a. Ta có:

f(x) = -2x2 - 3x3 - 5x + 5x3 - x + x2 + 4x + 3 + 4x2

= 2x3 + 3x2 - 2x + 3 (0.5 điểm)

g(x) = 2x2 - x3 + 3x + 3x3 + x2 - x - 9x + 2

= 2x3 + 3x2 - 7x + 2 (0.5 điểm)

h(x)=5x+1

nghiệm_của_đa_thức_h(x)_là_-1/5

a)h(x)=f(x)-g(x)

        =(2x³ +3x² -2x +3)-(2x³ +3x² -7x +2)

        =2x³ + 3x² - 2x +3 - 2x³ -3x² + 7x -2

        =5x+1

b)h(x)=5x+1=0

=>5x=-1

    x=\(\frac{-1}{5}\)

a) \(=\left(x-2\right)^2\)

b) \(=\left(2x+1\right)^2\)

c) \(=\left(4x-3y\right)\left(4x+3y\right)\)

d) \(=\left(4-x-3\right)\left(4+x+3\right)=\left(1-x\right)\left(x+7\right)\)

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

f) \(=\left(x-y\right)\left(x^2+xy+y^2\right)\)

g) \(=\left(x+3\right)\left(x^2-3x+9\right)\)

h) \(=\left(x+2\right)^3\)

i) \(=\left(1-x\right)^3\)

a/ $=(x-2)^2$

b/ $=(2x+1)^2$

c/ $=(4x-3y)(4x+3y)$

d/ $=(1-x)(x+7)$

e/ $=(-x+1)(5x-1)$

f/ $=(x-y)(x^2+xy+y^2)$

g/ $=(3+x)(9-3x+x^2)$

h/ $=(x+2)^3$

i/ $=(1-x)^3$

a: \(x^2-4x+4=\left(x-2\right)^2\)

b: \(4x^2+4x+1=\left(2x+1\right)^2\)

g: \(x^3+27=\left(x+3\right)\left(x^2-3x+9\right)\)

b: 1/2x-4=0

=>1/2x=4

hay x=8

a: x+7=0

=>x=-7

e: 4x²-81=0

=>(2x-9)(2x+9)=0

=>x=9/2 hoặc x=-9/2

g: x²-9x=0

=>x(x-9)=0

=>x=0 hoặc x=9

a)\(x+7=0=>x=-7\)

b)\(\dfrac{1}{2}x-4=0=>\dfrac{1}{2}x=4=>x=8\)

c)\(-8x+20=0=>-8x=-20=>x=\dfrac{5}{2}\)

d)\(x^2-100=0=>x^2=100=>\left[{}\begin{matrix}x=10\\x=-10\end{matrix}\right.\)

e)\(4x^2-81=0=>4x^2=81=>x^2=\dfrac{81}{4}=>\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{9}{2}\end{matrix}\right.\)

f)\(x^2-7=0=>x^2=7=>x=\sqrt{7}\)

g)\(x^2-9x=0=>x\left(x-9\right)=0=>\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\)

a: x+7=0

nên x=-7

b: x-4=0

nên x=4

c: -8x+20=0

=>-8x=-20

hay x=5/2

d: x²-100=0

=>(x-10)(x+10)=0

=>x=10 hoặc x=-10

a) x +7 =0

=>x = -7

b) x - 4 =0=>x = 4

c) -8x + 20 = 0 =>-8x =-20 =>\(x=-\dfrac{20}{-8}=\dfrac{5}{2}\)

d)\(x^2-100=0=>x^2=100>\left[{}\begin{matrix}x=10\\x=-10\end{matrix}\right.\)

e)\(4x^2-81=0=>4x^2=81=>x^2=\dfrac{81}{4}=>\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{9}{2}\end{matrix}\right.\)

f)\(x^2-7=0=>x^2=7=>x=\sqrt{7}\)

g)\(x^2-9x=0=>x\left(x-9\right)=0=>\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\)

H)\(x^3+3x=0=>x\left(x^2 +3\right)=0=>\left[{}\begin{matrix}x=0\\x^2=-3\left(vl\right)\end{matrix}\right.\)

\(a,\left(x^2+4\right)^2-16x^2=\left(x^2+4\right)-\left(4x\right)^2=\left(x^2+4-4x\right).\left(x^2+4+4x\right)=\left(x-2\right)^2.\left(x+2\right)^2\)

\(b,\left(x+3\right)^3-8x^3=\left(x+3\right)^3-\left(2x\right)^3=\left(x+3-2x\right).\left[x^2+\left(x+3\right).2x+\left(2x\right)^2\right]=\left(3-x\right).\left(x^2+2x^2+6x+4x^2\right)\)

\(c,\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2=\left(4x^2-3x-18-4x^2-3x\right).\left(4x^2-3x-18+4x^2+3x\right)=\left(-6x-18\right).\left(8x^2-18\right)\)

b. h(x) = (2x3 + 3x2 - 2x + 3) - (2x3 + 3x2 - 7x + 2)

= 2x3 + 3x2 - 2x + 3 - 2x3 - 3x2 + 7x - 2

= 5x + 1 (0.5 điểm)

g(x) = (2x3 + 3x2 - 2x + 3) + (2x3 + 3x2 - 7x + 2)

= 2x3 + 3x2 - 2x + 3 + 2x3 + 3x2 - 7x + 2

= 4x3 + 6x2 - 9x + 5 (0.5 điểm)