\(a\\ -5x^2+3x.\left(x+2\right)=-5x^2+3x^2+6x=-2x^2+6x\\ b\\ -2x.\left(1-x^2\right)-2x^3=-2x+2x^3-2x^3=-2x\\ c\\ 4x.\left(x-1\right)-4.\left(x^2+2x-1\right)\\ =4x^2-4x-4x^2-8x+4=-12x+4\)

\(d\\ 6x^3-2x^2.\left(-x^2-3x\right)=6x^3+2x^4+6x^3=2x^4+12x^3\\ e\\ 3x.\left(x-1\right)-\left(1+2x\right).5x\\ =3x^2-3x-5x-10x^2=-7x^2-8x\\ f\\ -5x^2-\left(x-6\right).\left(-2x^2\right)=-5x^2+2x^3-12x^2=2x^3-17x^2\)

Đúng rồi bạn nhé.

cảm ơn b

a, \(\left|2x-3\right|=\left|3x-7\right|\)

\(\Rightarrow\orbr{\begin{cases}2x-3=3x-7\\2x-3=7-3x\end{cases}\Rightarrow}\orbr{\begin{cases}-x=-4\\5x=10\end{cases}\Rightarrow}\orbr{\begin{cases}x=2\\x=2\end{cases}\Rightarrow}x=2\)

b, \(\left|7x-1\right|-\left|2x-5\right|=0\)

\(\left|7x-1\right|=\left|2x-5\right|\)

\(\Rightarrow\orbr{\begin{cases}7x-1=2x-5\\7x-1=5-2x\end{cases}\Rightarrow}\orbr{\begin{cases}5x=-4\\9x=6\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{-4}{5}\\x=\frac{2}{3}\end{cases}}\)

c, \(\left|3x-1\right|+\left|4+3x\right|=0\)

Vì \(\left|3x-1\right|\ge0\); \(\left|4+3x\right|\ge0\)

\(\Rightarrow\left|3x-1\right|+\left|4+3x\right|\ge0\)

Dấu " = " xảy ra <=> \(\hept{\begin{cases}3x-1=0\\4+3x=0\end{cases}\Rightarrow}\hept{\begin{cases}3x=1\\3x=-4\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1}{3}\\x=\frac{-4}{3}\end{cases}}\)(loại)

d, 2x + 1 = 25 => 2x = 24 => x = 12

đề là thế này? 

(2x + 1)² = 25

\(\Rightarrow\orbr{\begin{cases}2x+1=5\\2x+1=-5\end{cases}\Rightarrow}\orbr{\begin{cases}2x=4\\2x=-6\end{cases}\Rightarrow}\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)

{2x-3=3x-7

{2x-3=7-3x

a. \(\left|2-x\right|+\dfrac{3}{4}=6,75\Leftrightarrow\left|2-x\right|=6,75-\dfrac{3}{4}=6\Leftrightarrow\left[{}\begin{matrix}2-x=6\\2-x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=8\end{matrix}\right.\)

Vậy x=-4 hoặc x=8

thank

tiếp đi bạn

b) \(\left|4-7x\right|-\dfrac{3}{2}:5=\left|-1\dfrac{1}{3}\right|\)

\(\left|4-7x\right|-\dfrac{3}{10}=\dfrac{4}{3}\)

\(\left|4-7x\right|=\dfrac{49}{30}\) (*)

+) Nếu 4 - 7x \(\ge\) 0 \(\Rightarrow x\le\dfrac{4}{7}\)

PT (*) \(\Leftrightarrow4-7x=\dfrac{49}{30}\)

\(-7x=-\dfrac{71}{30}\)

x = \(\dfrac{71}{210}\) (t/m)

+) Nếu \(4-7x< 0\Rightarrow x>\dfrac{4}{7}\)

Pt (*) \(\Leftrightarrow-4+7x=\dfrac{49}{30}\)

x = \(\dfrac{169}{210}\) (t/m)

Vậy x=\(\dfrac{71}{210}\) hoặc x = \(\dfrac{169}{210}\)

a: 4x+9=0

=>4x=-9

=>x=-9/4

b: -5x+6=0

=>-5x=-6

=>x=6/5

c: 7-2x=0

=>2x=7

=>x=7/2

d: 2x+5=0

=>2x=-5

=>x=-5/2

e: 2x+6=0

=>2x=-6

=>x=-3

g: 3x-1/4=0

=>3x=1/4

hay x=1/12

Đk : x khác 0 và -1/2

=> (3x-1).(2x+1) = (2x-1).3x

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

=> 6x^2+x-1-(6x^2-3x) = 0

=> 6x^2+x-1-6x^2+3x = 0

=> 4x-1 = 0

=> 4x=1

=> x=1/4

Vậy x=1/4

Tk mk nha

a: \(\Leftrightarrow\left\{{}\begin{matrix}x>=-2\\\left(3x+8+2x+4\right)\left(3x+8-2x-4\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-2\\\left(5x+12\right)\left(x+4\right)=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

b: \(\Leftrightarrow\left|4x+2\right|=x+15\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-15\\\left(4x+2+x+15\right)\left(4x+2-x-15\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-15\\\left(5x+17\right)\left(3x-13\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{17}{5};\dfrac{13}{3}\right\}\)

c: =>3x+7>=0

hay x>=-7/3

d: =>|2x-5|=-2x+5

=>2x-5<=0

hay x<=5/2

hehehehehehehhehehehehehheheheheheheheehhehehhehehehehehhehehehhehehehhehheehehhehehhe

a) Ta có: \(5x^2-3x\left(x+2\right)\)

\(=5x^2-3x^2-6x\)

\(=2x^2-6x\)

b) Ta có: \(3x\left(x-5\right)-5x\left(x+7\right)\)

\(=3x^2-15x-5x^2-35x\)

\(=-2x^2-50x\)

c) Ta có: \(3x^2y\left(2x^2-y\right)-2x^2\left(2x^2y-y^2\right)\)

\(=3x^2y\left(2x^2-y\right)-2x^2y\left(2x^2-y\right)\)

\(=x^2y\left(2x^2-y\right)=2x^4y-x^2y^2\)

d) Ta có: \(3x^2\left(2y-1\right)-\left[2x^2\cdot\left(5y-3\right)-2x\left(x-1\right)\right]\)

\(=6x^2y-3x^2-\left[10x^2y-6x^2-2x^2+2x\right]\)

\(=6x^2y-3x^2-10x^2y+6x^2+2x^2-2x\)

\(=-4x^2y+5x^2-2x\)

e) Ta có: \(4x\left(x^3-4x^2\right)+2x\left(2x^3-x^2+7x\right)\)

\(=4x^4-16x^3+4x^4-2x^3+14x^2\)

\(=8x^4-18x^3+14x^2\)

f) Ta có: \(25x-4\left(3x-1\right)+7x\left(5-2x^2\right)\)

\(=25x-12x+4+35x-14x^3\)

\(=-14x^3+48x+4\)

Câu a:

2.(3\(x\) - \(\frac12\)) - 2\(x\) = \(\frac12\).(2\(x\) - 3)

6\(x\) - 1 - 2\(x\) = \(x\) - \(\frac32\)

6\(x\) - 2\(x\) - \(x\) = 1 - \(\frac32\)

4\(x\) - \(x\) = - \(\frac12\)

3\(x\) = - \(\frac12\)

\(x\) = - \(\frac12\) : 3

\(x=-\frac16\)

Vậy \(x=-\frac16\)

Câu b:

(2\(x\) - \(\frac35\))\(^2\) = \(\frac{4}{25}\)

(2\(x-\frac35\))\(^2\) = \(\left(\frac{2}{25}\right)\)\(^2\)

2\(x\) - \(\frac35\) = \(\frac25\) hoặc 2\(x\) - \(\frac35\) = - \(\frac25\)

TH: 2\(x\) - \(\frac35\) = \(\frac25\)

2\(x\) = \(\frac25+\frac35\)

2\(x\) = 1

\(x=\frac12\)

2\(x\) - \(\frac35\) = - \(\frac25\)

2\(x\) = - \(\frac25\) + \(\frac35\)

2\(x\) = \(\frac15\)

\(x\) = \(\frac{13}{25}\) : 2

\(x\) = \(\frac15\)

Vậy \(x\) ∈ {1/5; 1/2}