Thêm nữa câu a) Tính: M(x) + N(x)+ P(x)

B) Tính M(x) - N (x) - P(x)

ok rồi giúp mình với nha

\(P\left(x\right)=3x^2+7+2x^4-3x^2-4-5x+2x^3\)

\(=2x^4+2x^3+\left(3x^2-3x^2\right)-5x-4+7\)

\(=2x^4+2x^3-5x+3\)

\(Q\left(x\right)=-3x^3+2x^2-x^4+x+x^3+4x-2+5x^4\)

\(=\left(5x^4-x^4\right)+\left(-3x^3+x^3\right)+2x^2+\left(x+4x\right)-2\)

\(=4x^4-2x^3+2x^2+5x-2\)

a) ( x - 3 )² - 4 = 0

<=> ( x - 3 )² - 2² = 0

<=> ( x - 3 - 2 )( x - 3 + 2 ) = 0

<=> ( x - 5 )( x - 1 ) = 0

<=> x = 5 hoặc x = 1

b( 2x + 3 )² - ( 2x + 1 )( 2x - 1 ) = 22

<=> 4x² + 12x + 9 - ( 4x² - 1 ) = 22

<=> 4x² + 12x + 9 - 4x² + 1 = 22

<=> 12x + 10 = 22

<=> 12x = 12

<=> x = 1

c) ( 4x + 3 )( 4x - 3 ) - ( 4x - 5 )² = 16

<=> 16x² - 9 - ( 16x² - 40x + 25 ) = 16

<=> 16x² - 9 - 16x² + 40x - 25 = 16

<=> 40x - 34 = 16

<=> 40x = 50

<=> x = 50/40 = 5/4

d) x³ - 9x² + 27x - 27 = -8

<=> ( x - 3 )³ = -8

<=> ( x - 3 )³ = (-2)³

<=> x - 3 = -2

<=> x = 1 

e) ( x + 1 )³ - x²( x + 3 ) = 2

<=> x³ + 3x² + 3x + 1 - x³ - 3x² = 2

<=> 3x + 1 = 2

<=> 3x = 1

<=> x = 1/3

f) ( x - 2 )³ - x( x - 1 )( x + 1 ) + 6x² = 5

<=> x³ - 6x² + 12x - 8 - x( x² - 1 ) + 6x² = 5

<=> x³ + 12x - 8 - x³ + x = 5

<=> 13x - 8 = 5

<=> 13x = 13

<=> x = 1

a) \(\left(x-3\right)^2-4=0\)

=> \(\left(x-3\right)^2-2^2=0\)

=> \(\left(x-3-2\right)\left(x-3+2\right)=0\)

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

=> \(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)

b) \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)

=> \(\left(2x+3\right)^2-\left[\left(2x\right)^2-1^2\right]=22\)

=> \(\left(2x+3\right)^2-\left(4x^2-1\right)=22\)

=> \(\left(2x\right)^2+2\cdot2x\cdot3+3^2-4x^2+1=22\)

=> \(4x^2+12x+9-4x^2+1=22\)

=> \(12x+9+1=22\)

=> \(12x+10=22\)

=> 12x = 12

=> x = 1

c) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=16\)

=> \(\left(4x\right)^2-3^2-\left[\left(4x\right)^2-2\cdot4x\cdot5+5^2\right]=16\)

=> \(16x^2-9-\left(16x^2-40x+25\right)=16\)

=> \(16x^2-9-16x^2+40x-25=16\)

=> \(-9+40x-25=16\)

=> \(40x=16+25-\left(-9\right)=16+25+9=50\)

=> x = 50/40 = 5/4

d) \(x^3-9x^2+27x-27=-8\)

=> \(x^3-3\cdot x^2\cdot3+3\cdot x\cdot3^2-3^3=8\)

=> \(\left(x-3\right)^3=-8\)

=> \(\left(x-3\right)^3=\left(-2\right)^3\)

=> x - 3  = -2 => x = 1

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

=> \(x^3+3x^2+3x+1-x^3-3x^2=2\)

=> \(3x+1=2\)

=> \(3x=1\)=> x = 1/3

f) \(\left(x-2\right)^3-x\left(x-1\right)\left(x+1\right)+6x^2=5\)

=> \(x^3-3\cdot x^2\cdot2+3\cdot x\cdot2^2-2^3-x\left(x^2-1\right)+6x^2=5\)

=> \(x^3-6x^2+12x-8-x^3+x+6x^2=5\)

=> \(\left(12x+x\right)-8=5\)

=> 13x  = 13

=> x = 1

a) (x+3)^2-4=0

=>(x+3)^2 = 4

=>(x+3)^2 = 2^2 = (-2)^2

=>x+3 = 2 hoặc -2

=> x= -1 hoặc -5

giúp với trí ơi

giúp nốt trí ui

a) \(4x^2-2x-y^2-y=\left(4x^2-y^2\right)-\left(2x+y\right)\)

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

b) \(9x^2-25y^2-6x+10y=\left(3x-5y\right)\left(3x-5y\right)-2\left(3x-5y\right)\)

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

c) \(x^3-2x^2+2x-1=x^3-1-2x^2+2x\)

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

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

d) \(x^4+2x^3-4x-4=x^4-4+2x^3-4x\)

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

`#3107`

`a.`

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

`b.`

`9x^4y^3 + 3x^2y^4 = 3x^2y^2(3x^2y + y^2)`

`c.`

`x^3 - 2x^2 + 5x`

`= x(x^2 - 2x + 5)`

a) 4x² - 6x

= 2x(2x - 3)

b) 9x⁴y³ + 3x²y⁴

= 3x²y³(3x² + 3y)

c) x³ - 2x² + 5x

= x(x² - 2x + 5)

ko có biết

6, x mũ 4 - 4x mũ 3 - 8x mũ 2 + 8x =x (x+2) (x^2-6x+4)

8, x mũ 4 + 2x mũ 3 + x mũ 2 - y mũ 2  = -(y-x^2-x) (y+x^2+x)

10, 4x mũ 2 ( x + y ) -x - y  = (2x-1) (2x+1) (y+x)