Question

Giúp với

Answer 1

Câu 1:

a: Ta có: -4x+5=9

\(\Leftrightarrow-4x=4\)

hay x=-1

b: Ta có: \(\dfrac{1}{x+3}+\dfrac{2}{x\left(x+3\right)}=\dfrac{3}{x}\)

Suy ra: \(x+2=3x+9\)

\(\Leftrightarrow-2x=7\)

hay \(x=-\dfrac{7}{2}\left(nhận\right)\)

Answer 2

-4x + 5 = 9
-4x -4 = 0
-4 ( x - 1 ) = 0
x-1=0
x=1

Answer 3

-3x+5>= 7
-3x>=2
x=<-2/3

Answer 4

Câu 2: 

c: Ta có: \(-3x+5\ge7\)

\(\Leftrightarrow-3x\ge2\)

hay \(x\le-\dfrac{2}{3}\)

d: Ta có: \(\left(x-3\right)\left(x+3\right)< \left(x-2\right)^2+3\)

\(\Leftrightarrow x^2-9-x^2+4x-4< 3\)

\(\Leftrightarrow4x< 16\)

hay x<4