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

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

\(=8x^2+2x-3\)

a: Ta có: \(\left(x+5\right)^2-4x\left(2x+3\right)^2-\left(2x-1\right)\left(x+3\right)\left(x-3\right)\)

\(=x^2+10x+25-4x\left(4x^2+12x+9\right)-\left(2x-1\right)\left(x^2-9\right)\)

\(=x^2+10x+25-16x^3-48x^2-36x-2x^3+18x+x^2-9\)

\(=-18x^3-46x^2-8x+16\)

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

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

1) `2x(3x-1)-(2x+1)(x-3)`

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

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

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

`=3x^2-9x-4x^2+4x-2x+2`

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

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

`=3x^2-15x-(x^2-4x+4)-(4x^2-9)`

`=3x^2-15x-x^2+4x-4-4x^2+9`

`=-2x^2-11x+5`

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

`=4x^2-12x+9+2x^2+8x-x-4`

`=6x^2-5x+5`

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.