a) `3x+5 =0`

`3x=-5`

`x=-5/3`

`b) -4x+8=0`

`-4x =-8`

`x=2`

`c) 3x -6=0`

`3x=6`

`x=2`

`d)x^2 +x =0`

`x(x+1) =0`

`=>[(x=0),(x=-1):}`

`e) x^2 -4 =0`

`x^2 =4`

`=> x = +-2`

`f) x^3 -27 =0`

`x^3 =27`

`=> x=3`

`g) 3x^2 +4 =0`

`3x^2 =-4`

`x^2 =-4/3(vô-lí)`

=> Đa thức ko có nghiệm

h) `x^3 -4x =0`

`x(x^2 -4) =0`

`=>[(x=0),(x^2=4 => x=+-2):}`

i) `2x^3 -32x =0`

`2x(x^2 -16)=0`

`=>[(2x=0),(x^2=16):}`

`=>[(x=0),(x=+-4):}`

Phần nào bạn ko nhìn thấy thì bảo mk nhé

`@` `\text {Ans}`

`\downarrow`

Gửi c!

Bài 1: 

a) \(3x^2\left(2x^3-x+5\right)-6x^5-3x^3+10x^2\)

\(=6x^5-3x^3+10x^2-6x^5-3x^3+10x^2\)

\(=10x^2+10x^2\)

\(=20x^2\)

b) \(-2x\left(x^3-3x^2-x+11\right)-2x^4+3x^3+2x^2-22x\)

\(=-2x^4+6x^3+2x^2-22x-2x^4+3x^3+2x^2-22x\)

\(=-4x^4+9x^3+4x^2-44x\)

4:

a: =>1/4x^2-1/4x^2+2x=-14

=>2x=-14

=>x=-7

b: =>2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20

=>3x^2-12x-2=3x^2-17x+20

=>5x=22

=>x=22/5

`@` `\text {dnv4510}`

`A)`

`P(x)+Q(x)=`\((2x^4+3x^2-3x^2+6)+(x^4+x^3-x^2+2x+1)\)

`= 2x^4+3x^2-3x^2+6+x^4+x^3-x^2+2x+1`

`= (2x^4+x^4)+x^3+(3x^2-3x^2-x^2)+2x+(6+1)`

`= 3x^4+x^3-x^2+2x+7`

`B)`

`P(x)+M(x)=2Q(x)`

`-> M(x)= 2Q(x) - P(x)`

`2Q(x)=2(x^4+x^3-x^2+2x+1)`

`= 2x^4+2x^3-2x^2+4x+2`

`-> 2Q(x)-P(x)=(2x^4+2x^3-2x^2+4x+2)-(2x^4+3x^2-3x^2+6)`

`= 2x^4+2x^3-2x^2+4x+2-2x^4-3x^2+3x^2-6`

`= (2x^4-2x^4)+2x^3+(-2x^2-3x^2+3x^2)+4x+(2-6)`

`= 2x^3-2x^2+4x-4`

Vậy, `M(x)=2x^3-2x^2+4x-4`

`C)`

Thay `x=-4`

`M(-4)=2*(-4)^3-2*(-4)^2+4*(-4)-4`

`= 2*(-64)-2*16-16-4`

`= -128-32-16-4`

`= -180`

`->` `x=-4` không phải là nghiệm của đa thức.

Bài 2 : 

a, \(x^2-4x+4+1=\left(x-2\right)^2+1\ge1\)

Dấu ''='' xảy ra khi x = 2 

b, Ta có \(\left(x+1\right)^2+10\ge10\Rightarrow\dfrac{-100}{\left(x+1\right)^2+10}\ge-\dfrac{100}{10}=-10\)

Dấu ''='' xảy ra khi x = -1 

 Bài 1 : 

a, Ta có \(A\left(x\right)=x^2-4x+4=0\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x=2\)

b, \(B\left(x\right)=x^2\left(2x+1\right)+\left(2x+1\right)=\left(x^2+1>0\right)\left(2x+1\right)=0\Leftrightarrow x=-\dfrac{1}{2}\)

c, \(C\left(x\right)=\left|2x-3\right|=\dfrac{1}{3}\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{1}{3}+3=\dfrac{10}{3}\\2x=-\dfrac{1}{3}+3=\dfrac{8}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)

b: \(=\dfrac{2x^4-2x^3-2x^2-3x^3+3x^2+3x+x^2-x-1}{x^2-x-1}\)

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

Thu gọn Q(x) = x4 + 7x2 + 1

Khi đó R(x) = Q(x) - P(x) = 4x2 + 3x + 2. Chọn A

\(a) f ( x ) = 2 x ^4 + 3 x ^2 − x + 1 − x ^2 − x ^4 − 6 x ^3\)

\(= ( 2 x ^4 − x ^4 ) − 6 x ^3 + ( 3 x ^2 − x ^2 ) − x + 1\)

\(= x ^4 − 6 x ^3 + 2 x ^2 − x + 1\)

\(g ( x ) = 10 x ^3 + 3 − x ^4 − 4 x ^3 + 4 x − 2 x ^2\)

\(= − x ^4 + ( 10 x ^3 − 4 x ^3 ) − 2 x ^2 + 4 x + 3\)

\(= − x ^4 + 6 x ^3 − 2 x ^2 + 4 x + 3\)

\(b) f ( x ) + g ( x ) = x ^4 − 6 x ^3 + 2 x ^2 − x + 1 − x ^4 + 6 x ^3 − 2 x ^2 + 4 x + 3\)

\(= ( x ^4 − x ^4 ) + ( − 6 x ^3 + 6 x ^3 ) + ( 2 x ^2 − 2 x ^2 ) + ( − x + 4 x ) + ( 1 + 3 )\)

\(= 3 x + 4\)

c)Có \(h ( x ) = f ( x ) + g ( x ) = 3 x + 4\)

\(Cho h ( x ) = 0 ⇒ 3 x + 4 = 0\)

\(⇒ 3 x = − 4\) 

\(⇒ x = − \frac{4 }{3} \)

Vậy  \(x=-\frac{4}{3}\) là nghiệm của \(h ( x ) \)

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