`x^2+2x+3>2`

`<=>x^2+2x+1>0`

`<=>(x+1)^2>0`

`<=>x+1 ne 0`

`<=>x ne -1`

`(x+5)(3x^2+2)>0`

Vì `3x^2+2>=2>0`

`=>x+5>0<=>x>-5`

c) Ta có: \(21x-10x^2+9< 0\)

\(\Leftrightarrow10x^2-21x-9>0\)

\(\Leftrightarrow x^2-\dfrac{21}{10}x-\dfrac{9}{10}>0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{21}{20}+\dfrac{441}{400}>\dfrac{801}{400}\)

\(\Leftrightarrow\left(x-\dfrac{21}{20}\right)^2>\dfrac{801}{400}\)

\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{3\sqrt{89}+21}{20}\\x< \dfrac{-3\sqrt{89}+21}{20}\end{matrix}\right.\)

a, \(\left|2x+1\right|=3\)

=> 2x + 1 = 3 hoặc 2x + 1 = -3

=> 2x = 3 - 1 hoặc 2x = -3 - 1

=> 2x = 2 hoặc 2x = -4

=> x = 1 hoặc x = -2

b, \(2\frac{1}{2}x+1\frac{1}{2}=2\frac{2}{3}\)

=> \(\frac{5}{2}x+\frac{3}{2}=\frac{8}{3}\)

=> \(\frac{5}{2}x=\frac{8}{3}-\frac{3}{2}\)

=> \(\frac{5}{2}x=\frac{16-9}{6}\)

=> \(\frac{5}{2}x=\frac{7}{6}\)

=> \(x=\frac{7}{6}:\frac{5}{2}=\frac{7}{6}\cdot\frac{2}{5}=\frac{7}{3}\cdot\frac{1}{5}=\frac{7}{15}\)

c, \(3\cdot5^{x-3}+1=16\)

=> 3 . 5^x-3 = 16 - 1

=> 3 . 5^x-3 = 15

=> 5^x-3 = 15 : 3

=> 5^x-3 = 5

=> x - 3 = 5 : 5

=> x - 3 = 1

=> x = 1 + 3 = 4

d, \((x-1)^2=25\)

=> \((x-1)^2=5^2\)

=> x - 1 = 5 hoặc x - 1 = -5

=> x = 6 hoặc x = -4

e, \((-2)^2+\left|3x+1\right|=(-28)\cdot7\)

=> 4 + |3x + 1| = -196

=> |3x + 1| = -196 - 4 = -200

=> |3x + 1| = -200

Không thỏa mãn điều kiện

Gía trị biểu thức không phụ thuộc vào biến nghĩa là với mọi x, biểu thức đó có giá trị là 1 số thực.Ta có :

A = 2x(x - 1) - x(2x + 1) - (3 - 3x) = 2x² - 2x - 2x² - x - 3 + 3x = (2x² - 2x²) + (3x - 2x - x) - 3 = -3

B = 2x(x - 3) - (2x - 2)(x - 2) = 2x² - 6x - 2x² + 4x + 2x - 4 = (2x² - 2x²) + (4x + 2x - 6x) - 4 = -4

C = (3x - 5)(2x + 11) - (2x + 3)(3x + 7) = 6x² + 33x - 10x - 55 - 6x² - 14x - 9x - 21 = (6x² - 6x²) + (33x - 10x - 14x - 9x) - 55 - 21 = -76      = D = (2x + 11)(3x - 5) - (2x + 3)(3x + 7)

Vậy với mọi x , (A,B,C,D) = (-3;-4;-76;-76) => đpcm 

D = 

\(a,3x-2\left(x-3\right)=0\\ \Leftrightarrow3x-2x+6=0\\ \Leftrightarrow x=-6\\ b,\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\\ \Leftrightarrow2x^2+2x-3x-3=2x^2-x+10x-5\\ \Leftrightarrow2x^2-x-3=2x^2+9x-5\\ \Leftrightarrow10x-2=0\\ \Leftrightarrow x=\dfrac{1}{5}\\ c,ĐKXĐ:x\ne\pm1\\ \dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\\ \Leftrightarrow\dfrac{2x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x-x^2+1}{\left(x+1\right)\left(x-1\right)}=0\)

\(\Rightarrow3x+1=0\\ \Leftrightarrow x=-\dfrac{1}{3}\left(tm\right)\)

\(d,\left(2x+3\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x-5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\\ e,ĐKXĐ:x\ne\pm2\\ \dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\\ \Leftrightarrow\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-22}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\dfrac{x^2-4x+4-3x-6-2x+22}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow x^2-9x+20=0\\ \Leftrightarrow\left(x^2-5x\right)-\left(4x-20\right)=0\\ \Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)

b. 2x³-3x²+3x-1=2x³-x²-2x²+x+2x-1

= x²(2x-1)-x(2x-1)+(2x-1)

=(2x-1)(x²-x-1)

c. 3x³-14x²+4x+3= 3x³+x²-15x²-5x+9x+3

=x²(3x+1)-5x(3x-1)+3(3x+1)

=(3x+1)(x²-5x+3)

bài 1:
a 2x(x-5)-2x^2=20
<=>2x^2-10x-2x^2=20
<=>-10x=20
<=>x=-2
v....
b x^2-2x+1=0
<=>(x-1)^2=0
<=>x-1=0
<=>x=1
v...
bài 3

A=x-x^2+1=-(x^2-x-1)=-(x^2-2*x*1/2+1/4-5/4)=-(x-1/2)^2+5/4<=5/4
dấu bằng xảy ra <=>x=1/2
bài 2 mình ko biết làm sorry cậu

tran thu phuong cảm ơn bn nhá.

Ai giúp tớ câu 2 đi