\(\frac{5x^2-30x+50+3}{x^2-6x+10}=\frac{5\left(x^2-6x+10\right)}{x^2-6x+10}+\frac{3}{\left(x-3\right)^2+1}=5+\frac{3}{\left(x-3\right)^2+1}>5\)

Bài 1:

\(D=\dfrac{5x^2-30x+53}{x^2-6x+10}=\dfrac{5\left(x^2-6x+10\right)+3}{x^2-6x+10}=5+\dfrac{3}{x^2-6x+10}\)

\(=5+\dfrac{3}{\left(x-3\right)^2+1}\)

Ta có: \(\left(x+3\right)^2+1\ge1\Rightarrow\dfrac{3}{\left(x-3\right)^2+1}\le3\)

\(\Rightarrow D\le3+5=8\)

Vậy max D= 8 <=> x=3

Bài 2: 

\(8\left(x-3\right)^3+x^3=6x^2-12x+8\)

\(\Leftrightarrow\left[2\left(x-3\right)^3\right]=-x^3+3.2x^2-3.2^2x+2^3\)

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

\(\Leftrightarrow2x-6=2-x\)

\(\Leftrightarrow3x=8\Leftrightarrow x=\dfrac{8}{3}\)

Vậy tập nghiệm : \(S=\left\{\dfrac{8}{3}\right\}\)

a/ ĐK: $x\ne -5$

$\dfrac{6x^2+30x}{4}=\dfrac{6x(x+5)}{4}=\dfrac{3x(x+5)}{2}$ 

Đề này sai

b/ ĐK: $x\ne \pm 1$

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

$\to$ ĐPCM

Câu a sai đề nhé.

a, Xét \(VT=\dfrac{3x\left(x+5\right)}{2\left(x+5\right)}=\dfrac{3x}{2}\)

\(VP=\dfrac{6x^2+30x}{4}=\dfrac{6x\left(x+5\right)}{4}=\dfrac{3x\left(x+5\right)}{2}\)

Vậy \(VT\ne VP\)hay đpcm ko xảy ra 

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

Vậy ta có đpcm 

\(x=\dfrac{3+\sqrt{5}}{2}\Rightarrow2x-3=\sqrt{5}\Rightarrow4x^2-12x+9=5\)

\(\Rightarrow4x^2-12x+4=0\Rightarrow x^2-3x+1=0\)

\(\Rightarrow P=\left[10\left(x^2-3x+1\right)+1\right]^2+\dfrac{\left[2\left(x^2-3x+1\right)+1\right]^{10}}{x^3\left(x^2-3x+1\right)-1}=1^2+\dfrac{1^2}{0-1}=...\)

a:=>3x=15

=>x=5

b: =>8-11x<52

=>-11x<44

=>x>-4

c: \(VT=\left(\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2x-6}+\dfrac{x}{6-x}\)

\(=\dfrac{12x-36}{2x-6}\cdot\dfrac{1}{x-6}-\dfrac{x}{x-6}=\dfrac{6}{x-6}-\dfrac{x}{x-6}=-1\)

a ) \(\dfrac{x^2+3x+2}{3x+6}=\dfrac{\left(x+1\right)\left(x+2\right)}{3\left(x+2\right)}=\dfrac{x+1}{3}\) (1)

\(\dfrac{2x^2+x-1}{6x-3}=\dfrac{\left(2x-1\right)\left(x+1\right)}{3\left(2x-1\right)}=\dfrac{x+1}{3}\) (2)

Từ (1) ; (2) \(\Rightarrow\dfrac{x^2+3x+2}{3x+6}=\dfrac{2x^2+x-1}{6x-3}\) (đpcm)

b ) \(\dfrac{15x-10}{3x^2+3x-\left(2x+2\right)}=\dfrac{5\left(3x-2\right)}{\left(3x-2\right)\left(x+1\right)}=\dfrac{5}{x+1}\) (3)

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

Từ (3) và (4) \(\Rightarrow\dfrac{15x-10}{3x^2+3x-\left(2x+2\right)}=\dfrac{5x^2-5x+5}{x^3+1}\) (đpcm)

a) \(\dfrac{x^2+3x+2}{3x+6}=\dfrac{x^2+x+2x+2}{3\left(x+2\right)}=\dfrac{\left(x^2+x\right)+\left(2x+2\right)}{3\left(x+2\right)}=\dfrac{x\left(x+1\right)+2\left(x+1\right)}{3\left(x+2\right)}=\dfrac{\left(x+1\right)\left(x+2\right)}{3\left(x+2\right)}=\dfrac{x+1}{3}\left(1\right)\) \(\dfrac{2x^2+x-1}{6x-3}=\dfrac{2x^2+2x-x-1}{3\left(2x-1\right)}=\dfrac{2x\left(x+1\right)-\left(x+1\right)}{3\left(2x-1\right)}=\dfrac{\left(2x-1\right)\left(x+1\right)}{3\left(2x-1\right)}=\dfrac{x+1}{3}\left(2\right)\) Từ (1)và (2)=> \(\dfrac{x^2+3x+2}{3x+6}=\dfrac{2x^2+x-1}{6x-3}\) b)\(\dfrac{15x-10}{3x^2+3x-\left(2x+2\right)}=\dfrac{5\left(3x-2\right)}{3x\left(x+1\right)-2\left(x+1\right)}=\dfrac{5\left(3x-2\right)}{\left(3x-2\right)\left(x+1\right)}=\dfrac{5}{x+1}\left(3\right)\) \(\dfrac{5x^2-5x+5}{x^3+1}=\dfrac{5\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{5}{x+1}\left(4\right)\) Từ (3) và (4) => \(\dfrac{15x-10}{3x^2+3x-\left(2x+2\right)}=\dfrac{5x^2-5x+5}{x^3+1}\)

a, \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)

\(\Leftrightarrow\left(\dfrac{59-x}{49}+1\right)+\left(\dfrac{57-x}{43}+1\right)+\left(\dfrac{55-x}{45}+1\right)+\left(\dfrac{53-x}{47}+1\right)+\left(\dfrac{51-x}{49}+1\right)=0\)

\(\Leftrightarrow\dfrac{100-x}{45}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}+\dfrac{100-x}{49}=0\)

\(\Leftrightarrow\left(100-x\right).\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\right)=0\)

Mà \(\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\right)\ne0\)

\(\Rightarrow100-x=0\)

\(\Rightarrow x=100\)

Vậy \(S=\left\{100\right\}\)

b, \(6x^2-5x+3=2x-3x\left(3-2x\right)\)

\(\Leftrightarrow6x^2-5x+3=2x-9x+6x^2\)

\(\Leftrightarrow6x^2-5x+3=-7x+6x^2\)

\(\Leftrightarrow6x^2-5x+3+7x-6x^2=0\)

\(\Leftrightarrow2x+3=0\)

\(\Leftrightarrow2x=-3\)

\(\Leftrightarrow x=\dfrac{-3}{2}\)

Vậy \(S=\left\{\dfrac{-3}{2}\right\}\)

c,\(\dfrac{2\left(x-4\right)}{4}-\dfrac{3+2x}{10}=x+\dfrac{1-x}{5}\)

\(\Leftrightarrow\dfrac{10x-40}{20}-\dfrac{6+4x}{20}=\dfrac{20x}{20}+\dfrac{4-4x}{20}\)

\(\Leftrightarrow\dfrac{6x-46}{20}=\dfrac{16x+4}{20}\)

\(\Leftrightarrow6x-46=16x+4\)

\(\Leftrightarrow6x-46-16x-4=0\)

\(\Leftrightarrow-10x-50=0\)

\(\Leftrightarrow-10x=50\)

\(\Leftrightarrow x=-5\)

Vậy \(S=\left\{-5\right\}\)

\(=5x^2-3x-x^3+x^2+x^3-6x^2-10+3x=0\)