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

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

3, 2x(x^2-8x+16)-(x+5)(x^2-4)+2(x^2+10x+25)-x+1

=2x^3-16x^2+32x-(x^3-4x+5x^2-20)+2x^2+20x+50-x+1

=2x^3-16x^2+32x-x^3+4x-5x^2+20+2x^2+20x+50-x+1

=x^3-19x^2+55x+71

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

\(\Rightarrow x+2-2=0\)

\(\Rightarrow x=0\)

\(\left(x+3\right)+1=7\)

\(\Rightarrow x+3+1=7\)

\(\Rightarrow x+4=7\)

\(\Rightarrow x=3\)

\(\left(3x-4\right)+4=12\)
\(\Rightarrow3x-4+4=12\)

\(\Rightarrow3x=12\)

\(\Rightarrow x=4\)

\(\left(5x+4\right)-1=13\)

\(\Rightarrow5x+4-1=13\)

\(\Rightarrow5x+3=13\)

\(\Rightarrow5x=10\)

\(\Rightarrow x=2\)

\(\left(4x-8\right)-3=5\)

\(\Rightarrow4x-8-3=5\)

\(\Rightarrow4x-11=5\)

\(\Rightarrow4x=16\)

\(\Rightarrow x=4\)

\(8-\left(2x+4\right)=2\)

\(\Rightarrow8-2x-4=2\)

\(\Rightarrow4-2x=2\)

\(\Rightarrow2x=2\)

\(\Rightarrow x=1\)

\(7+\left(5x+2\right)=14\)

\(\Rightarrow7+5x+2=14\)

\(\Rightarrow9+5x=14\)

\(\Rightarrow5x=5\)

\(\Rightarrow x=1\)

\(5-\left(3x-11\right)=1\)

\(\Rightarrow5-3x+11=1\)

\(\Rightarrow16-3x=1\)

\(\Rightarrow3x=15\)

\(\Rightarrow x=5\)

Ít thôi -..-

a) ( 3x + 2 )( 2x + 9 )  - ( x + 3 )( 6x + 1 ) = ( x + 1 )² - ( x + 2 )( x - 2 )

<=> 6x² + 31x + 18 - ( 6x² + 19x + 3 ) = x² + 2x + 1 - ( x² - 4 )

<=> 6x² + 31x + 18 - 6x² - 19x - 3 = x² + 2x + 1 - x² + 4

<=> 12x + 15 = 2x + 5

<=> 12x - 2x = 5 - 15

<=> 10x = -10

<=> x = -1

b) ( 2x + 3 )( x - 4 ) + ( x - 5 )( x - 2 ) = ( 3x - 5 )( x - 4 )

<=> 2x² - 5x - 12 + x² - 7x + 10 = 3x² - 17x + 20

<=> 3x² - 12x - 2 = 3x² - 17x + 20

<=> 3x² - 12x - 3x² + 17x = 20 + 2

<=> 5x = 22

<=> x = 22/5

c) ( x + 2 )³ - ( x - 2 )³ - 12x( x - 1 ) = -8

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

<=>  x³ + 6x² + 12x + 8 - x³ + 6x² - 12x + 8 - 12x² + 12x = -8

<=> 12x + 16 = -8

<=> 12x = -24

<=> x = -2

d) ( 3x - 1 )² - 5( x + 1 ) + 6x - 3.2x + 1 - ( x - 1 )² = 16

<=> 9x² - 6x + 1 - 5x - 5 + 6x - 6x + 1 - ( x² - 2x + 1 ) = 16

<=> 9x² - 11x - 3 - x² + 2x - 1 = 16

<=> 8x² - 9x - 4 = 16

<=> 8x² - 9x - 4 - 16 = 0

<=> 8x² - 9x - 20 = 0

( Đến đây bạn có hai sự lựa chọn : 1 là vô nghiệm

                                                         2 là nghiệm vô tỉ =) )

a) (3x + 2)(2x + 9) - (x + 3)(6x + 1) = (x + 1)² - (x + 2)(x - 2)

=> 3x(2x + 9) + 2(2x + 9) - x(6x + 1) - 3(6x + 1) = x² + 2x + 1 - x(x - 2) - 2(x - 2)

=> 6x² + 27x + 4x + 18 - 6x² - x - 18x - 3 = x² + 2x + 1 - x² + 2x - 2x + 4

=> (6x² - 6x²) + (27x + 4x - x - 18x) + (18 - 3) = (x² - x²) + (2x + 2x - 2x) + (1 + 4)

=> 12x + 15 = 2x + 5

=> 12x + 15  - 2x - 5 = 0

=> 10x + 10 = 0

=> 10x = -10 => x = -1

b) (2x + 3)(x - 4) + (x - 5)(x - 2) = (3x - 5)(x - 4)

=> 2x(x - 4) + 3(x - 4) + x(x - 2) - 5(x - 2) = 3x(x - 4) - 5(x - 4)

=> 2x² - 8x + 3x - 12 + x² - 2x - 5x + 10 = 3x² - 12x - 5x + 20

=> (2x² + x²) + (-8x + 3x - 2x - 5x) + (-12 + 10) = 3x² - 17x + 20

=> 3x² - 12x - 2 = 3x² - 17x + 20

=> 3x² - 12x - 2 - 3x² + 17x - 20 = 0

=> (3x² - 3x²) + (-12x + 17x) + (-2 - 20) = 0

=> 5x - 22 = 0

=> 5x = 22 => x = 22/5

c) (x + 2)³ - (x - 2)³ - 12x(x - 1) = -8

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

=> x³ + 6x² + 12x + 8 -x³ + 6x² - 12x + 8 - 12x² + 12x = -8

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

=> 12x + 16 = -8

=> 12x = -24

=> x = -2

Còn bài cuối làm nốt

\(\text{a, 3(x+1)+4x=10}\)

\(\Rightarrow3x+3+4x=10\)

\(\Rightarrow7x+3=10\)

\(\Rightarrow7x=10-3=7\)

\(\Rightarrow x=1\)

c, x+1/10+x+2/9=x+3/8+x+4/7

=> (x+1/10 +1) +(x+2/9 +1)= ( x+3/8 +1) +(x+4/7 +1)

=> x+11/10 + x+11/9 = x+11/8 + x+11/7

...............

a) \(3\left(x+1\right)+4x=10\)

\(\Rightarrow3x+3+4x=10\)

\(\Rightarrow3x+4x=10-3\)

\(\Rightarrow7x=7\)

\(\Rightarrow x=7\)

tải photomath về bn

thank you nha 

Tui chưa nháp nhưng câu 1 thử nhân hết ra coi triệt tiêu bớt đc ko, mà chắc chắn là nhân ra sẽ mất cái 27x^3 rồi nên thành pt bậc 2 giải vô tư nhé, câu 2 tách hết ra cx lm đc vì nó là pt bậc 2 

câu 3 tách thành (x+3)(x^2-7x+9)=0 có pt bậc 2 nên ok r

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

<=> 27x³ - 8 - 27x³ + 1 =  x - 4

<=> x - 4 = -7

<=> x = -3

Vậy S = {-3}

9(2x + 1) = 4(x - 5)²

<=> 4(x² - 10x + 25) - 18x - 9 = 0

<=>4x² - 40x + 100 - 18x - 9 = 0

<=> 4x² - 58x + 91 = 0

<=> (4x² - 58x + 210,25) - 119,25 = 0

<=> (2x - 14,5)² = 119,25

<=> \(\orbr{\begin{cases}2x-14,5=\sqrt{119,25}\\2x-14,5=-\sqrt{119,25}\end{cases}}\)

<=> \(\orbr{\begin{cases}x=\frac{29+3\sqrt{53}}{4}\\x=\frac{29-3\sqrt{53}}{4}\end{cases}}\)

Vậy S = {...}

x³ - 4x²  - 12x + 27 = 0

<=> (x³ + 3x²) - (7x² + 21x) + (9x + 27) = 0

<=> x²(x + 3) - 7x(x + 3) + 9(x + 3) = 0

<=> (x² - 7x + 9)(x + 3) = 0

<=> \(\orbr{\begin{cases}x-7x+9=0\\x+3=0\end{cases}}\)

<=> \(\orbr{\begin{cases}\left(x^2-7x+12,25\right)-3,25=0\\x=-3\end{cases}}\)

<=> \(\orbr{\begin{cases}\left(x-3,5\right)^2=3,25\\x=-3\end{cases}}\)

<=> \(\orbr{\begin{cases}x-3,5=\sqrt{3,25}\\x-3,5=-\sqrt{3,25}\end{cases}}\)

hoặc x = -3

<=>  \(\orbr{\begin{cases}x=\frac{7+\sqrt{13}}{2}\\x=\frac{7-\sqrt{13}}{2}\end{cases}}\)

hoặc x = -3

Vậy S = {...}