\(x\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\)

<=> \(\left[x\left(x+1\right)\right]\left[\left(x-1\right)\left(x+2\right)\right]-24=0\)

<=> \(\left(x^2+x\right)\left(x^2+2x-x-2\right)-24=0\)

<=> \(\left(x^2+x\right)\left(x^2+x-2\right)-24=0\)

Đặt t = x² + x 

<=> t(t - 2) - 24 = 0

<=> t² - 2t - 24 = 0

<=> t² - 6t + 4t - 24 = 0

<=> (t + 4)(t - 6) = 0

<=> \(\orbr{\begin{cases}x^2+x+4=0\\x^2+x-6=0\end{cases}}\)

<=> \(\orbr{\begin{cases}\left(x^2+x+\frac{1}{4}\right)+\frac{15}{4}=0\\x^2+3x-2x-6=0\end{cases}}\)

<=> \(\orbr{\begin{cases}\left(x+\frac{1}{2}\right)^2+\frac{15}{4}=0\left(ktm\right)\\\left(x-2\right)\left(x+3\right)=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)

Vậy S = {2; -3}

(lưu ý: thay "ktm" thành vô lý và giải thích thêm)

\(\left(x+3\right)^4+\left(x+5\right)^4=2\)

<=> (x + 4 - 1)⁴ + (x + 4 + 1)⁴ - 2 = 0

Đặt y = x + 4

<=> (y - 1)⁴ + (y + 1)⁴ - 2 = 0

<=> y⁴ - 4y³ + 6y² - 4y + 1 + y⁴ + 4y³ + 6y² + 4y + 1 - 2 = 0

<=> 2y⁴ + 12y² = 0

<=> 2y²(y² + 6) = 0

<=> \(\orbr{\begin{cases}y^2=0\\y^2+6=0\left(ktm\right)\end{cases}}\)

<=> y = 0

<=> x + 4 = 0

<=> x = -4

Vậy S = {-4}

\(\frac{x^2+x+4}{2}+\frac{x^2+x+7}{3}=\frac{x^2+x+13}{5}+\frac{x^2+x+16}{6}\)

<=> \(\frac{x^2+x+4}{2}-3+\frac{x^2+x+7}{3}-3=\frac{x^2+x+13}{5}-3+\frac{x^2+x+16}{6}-3\)

<=> \(\frac{x^2+x+4-6}{2}+\frac{x^2+x+7-9}{3}=\frac{x^2+x+13-15}{5}+\frac{x^2+x+16-18}{6}\)

<=> \(\frac{x^2+x-2}{2}+\frac{x^2+x-2}{3}=\frac{x^2+x-2}{5}+\frac{x^2+x-2}{6}\)

<=> \(\left(x^2+2x-x-2\right)\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{5}-\frac{1}{6}\right)=0\)

<=> (x + 2)(x - 1) = 0 (do \(\frac{1}{2}+\frac{1}{3}-\frac{1}{5}-\frac{1}{6}\ne0\))

<=> \(\orbr{\begin{cases}x+2=0\\x-1=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-2\\x=1\end{cases}}\)

Vậy S = {-2; 1}

câu cuối: + 3 vào sau các phân số của pt như trên

\(a)\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}=\frac{-3}{4}\left(x\ne-3;x\ne2\right)\)

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

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

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

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

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

\(\Leftrightarrow\frac{x-4}{x-2}=\frac{-3}{4}\)

<=> 4x-16=-3x+6

<=> 4x-16+3x-6=0

<=> 7x-22=0

<=> 7x=22

<=> \(x=\frac{22}{7}\)(TMĐK)
 

\(x\ne\pm2\)

Đặt \(\left\{{}\begin{matrix}\frac{x+3}{x-2}=a\\\frac{x-3}{x+2}=b\end{matrix}\right.\) phương trình trở thành:

\(a^2+6b^2=7ab\)

\(\Leftrightarrow a^2-7ab+6b^2=0\)

\(\Leftrightarrow a^2-ab-6ab+6b^2=0\)

\(\Leftrightarrow a\left(a-b\right)-6b\left(a-b\right)=0\)

\(\Leftrightarrow\left(a-6b\right)\left(a-b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=6b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\frac{x+3}{x-2}=\frac{x-3}{x+2}\\\frac{x+3}{x-2}=\frac{6\left(x-3\right)}{x+2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x+3\right)\left(x+2\right)=\left(x-3\right)\left(x-2\right)\\\left(x+3\right)\left(x+2\right)=6\left(x-3\right)\left(x-2\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=-5x\\x^2-7x+6=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=6\end{matrix}\right.\)

a) \(\frac{x+5}{4}-\frac{2x-3}{3}=\frac{6x-1}{8}+\frac{2x-1}{12}\)

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

<=> \(-\frac{4}{3}x=-\frac{59}{24}\)

<=> \(x=\frac{59}{32}\)

Vậy S = { 59/32}

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

<=> \(\frac{x^2+14x+40}{12}-\frac{-x^2-2x+8}{4}=\frac{x^2+8x-20}{3}\)

<=> \(\left(\frac{x^2}{12}+\frac{x^2}{4}-\frac{x^2}{3}\right)+\left(\frac{14}{12}x+\frac{2}{4}x-\frac{8}{3}x\right)=-\frac{20}{8}+\frac{8}{4}-\frac{40}{12}\)

<=> \(-x=-8\)

<=> x = 8 

Vậy S = { 8 }

Cái này bạn đặt x+3/x-2 = a 

x-3/x+2 = b

=> x^2-9/x^2-4 = ab

Ta có : a^2 - 7ab + 6b^2 = 0

<=> a^2 - 6ab - ab + 6b^2 = 0

PT đa thức thành nhân tử là xong :D 

a)Ta có: \(\frac{-2}{5}+\frac{6}{5}.\left(y-\frac{2}{3}\right)=\frac{-4}{15}\)

\(\Rightarrow\frac{6}{5}.\left(y-\frac{2}{3}\right)=\frac{-4}{15}-\frac{-2}{15}\)

\(\Rightarrow\frac{6}{5}.\left(y-\frac{2}{3}=\right)\frac{-2}{5}\)

\(\Rightarrow y-\frac{2}{3}=\frac{-2}{5}:\frac{6}{5}=\frac{-1}{3}\)

\(\Rightarrow y=\frac{-1}{3}+\frac{2}{3}=\frac{1}{3}\)

Vậy x = \(\frac{1}{3}\)

b) Ta có: \(\frac{-2}{5}+\frac{2}{3}x+\frac{1}{6}x=\frac{-4}{15}\)

        \(\Rightarrow\frac{-2}{5}+x.\left(\frac{2}{3}+\frac{1}{6}\right)=\frac{-4}{15}\)

        \(\Rightarrow x.\frac{5}{6}=\frac{-4}{15}-\frac{-2}{15}\)

         \(x.\frac{5}{6}=\frac{-2}{15}\)

\(\Rightarrow x=\frac{-2}{15}:\frac{5}{6}=\frac{-4}{25}\)

Vậy x = \(\frac{-4}{25}\)

c) Ta có: \(\frac{3}{2}x+\frac{-2}{5}-\frac{2}{3}.x=\frac{-4}{15}\)

\(\Rightarrow\frac{3}{2}x-\frac{2}{3}x+\frac{-2}{5}=\frac{-4}{15}\)

\(\Rightarrow x.\left(\frac{3}{2}-\frac{2}{4}\right)=\frac{-4}{15}-\frac{-2}{15}\)

\(\Rightarrow x.\frac{5}{6}=\frac{-2}{15}\)

\(\Rightarrow x=\frac{-2}{15}:\frac{5}{6}=\frac{-4}{25}\)

Vậy x = \(\frac{-4}{25}\)

Ủng hộ tớ nha m.n

a) \(\hept{\begin{cases}3\left(x+1\right)+2\left(x+2y\right)=4\\4\left(x+1\right)-\left(x+2y\right)=9\end{cases}}\Leftrightarrow\hept{\begin{cases}3\left(x+1\right)+2\left(x+2y\right)=4\\8\left(x+1\right)-2\left(x+2y\right)=18\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}11\left(x+1\right)=22\\3\left(x+1\right)+2\left(x+2y\right)=4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\4y+8=4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-1\end{cases}}\)

b) ĐK : y khác 0

\(\hept{\begin{cases}x+\frac{1}{y}=-\frac{1}{2}\\2x-\frac{3}{y}=-\frac{7}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}3x+\frac{3}{y}=-\frac{3}{2}\\2x-\frac{3}{y}=-\frac{7}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}5x=-5\\3x+\frac{3}{y}=-\frac{3}{2}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=-1\\-3+\frac{3}{y}=-\frac{3}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\\frac{3}{y}=\frac{3}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\y=2\left(tm\right)\end{cases}}\)

c) Ta có: \(\left\{{}\begin{matrix}\dfrac{x+2}{x+1}+\dfrac{2}{y-2}=6\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x+1}+\dfrac{10}{y-2}=25\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{11}{y-2}=22\\\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-2=\dfrac{1}{2}\\\dfrac{1}{x+1}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1=1\\y-2=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{5}{2}\end{matrix}\right.\)