Bài 2: 

a: =>x-45=-20

hay x=25

b: =>35+30-5x=-12+112

=>65-5x=100

=>5x=-35

hay x=-7

c: =>x-124=1000

hay x=1124

d: \(\Leftrightarrow46-\left(3x-2\right)^3=-18\)

\(\Leftrightarrow\left(3x-2\right)^3=64\)

=>3x-2=4

hay x=2

Bài 2: 

a: =>x-45=-20

hay x=25

b: =>35+30-5x=-12+112=100

=>65-5x=100

=>5x=-35

hay x=-7

c: =>x-124=1000

hay x=1124

d: \(\Leftrightarrow46-\left(3x-2\right)^3=-18\)

\(\Leftrightarrow\left(3x-2\right)^3=64\)

=>3x-2=4

hay x=2

sao bạn ko làm bài 1 cho mình zậy

Ta có: \(\frac{2}{x^2-1}+Q=\frac{6}{x-3}-\frac{2x^2}{1-x^2}\)

=>\(Q+\frac{2}{x^2-1}=\frac{6}{x-3}+\frac{2x^2}{x^2-1}\)

=>\(Q=\frac{6}{x-3}+\frac{2x^2-2}{x^2-1}=\frac{6}{x-3}+2=\frac{6+2x-6}{x-3}=\frac{2x}{x-3}\)

Câu 1:

\(\left(x-2\right)\left(x^2+2x+4\right)+25x=x\left(x+5\right)\left(x-5\right)+8\)

\(\Leftrightarrow x^3-8+25x=x\left(x^2-25\right)+8\)

\(\Leftrightarrow x^3-8+25x=x^3-25x+8\)

\(\Leftrightarrow x^3-8+25x-x^3+25x-8=0\)

\(\Leftrightarrow50x-16=0\)

\(\Leftrightarrow50x=16\)

\(\Leftrightarrow x=\dfrac{8}{25}\)

Câu 2 :

\(\dfrac{x+5}{4}+\dfrac{3+2x}{3}=\dfrac{6x-1}{3}-\dfrac{1-2x}{12}\)

<=> \(\dfrac{3\left(x+5\right)}{12}+\dfrac{4\left(3+2x\right)}{12}=\dfrac{4\left(6x-1\right)}{12}-\dfrac{1-2x}{12}\)

<=>\(\dfrac{3x+15+12+8x}{12}=\dfrac{24x-4-1+2x}{12}\)

<=> 3x + 15 + 12 + 8x = 24x - 4 - 1 +2x

<=> 11x+27 = 26x -5

<=> ( 26x - 5 ) - ( 11x + 27 ) = 0

<=> 15x - 32 = 0

<=> 15x = 32

<=> x = \(\dfrac{32}{15}\)

bài này cần có đkxđ 

a: =>4/3x=7/9-4/9=1/3

=>x=1/4

b: =>5/2-x=9/14:(-4/7)=-9/8

=>x=5/2+9/8=29/8

c: =>3x+3/4=8/3

=>3x=23/12

hay x=23/36

d: =>-5/6-x=7/12-4/12=3/12=1/4

=>x=-5/6-1/4=-10/12-3/12=-13/12

em moi lop 4 mà

Sửa đề: \(19 + 3x + 4\sqrt{-x^2 - x + 6} = 10\sqrt{2 - x} + 10\sqrt{x + 3}\) (1)

ĐKXĐ: -3<=x<=2

Đặt \(u=\sqrt{2-x};v=\sqrt{x+3}\) (Điều kiện: u>0; v>0)

\(u^2+v^2=2-x+3+x=5\)

\(uv=\sqrt{\left(2-x\right)\left(x+3\right)}=\sqrt{2x+6-x^2-3x}=\sqrt{-x^2-x+6}\)

\(v^2=x+3\)

=>\(x=v^2-3\)

=>\(3x=3v^2-9\)

(1) sẽ trở thành: \(19 + (3v^2 - 9) + 4uv = 10u + 10v\)

=>\(10+3v^2+4uv-10u-10v=0\)

=>\(u(4v-10)=-3v^2+10v-10=-3\left(v^2-\frac{10}{3}v+\frac{10}{3}\right)=-3\left(v^2-2\cdot v\cdot\frac53+\frac{25}{9}+\frac59\right)=-3\left(v-\frac53\right)^2-\frac53<0\) ∀v

=>u(4v-10)<0

=>4v-10<0

=>4v<10

=>v<5/2

=>\(u = \frac{3v^2 - 10v + 10}{10 - 4v}\)

=>\(5 - v^2 = \left( \frac{3v^2 - 10v + 10}{10 - 4v} \right)^2\)

=>\((5 - v^2)(10 - 4v)^2 = (3v^2 - 10v + 10)^2\)

=>\((5 - v^2)(16v^2 - 80v + 100) = 9v^4 - 60v^3 + 160v^2 - 200v + 100\)

=>\(-16v^4 + 80v^3 - 20v^2 - 400v + 500 = 9v^4 - 60v^3 + 160v^2 - 200v + 100\)

=>\(25v^4 - 140v^3 + 180v^2 + 200v - 400 = 0\)

=>\(5v^4 - 28v^3 + 36v^2 + 40v - 80 = 0\)

=>\((v - 2)(5v^3 - 18v^2 + 40) = 0\)

=>v-2=0

=>v=2

=>x+3=4

=>x=1(nhận)

1) 2x – (3 – 5x) = 4( x +3)

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

<=>2x-3+5x-4x-12=0

<=>3x-15=0

<=>x=5

2) 5(2x-3) - 4(5x-7) =19 - 2(x+11)

<=>10x-15-20x+28=19-2x-22

<=>10x-15-20x+28-19+2x+22=0

<=>-8x+16=0

<=>x=2

Dễ thế mà không làm được

$\left|\frac54x-\frac72\right|=\frac58+\frac35$

$\left|\frac54x-\frac72\right|=\frac{49}{40}$

$\frac54x-\frac72=\pm\frac{49}{40}$

TH1:

$\frac54x-\frac72=\frac{49}{40}$

$\frac54x=\frac{189}{40}$

$x=\frac{189}{50}$

TH2:

$\frac54x-\frac72=-\frac{49}{40}$

$\frac54x=\frac{91}{40}$

$x=\frac{91}{50}$

Vậy $\boxed{x=\frac{91}{50};\frac{189}{50}}$.