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

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

\(1,4x\left(1-x\right)-8=1-\left(4x^2+3\right)\\ \Leftrightarrow4x-4x^2-8=1-4x^2-3\\ \Leftrightarrow4x-4x^2-8-1+4x^2+3=0\\ \Leftrightarrow4x-6=0\\ \Leftrightarrow x=\dfrac{3}{2}\)

\(2,\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\\ \Leftrightarrow\left(2-3x\right)\left(x+11\right)-\left(2-3x\right)\left(5x-2\right)=0\\ \Leftrightarrow\left(2-3x\right)\left(x+11-5x+2\right)=0\\ \Leftrightarrow\left(2-3x\right)\left(-4x+13\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{13}{4}\end{matrix}\right.\)

\(a,3x-2\left(x-3\right)=0\\ \Leftrightarrow3x-2x+6=0\\ \Leftrightarrow x=-6\\ b,\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\\ \Leftrightarrow2x^2+2x-3x-3=2x^2-x+10x-5\\ \Leftrightarrow2x^2-x-3=2x^2+9x-5\\ \Leftrightarrow10x-2=0\\ \Leftrightarrow x=\dfrac{1}{5}\\ c,ĐKXĐ:x\ne\pm1\\ \dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\\ \Leftrightarrow\dfrac{2x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x-x^2+1}{\left(x+1\right)\left(x-1\right)}=0\)

\(\Rightarrow3x+1=0\\ \Leftrightarrow x=-\dfrac{1}{3}\left(tm\right)\)

\(d,\left(2x+3\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x-5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\\ e,ĐKXĐ:x\ne\pm2\\ \dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\\ \Leftrightarrow\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-22}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\dfrac{x^2-4x+4-3x-6-2x+22}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow x^2-9x+20=0\\ \Leftrightarrow\left(x^2-5x\right)-\left(4x-20\right)=0\\ \Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)

Đề bài mình viết thiếu là CM biểu thức sau không phụ thuộc vào x ( nghĩa là kết quả phải ra số tự nhiên không có x ) 

\(A=\left(2x+1\right)\left(x-1\right)-2x\left(x+2\right)-5\left(-x+3\right)+4\)

\(=2x^2-2x+x-1-2x^2-4x+5x-15+4\)

\(=-12\left(đpcm\right)\)

\(B=\left(4x+3\right)\left(2x-5\right)-\left(8x+1\right)\left(x+3\right)+13\left(3x+1\right)+2\)

\(=8x^2-20x+6x-15-\left(8x^2+24x+x+3\right)+39x+13+2\)

\(=-3\left(đpcm\right)\)

\(\left(2x-4\right)\left(1-3x\right)=0\)

<=>  \(2\left(x-2\right)\left(1-3x\right)=0\)

<=>  \(\orbr{\begin{cases}x-2=0\\1-3x=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=2\\x=\frac{1}{3}\end{cases}}\)

Vậy....

\(\left(2x-4\right)\left(1-3x\right)=0\)

\(\Rightarrow2x-4=0\)hoặc\(1-3x=0\)

\(TH1:2x-4=0\)

                     \(2x=0+4\)

                     \(2x=4\)

                       \(x=4:2\)

                       \(x=2\)

\(TH2:1-3x=0\)

                    \(3x=1-0\)

                   \(3x=1\)

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

Vậy:\(x=2\)hoặc \(x=\frac{1}{3}\)

a. (2x-4)(1-3x)=0

<=> 2x-4=0 hoặc 1-3x=0

<=> x=2 hoặc x=1/3

b. (x-3)(x²-3x+2)=0

<=> x-3=0 hoặc x²-3x+2=0

<=> x=3 hoặc x=2 hoặc x=1

c. (x-1)(5x+3)=(2x-7)(x-1)

<=> (x-1)(3x+10)=0

<=> x-1=0 hoặc 3x+10=0

<=> x=1 hoặc x=-10/3

\(\left(-3x-2\right)^2+\left(3x+5\right)\left(5-3x\right)=-7\)

\(\Leftrightarrow9x^2+12x+4+15x-9x^2+25-15x=-7\)

\(\Leftrightarrow12x+36=0\Leftrightarrow x=-3\)

\(\left(x+2\right)\left(x^2+2x+2\right)-x\left(x-8\right)^2=\left(4x-3\right)\left(4x+3\right)\)

\(\Leftrightarrow x^3+2x^2+2x+2x^2+4x+4-x\left(x^2-16x+64\right)=16x^2-9\)

\(\Leftrightarrow x^3+4x^2+6x+4-x^3+16x^2-64=16x^2-9\)

\(\Leftrightarrow4x^2+6x-51=0\)

\(\cdot\Delta=6^2-4.4.\left(-51\right)=852\)

Vậy pt có 2 nghiệm phân biệt

\(x_1=\frac{-6+\sqrt{852}}{8}\);\(x_2=\frac{-6-\sqrt{852}}{8}\)

a) x+3=12

x=12-3

x=9

b)(x-3):2=5¹⁴:5¹²

=>(x-3):2=5²

=>(x-3):2=25

=>x-3=25.2

=>x-3=50

=>x=50+3

=>x=53

c)4x+3x=30-20:10

=>x(4+3)=30-2

=>7x=28

=>x=28:7

=>x=4

d)2x-138=2³.3²

=>2x-138=8.9

=>2x-138=72

=>2x=72+138

=>2x=210

=>x=210:2

=>x=105

a) x + 3 = 12

x = 12 - 3

x = 9

b) ( x - 3 ) : 2 = 5¹⁴ : 5¹²

( x - 3 ) : 2 = 5^14-12

( x - 3 ) : 2 = 5²

( x - 3 ) : 2 = 25

x - 3 = 50

x = 53

c) 4x + 3x = 30 - 20 : 10

7x = 28

x = 4

d) 2x - 138 = 2³ x 3²

2x - 138 = 8 x 9

2x - 138 = 72

2x = 210

x = 105

a) 3x = 7y ⇒ x/7 = y/3

⇒ x/7 = 2y/6

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

x/7 = 2y/6 = (x - 2y)/(7 - 6) = 2/1 = 2

x/7 = 2 ⇒ x = 2.7 = 14

y/3 = 2 ⇒ y = 2.3 = 6

Vậy x = 14; y = 6

b) x/2 = y/3 ⇒ x/6 = y/9 (1)

x/3 = z/4 ⇒ x/6 = z/8 (2)

Từ (1) và (2) ⇒ x/6 = y/9 = z/8

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

x/6 = y/9 = z/8 = (x + y - z)/(6 + 9 - 8) = 7/7 = 1

x/6 = 1 ⇒ x = 1.6 = 6

y/9 = 1 ⇒ y = 1.9 = 9

z/8 = 1 ⇒ z = 1.8 = 8

Vậy x = 6; y = 9; z = 8

c) x/2 = y/3 ⇒ x/10 = y/15 ⇒ 2x/20 = y/15 (3)

y/5 = z/4 ⇒ y/15 = z/12 (4)

Từ (3) và (4) ⇒ 2x/20 = y/15 = z/12

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

2x/20 = y/15 = z/12 = (2x - y + z)/(20 - 15 + 12) = 17/17 = 1

2x/20 = 1 ⇒ x = 1.20 : 2 = 10

y/15 = 1 ⇒ y = 1.15 = 15

z/12 = 1 ⇒ z = 1.12 = 12

Vậy x = 10; y = 15; z = 12