a) \(\left(3x+2\right)\left(3x+2\right)-\left(3x+1\right)^2=\left(3x+2\right)^2-\left(3x+1\right)^2=\left(3x+2-3x-1\right)\left(3x+2+3x+1\right)=1.\left(6x+3\right)=6x+3\)

b) \(\left(1-4x^2\right)\left(1+4x^2\right)-\left(2x+3\right)^2=1-16x^4-4x^2-12x-9=-16x^4-4x^2-12x-8\)

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

\(=9x^4+12x+4-9x^2-6x-1\)

=6x+3

b: \(\left(1-4x^2\right)\left(1+4x^2\right)-\left(2x+3\right)^2\)

\(=1-16x^4-4x^2-12x-9\)

\(=-16x^4-4x^2-12x-8\)

a/ 2x\(^{^{ }3}\)-3\(^{^{ }3}\)-2x\(^3\)-1\(^{^{ }3}\)=-28

b/x\(^{^{ }3}\)+2\(^{^{ }3}\)-x\(^3\)+2=10

c/3x\(^3\)+5\(^3\)-3x(3x\(^2\)-1)=3x\(^3\)+5\(^3\)-3x\(^3\)+3x=125+3x

d/ x\(^6\)-(x\(^3\)+1)(x\(^2\)-x+1)= x\(^6\)-(x\(^6\)-x\(^4\)+x\(^3\)+x\(^2\)-x+1)=x\(^4\)-x\(^3\)-x\(^2\)+x-1

Ta có: \(2x\left(3x-1\right)-\left(2x+1\right)\left(x-3\right)\)

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

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

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

Ta có: \(3\left(x^2-2x\right)-\left(4x+2\right)\left(x-1\right)\)

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

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

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

\(2x\left(3x-1\right)-\left(2x+1\right)\left(x-3\right)=6x^2-2x-2x^2+5x+3=4x^2+3x+3\)

\(3\left(x^2-2x\right)-\left(4x+2\right)\left(x-1\right)=3x^2-6x-4x^2+2x-2=-x^2-4x-2\)

cậu xem câu hỏi của Quỳnh Phạm nha

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

A = \(\left(3x-1+2x+1\right)^2\)

A)

<=>(3x)^2−2×3x+1+2(3x−1)(2x+1)+(2x+1)^2

<=>(3x)^2−2×3x+1+(6x−2)(2x+1)+(2x+1)^2

<=>(3x)^2−2×3x+1+12x^2+6x−4x−2+(2x+1)^2

<=>(3x)^2−2×3x+1+12x^2+6x−4x−2+(2x)^2+2×2x+1

<=>32x^2−2×3x+1+12x^2+6x−4x−2+(2x)^2+2×2x+1

<=>9x^2−2×3x+1+12x^2+6x−4x−2+(2x)^2+2×2x+1

<=>9x^2−2×3x+1+12x^2+6x−4x−2+2^2x^2+2×2x+1

<=>9x^2−2×3x+1+12x^2+6x−4x−2+4x^2+2×2x+1

<=>9x^2−6x+1+12x^2+6x−4x−2+4x^2+2×2x+1

<=>9x^2−6x+1+12x^2+6x−4x−2+4x^2+4x+1

<=>(9x^2+12x^2+4x^2)+(−6x+6x−4x+4x)+(1−2+1)

<=> 25x^2

B)

<=>2x(4x^2−6x+9)+3(4x^2−6x+9)+8(1−x)(1+x+x^2)

<=>8x^3−12x^2+18x+3(4x^2−6x+9)+8(1−x)(1+x+x^2)

<=>8x^3−12x^2+18x+12x^2−18x+27+8(1−x)(1+x+x^2)

<=>8x^3−12x^2+18x+12x^2−18x+27+(8−8x)(1+x+x^2)

<=>8x^3−12x^2+18x+12x^2−18x+27+8(1+x+x^2)−8x(1+x+x^2)

<=>8x^3−12x^2+18x+12x^2−18x+27+8+8x+8x^2−8x(1+x+x^2)

<=>8x^3−12x^2+18x+12x^2−18x+27+8+8x+8x^2−(8x+8x2+8x^3)

<=>8x^3−12x^2+18x+12x^2−18x+27+8+8x+8x^2−8x−8x^2−8x^3

<=>(8x^3−8x^3)+(−12x^2+12x^2+8x^2−8x^2)+(18x−18x+8x−8x)+(27+8)

<=> 35

(3x+4)² +(4x -1 )² + 2x + 5(2x-5)

= 9x²+24x+16 + 16x²-8x+1 + 2x + 10x - 25

= 25x² + 28x - 8

b. (2x+1)(4x² -2x + 1 ) + ( 2-3x ) ( 4+ 6x + 9x² )

= 8x³ + 1 + 8-27x³

= -19x³ +9

Đặt `A=(1-3x)/(2x)+(3x-2)/(2x-1)+(3x-2)/(2x-4x^2)`

`=(2x(3x-2))/(2x(2x-1))-((3x-1)(2x-1))/(2x(2x-1))-(3x-2)/(2x(2x-1))`

`=(6x^2-4x-6x^2+5x-1-3x+2)/(2x(2x-1))`

`=(-2x+1)/(2x(2x-1))`

`=-1/(2x)`

`2x=1/(483)`

`=>A=-1/(1/483)=-483`

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

⇔\(\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}+\dfrac{3x-2}{2x\left(1-2x\right)}\)

⇔\(\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}-\dfrac{(3x-2)}{2x\left(2x-1\right)}\)

ĐKXĐ: \(\left\{{}\begin{matrix}2x\ne0\\2x-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\2x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne\dfrac{1}{2}\end{matrix}\right.\)

MTC: 2x(2x-1)

\(\dfrac{1-3x\left(2x-1\right)}{2x\left(2x-1\right)}+\dfrac{(3x-2)(2x)}{(2x-1)(2x)}-\dfrac{(3x-2)}{2x\left(2x-1\right)}\)

\(\Rightarrow1-3x\left(2x-1\right)+\left(3x-2\right)\left(2x\right)-\left(3x-2\right)\)

\(\Leftrightarrow1-6x^2+3x+6x^2-4x-3x+2\\ \Rightarrow-4x+3\)

Thay \(x=\dfrac{1}{966}\)vào biểu thức trên

ta có -4x+3= \(-4\times\dfrac{1}{966}+3=\dfrac{-4}{966}+3=\dfrac{-2}{483}+3=\dfrac{-2}{483}+\dfrac{1449}{483}=\dfrac{1447}{483}\)

a: \(P\left(x\right)=2x^3+x^2+x+2\)

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

b: \(P\left(-1\right)=2\cdot\left(-1\right)+1-1+2=0\)

\(Q\left(-1\right)=-1+1-1+1=0\)

Do đó: x=-1 là nghiệm chung của P(x), Q(x)

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

\(P\left(x\right)=2x^3+x^2+x+2\)

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

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

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

\(P\left(-1\right)=0\)

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

\(Q\left(-1\right)=0\)

Vậy x = -1  là nghiệm của P(x),Q(x)

a, `P(x) = 2x^3 + x^2 + x + 2`.

`Q(x) = x^3 + x^2 + x + 1`.

`P(-1) = 0`

`Q(-1) = 0`

`=>` `-1` là nghiệm chung của `2` đa thức trên.