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\(\Leftrightarrow\frac{2x}{x^2-3x+12}+\frac{6x}{x^2+2x+12}=1\)

\(\Leftrightarrow\frac{2}{x+\frac{12}{x}-3}+\frac{6}{x+\frac{12}{x}+2}=1\)

Đặt \(x+\frac{12}{x}-3=t\)

\(\Rightarrow\frac{2}{t}+\frac{6}{t+5}=1\Leftrightarrow2\left(t+5\right)+6t=t\left(t+5\right)\)

\(\Leftrightarrow t^2-3t-10=0\Rightarrow\left[{}\begin{matrix}t=5\\t=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{12}{x}-3=-2\\x+\frac{12}{x}-3=5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-x+12=0\\x^2-8x+12=0\end{matrix}\right.\) (casio)

a) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

Vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne=\)

Nên x + 1 = 0 => x = -1

b) \(\frac{x+1}{14}+\frac{x+2}{13}=\frac{x+3}{12}+\frac{x+4}{11}\)

\(\Leftrightarrow\frac{x+1}{14}+1+\frac{x+2}{13}+1=\frac{x+3}{12}+1+\frac{x+4}{11}+1\)

\(\Leftrightarrow\frac{x+15}{14}+\frac{x+15}{13}=\frac{x+15}{12}+\frac{x+15}{11}\)

\(\Leftrightarrow\frac{x+15}{14}+\frac{x+15}{13}-\frac{x+15}{12}-\frac{x+15}{11}=0\)

\(\Leftrightarrow\left(x+15\right)\left(\frac{1}{14}+\frac{1}{13}-\frac{1}{12}-\frac{1}{11}\right)=0\)

Vì \(\frac{1}{14}+\frac{1}{13}-\frac{1}{12}-\frac{1}{11}\ne0\)

Nên x  +15 = 0 => x = -15

a,\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Rightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)=\left(x+1\right).\left(\frac{1}{13}+\frac{1}{14}\right)\)

\(\Rightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)-\left(x+1\right).\left(\frac{1}{13}+\frac{1}{14}\right)=0\)

\(\Rightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

Vì \(\frac{1}{10}>\frac{1}{13};\frac{1}{11}>\frac{1}{14}\Rightarrow\frac{1}{10}+\frac{1}{11}>\frac{1}{13}+\frac{1}{14}\Rightarrow\frac{1}{10}+\frac{1}{11}+\frac{1}{12}>\frac{1}{13}+\frac{1}{14}\)

\(\Rightarrow\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}>0\)

\(\Rightarrow x+1=0\Rightarrow x=-1\)

b, Bạn cộng thêm 1 vào \(\frac{x+1}{14};\frac{x+1}{13};\frac{x+1}{12};\frac{x+1}{11}\)Mội bên phân số 1 đơn vị rồi áp dụng như bài 1

\(a)\) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Leftrightarrow\)\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

\(\Leftrightarrow\)\(\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

Vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\)

Nên \(x+1=0\)

\(\Rightarrow\)\(x=-1\)

Vậy \(x=-1\)

Chúc bạn học tốt ~ 

\(pt\Leftrightarrow\frac{2x}{x^2-3x+12}+\frac{6x}{x^2+2x+12}=1\)

\(\Leftrightarrow\frac{2}{x-3+\frac{12}{x}}+\frac{6}{x+2+\frac{12}{x}}=1\)

Đặt \(x+\frac{12}{x}=t\)

Khi đó:

\(pt\Leftrightarrow\frac{2}{t-3}+\frac{6}{t+2}=1\Leftrightarrow2t+4+6t-18=t^2-t-6\)

\(\Leftrightarrow t^2-t-6=8t-14\)

\(\Leftrightarrow t^2-9t+8=0\)

\(\Leftrightarrow\left(t-8\right)\left(t-1\right)=0\)

\(\Leftrightarrow x+\frac{12}{x}=8;x+\frac{12}{x}=1\)

Thôi,bí rồi

Sao khó vậy? Mình còn chưa học.

Cậu cộng cả 2 pt zao

\(\int^{x\left(\frac{1}{y}-\frac{1}{y+12}\right)=1}_{x\left(\frac{1}{y-12}-\frac{1}{y}\right)=2}\Leftrightarrow\int^{\frac{1}{y}-\frac{1}{y+12}=\frac{1}{x}}_{\frac{1}{y-12}-\frac{1}{y}=\frac{2}{x}}\Leftrightarrow\int^{\frac{2}{y}-\frac{2}{y+12}=\frac{2}{x}\left(1\right)}_{\frac{1}{y-12}-\frac{1}{y}=\frac{2}{x}\left(2\right)}\)

Lấy vế trừ vế của pt (1) và (2) ta có 

\(\frac{2}{y}-\frac{2}{y+12}-\frac{1}{y-12}+\frac{1}{y}=0\)

\(\Leftrightarrow\frac{3}{y}-\frac{2}{y+12}-\frac{1}{y-12}=0\Leftrightarrow3\left(y+12\right)\left(y-12\right)-2y\left(y-12\right)-y\left(y+12\right)=0\)

Rút gọn giải pt bậc 2 sau thay trở lại tìm x