a) \(f\left(-2\right)=2.\left(-2\right)^2+3.\left(-2\right)-5\)

\(f\left(-2\right)=8-6-5=-3\)

b) \(f\left(x\right)+g\left(x\right)\)

\(=2x^2+3x-5+3x^2+2x-7\)

\(=5x^2+5x-12\)

       \(5.13.21+7.65.2+65\)

\(=5.21.13+7.2.65+65.1\)

\(=110.13+11.65\)

\(=1430+715\)

\(=2145\)

1111² + 1234321

bài 1: 

a) ta có: 3x + 5 = (3(x+1)+2)\(⋮\)(x+1)

vì  (3(x+1)\(⋮\)(x+1) nên 2 \(⋮\)(x+1) => (x+1) \(\in\)Ư(2) => (x+1) \(\in\)\(\xi\)-2;-1;1;2  \(\xi\)=> x \(\in\)\(\xi\)-3; -2; 0; 1  \(\xi\)

vậy, x= -3; -2; 0; 1

x³ + 3x - 4 = 0

=> x³ - x² + x² - x + 4x - 4 = 0

=> x² ( x - 1 ) + x ( x - 1 ) + 4 ( x - 1 ) = 0

=> ( x² + x + 4 ) ( x - 1 ) = 0

\(\Leftrightarrow\orbr{\begin{cases}x^2+x+4=0\\x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x^2+x=-4\left(loai\right)\\x=1\end{cases}}\)

c. - x ( x + 3 ) + 2 = ( 4x + 1 ) ( x - 1 ) + 2x

<=> - x² - 3x + 2 = 4x² - x - 1

<=> 4x² - x - 1 + x² + 3x - 2 = 0

<=> 5x² + 2x - 3 = 0

<=> ( 5x² + 5x ) - ( 3x + 3 ) = 0

<=> 5x ( x + 1 ) - 3 ( x + 1 ) = 0

<=> ( 5x - 3 ) ( x + 1 ) = 0 

\(\Leftrightarrow\orbr{\begin{cases}5x-3=0\\x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=-1\end{cases}}\)

d. ( 2x + 3 ) ( x - 3 ) - ( x - 3 ) ( x + 1 ) = ( 2 - x ) ( 3x + 1 ) + 3

<=> ( x - 3 ) ( 2x + 3 - x - 1 ) = - 3x² + 5x + 5

<=> x² - x - 6 = - 3x² + 5x + 5

<=>  - 3x² + 5x + 5 - x² + x + 6 = 0

<=> - 4x² + 6x + 11 = 0

\(\Leftrightarrow x=\frac{6\pm\sqrt{\left(-6\right)^2-4\left(4.\left(-11\right)\right)}}{2.4}\)( xài công thức bậc 2 )

\(\Leftrightarrow x=\frac{6\pm2\sqrt{53}}{8}\Leftrightarrow x=\frac{3\pm\sqrt{53}}{4}\)

Vậy \(x=\frac{3+\sqrt{53}}{4};x=\frac{3-\sqrt{53}}{4}\)