5-(x +8)=3x + 3(x-9)

<=> 5 - x - 8 = 3x + 3x - 27

<=> -x - 3x - 3x = -27 + 8 - 5

<=> -7x = -24

<=> x = 24/7

e. 3x - 18 + x= 12-(5x + 3)

<=> 3x - 18 + x = 12 - 5x - 3

<=> 3x + x - 5x = 12 - 3 + 18

<=> -x = 27

<=> x = - 27

Bài 1: Giải các phương trình sau:

a) 3(2,2-0,3x)=2,6 + (0,1x-4)

<=> 6.6 - 0.9x = 2,6 + 0,1x - 4

<=> - 0.9x - 0,1x = -6.6 -1,4

<=> -x = -8

<=> x = 8

Vậy x = 8

b) 3,6 -0,5 (2x+1) = x - 0,25(22-4x)

<=> 3,6 - x - 0,5 = x - 5,5 + x

<=> - x - 3,1 = -5,5

<=> - x = -2.4

<=> x = 2.4

Vậy  x = 2.4

a: =>4x-2x-2-3x-2=0

=>-x-4=0

=>x=-4

b: =>x+2-2x-2+x=0

=>0x=0(luôn đúng)

d: =>3x=3

hay x=1

e: =>2x=1

hay x=1/2

f: =>4x=-4

hay x=-1

g: =>3x=-3

hay x=-1

c: =>2x+3-5-4+x=0

=>3x-6=0

=>x=2

d: =>3x=3

hay x=1

e: =>2x=1

hay x=1/2

f: =>4x=-4

hay x=-1

g: =>3x=-3

hay x=-1

\(a,4x-2\left(x+1\right)=3x+2\\ \Leftrightarrow4x-2x-2-3x-2=0\\ \Leftrightarrow-x-4=0\\ \Leftrightarrow x+4=0\\ \Leftrightarrow x=-4\)

Vậy pt có tập nghiệm \(S=\left\{-4\right\}\)

\(b,x+2-2\left(x+1\right)=-x\\ \Leftrightarrow x+2-2x-2+x=0\\ \Leftrightarrow0=0\)

Vậy pt có tập nghiệm \(S=R\)

\(c,2\left(x+3\right)-5=4-x\\ \Leftrightarrow2x+6-5-4+x=0\\ \Leftrightarrow3x-3=0\\ \Leftrightarrow3x=3\\ \Leftrightarrow x=1\)

Vậy pt có tập nghiệm \(S=\left\{1\right\}\)

\(d,3x-2=1\\ \Leftrightarrow3x=3\\ \Leftrightarrow x=1\)

Vậy pt có tập nghiệm \(S=\left\{1\right\}\)

\(e,2x-1=0\\ \Leftrightarrow2x=1\\ \Leftrightarrow x=\dfrac{1}{2}\)

Vậy pt có tập nghiệm \(S=\left\{\dfrac{1}{2}\right\}\)

\(f,4x+3=-1\\ \Leftrightarrow4x=-4\\ \Leftrightarrow x=-1\)

Vậy pt có tập nghiệm \(S=\left\{-1\right\}\)

\(g,3x+2=-1\\ \Leftrightarrow3x=-3\\ \Leftrightarrow x=-1\)

Vậy pt có tập nghiệm \(S=\left\{-1\right\}\)

a: =>3,6-1,7x=2,3-1,4-4=0,9-4=-3,1

=>1,7x=6,7

hay x=67/17

b: \(\Leftrightarrow30\left(5x+4\right)-15\left(3x+5\right)=24\left(4x+9\right)-40\left(x-9\right)\)

=>150x+120-45x-75=96x+216-40x+360

=>105x+45=56x+576

=>49x=531

hay x=531/49

a) Ta có: \(2\left(3x+1\right)-4\left(5-2x\right)>2\left(4x-3\right)-6\)

\(\Leftrightarrow6x+2-20+8x>8x-6-6\)

\(\Leftrightarrow14x-18-8x+12>0\)

\(\Leftrightarrow6x-6>0\)

\(\Leftrightarrow6x>6\)

hay x>1

Vậy: S={x|x>1}

b) Ta có: \(9x^2-3\left(10x-1\right)< \left(3x-5\right)^2-21\)

\(\Leftrightarrow9x^2-30x+3< 9x^2-30x+25-21\)

\(\Leftrightarrow9x^2-30x+3-9x^2+30x-4< 0\)

\(\Leftrightarrow-1< 0\)(luôn đúng)

Vậy: S={x|\(x\in R\)}

a) \(x^2-4x+4=25\\ \Rightarrow\left(x-2\right)^2=25\\ \Rightarrow\left[{}\begin{matrix}x-2=-5\\x-2=5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)

b) \(\left(5-2x\right)^2-16=0\\ \Rightarrow\left(5-2x\right)^2=16\\ \Rightarrow\left[{}\begin{matrix}5-2x=-4\\5-2x=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4,5\\0,5\end{matrix}\right.\)

c) \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\\ \Rightarrow\left(x-3\right)^3-\left(x-3\right)^3+9\left(x+1\right)^2=15\\ \Rightarrow9\left(x+1\right)^2=15\\ \Rightarrow\left(x+1\right)^2=\dfrac{5}{3}\\ \Rightarrow\left[{}\begin{matrix}x+1=-\sqrt{\dfrac{5}{3}}\\x+1=\sqrt{\dfrac{5}{3}}\end{matrix}\right.\)

   \(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3+\sqrt{15}}{3}\\x=\dfrac{-3+\sqrt{15}}{3}\end{matrix}\right.\)

a)\(\Leftrightarrow\)\(x^2-4x-21=0\)

\(\Leftrightarrow\)\(x^2-7x+3x-21=0\)

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

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

\(\Leftrightarrow\)\(\left[\begin{array}{} x=7\\ x=-3 \end{array} \right.\)

b)\(\Leftrightarrow\)\((5-2x)^2-4^2=0\)

\(\Leftrightarrow\)\((5-2x-4)(5-2x+4)=0\)

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

\(\Leftrightarrow\)\(\left[\begin{array}{} x=\dfrac{1}{2}\\ x=\dfrac{9}{2} \end{array} \right.\)

\(a.\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)

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

\(\Leftrightarrow x-1=3x-2\)

\(\Leftrightarrow2x=1\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

c: =>x-3=0

hay x=3

d: \(\Leftrightarrow\left(3x-1\right)\cdot\left(x^2+2-7x+10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)

hay \(x\in\left\{\dfrac{1}{3};3;4\right\}\)