\(\left(+\right)A=\left(2x+3\right)\left(4x^2-6x+9\right)-2\left(4x^3-1\right).\)

\(A=8x^3-6x^2-18x+27-8x^3+2\)

\(A=6x^2-18x+29\)

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

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

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

\(B=-6\)

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

a: A=x+3+|x-3|

=x+3+3-x(x<=3)

=6

b:\(B=\sqrt{x^2+4x+4}-\sqrt{x^2}\)

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

=x+2-x=2

c: \(C=\dfrac{\sqrt{x^2-2x+1}}{x-1}\)

\(=\dfrac{\left|x-1\right|}{x-1}=\dfrac{x-1}{x-1}=1\)

\(a,\dfrac{\left(x-1\right)^2}{x^2-1}=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{x+1}\\ b,\dfrac{x^2-16}{4x-x^2}=\dfrac{\left(x-4\right)\left(x+4\right)}{x\left(4-x\right)}=\dfrac{-\left(4-x\right)\left(x+4\right)}{x\left(4-x\right)}=\dfrac{-\left(x+4\right)}{x}\\ c,\dfrac{x^2+6x+9}{2x+6}=\dfrac{\left(x+3\right)^2}{2\left(x+3\right)}=\dfrac{x+3}{2}\)

\(d,\dfrac{x^2+x}{x^2+4x+3}=\dfrac{x\left(x+1\right)}{\left(x^2+x\right)+\left(3x+3\right)}=\dfrac{x\left(x+1\right)}{x\left(x+1\right)+3\left(x+1\right)}=\dfrac{x\left(x+1\right)}{\left(x+1\right)\left(x+3\right)}=\dfrac{x}{x+3}\)

\(e,\dfrac{x^2-x+1}{x^3+1}=\dfrac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x+1}\\ f,\dfrac{\left(x+y\right)^2-z^2}{x+y+z}=\dfrac{\left(x+y-z\right)\left(x+y+z\right)}{x+y+z}=x+y-z\)

mình chỉ biết làm một nửa k biết có đứng k bạn có chắc đề bài đúng k

5x^2 - 1^2 - (2x^3-3^3)= (5x^2-1x^2)-(2x^3-3^3) hdt số 3 và số 7

\(=8x^3-27-8x^3+50=23\)

câu c (x+3)(x^2-3x+9)-(54+x^3)=x^3+27-54-x^3

=27