321 x 3 + 321 x 5 + 321 x 2

= 321 x (3 + 5 + 2)

= 321 x 10

= 3 210.

321 x 3 + 321 x 5 + 321 x 2

= 321 x (3 + 5 + 2)

= 321 x 10

= 3210

\(C,\dfrac{1}{2}.\dfrac{3}{4}+\dfrac{1}{2}.\dfrac{1}{4}=\dfrac{1}{2}.\left(\dfrac{3}{4}+\dfrac{1}{4}\right)\)

\(=\dfrac{1}{2}.\dfrac{4}{4}=\dfrac{1}{2}.1=\dfrac{1}{2}\)

\(D,\dfrac{11}{3}.\dfrac{26}{7}-\dfrac{26}{7}.\dfrac{8}{3}=\dfrac{26}{7}.\left(\dfrac{11}{3}-\dfrac{8}{3}\right)\)

\(=\dfrac{26}{7}.\dfrac{3}{3}=\dfrac{26}{7}.1=\dfrac{26}{7}\)

\(4\left(x+1\right)\left(-x+2\right)+\left(2x-1\right)\left(2x+3\right)=-11\)

\(\text{⇔}-4x^2+4x+8+4x^2+4x-3=-11\)

\(\text{⇔}8x+5=-11\) 

\(\text{⇔}8x=-16\)

\(\text{⇔}x=-2\)

Vậy: \(x=-2\)

==========

\(\left(2x+4\right)\left(3x+1\right)\left(x-2\right)-\left(-3x^2+1\right)\left(-2x+\dfrac{2}{3}\right)=-\dfrac{26}{3}\)

\(\text{⇔}6x^3+2x^2-24x-8-6x^3-2x^2-2x+\dfrac{2}{3}=-\dfrac{26}{3}\)

\(\text{⇔}-26x-\dfrac{22}{3}=-\dfrac{26}{3}\)

\(\text{⇔}-26x=-\dfrac{4}{3}\)

\(\text{⇔}x=\dfrac{2}{39}\)

\(=\dfrac{3}{2}\cdot\dfrac{4}{2}\cdot\dfrac{5}{3}\cdot\dfrac{6}{4}\cdot...\cdot\dfrac{27}{25}\cdot\dfrac{28}{26}\cdot\dfrac{29}{27}\)

\(=\dfrac{1}{4}\cdot28\cdot29=7\cdot29=203\)

a) (x - 1)³ + 3(x + 1)² = (x² - 2x + 4)(x + 2)

x³ - 3x² + 3x - 1 + 3(x² + 2x+ 1) = x³ + 8

\(\Rightarrow\)x³ - 3x² + 3x - 1 + 3x² + 6x - x³ - 8 = 0

\(\Rightarrow\) 9x - 9 = 0

\(\Rightarrow\) 9x = 9

\(\Rightarrow\) x = 1

b) (2x - 1)(x + 3) - x(3 + 2x) = 26

2x² + 6x - x - 3 - 3x - 2x² = 26

2x - 3 = 26

\(\Rightarrow\) 2x = 29

\(\Rightarrow\) x = 14.5

a) ( x - 1)³ + 3(x+1)² = (x² - 2x + 4 )( x+ 2)

=>x³-3x²+3x-1+3(x²+2x+1)=x3+8

=>x³-3x²+3x-1+3x²+6x+3-x³-8=0

=>(x³-x³)+(-3x²+3x²)+(3x+6x)+(-1+3-8)=0

=>9x-6=0

=>9x=6

=>x=\(\dfrac{2}{3}\)

b) (2x-1)(x+3)-x(3+2x)=26

=>2x²+6x-x-3-3x-2x²=26

=>(2x²-2x²)+(6x-x-3x)-3=26

=>2x-3=26

=>2x=29

=>x=\(\dfrac{29}{2}\)

vậy x=\(\dfrac{29}{2}\)

a: Ta có: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2+3x^2=-33\)

\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+6x^2+12x+1+3x^2=-33\)

\(\Leftrightarrow39x=-34\)

hay \(x=-\dfrac{34}{39}\)

b: Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x-2\right)\left(x+2\right)=1\)

\(\Leftrightarrow x^3-27-x^3+4x=1\)

\(\Leftrightarrow4x=28\)

hay x=7

c: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x-3\right)\left(x+3\right)=26\)

\(\Leftrightarrow x^3+8-x^3+9x=26\)

\(\Leftrightarrow x=2\)

a/\(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x+3\right)\left(x-3\right)=26\)

↔ \(x^3+2^3\)\(-x\left(x^2-3^2\right)\)= 26

↔\(x^3+8-x^3+9x=26\)

↔\(9x=18\leftrightarrow x=2\)

Vậy x=2

b/\(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x-4\right)\left(x+4\right)=21\)

\(\Leftrightarrow x^3-3^3-x\left(x^2-4^2\right)=21\)

\(\Leftrightarrow x^3-9-x^3+16x=21\)

\(\Leftrightarrow16x=30\)

\(\Leftrightarrow x=\frac{15}{8}\)

Vậy \(x=\frac{15}{8}\)

c/\(\left(2x-1\right)\left(4x^2+2x+1\right)-4x\left(2x^2-3\right)=23\)

↔\(\left(2x\right)^3-1^3-4x\left(2x^2-3\right)=23\)

↔\(8x^3-1-8x^3+12x=23\)

↔\(12x=24\leftrightarrow x=2\)

Vậy x=2

a, (x + 2)(x² - 2x + 4 ) - x(x + 3)(x - 3) = 26

<=> x³ + 8 - x(x² - 9) = 26

<=> x³ + 8 - x³ + 9x = 26

<=> 9x - 18 = 0

<=> 9x = 18

<=> x = 2
b, (x - 3)(x² + 3x + 9) - x(x - 4)(x + 4) = 21

<=> x³ - 27 - x(x² - 16) = 21

<=> x³ - 27 - x³ + 16x = 21

<=> 16x - 48 = 0

<=> 16x = 48

<=> x = 3
c, (2x - 1)(4x² + 2x + 1) - 4x(2x² - 3) = 23

<=> 8x³ - 1 - 8x³ + 12x = 23

<=> 12x - 24 = 0

<=> 12x = 24

<=> x = 2

\(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x+3\right)\left(x-3\right)=26\)

\(< =>x^3-2x^2+4x+2x^2-4x+8-x\left(x^2-9\right)-26=0\)

\(< =>x^3+8-x^3+9x-26=0\)

\(< =>9x-18=0< =>x=2\)

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

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

\(=2x^2-2x+13\)

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

\(=\left(4-x-x+3\right)\left(4-x+x-3\right)\)

\(=-2x+7\)

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

\(=x^2-25-x^2-10x-25\)

=-10x-50

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

\(=x^2-6x+9-x^2+16\)

=-6x+25

5) Ta có: \(\left(y^2-6y+9\right)-\left(y-3\right)^2\)

\(=y^2-6y+9-y^2+6y-9\)

=0

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

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

=12x+18