a: Ta có: \(x\left(2x-3\right)-\left(2x-1\right)\left(x+5\right)=17\)

\(\Leftrightarrow2x^2-3x-2x^2-10x+x+5=17\)

\(\Leftrightarrow-12x=12\)

hay x=-1

a: \(\frac{2x+2}{5}+\frac{3}{10}<\frac{3x-2}{4}\)

=>\(\frac{4\left(2x+2\right)}{20}+\frac{6}{20}<\frac{5\left(3x-2\right)}{20}\)

=>4(2x+2)+6<5(3x-2)

=>8x+8+6<15x-10

=>8x+14<15x-10

=>-7x<-24

=>\(x>\frac{24}{7}\)

b: \(\frac{2+x}{3}<\frac{3+2x}{5}\)

=>\(\frac{x+2}{3}-\frac{2x+3}{5}<0\)

=>\(\frac{5\left(x+2\right)-3\left(2x+3\right)}{15}<0\)

=>5(x+2)-3(2x+3)<0

=>5x+10-6x-9<0

=>-x+1<0

=>-x<-1

=>x>1

d: \(1+\frac{3\left(x+1\right)}{10}>\frac{x-2}{5}\)

=>\(\frac{10+3\left(x+1\right)}{10}>\frac{2\left(x-2\right)}{10}\)

=>3(x+1)+10>2(x-2)

=>3x+3+10>2x-4

=>3x+13>2x-4

=>3x-2x>-4-13

=>x>-17

e: \(\frac{2x-7}{6}\ge\frac{3x-7}{2}\)

=>\(\frac{2x-7}{6}\ge\frac{3\left(3x-7\right)}{6}\)

=>2x-7>=3(3x-7)

=>2x-7>=9x-21

=>-7x>=-14

=>x<=2

f: \(\frac{2x-1}{3}>\frac{3x+1}{2}\)

=>\(\frac{2x-1}{3}-\frac{3x+1}{2}>0\)

=>\(\frac{2\left(2x-1\right)-3\left(3x+1\right)}{6}>0\)

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

=>4x-2-9x-3>0

=>-5x-5>0

=>5x+5<0

=>5x<-5

=>x<-1

- Thay lần lượt xo vào từng phương trình trên ta được kết quả sau :

 +, Phương trình nhận xo là nghiệm : a, b, c, d, e .

`2x+10=|3x-2|(x>=-5)`

`<=>`$\left[ \begin{array}{l}2x+10=3x-2\\2x+10=2-3x\end{array} \right.$

`<=>` $\left[ \begin{array}{l}x=12\\5x=-8\end{array} \right.$

`<=>` $\left[ \begin{array}{l}x=12(tm)\\x=-\dfrac85(tm)\end{array} \right.$

Vậy `S={1,-8/5}`

`2x+10=|3x-2|(x>=-5)`

`<=>`$\left[ \begin{array}{l}2x+10=3x-2\\2x+10=2-3x\end{array} \right.$

`<=>` $\left[ \begin{array}{l}x=12\\5x=-8\end{array} \right.$

`<=>` $\left[ \begin{array}{l}x=12(tm)\\x=-\dfrac85(tm)\end{array} \right.$

Vậy `S={12,-8/5}`

c: =>2x+4>=2x+2-3

=>4>=-1(luôn đúng)

a: 5x+10>3x+3

=>2x>-7

=>x>-7/2

bạn coi lại đề nhé!

1/ ( x-1) (2x+1) =0

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\2x+1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-0,5\end{matrix}\right.\)

2/ x (2x-1) (3x+15) =0

\(\Rightarrow\left[{}\begin{matrix}x=0\\2x-1=0\\3x+15=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=0,5\\x=-5\end{matrix}\right.\)

3/ (2x-6) (3x+4).x=0

\(\Rightarrow\left[{}\begin{matrix}2x-6=0\\3x+4=0\\x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{4}{3}\\x=0\end{matrix}\right.\)

4/ (2x-10)(x²+1)=0

\(\Rightarrow\left[{}\begin{matrix}2x-10=0\\x^2+1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x^2=-1\left(loại\right)\end{matrix}\right.\)

5/ (x²+3) (2x-1) =0

\(\Rightarrow\left[{}\begin{matrix}x^2+3=0\\2x-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x^2=-3\left(loại\right)\\x=0,5\end{matrix}\right.\)

6/ (3x-1) (2x² +1)=0

\(\Rightarrow\left[{}\begin{matrix}3x-1=0\\2x^2+1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x^2=-0,5\left(loại\right)\end{matrix}\right.\)

1: Ta có: \(\left(x-1\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)

2: Ta có: \(x\left(2x-1\right)\left(3x+15\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-1=0\\3x+15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-5\end{matrix}\right.\)

3: Ta có: \(\left(2x-6\right)\left(3x+4\right)x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\3x+4=0\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{4}{3}\\x=0\end{matrix}\right.\)

Thế mày làm đi

cho ít thôi thì làm

Bài 1:

\(a,=6x^2+6x\\ b,=15x^3-10x^2+5x\\ c,=6x^3+12x^2\\ d,=15x^4+20x^3-5x^2\\ e,=2x^2+3x-2x-3=2x^2+x-3\\ f,=3x^2-5x+6x-10=3x^2+x-10\)

Bài 2:

\(a,\Leftrightarrow3x^2+3x-3x^2=6\\ \Leftrightarrow3x=6\Leftrightarrow x=2\\ b,\Leftrightarrow6x^2+3x-6x^2+9x-2x-3=10\\ \Leftrightarrow10x=13\Leftrightarrow x=\dfrac{13}{10}\)

Giúp em với ạ

(3x-5)(2x+1)-(2x-1)^2-2x(x-2)-x+10=4

=>6x^2+3x-10x-5-(4x^2-4x+1)-2x^2+4x-x+10=4

=>(6x^2-4x^2-2x^2)+(3x-10x+4x+4x-x)+(-5-1+10)=4

=>4=4