1) Ta có : 2x² + 3x - 5

= 2x² - 2x + 5x - 5

= 2x(x - 1) + 5(x - 1)

= (x - 1) (2x + 5) 

3) x² + x - 6

= x² + 2x - 3x - 6

= x(x + 2) - (3x + 6)

= x(x + 2) - 3(x + 2)

= (x - 3)(x + 2) 

a) \(C=\left(\dfrac{x}{x^2-x-6}-\dfrac{x-1}{3x^2-4x-15}\right):\dfrac{x^4-2x^2+1}{3x^2+11x+10}\cdot\left(x^2-2x+1\right)\) (ĐK: \(x\ne-\dfrac{5}{3};x\ne3;x\ne-2;x\ne1\))

\(C=\left[\dfrac{x}{\left(x-3\right)\left(x+2\right)}-\dfrac{x-1}{\left(x-3\right)\left(3x+5\right)}\right]:\dfrac{\left(x^2-1\right)^2}{\left(3x+5\right)\left(x+2\right)}\cdot\left(x-1\right)^2\)

\(C=\left[\dfrac{x\left(3x+5\right)}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}-\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x-3\right)\left(3x+5\right)\left(x+2\right)}\right]\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x^2-1\right)^2\left(x-1\right)^2}\)

\(C=\dfrac{3x^2+5x-x^2-2x+x+2}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x^2-1\right)^2\left(x-1\right)^2}\)

\(C=\dfrac{2x^2+4x+2}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x+1\right)^2\left(x-1\right)^4}\)

\(C=\dfrac{2\left(x+1\right)^2}{\left(3x+5\right)\left(x-3\right)\left(x+2\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x+1\right)^2\left(x-1\right)^4}\)

\(C=\dfrac{2}{\left(x-1\right)^4\left(x-3\right)}\)

b) Thay x = 2003 ta có: 

\(C=\dfrac{2}{\left(2003-1\right)^4\left(2003-3\right)}=\dfrac{2}{2002^4\cdot2000}=\dfrac{1}{2002^4\cdot1000}\)

c) \(C>0\) khi: 

\(\dfrac{2}{\left(x-1\right)^4\left(x-3\right)}>0\) mà: \(\left\{{}\begin{matrix}2>0\\\left(x-1\right)^4>0\end{matrix}\right.\)

\(\Leftrightarrow x-3>0\)

\(\Leftrightarrow x>3\) (đpcm) 

20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)

Vậy...

1) 16 - 8x = 0 ⇔ 8(2 - x) = 0⇔ 2 - x = 0 ⇔ x = 2

Vậy phương trình có nghiệm là x = 2

x²-4x+7 = 0 ⇔ x² -4x + 4 + 3 = 0 

⇔ (x-2)²+3=0 ⇔ (x-2)²=-3 (vô lí)

Vậy pt vô nghiệm

*Chứng minh phương trình \(x^2-4x+7=0\) vô nghiệm

Ta có: \(x^2-4x+7=0\)

\(\Leftrightarrow x^2-4x+4+3=0\)

\(\Leftrightarrow\left(x-2\right)^2+3=0\)

mà \(\left(x-2\right)^2+3\ge3>0\forall x\)

nên \(x\in\varnothing\)(đpcm)

a: 11x+4=-3/2

=>\(11x=-\dfrac{3}{2}-4=-\dfrac{11}{2}\)

=>\(x=-\dfrac{1}{2}\)

b: \(x^2-9+2\left(x-3\right)=0\)

=>\(\left(x-3\right)\left(x+3\right)+2\left(x-3\right)=0\)

=>\(\left(x-3\right)\left(x+3+2\right)=0\)

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

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

c: \(\dfrac{x-3}{5}+\dfrac{1+2x}{3}=6\)

=>\(\dfrac{3\left(x-3\right)+5\left(2x+1\right)}{15}=6\)

=>\(3x-9+10x+5=90\)

=>13x-4=90

=>13x=94

=>\(x=\dfrac{94}{13}\)

d: \(\dfrac{2}{x+1}-\dfrac{1}{x-2}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\)(ĐKXĐ: \(x\notin\left\{-1;2\right\}\))

=>\(\dfrac{2\left(x-2\right)-\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{3x-11}{\left(x-2\right)\left(x+1\right)}\)

=>3x-11=2x-4-x-1

=>3x-11=x-5

=>2x=6

=>x=3(nhận)

f)  11.(x+2)=5.x+40

=> 11x+22=5x+40

=> 6x=18

=> x=18:6=3

g)11.(x-6)=4.x+11

=> 11.x+(-66)=4x+11

=> 7x=77

=> x=77:7=11

h) (3x-4):x+10=12

=> 3x:x-4:x=12-10=2

=> 3-4:x=2

=> 4:x=3-2=1

=> x=4:1=4

k)(5x+7):x+20=26

=> 5x:x+7:x=26-20=6

=> 5+7:x=6

=> => 7:x=6-5=1

=> x=7:1=7

m) x.1999-x=1999.1997+1999

=> x.1998=1999.(1997+1)

=> x.1998=1999.1998

=> x=1999

**** cho mik nha Nguyễn Thị Thảo Ly

Dài thế này chả ai dám trả lời :D

đúng đó đăng ít một thui bạn ơi

Đổi x thành a, ta có:

f. 11 x ( a + 2)         = 5 x a + 40

11 x a + 11 x 2       = 5 x a + 40

11 x a + 22             = 5 x a + 40

6 x a + 5 x a + 22   = 5 x a + 40

6 x a + 22              = 40(Cùng bỏ 5 x a ở hai vế)

6 x a                      = 40 - 22

6 x a                      = 18

                         a   = 18 : 6

                         a   = 3

g. 11 x ( a - 6 )        = 4 x a + 11

11 x a - 11 x 6         = 4 x a + 11

4 x a + 7 x a - 66     = 4 x a + 11

7 x a - 66                = 11(Cùng bỏ đi 4 x a ở 2 vế)

7 x a                       = 11 + 66

7 x a                       = 77

a                            = 77 : 7

                         a   = 11

h. ( 3 x a - 4 ) : a + 10 = 12

3 x a - 4                     = ( 12 - 10 ) x a

3 x a - 4                     = 2 x a

2 x a + a                    = 2 x a + 4

                           a    = 4(Cùng bỏ đi 2 x a)

k. Tương tự h

m. a x 1999 - a = 1999 x 1997 + 1999

     a x 1998      = 1999 x 1998

=> a = 1999

A=1999

\(x+\dfrac{2}{3}=\dfrac{9}{11}\Rightarrow x=\dfrac{9}{11}-\dfrac{2}{3}=\dfrac{5}{33}\\ x-\dfrac{3}{10}=\dfrac{4}{15}\Rightarrow x=\dfrac{4}{15}+\dfrac{3}{10}=\dfrac{17}{30}\\ x\times\dfrac{1}{7}=\dfrac{5}{6}\Rightarrow x=\dfrac{5}{6}:\dfrac{1}{7}=\dfrac{35}{6}\\ x:\dfrac{3}{5}=\dfrac{1}{6}\Rightarrow x=\dfrac{1}{6}\times\dfrac{3}{5}=\dfrac{1}{10}\)