a,\(2x-5=3x+15\)

\(3x-2x=-5-15\)

\(x=-20\)

b,\(\frac{2}{x-1}=\frac{6}{x+1}\)

\(2x+2=6x-6\)

\(4x=8\)

\(x=2\)

\(\frac{2x+1}{x-1}=\frac{5.\left(x-1\right)}{x+1}\)

\(\frac{2x+1}{x-1}=\frac{5x-5}{x+1}\)

\(2x^2+3x+1=5x^2-10+5\)

\(3x^2-3x=10-5+1=6\)

\(3x.\left(x-1\right)=6\)

\(x.\left(x-1\right)=3\)

Lập bảng

bạn ghi rõ đề ra được không

a: \(=\left(\dfrac{-\left(x+2\right)}{x-2}-\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x+2}\right)\cdot\dfrac{2x^2-x^3}{x^2-3x}\)

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

\(=\dfrac{-4x^2-8x}{x+2}\cdot\dfrac{-x}{x-3}\)

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

b: \(=\dfrac{2x-1}{2x+1}:\left(2x-1+\dfrac{2-4x}{2x+1}\right)\)

\(=\dfrac{2x-1}{2x+1}:\dfrac{4x^2-1+2-4x}{2x+1}\)

\(=\dfrac{2x-1}{4x^2-4x+1}=\dfrac{1}{2x-1}\)

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

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

\(=\dfrac{-x}{x-1}\cdot\dfrac{x-1}{x\left(x-2\right)}=\dfrac{-1}{x-2}\)

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)(

1) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

Ta có: \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}\)

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

Suy ra: \(x^2+2x+1-\left(x^2-2x+1\right)=4\)

\(\Leftrightarrow x^2+2x+1-x^2+2x-1=4\)

\(\Leftrightarrow4x=4\)

hay x=1(loại)

Vậy: \(S=\varnothing\)

2) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{x+2}{x-2}+\dfrac{x}{x+2}=2\)

\(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2\left(x^2-4\right)}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(x^2+4x+4+x^2-2x=2x^2-8\)

\(\Leftrightarrow2x^2+2x+4-2x^2-8=0\)

\(\Leftrightarrow2x-4=0\)

\(\Leftrightarrow2x=4\)

hay x=2(loại)

Vậy: \(S=\varnothing\)

b: =>2x^2-8x+3x-12+x^2-7x+10=3x^2-17x+20

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

=>5x=22

=>x=22/5

c: =>24x^2+16x-9x-6-4x^2-16x-7x-28=20x^2-4x+5x-1

=>-16x-34=x-1

=>-17x=33

=>x=-33/17

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

=>5x=-3

=>x=-3/5

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

=>-6x+8=4

=>-6x=-4

=>x=2/3

f: =>4(x^2+4x-5)-x^2-7x-10=3x^2+3x-6

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

=>6x=16

=>x=8/3

a. (x - 2)(x + 2) - (x - 3)² = 9

<=> x² - 2² - (x - 3)² = 3²

<=> x - 2 - (x - 3) = 3

<=> x - 2 - x + 3 = 3

<=> x - x = 3 - 3 + 2

<=> 0 = 2 (Vô lí)

Vậy nghiệm của PT là S = \(\varnothing\)

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

\(\Leftrightarrow x^3+x-x^2-1-x^3-1=2x-x^2\)

\(\Leftrightarrow-x^2+x-2-2x+x^2=0\)

\(\Leftrightarrow-x=2\)

hay x=-2

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