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

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

\(\Rightarrow x^2+x-12+x^2-x-2=2x^2+x-10\Leftrightarrow x=-4\)

\(\frac{x+4}{2x^2-5x+2}+\frac{x+1}{2x^2-7x+3}=\frac{2x+5}{2x^2-7x+3}\)

\(\Rightarrow\frac{x+4}{2x^2-5x+2}=\frac{2x-5}{2x^2-7x+3}-\frac{x+1}{2x^2-7x+3}\)

\(\Rightarrow\frac{x+4}{2x^2-5x+2}=\frac{x+4}{2x^2-7x+3}\)

TH1:\(x+4\ne0\)

\(\Rightarrow2x^2-5x+2=2x^2-7x+3\)

\(\Rightarrow-5x+2=-7x+3\)

\(\Rightarrow2x=1\)

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

TH2:\(x+4=0\)

\(\Rightarrow x=-4\)

\(\Leftrightarrow\frac{5\left(x+5\right)-3\left(x-3\right)}{15}=\frac{5\left(x+5\right)-3\left(x-3\right)}{\left(x-3\right)\left(x+5\right)}\)

\(\Leftrightarrow\frac{2x+34}{15}=\frac{2x+34}{x^2+2x-15}\Leftrightarrow\orbr{\begin{cases}2x+34=0\\x^2+2x-15=15\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-17\\x^2+2x-30=0\end{cases}}\)

Từ đó tìm được \(S=\left\{-17;\sqrt{31}-1;-\sqrt{31}-1\right\}\)

a) Ta có: x⁴ - x³ + 2x² - x + 1 = 0

=> (x⁴ + 2x² + 1) - x(x² + 1) = 0

=> (x² + 1)² - x(x² + 1) = 0

=> (x² + 1)(x² - x + 1) = 0

=> (x² + 1)[(x² - x + 1/4) + 3/4] = 0

=> (x²+  1 )[(x - 1/2)² + 3/4] = 0

=> pt vô nghiệm (vì x² + 1 > 0; (x - 1/2)² + 3/4 > 0)

b) Ta có: x³ + 2x² - 7x + 4 = 0

=> (x³ - x) + (2x² - 6x + 4) = 0

=> x(x² - 1) + 2(x² - 3x + 2) = 0

=> x(x - 1)(x + 1) + 2(x² - 2x - x + 2) = 0

=> x(x - 1)(x + 1) + 2(x - 2)(x - 1) = 0

=> (x - 1)(x² + x + 2x - 4) = 0

=> (x - 1)(x² + 3x - 4) = 0

=> (x - 1)(x²  + 4x - x - 4) = 0

=> (x - 1)(x + 4)(x - 1) = 0

=> (x - 1)²(x + 4) = 0

=> \(\orbr{\begin{cases}x-1=0\\x+4=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=1\\x=-4\end{cases}}\)

a) \(x^4-x^3+2x^2-x+1=0\)

\(\Leftrightarrow\left(x^4+2x^2+1\right)-x\left(x^2+1\right)=0\)

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

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

\(\Leftrightarrow\left(x^2+1\right)\left[\left(x^2-x+\frac{1}{4}\right)+\frac{3}{4}\right]=0\)

\(\Leftrightarrow\left(x^2+1\right)\left[\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\right]=0\)

Ta có: \(\hept{\begin{cases}x^2+1>0\forall x\\\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\end{cases}}\)

\(\Rightarrow\)Phương trình vô nghiệm

Vậy không có giá trị x thỏa mãn đề bài
 

b) \(x^3+2x^2-7x+4=0\)

\(\Leftrightarrow\left(x^3-x\right)+\left(2x^2-6x+4\right)=0\)

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

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

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

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

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

\(\Leftrightarrow\left(x-1\right)\left[x^2+x+2x-4\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^2+3x-4\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^2+4x-x-4\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left[x\left(x+4\right)-\left(x+4\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)^2\left(x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^2=0\\x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+4=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=-4\end{cases}}}\)

Vậy x=1; x=-4

a) 7x - 35 = 0

<=> 7x = 0 + 35

<=> 7x = 35

<=> x = 5

b) 4x - x - 18 = 0

<=> 3x - 18 = 0

<=> 3x = 0 + 18

<=> 3x = 18

<=> x = 5

c) x - 6 = 8 - x

<=> x - 6 + x = 8

<=> 2x - 6 = 8

<=> 2x = 8 + 6

<=> 2x = 14

<=> x = 7

d) 48 - 5x = 39 - 2x

<=> 48 - 5x + 2x = 39

<=> 48 - 3x = 39

<=> -3x = 39 - 48

<=> -3x = -9

<=> x = 3

có bị viết nhầm thì thông cảm nha!

\(a,ĐKXĐ:x\ne\pm\frac{1}{2}\)

Ta có: \(\frac{2}{2x+1}-\frac{3}{2x-1}=\frac{4}{4x^2-1}\)

\(\Leftrightarrow2\left(2x-1\right)-3\left(2x+1\right)=4\)

\(\Leftrightarrow4x-2-6x-3=4\)

\(\Leftrightarrow-2x=9\)

\(\Leftrightarrow x=-\frac{9}{2}\)(Tm ĐKXĐ)

Vậy pt có nghiệm duy nhất \(x=-\frac{9}{2}\)

\(b,ĐKXĐ:x\ne\pm1;-3\)

Ta có: \(\frac{2x}{x+1}+\frac{18}{x^2+2x-3}=\frac{2x-5}{x+3}\)

\(\Leftrightarrow\frac{2x}{x+1}+\frac{18}{\left(x-1\right)\left(x+3\right)}=\frac{2x-5}{x+3}\)

\(\Leftrightarrow2x\left(x-1\right)\left(x+3\right)+18\left(x+1\right)=\left(2x-5\right)\left(x-1\right)\left(x+1\right)\)

\(\Leftrightarrow2x\left(x^2+2x-3\right)+18x+18=\left(2x-5\right)\left(x^2-1\right)\)

\(\Leftrightarrow2x^3+4x^2-6x+18x+18=2x^3-2x-5x^2+5\)

\(\Leftrightarrow9x^2+14x+13=0\)

\(\Leftrightarrow\left(9x^2+14x+\frac{49}{9}\right)+\frac{68}{9}=0\)

\(\Leftrightarrow\left(3x+\frac{7}{3}\right)^2+\frac{68}{9}=0\)

Pt vô nghiệm 

\(c,ĐKXĐ:x\ne1\)

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

\(\Leftrightarrow x^2+x+1+2x^2-5=x-1\)

\(\Leftrightarrow3x^2=3\)

\(\Leftrightarrow x^2=1\)

\(\Leftrightarrow x=\pm1\)

Kết hợp vs ĐKXĐ được x = -1

Vậy pt có nghiệm duy nhất x = -1

làm lần lượt nha(bài nào k bt bỏ qua)

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

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

\(\Rightarrow-2x-5=4\)

\(\Rightarrow-2x=9\)

\(\Rightarrow x=\frac{9}{-2}\)

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

\(\Rightarrow\frac{3}{x-5}=-1\)

\(\Rightarrow3=-x+5\)

\(\Leftrightarrow x+3=5\)

\(\Rightarrow x=2\)