`Answer:`

\(\frac{\left(m^2+1\right).x+1-2m^2}{x-5}\)\(=2m\left(ĐKXĐ:x\ne5\right)\)

`=>(m^2+1).x+1-2m^2=2m.(x-5)`

`<=>(m^2-2m+1).x=2m^2-10m-1`

`<=>(m-1)^2``x=2m^2-10m-1(1)`

Phương trình `(1)` có nghiệm duy nhất `<=>m\ne1`

Lúc đó nghiệm của phương trình `(1)` sẽ là \(x=\frac{2m^2-10m-1}{\left(m-1\right)^2}\)

Ta xét hiệu:

 \(x-3=\frac{2m^2-10m-1}{\left(m-1\right)^2}-3\)

\(=\frac{2m^2-10m-1-3\left(m-1\right)^2}{\left(m-1\right)^2}\)

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

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

\(\le0;\forall m\ne1\)

\(\Rightarrow x-3\le0;\forall m\ne1\)

\(\Rightarrow x\le3;\forall m\ne1\)

Dấu "=" xảy ra khi `m+2=0<=>m=-2`

tt

liuliu

\(\dfrac{x-2}{40}-\dfrac{2-x}{102}+1=1-\dfrac{2x-4}{2022}\\ \Leftrightarrow\dfrac{x-2}{40}+\dfrac{x-2}{102}+\dfrac{x-2}{1011}=0\\ \Leftrightarrow\left(x-2\right)\left(\dfrac{1}{40}+\dfrac{1}{102}+\dfrac{1}{1011}\right)=0\\ \Leftrightarrow x-2=0\left(vì.\dfrac{1}{40}+\dfrac{1}{102}+\dfrac{1}{1011}\ne0\right)\\ \Leftrightarrow x=2\)

\(\Leftrightarrow4x^2-4xy+4y^2=12\)

\(\Leftrightarrow\left(2x-y\right)^2+3y^2=12\)

\(\Leftrightarrow3y^2=12-\left(2x-y\right)^2\le12\)

\(\Rightarrow y^2\le4\Rightarrow y^2=\left\{0;1;4\right\}\)

\(\Rightarrow y=\left\{-2;-1;0;1;2\right\}\)

Thế vào pt ban đầu tính x tương ứng

a, \(\Rightarrow x-5\ne0\Leftrightarrow x\ne5\)

b, mà \(\left(5x-3\right)^2\ge0\forall x\)

 \(\Rightarrow5x-3=0\Leftrightarrow x=\dfrac{3}{5}\)

(x + 1) + (x + 2)  + (x + 3) = 4x

x + 1 + x + 2 + x + 3 = 4x

(x + x + x) + (1 + 2 + 3) = 4x

3x + 6 = 4x

3x - 4x = -6

- x = -6

   x = \(\frac{-6}{-1}\)

   x = 6

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

\(\Leftrightarrow12.\left(\frac{x}{3}-x+1\right)=12.\left(\frac{x}{4}-\frac{x+1}{3}\right)\)

\(\Leftrightarrow4x-12x+12=3x-4.\left(x+1\right)\)

\(\Leftrightarrow4x-12x+12=3x-4x-4\)

\(\Leftrightarrow4x-12x-3x+4x=\left(-12\right)-4\)

\(\Leftrightarrow\left(-7\right)x=-16\)

\(\Leftrightarrow x=\frac{16}{7}\)

Vậy \(x=\frac{16}{7}\)

\(4x-12x+12=3x-4\left(x+1\right)\Leftrightarrow-8x+12=-x-4\)

\(\Leftrightarrow-7x=-16\Leftrightarrow x=\dfrac{16}{7}\)