\(1,A=\left(3x+7\right)\left(2x+3\right)-\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\\ =6x^2+23x+21-2x-3-6x^2-23x+55\\ =73-2x\left(đề.sai\right)\\ B=x^4+x^3-x^2-2x^2-2x+2-x^4-x^3+3x^2+2x\\ =2\\ 2,\\ a,\Leftrightarrow30x^2+18x+3x-30x^2=7\\ \Leftrightarrow21x=7\Leftrightarrow x=\dfrac{1}{3}\\ b,\Leftrightarrow-63x^2+78x-15+63x^2+x-20=44\\ \Leftrightarrow79x=79\Leftrightarrow x=1\\ c,\Leftrightarrow\left(x+5\right)\left(x^2+3x+2\right)-x^3-8x^2=27\\ \Leftrightarrow x^3+3x^2+2x+5x^2+15x+10-x^3-8x^2=27\\ \Leftrightarrow17x=17\Leftrightarrow x=1\)

\(d,\Leftrightarrow7x-2x^2-3+x^2+x-6=-x^2-x+2\\ \Leftrightarrow9x=11\Leftrightarrow x=\dfrac{11}{9}\)

Bài 1.

Bài 1: 

b: \(3x-6=x^2-16\)

\(\Leftrightarrow x^2-3x-10=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

 giúp mình nhanh với 

f/ \(3xy\left(x+y\right)-\left(x+y\right)\left(x^2+y^2+2xy\right)+y^3=27\)

\(3x^2y+3xy^2-\left(x+y\right)\left(x+y\right)^2+y^3=27\)

\(3x^2y+3xy^3-\left(x+y\right)^3+y^3=27\)

\(3x^2y+3xy^3-\left(x^3+3x^2y+3xy^2+b^3\right)+y^3=27\)

\(-x^3=27\)

\(x=-3\)

Bài 1:

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

\(6x-9+4-2x=-3\)

\(4x=-2\)

\(x=-\frac{1}{2}\)

b/ \(2x\left(x^2-2\right)+x^2\left(1-2x\right)-x^2=-12\)

\(2x^3-4x+x^2-2x^3-x^2=-12\)

\(-4x=-12\)

\(x=\frac{1}{3}\)

Bài 2:

a/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)

\(6x^2+9x-6x^2-15x+4x+10=8\)

\(-2x=8\)

\(x=-4\)

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

\(4x^2-4x-3x^2+15-x^2=-7\)

\(-4x=-22\)

\(x=\frac{11}{2}\)

c/ \(2\left(3x-1\right)\left(2x+5\right)-6\left(2x-1\right)\left(x+2\right)=-6\)

\(6x-2\left(2x+5\right)-12x+6\left(x+2\right)=-6\)

\(6x-4x-10-12x+6x+12=-6\)

\(-4x=-8\)

\(x=2\)

Bài 5 :

S = 1 + 3 - 5 - 7 + 9 + 11 - ... - 397 - 399

S = 1 + (3 - 5 - 7 + 9) + (11 - 13 - 15 + 17) + ... + (387 - 389 - 391 + 393) + (395 - 397 - 399)

S = 1 + 0 + 0 + ... + 0 + (- 401)

S = 1 - 401

S = - 400

Bài 5

A= 1+3-5-7+9+11-13-15+...-397-399

A= ( 1+3-5-7)+( 9+11-13-15)+...+( 393+395-397-399)

A= -8 -8 -...-8

A = -8.50 ( từ 1 đến 399 có 200 số, chia làm 4 cặp)

A= -400

Bài 6: 

b) Ta có: x(3x+9)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\3x=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)

Vậy: S={0;-3}

c) Ta có: 52-(x+21)=19-x

\(\Leftrightarrow52-x-21-19+x=0\)

\(\Leftrightarrow12=0\)(vô lý)

Vậy: \(x\in\varnothing\)

Rút gọn hết ta được :

a/ 41x - 17 = -21

=> 41x = -4 => x = 4/41

b/ 34x - 17 = 0 

=> 34x = 17

=> x = 17/34 = 1/2

c/ 19x + 56 = 52 

=> 19x = -4

=> x = -4/19

d/ 20x² - 16x - 34 = 10x² + 3x - 34

=> 10x² - 19x = 0

=> x(10x - 19) = 0

=> x = 0 

hoặc 10x - 19 = 0 => 10x = 19 => x = 19/10

Vậy x = 0 ; x = 19/10

Rút gọn hết ta được :

a/ 41x - 17 = -21

=> 41x = -4 => x = 4/41

b/ 34x - 17 = 0

=> 34x = 17

=> x = 17/34 = 1/2

c/ 19x + 56 = 52

=> 19x = -4

=> x = -4/19

d/ 20x 2 - 16x - 34 = 10x 2 + 3x - 34

=> 10x 2 - 19x = 0

=> x(10x - 19) = 0

=> x = 0 hoặc 10x - 19 = 0

=> 10x = 19

=> x = 19/10

Vậy x = 0 ; x = 19/10 

a) ( 6x - 3 ) ( 2x + 4 ) + ( 4x - 1 ) ( 5 - 3x ) = -21

<=> 12x² + 24x - 6x - 12 + 20x - 12x² - 5 + 3x = -21

<=> 41x = -21 + 12 + 5 

<=> 41x = -4

<=> x = -4/41