1, \(\left(x-1\right)\left(x+2\right)-\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left[x+2-\left(x-1\right)\right]=0\)

\(\Leftrightarrow3\left(x-1\right)=0\Leftrightarrow x=1\)

2, \(\left(x-2\right)^2-3\left(x-2\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left[x-2-3\left(x+1\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(-2x-5\right)=0\Leftrightarrow x=-\dfrac{5}{2};x=2\)

3, \(\left(5-2x\right)\left(2x+7\right)=4x^2-25=\left(2x-5\right)\left(2x+5\right)\)

\(\Leftrightarrow\left(5-2x\right)\left(2x+7\right)+\left(5-2x\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\left(5-2x\right)\left(2x+7+2x+5\right)=0\Leftrightarrow\left(4x+12\right)\left(5-2x\right)=0\Leftrightarrow x=-3;x=\dfrac{5}{2}\)

1) Ta có: \(\left(x-1\right)\left(x+2\right)-\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2-x+1\right)=0\)

\(\Leftrightarrow x-1=0\)

hay x=1

2) Ta có: \(\left(x-2\right)^2-3\left(x-2\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-2-3x-3\right)=0\)

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

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

a) \(\dfrac{x}{3}=\dfrac{4}{12}\Rightarrow x=\dfrac{4}{12}\cdot3=\dfrac{12}{12}=1\)

b) \(\dfrac{x-1}{x-2}=\dfrac{3}{5}\) (Điều kiện : \(x\ne2\))

\(\Rightarrow5\left(x-1\right)=3\left(x-2\right)\)

\(\Leftrightarrow5x-5=3x-6\Leftrightarrow5x-3x=-6+5\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)

c) \(2x:6=\dfrac{1}{4}\Leftrightarrow2x=\dfrac{1}{4}\cdot6=\dfrac{6}{4}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{3}{2}:2=\dfrac{3}{2}\cdot\dfrac{1}{2}=\dfrac{3}{4}\)

d) \(\dfrac{x^2+x}{2x^2+1}=\dfrac{1}{2}\)

\(\Rightarrow2\left(x^2+x\right)=2x^2+1\)

\(\Leftrightarrow2x^2+2x=2x^2+1\)

\(\Leftrightarrow2x^2+2x-2x^2=1\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\).

Bạn nên viết lại đề bài cho sáng sủa, rõ ràng để người đọc dễ hiểu hơn.

f: =>4(x^2+4x-5)-x^2-7x-10=3(x^2+x-2)

=>4x^2+16x-20-x^2-7x-10-3x^2-3x+6=0

=>6x-24=0

=>x=4

e: =>8x+16-5x^2-10x+4(x^2-x-2)=4-x^2

=>-5x^2-2x+16+4x^2-4x-8=4-x^2

=>-6x+8=4

=>-6x=-4

=>x=2/3

d: =>2x^2+3x^2-3=5x^2+5x

=>5x=-3

=>x=-3/5

b: =>2x^2-8x+3x-12+x^2-7x+10=3x^2-12x-5x+20

=>-12x-2=-17x+20

=>5x=22

=>x=22/5

Chúng ta sẽ giải từng phương trình một:

a. Đặt �=�2−2y=x2−2, ta có: (−2+�2)(�2−2)4=1(−2+x2)(

a)2x.(x+3)-3.(x^2+1)=x+1-x.(x-2)

<=> 2x² + 6x - 3x² - 3 = x - 1 - x² + 2x

<=> 2x² + 6x - 3x² - x + x² - 2x = -1 +3

<=> 3x = 2

<=> x = 2/3

b)(x+2).(x-2)-(x-3).(x+5)=0

<=> x² - 4 - x² - 5x - 3x - 15 = 0

<=> -5x - 3x = 4 + 15

<=> -8x = 19

<=> x = -19/8

Phần c tương tự ạ

\(a,\Leftrightarrow x^3-8-x\left(x^2-9\right)=1\\ \Leftrightarrow x^3-8-x^3+9x=1\\ \Leftrightarrow9x=9\Leftrightarrow x=1\\ b,\Leftrightarrow8x^3+12x^2+6x+1-8x^3 +12x^2-6x+1-24x^2+24x-1=0\Leftrightarrow1=0\Leftrightarrow x\in\varnothing\)

a) \(\Leftrightarrow x^3-8-x^3+9x=1\)

\(\Leftrightarrow9x=9\Leftrightarrow x=1\)

b) \(\Leftrightarrow8x^3+12x^2+6x+1-8x^3+12x^2-6x+1-24x^2+24x-6=5\)

\(\Leftrightarrow24x=9\Leftrightarrow x=\dfrac{3}{8}\)

Câu 1: C

Câu 2: A

Câu 3: C

c, viết thiếu kìa :))

\(a,\left(2x-1\right)-\left(5x+1\right)=\left(x+3\right)-\left(x-2\right)\)

\(\Leftrightarrow2x-1-5x-1=x+3-x+2\)

\(\Leftrightarrow-3x-2=5\Leftrightarrow-3x=7\Leftrightarrow x=\frac{7}{3}\)

\(b,\left(2x-3\right)-\left(x-5\right)=\left(x+2\right)-\left(x-1\right)\)

\(\Leftrightarrow2x-3-x+5=x+2-x+1\Leftrightarrow x+2=3\Leftrightarrow x=1\)

\(c,2\left(x-1\right)-5\left(x+2\right)=0\)

\(\Leftrightarrow2x-2-5x-10=0\Leftrightarrow-3x-12=0\Leftrightarrow-3x=12\Leftrightarrow x=-4\)