ai giup mik dc ko ak pls mik can gap

\(a,A=\dfrac{5-3}{5+2}=\dfrac{2}{7}\\ b,B=\dfrac{3x-9+2x+6-3x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ c,C=AB=\dfrac{x-3}{x+2}\cdot\dfrac{2}{x-3}=\dfrac{2}{x+2}\\ C=-\dfrac{1}{3}\Leftrightarrow x+2=-6\Leftrightarrow x=-8\left(tm\right)\)

\(b,4x^2-x-5=0\)

\(\Leftrightarrow4x^2-5x+4x-5=0\)

\(\Leftrightarrow x\left(4x-5\right)+4x-5=0\)

\(\Leftrightarrow\left(4x-5\right)\left(x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{5}{4}\end{cases}}\)

Bài 2

\(a,x^3+5x^2+3x-9\)

\(\Leftrightarrow x^3-x^2+6x^2-6x+9x-9\)

\(\Leftrightarrow x^2\left(x-1\right)+6x\left(x-1\right)+9\left(x-1\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+6x+9\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x+3\right)^2\)

b,\(x^3-7x-6\)

\(\Leftrightarrow x^3-3x^2+3x^2-9x+2x-6\)

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

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

\(\Leftrightarrow\left(x-3\right)\left(x+1\right)\left(x+2\right)\)

c,\(3x^3-7x^2+17x-5\)

\(\Leftrightarrow3x^3-x^2-6x^2+2x+15x-5\)

\(\Leftrightarrow x^2\left(3x-1\right)-2x\left(3x-1\right)+5\left(3x-1\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2-2x+5\right)\)

\(4x^2-x-5=0\)

<=>  \(4x^2+4x-5x-5=0\)

<=>   \(4x\left(x+1\right)-5\left(x+1\right)=0\)

<=>  \(\left(x+1\right)\left(4x-5\right)=0\)

tự lm tiếp

Bài 1:

a)\(28x^3+15x^2+75x+125=0\)

\(\Leftrightarrow\left(4x+5\right)\left(7x^2-5x+25\right)=0\)

Dễ thấy: \(7x^2-5x+25=7\left(x-\frac{5}{14}\right)^2+\frac{675}{28}>0\)

\(\Rightarrow4x+5=0\Rightarrow x=-\frac{5}{4}\)

b)\(4x^2-x-5=0\)

\(\Leftrightarrow\left(x+1\right)\left(4x-5\right)=0\)

\(\Rightarrow x=-1;x=\frac{5}{4}\)

Bài 2:

a)\(x^3+5x^2+3x-9\)

\(=\left(x-1\right)\left(x+3\right)^2\)

b)\(x^3-7x-6\)

\(=\left(x-3\right)\left(x+1\right)\left(x+2\right)\)

c)\(3x^3-7x^2+17x-5\)

\(=\left(3x-1\right)\left(x^2-2x+5\right)\)

b)(x-2)3-(x-3)(x2+3x+9)+6(x+1)2=49

(=) x³- 6x² +12 x -8 - ( x3 - 27 ) + 6( x² + 2x +1)

(=) x³ - 6x² +12x -8 - x³ +27 + 6x² +12x +6

(=) 24x + 25 = 49

(=) 24x = 49 - 25 = 24

(=) x = 24/24 =1

x^3 - 3x^2 + 3x - 1 = 0

( x- 1)^3    = 0 

=> x -1 = 0

=> x = 1