a: Ta có: \(\left(x+1\right)^3-\left(x+2\right)\left(x-1\right)^2-3\left(x-3\right)\left(x+3\right)=5\)

\(\Leftrightarrow x^3+3x^2+3x+1-\left(x+2\right)\left(x^2-2x+1\right)-3\left(x^2-9\right)=5\)

\(\Leftrightarrow x^3+3x^2+3x+1-\left(x^3-2x^2+x+2x^2-4x+2\right)-3\left(x^2-9\right)=5\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x-2-3x^2+9=5\)

\(\Leftrightarrow6x=-3\)

hay \(x=-\dfrac{1}{2}\)

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

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

\(\Leftrightarrow2x^3+6x=2x^3+24x\)

\(\Leftrightarrow x=0\)

c: Ta có: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-1=-10\)

\(\Leftrightarrow12x=-11\)

hay \(x=-\dfrac{11}{12}\)

Ý bạn là \(f(x)=\frac{1}{9+x^2}+\frac{3}{9+x^2}\) hay thế nào? Bạn cần viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo để được hỗ trợ tốt hơn).

a) \(x-\dfrac{3}{4}=6\times\dfrac{3}{8}\)

\(x-\dfrac{3}{4}=\dfrac{9}{4}\)

=> \(x=\dfrac{9}{4}+\dfrac{3}{4}=3\)

b) \(\dfrac{7}{8}:x=3-\dfrac{1}{2}\)

\(\dfrac{7}{8}:x=\dfrac{5}{2}\)

=> \(x=\dfrac{7}{8}:\dfrac{5}{2}=\dfrac{7}{20}\)

c) \(x+\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{3}{4}\)

\(x+\dfrac{1}{6}=\dfrac{3}{4}\)

=> \(x=\dfrac{3}{4}-\dfrac{1}{6}=\dfrac{7}{12}\)

d) \(\dfrac{3}{2}\times\dfrac{4}{5}-x=\dfrac{2}{3}\)

\(\dfrac{6}{5}-x=\dfrac{2}{3}\)

=> \(x=\dfrac{6}{5}-\dfrac{2}{3}=\dfrac{8}{15}\)

e) \(x\times3\dfrac{1}{3}=3\dfrac{1}{3}:4\dfrac{1}{4}\)(?)

\(x\times\dfrac{10}{3}=\dfrac{40}{51}\)

=> \(x=\dfrac{40}{51}:\dfrac{10}{3}=\dfrac{4}{17}\)

f) \(5\dfrac{2}{3}:x=3\dfrac{2}{3}-2\)

\(\dfrac{17}{3}:x=\dfrac{5}{3}\)

=> \(x=\dfrac{17}{3}:\dfrac{5}{3}=\dfrac{17}{5}\)

a: =>x-3/4=18/8=9/4

=>x=9/4+3/4=12/4=3

b: =>7/8:x=5/2

=>x=7/8:5/2=7/8*2/5=14/40=7/20

c: x+1/2*1/3=3/4

=>x+1/6=3/4

=>x=3/4-1/6=9/12-2/12=7/12

d: =>12/10-x=2/3

=>6/5-x=2/3

=>x=6/5-2/3=18/15-10/15=8/15

e: =>x*10/3=10/3:17/4=10/3*4/17

=>x=4/17

f: =>17/3:x=13/3-5/2=26/6-15/6=11/6

=>x=17/3:11/6=17/3*6/11=34/11

a: =>x-2/5=3/4:1/3=3/4*3=9/4

=>x=9/4+2/5=45/20+8/20=53/20

b: =>x-2/3=7/3:4/5=7/3*5/4=35/12

=>x=35/12+2/3=43/12

c: 1/3(x-2/5)=4/5

=>x-2/5=4/5*3=12/5

=>x=12/5+2/5=14/5

d: =>2/3x-1/3-1/4x+1/10=7/3

=>5/12x-7/30=7/3

=>5/12x=7/3+7/30=77/30

=>x=77/30:5/12=154/25

e: \(\Leftrightarrow x\cdot\dfrac{3}{7}-\dfrac{2}{7}+\dfrac{1}{2}-\dfrac{5}{4}x+\dfrac{5}{2}=0\)

=>\(x\cdot\dfrac{-23}{28}=\dfrac{2}{7}-3=\dfrac{-19}{7}\)

=>x=19/7:23/28=76/23

f: =>1/2x-3/2+1/3x-4/3+1/4x-5/4=1/5

=>13/12x=1/5+3/2+4/3+5/4=257/60

=>x=257/65

i: =>x^2-2/5x-x^2-2x+11/4=4/3

=>-12/5x=4/3-11/4=-17/12

=>x=17/12:12/5=85/144

Bài dài quá, lần sau chia nhỏ câu hỏi nhé!!!!!

đúng vậy

1) Do x ∈ Z và 0 < x < 3

⇒ x ∈ {1; 2}

2) Do x ∈ Z và 0 < x ≤ 3

⇒ x ∈ {1; 2; 3}

3) Do x ∈ Z và -1 < x ≤ 4

⇒ x ∈ {0; 1; 2; 3; 4}

a: =>9x^2+12x+4-9x^2+12x-4=5x+38

=>24x=5x+38

=>19x=38

=>x=2

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

=>-2x+1=-x

=>-x=-1

=>x=1

f: =>x^3-6x^2+12x-8+9x^2-1=x^3+3x^2+3x+1

=>12x-9=3x+1

=>9x=10

=>x=10/9

b: \(\Leftrightarrow3x^2-12x+12+9x-9=3x^2+3x-9\)

=>-3x+3=3x-9

=>-6x=-12

=>x=2

`a)(1-1/2)xx(1-1/3)xx(1-1/4)xx(1-1/5)`

`=1/2xx2/3xx3/4xx4/5`

`=[1xx2xx3xx4]/[2xx3xx4xx5]`

`=1/5`

`b)(1-3/4)xx(1-3/7)xx(1-3/10)xx(1-3/13)xx .... xx(1-3/97)xx(1-3/100)`

`=1/4xx4/7xx7/10xx10/13xx .... xx94/97xx97/100`

`=[1xx4xx7xx10xx...xx94xx97]/[4xx7xx10xx13xx....xx97xx100]`

`=1/100`

\(A=2x^3+3x^2-3-5x^2-5x=2x^3-2x^2-5x-3\\ B=125-150x+60x^2-8x^3-25+9x^2=-8x^3+69x^2-150x+100\\ C=\left(3x+1-2x+1\right)\left(3x+1+2x-1\right)=5x\left(x+2\right)=5x^2+10x\\ D=\left(2x+1-3+x\right)^2=\left(3x-2\right)^2=9x^2-12x+4\\ E=x^3-6x^2+12x-8-x^3+x+6x^2-18x=-5x-8\\ F=x^3-3x^2+3x-1-3+3x^2-x^3+1-3x=-3\)

`e)(x+2)(x+3)=5-x+x(x-1)-2`

`<=>x^2+3x+2x+6=5-x+x^2-x-2`

`<=>7x=-3`

`<=>x=-3/7`

`f)(2x-3)(3-x)+(x-1)^2=1-(x+3)(x-3)`

`<=>6x-2x^2-9+3x+x^2-2x+1=1-x^2+9`

`<=>7x=17`

`<=>x=17/7`

`j)3(x+1)(x-1)=3(x^2+2x)+1`

`<=>3x^2-3=3x^2+6x+1`

`<=>6x=-4`

`<=>x=-2/3`