a: s=v*t

b: Theo đề, ta có: x/10-x/12=1/4

=>x/60=1/4

=>x=15

mong m.n giúp đỡ mik 

a) A= 3.(x²-2xy+y²)- 2. (x²+2xy+y²) - x²-y²

A= 3.x²-2xy+y²-2. x²+2xy+y²-x²-y²

a) \(3\left(2x-5\right)+125=134\)

\(\Leftrightarrow3\left(2x-5\right)=9\)

\(\Leftrightarrow2x-5=3\)

\(\Leftrightarrow2x=8\Leftrightarrow x=4\)

b) \(\left(2x+5\right)+\left(2x+3\right)+\left(2x+1\right)=27\)

\(\Leftrightarrow6x+9=27\)

\(\Leftrightarrow6x=18\Leftrightarrow x=3\)

d) \(27\left(x-27\right)-27=0\)

\(\Leftrightarrow27\left(x-27\right)=27\)

\(\Leftrightarrow x-27=1\Leftrightarrow x=28\)

3.(x+5)-2x=27
3.x+3.5-2x=27
3x+15-2x=27
x.(3-2)+15=27
x.  1    +15=27
  x       +15=27
  x             =27-15
  x             =12

Ta có: \(\sqrt{2x+7}-6=x\)

\(\Leftrightarrow\sqrt{2x+7}=x+6\)

\(\Leftrightarrow x^2+12x+36-2x-7=0\)

\(\Leftrightarrow x^2+10x+29=0\)(Vô lý)

Vậy: \(S=\varnothing\)

Điều kiện : x ≥ 0

\(\sqrt{2x+97}-6=x\text{⇔}\sqrt{2x+97}=x+6\\ \text{⇔}2x+97=x^2+12x+36\text{⇔}x^2+10x-61=0\\ \text{⇔}\left[{}\begin{matrix}x=-5+\sqrt{86}\\x=-5-\sqrt{86}\end{matrix}\right.\)

\(\sqrt{2x+97}-6=x\)

\(\Leftrightarrow\sqrt{2x+97}=x+6\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+6\ge0\\2x+97=\left(x+6\right)^2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-6\\x^2+10x-61=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-6\\\left[{}\begin{matrix}x=-5+\sqrt{86}\\x=-5-\sqrt{86}\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow x=-5+\sqrt{86}\)

Vậy..,

p

a. 12xy² - 8x²y = 4xy . (3y - 2x)

b. 3x + 3y - x² - xy = (3x + 3y) - (x² + xy) = 3 . (x + y) - x . (x + y) = (x + y)(3 - x)

GIUSP MIK VS MN ƠI

Bài 2:

a: =>2x^2-8+3x-2x^2=15

=>3x=23

=>x=23/3

b: \(\Leftrightarrow\left(x+3\right)\left(x-5\right)=0\)

=>x=5 hoặc x=-3

c: =>x^2+7x-30=0

=>(x+10)(x-3)=0

=>x=3 hoặc x=-10