\(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\Leftrightarrow\frac{x+1}{35}+1+\frac{x+3}{33}+1=\frac{x+5}{31}+1+\frac{x+7}{29}+1\)

\(\frac{x+36}{35}+\frac{x+36}{33}=\frac{x+36}{31}+\frac{x+36}{29}\Leftrightarrow\left(x+36\right)\left(\frac{1}{35}+\frac{1}{33}-\frac{1}{31}-\frac{1}{29}\right)=0\)

mà \(\frac{1}{35}+\frac{1}{33}-\frac{1}{31}-\frac{1}{29}\ne0\) vậy \(x+36=0\Rightarrow x=-36\)

\(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)

\(\Leftrightarrow\frac{x+1}{35}+1+\frac{x+3}{33}+1=\frac{x+5}{31}+1+\frac{x+7}{29}+1\)

\(\Leftrightarrow\frac{x+36}{35}+\frac{x+36}{33}-\frac{x+36}{31}-\frac{x+36}{29}=0\)

\(\Leftrightarrow\left(x+36\right)\left(\frac{1}{35}+\frac{1}{33}-\frac{1}{31}-\frac{1}{29}\ne0\right)=0\)

\(\Leftrightarrow x=-36\)

\(\frac{\left(x-2\right)\left(x+10\right)}{3}-\frac{\left(x+4\right)\left(x+10\right)}{12}=\frac{\left(x-2\right)\left(x+4\right)}{4}\)

<=> \(\frac{4\left(x^2+10x-2x-20\right)-\left(x^2+10x+4x+40\right)}{12}=\frac{3\left(x^2+4x-2x-8\right)}{12}\)

=> \(4x^2+40x-8x-80-x^2-10x-4x-40=3x^2+12x-6x-24\)

<=> \(4x^2+40x-8x-80-x^2-10x-4x-40-3x^2-12x+6x+24=0\)

<=> \(12x-96=0\)

<=>\(12x=96\)

<=>\(x=8\)

để \(\frac{7}{x^2-x+1}\in Z\Leftrightarrow x^2-x+1\inƯ_7=\left\{\pm1;\pm7\right\}\)

nếu \(x^2-x+1=-7\Leftrightarrow x^2-x+8=0\left(vo nghiem\right)\)

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

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

nếu \(x^2-x+1=7\Leftrightarrow x^2-x-6=0\Leftrightarrow\hept{\begin{cases}x=3\\x=-2\end{cases} }\)

vậy \(x\in\left\{-2,0,1,3\right\}\)

Để \(\frac{7}{x^2-x+1}\)ta có : \(x^2-x+1=x^2-x+\frac{1}{4}+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\)

hay \(7⋮\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

Xét từng trường hợp : 

TH1 : \(\left(x-\frac{1}{2}\right)^2+\frac{3}{4}=1\Leftrightarrow\left(x-\frac{1}{2}\right)^2=\frac{1}{4}\Leftrightarrow x-\frac{1}{2}=\pm\frac{1}{2}\)

\(\Leftrightarrow x_1=\frac{1}{2}+\frac{1}{2}=1;x_2=-\frac{1}{2}+\frac{1}{2}=0\)( chọn )

TH2 : \(\left(x-\frac{1}{2}\right)^2+\frac{3}{4}=-1\Leftrightarrow\left(x-\frac{1}{2}\right)^2=-\frac{7}{4}\)ko thỏa mãn 

tương tự 2 trường hợp còn lại 

\(\frac{2\left(x-4\right)}{3}+\frac{4\left(x+3\right)-x+1}{8}=\frac{3\left(2x-3\right)}{5}-7\)

<=> \(\frac{2x-8}{3}+\frac{4x+12-x+1}{8}-\frac{6x-9}{5}+7=0\)

<=> \(\frac{40\left(2x-8\right)}{120}+\frac{15\left(3x+13\right)}{120}-\frac{24\left(6x-9\right)}{120}+\frac{840}{120}=0\)

<=> \(\frac{80x-320}{120}+\frac{45x+195}{120}-\frac{144x-216}{120}+\frac{840}{120}=0\)

<=> \(\frac{80x-320+45x+195-144x+216+840}{120}=0\)

<=> \(\frac{-19x+931}{120}=0\)

<=> -19x + 931 = 0

<=> -19x = -931

<=> x = 49

{=}-19x+931=0

{=}-19x=-931

{=}x=49

( x - 3 )³ - 2( x - 1 ) = x( x - 2 )² - 5x²

<=> x³ - 9x² + 27x - 27 - 2x + 2 = x( x² - 4x + 4 ) - 5x²

<=> x³ - 9x² + 25x - 25 = x³ - 4x² + 4x - 5x²

<=> x³ - 9x² + 25x - 25 - x³ + 4x² - 4x + 5x² = 0

<=> 21x - 25 = 0

<=> 21x = 25

<=> x = 25/21

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

\(\Leftrightarrow x^3-9x^2+27x-27-2x+2=x\left(x^2-4x+4\right)-5x^2\)

\(\Leftrightarrow x^3-9x^2+25x-25=x^3-4x^2+4x-5x^2\)

\(\Leftrightarrow x^3-9x^2+25x-25-x^3+4x^2-4x+5x^2=0\)

\(\Leftrightarrow21x-25=0\Leftrightarrow x=\frac{25}{21}\)

x = 38/11 nha

h đúng cho mk nha

Tham khảo câu b xem

https://olm.vn/hoi-dap/detail/248753766595.html

Thêm điều kiện a ; b ; c dương nhé !!! 

\(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\)

\(\frac{a^2}{b^2}-\frac{a}{b}+\frac{b^2}{c^2}-\frac{b}{c}+\frac{c^2}{a^2}-\frac{c}{a}\ge0\)

\(\left(\frac{a^2}{b^2}-\frac{a}{b}+1\right)+\left(\frac{b^2}{c^2}-\frac{b}{c}+1\right)+\left(\frac{c^2}{a^2}-\frac{c}{a}+1\right)-3\ge0\)

\(\left(\frac{a}{b}-1\right)^2+\left(\frac{b}{c}-1\right)^2+\left(\frac{c}{a}-1\right)^2\ge3\)

hiển nhiên đúng vì 

\(\left(\frac{a}{b}-1\right)^2\ge0;\left(\frac{b}{c}-1\right)^2\ge0;\left(\frac{c}{a}-1\right)^2\ge0\)

mặt khác  \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge3\sqrt[3]{\frac{a}{b}\frac{b}{c}\frac{c}{a}}=3\)( cô si 3 số ) 

Dấu ''='' xảy ra <=> a = b = c ( đpcm )

quên hằng :< 

\(\frac{a^2}{b^2}-\frac{a}{b}+\frac{b^2}{c^2}-\frac{b}{c}+\frac{c^2}{a^2}-\frac{c}{a}\ge0\)

\(\left(\frac{a^2}{b^2}-2\frac{a}{b}+1\right)+\left(\frac{b^2}{c^2}-2\frac{b}{c}+1\right)+\left(\frac{c^2}{a^2}-2\frac{c}{a}+1\right)+\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)-3\ge0\)

\(\left(\frac{a}{b}-1\right)^2\ge0+\left(\frac{b}{c}-1\right)^2\ge0+\left(\frac{c}{a}+1\right)^2\ge0+\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\ge3\)

mà \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge3\sqrt[3]{\frac{a}{b}\frac{b}{c}\frac{c}{a}}=3\)( cô si 3 số ) 

Dấu ''='' xảy ra <=> a = b = c 

\(\frac{3x}{x^2-64}.\frac{x-8}{x^2-x}=\frac{3x\left(x-8\right)}{x\left(x-8\right)\left(x+8\right)\left(x-1\right)}\)

\(=\frac{3}{\left(x+8\right)\left(x-1\right)}\)