\(a)\frac{2x-1}{5x-10}\)    \(\text{Đ}K:x\ne2\)

\(\Leftrightarrow2x-1=0\)

\(\Leftrightarrow x=\frac{1}{2}(TM)\)

\(b)\frac{x^2-x}{2x}\)    \(\text{Đ}K:x\ne0\)

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

\(\Leftrightarrow x.(x-1)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0(lo\text{ại})\\x=1(TM)\end{cases}}\)

\(c)\frac{2x+3}{4x-5}\)      \(\text{Đ}K:x\ne\frac{5}{4}\)

\(\Leftrightarrow2x+3=0\)

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

\(d)\frac{(x-1).(x+2)}{(x-3).(x-1)}\)    \(\text{Đ}K:\hept{\begin{cases}x\ne3\\x\ne1\end{cases}}\)

\(\Leftrightarrow(x-1).(x+2)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1(l\text{oại})\\x=-2(TM)\end{cases}}\)

gửi cho 4 câu trc

dài vl

Tại vì nó được đề bài cho nên có nghĩa,k có nghĩa thì lm kiểu đếch j?

Đùa người ak 😡😡😡😡😡😡

ai giúp mình vớiiiii

a, ĐKXĐ:

9x^2 - 16 ≠ 0

=> (3x - 4)(3x + 4) ≠ 0

=> 3x - 4 ≠ 0 và 3x + 4 ≠ 0

=> 3x  ≠ 4 và 3x ≠ -4

=> x ≠ 4/3 hoặc x ≠ -4/3

b, ĐKXĐ:

x^2 - 5x + 6 ≠ 0

=> x^2 - 2x - 3x + 6 ≠ 0

=> x(x - 2) - 3(x - 2) ≠ 0

=> (x - 3)(x - 2) ≠ 0

=> x - 3 ≠ 0 và x - 2 ≠ 0

=> x ≠ 3 và x ≠ 2

c, ĐKXĐ : 

x^2 - 4x + 4 ≠ 0

=> (x - 2)^2 ≠ 0

=> x - 2 ≠ 0

=> x ≠ 2

a) \(\frac{1-x}{x+1}+3=\frac{2x+3}{x+1}\)

<=> 1 - x + 3(x + 1) = 2x + 3

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

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

<=> 4 = 3 (vô lý)

=> pt vô nghiệm

b) ĐKXĐ: \(x\ne1;x\ne2\)

\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)

<=> (x - 2)(2 - x) - 5(x + 1)(2 - x) = 15(x - 2)

<=> 2x - x² - 4 + 2x - 5x - 5x² + 10 = 15x - 30

<=> -x + 4x² - 14 = 15x - 30

<=> x - 4x² + 14 = 15x - 30 

<=> x - 4x² + 14 + 15x - 30 = 0

<=> 16x - 4x² - 16 = 0

<=> 4(4x - x² - 4) = 0

<=> -x² + 4x - 4 = 0

<=> x² - 4x + 4 = 0

<=> (x - 2)² = 0

<=> x - 2 = 0

<=> x = 2 (ktm)

=> pt vô nghiệm 

c) xem bài 4 ở đây: Câu hỏi của gjfkm

d) ĐKXĐ: \(x\ne1;x\ne2;x\ne3\)

\(\frac{x+4}{x^2-3x+2}+\frac{x+1}{x^2-4x+3}=\frac{2x+5}{x^2-4x+3}\)

<=> \(\frac{x+4}{\left(x-1\right)\left(x-2\right)}+\frac{x+1}{\left(x-1\right)\left(x-3\right)}=\frac{2x+5}{\left(x-1\right)\left(x-3\right)}\)

<=> (x + 4)(x - 3) + (x + 1)(x - 2) = (2x + 5)(x - 2)

<=> x² - 3x + 4x - 12 + x² - 2x + x - 2 = 2x² - 4x + 5x - 10

<=> 2x² - 14 = 2x² + x - 10

<=> 2x² - 14 - 2x² = x - 10

<=> -14 = x - 10

<=> -14 + 10 = x

<=> -4 = x

<=> x = -4

Đây là những bài cơ bản mà bạn!

bạn ấy muốn thách xem bạn nào đủ kiên nhẫn đánh hết chỗ này

a) Để \(\frac{2x+3}{4x-5}=0\)

=> 2x + 3 = 0

x = -3/2

b) Để \(\frac{\left(x-1\right)\left(x+2\right)}{x^2-4x+3}=\frac{\left(x-1\right).\left(x+2\right)}{\left(x-3\right).\left(x-1\right)}=\frac{x+2}{x-3}=0\)

=> x + 2 = 0=> x = -2

c) để \(\frac{x^2-1}{x^2-2x+1}=\frac{\left(x-1\right).\left(x+2\right)}{\left(x-1\right)^2}=\frac{x+2}{x-1}=0\)

=> x + 2 = 0 => x = -2 

d) để \(\frac{x^2-4}{x^2+3x-10}=\frac{\left(x+2\right).\left(x-2\right)}{\left(x-2\right).\left(x+5\right)}=\frac{x+2}{x+5}=0\)

=> ...

e) để \(\frac{x^3-16x}{x^3-3x^2-4x}=\frac{x.\left(x-4\right).\left(x+4\right)}{x.\left(x-4\right).\left(x+1\right)}=\frac{x+4}{x+1}=0\)

=> ....

a) \(\frac{x^2-16}{4x-x^2}=\frac{\left(x+4\right)\left(x-4\right)}{x\left(4-x\right)}\)

\(=\frac{\left(x+4\right)\left(x-4\right)}{-x\left(x-4\right)}=\frac{x+4}{-x}\)

b) \(\frac{x^2+4x+3}{2x+6}=\frac{x^2+3x+x+3}{2\left(x+3\right)}\)

\(=\frac{x\left(x+3\right)+\left(x+3\right)}{2\left(x+3\right)}\)

\(=\frac{\left(x+1\right)\left(x+3\right)}{2\left(x+3\right)}=\frac{x+1}{2}\)

c) \(\frac{\left(2x^2+2x\right)\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}\)

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

\(=\frac{2x\left(x-2\right)^2}{x\left(x+2\right)\left(x-2\right)}\)

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

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

d) \(\frac{x^3-x^2y+xy^2}{x^3+y^3}\)

\(=\frac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\frac{x}{x+y}\)

Bài 2: \(a,\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\) 

 \(\frac{5-3x}{x^2-9}=\frac{5-3x}{\left(x-3\right)\left(x+3\right)}=\frac{\left(5-3x\right)2x}{2x\left(x-3\right)\left(x+3\right)}\)

\(b,\frac{x+1}{x-x^2}=\frac{x+1}{x\left(1-x\right)}=-\frac{x+1}{x\left(x+1\right)}=-\frac{2\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)^2}\) 

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

\(c,\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\) 

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

\(\frac{6}{x-1}=\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

\(d,\frac{7}{5x}=\frac{7.2\left(2y-x\right)\left(2y+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)

\(\frac{4}{x-2y}=-\frac{4}{2y-x}=-\frac{4.2.5x\left(2x+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)

\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{2.5x.\left(2y-x\right)\left(2y+x\right)}\)