2:

a: =>x^2+3x-4x-12-(x^2-5x+x-5)=8

=>x^2-x-12-x^2+4x+5=8

=>3x-7=8

=>3x=15

=>x=5

b: =>3x^2+3x-2x-2-3x^2-21x=13

=>-20x=15

=>x=-3/4

c: =>x^2-25-x^2-2x=9

=>-2x=25+9=34

=>x=-17

d: =>x^3-1-x^3+3x=1

=>3x-1=1

=>3x=2

=>x=2/3

a) \(3\left(x^2-2x+1\right)+x\left(2-3x\right)=7\)

\(\Rightarrow3x^2-6x+3+2x-3x^2=7\)

\(\Rightarrow-4x+3=7\)

\(\Rightarrow-4x+3-7=0\)

\(\Rightarrow-4x-4=0\)

\(\Rightarrow-4\left(x+1\right)=0\)

\(\Rightarrow x+1=0\)

\(\Rightarrow x=-1\)

b) \(5\left(x-2\right)+2\left(x+3\right)=10\)

\(\Rightarrow5x-10+2x+6=10\)

\(\Rightarrow7x-4=10\)

\(\Rightarrow7x=10+4=14\)

\(\Rightarrow x=\dfrac{14}{7}=2\)

c) \(\left(x+1\right)\left(-3\right)+5\left(x-4\right)=-3\)

\(\Rightarrow-3x-3+5x-20=-3\)

\(\Rightarrow2x-23=-3\)

\(\Rightarrow2x=-3+23=20\)

\(\Rightarrow x=\dfrac{20}{2}=10\)

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

\(\Rightarrow2x-2-3x+x^2=x^2\)

\(\Rightarrow-x-2+x^2-x^2=0\)

\(\Rightarrow-x-2=0\)

\(\Rightarrow-x=2\)

\(\Rightarrow x=-2\)

đ) \(3x\left(x+5\right)-2\left(x+5\right)=3x^2\)

\(\Rightarrow3x^2+15x-2x-10=3x^2\)

\(\Rightarrow3x^2-3x^2+13x-10=0\)

\(\Rightarrow13x-10=0\)

\(\Rightarrow13x=10\)

\(\Rightarrow x=\dfrac{10}{13}\)

e) \(4x\left(x+2\right)+x\left(4-x\right)=3x^2+12\)

\(\Rightarrow4x^2+8x+4x-x^2=3x^2+12\)

\(\Rightarrow3x^2+12x=3x^2+12\)

\(\Rightarrow3x^2+12x-3x^2-12=0\)

\(\Rightarrow12\left(x-1\right)=0\)

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

f) \(\dfrac{1}{3}x\left(3x+6\right)-x\left(x-5\right)=9\)

\(\Rightarrow x^2+2x-x^2+5x=9\)

\(\Rightarrow7x=9\)

\(\Rightarrow x=\dfrac{9}{7}\)

a) (3x + 1)^2 - 2(3x + 1)(3x - 5) + (3x - 5)^2 

= 9x^2 + 6x + 1 - 18x^2 + 24x + 10 + 9x^2 - 30x + 25

= 36

b) (3x^2 - y)^2

= 9x^4 - 6x^2y + y^2

c) (3x + 5)^2 + (3x - 5)^2 - (3x + 2)(3x - 2)

= 9x^2 + 30x + 25 + 9x^2 - 30x + 25 - 9x^2 + 4

= 9x^2 + 54

d) 2x(2x - 1)^2 - 3x(x + 3)(x - 3) - 4x(x + 1)^2

= 8x^3 - 8x^2 + 2x - 3x^2 + 27x - 4x^3 - 8x^2 - 4x

= x^3 - 16x^2 + 25x

e) (x - 2)(x^2 + 2x + 4) - (x + 1)^2 + 3(x - 1)(x + 1)

= x^3 - 8 - x^2 - 2x - 1 + 3x^2 - 2

= x^3 + 2x^2 - 2x - 12

f) (x^4 - 5x^2 + 25)(x^2 + 5) - (2 + x^2)^2 + 3(1 + x^2)^2

= x^6 + 125 - 4 - 4x^2 - x^2 + 3 + 6x^2 + 3x^4

= x^6 + 2x^4 + 2x^2 + 124

1/ \(2\left(x-5\right)=\left(-x-5\right)\)

\(\Leftrightarrow2x-10=-x-5\)

\(\Leftrightarrow3x=5\)

\(\Leftrightarrow x=\dfrac{5}{3}\)

Vậy: \(S=\left\{\dfrac{5}{3}\right\}\)

==========

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

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

\(\Leftrightarrow-x=-7\)

\(\Leftrightarrow x=7\)

Vậy: \(S=\left\{7\right\}\)

==========

3/ \(4\left(x-5\right)-\left(3x-1\right)=x-19\)

\(\Leftrightarrow4x-20-3x+1=x-19\)

\(\Leftrightarrow0x=0\)

Vậy: \(S=\left\{x|x\text{ ∈ }R\right\}\) 

===========

4/ \(7-\left(x-2\right)=5\left(2-3x\right)\)

\(\Leftrightarrow7-x+2=10-15x\)

\(\Leftrightarrow14x=1\)

\(\Leftrightarrow x=\dfrac{1}{14}\)

Vậy: \(S=\left\{\dfrac{1}{14}\right\}\)

==========

5/ \(2x-\left(5-3x\right)=7x+1\)

\(\Leftrightarrow2x-5+3x=7x+1\)

\(\Leftrightarrow-2x=6\)

\(\Leftrightarrow x=-3\)

Vậy: \(S=\left\{-3\right\}\)

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Chúc bạn học tốt.

1. \(2\left(x-5\right)=-x-5\)

\(\Leftrightarrow3x=5\)

\(\Leftrightarrow x=\dfrac{5}{3}\)

Vậy \(S=\left\{\dfrac{5}{3}\right\}\)

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

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

\(\Leftrightarrow x=7\)

Vậy \(S=\left\{7\right\}\)

3. \(4\left(x-5\right)-\left(3x-1\right)=x-19\)

\(\Leftrightarrow4x-20-3x+1-x+19=0\)

\(\Leftrightarrow0x=0\)

Vậy \(S=\left\{x\in R\right\}\)

4. \(7-\left(x-2\right)=5\left(2-3x\right)\)

\(\Leftrightarrow7-x+2-10+15x=0\)

\(\Leftrightarrow14x-1=0\)

\(\Leftrightarrow x=\dfrac{1}{14}\)

Vậy \(S=\left\{\dfrac{1}{14}\right\}\)

4. \(2x-\left(5-3x\right)=7x+1\)

\(\Leftrightarrow2x-5+3x-7x-1=0\)

\(\Leftrightarrow-2x-6=0\)

\(\Leftrightarrow x=-3\)

Vậy \(S=\left\{-3\right\}\)

1: Ta có: \(2\left(x-5\right)=\left(-x-5\right)\)

\(\Leftrightarrow2x-10+x+5=0\)

\(\Leftrightarrow3x=5\)

hay \(x=\dfrac{5}{3}\)

2: Ta có: \(2\left(x+3\right)-3\left(x-1\right)=2\)

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

\(\Leftrightarrow-x=-7\)

hay x=7

3: Ta có: \(4\left(x-5\right)-\left(3x-1\right)=x-19\)

\(\Leftrightarrow4x-20-3x+1-x+19=0\)

\(\Leftrightarrow0x=0\)(luôn đúng

i: \(=\dfrac{x+1+x-18+x+2}{x-5}=\dfrac{3x-15}{x-5}=3\)

Bài 1:

\(i,\dfrac{x+1}{x-5}+\dfrac{x-18}{x-5}-\dfrac{x+2}{5-x}=\dfrac{x+1}{x-5}+\dfrac{x-18}{x-5}+\dfrac{x+2}{x-5}=\dfrac{x+1+x-18+x+2}{x-5}=\dfrac{3x-15}{x-5}=\dfrac{3\left(x-5\right)}{x-5}=3\)

\(j,\dfrac{3x\left(x-2\right)}{3x-2}+\dfrac{6x^2}{3x-2}-\dfrac{2\left(2-3x\right)}{2-3x}=\dfrac{3x^2-6x}{3x-2}+\dfrac{6x^2}{3x-2}+\dfrac{4-6x}{3x-2}=\dfrac{3x^2-6x+6x^2+4-6x}{3x-2}=\dfrac{9x^2-12x+4}{3x-2}=\dfrac{\left(3x-2\right)^2}{3x-2}=3x-2\)

\(n,\dfrac{2}{x}+\dfrac{3}{x-1}+\dfrac{1-4x}{x^2-x}=\dfrac{2\left(x-1\right)+3x+1-4x}{x\left(x-1\right)}=\dfrac{2x-2+3x+1-4x}{x\left(x-1\right)}=\dfrac{x-1}{x\left(x-1\right)}=\dfrac{1}{x}\)

Bài 2:

\(j,\dfrac{2}{3x}-\dfrac{1}{2x-2}-\dfrac{x-4}{6x-6x^2}=\dfrac{4\left(x-1\right)}{6x\left(x-1\right)}-\dfrac{3x}{6x\left(x-1\right)}-\dfrac{x-4}{6x\left(1-x\right)}=\dfrac{4x-4-3x+x-4}{6x\left(x-1\right)}=\dfrac{2x-8}{6x\left(x-1\right)}=\dfrac{2\left(x-4\right)}{6x\left(x-1\right)}=\dfrac{x-4}{3x\left(x-1\right)}\)

a: =>x^2-25-x^2-3x=10

=>-3x=35

=>x=-35/3

b: =>4x^2-9-4(x^2+4x+4)=5

=>4x^2-9-4x^2-16x-16-5=0

=>-16x-30=0

=>x=-15/8

c: =>9x^2+45x-9x^2+4=7

=>45x=3

=>x=1/15

d: =>x^3+3x^2+3x+1-x^3-3x^2+5x=8

=>8x=7

=>x=7/8

1 ) \(lim_{x\rightarrow+\infty}\dfrac{3x^2+5}{x^3-x+2}=lim_{x\rightarrow+\infty}\dfrac{\dfrac{3}{x}+\dfrac{5}{x^3}}{1-\dfrac{1}{x^2}+\dfrac{2}{x^3}}=0\)

2 ) \(lim_{x\rightarrow-\infty}\dfrac{2x^2\left(3x^2-5\right)^3\left(1-x\right)^5}{3x^{14}+x^2-1}\)  \(=lim_{x\rightarrow-\infty}\dfrac{\dfrac{2}{x}\left(3-\dfrac{5}{x^2}\right)^3\left(\dfrac{1}{x}-1\right)^5}{3+\dfrac{1}{x^{12}}-\dfrac{1}{x^{14}}}=0\)

3 ) \(lim_{x\rightarrow+\infty}\dfrac{3x-\sqrt{2x^2+5}}{x^2-4}=lim_{x\rightarrow+\infty}\dfrac{\left(7x^2-5\right)}{\left(3x+\sqrt{2x^2+5}\right)\left(x^2-4\right)}\)

\(=lim_{x\rightarrow+\infty}\dfrac{\dfrac{7}{x}-\dfrac{5}{x^3}}{\left(3+\sqrt{2+\dfrac{5}{x^2}}\right)\left(1-\dfrac{4}{x^2}\right)}=0\)

1: \(A=\left(-x+5\right)\left(x-2\right)+\left(x-7\right)\left(x+7\right)\)

\(=-x^2+2x+5x-10+x^2-49=7x-59\)

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

\(=9x^2+6x+1-9x^2+4=6x+5\)

=>7x-59=6x+5

=>x=64

2: \(A=\left(5x-1\right)\left(x+1\right)-2\left(x-3\right)^2\)

\(=5x^2+5x-x-1-2x^2+12x-9\)

\(=3x^2+16x-10\)

\(B=\left(x+2\right)\left(3x-1\right)-\left(x+4\right)^2+x^2-x\)

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

\(=3x^2-4x-18\)

=>16x-10=-4x-18

=>20x=-8

hay x=-2/5