Ta có :\(\frac{6}{x^2+2}+\frac{12}{x^2+8}=3-\frac{7}{x^2+3}\)

\(\Leftrightarrow\frac{6}{x^2+2}+\frac{12}{x^2+8}+\frac{7}{x^2+3}=3\)

\(\Leftrightarrow\left(\frac{6}{x^2+2}-1\right)+\left(\frac{12}{x^2+8}-1\right)+\left(\frac{7}{x^2+3}-1\right)=0\)

\(\Leftrightarrow\frac{4-x^2}{x^2+2}+\frac{4-x^2}{x^2+8}+\frac{4-x^2}{x^2+3}=0\)

\(\Leftrightarrow\left(4-x^2\right)\left(\frac{1}{x^2+2}+\frac{1}{x^2+8}+\frac{1}{x^2+3}\right)=0\)

Ta thấy : \(\frac{1}{x^2+2}+\frac{1}{x^2+8}+\frac{1}{x^2+3}\ne0\forall x\)

Do đó : \(4-x^2=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\) ( thỏa mãn )

Vậy : \(x\in\left\{-2,2\right\}\)

Câu hỏi của Trang - Toán lớp 7 | Học trực tuyến

ta thấy 6-2=4, 12-8=4, 7-3=4 nên 

6/(x^2+2)-1+12/(x^2+8)-1+7/(x^2+3-)=0

<=>(4-x^2)(1/(x^2+2)+1/(x^2+8)+1/(x^2+3))=0

=> x= 2 hoặc x=-2

Bạn làm rõ được không ạ , thiếu 3 mà bạn

\(\frac{6}{x^2+2}+\frac{12}{x^2+8}+\frac{7}{x^2+3}=3\)

\(\Leftrightarrow\left(\frac{6}{x^2+2}-1\right)+\left(\frac{12}{x^2+8}-1\right)+\left(\frac{7}{x^2+3}-1\right)=0\)

đến đay làm như trên

#)Giải :

a) x + 2x + 3x + ... + 100x = - 213

=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213 

=> 100x + 5049 = - 213 

<=> 100x = - 5262

<=> x = - 52,62

#)Giải :

b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)

\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{3}+\frac{1}{6}\)

\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{2}\)

\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}\right)x=\frac{1}{2}\)

\(\Rightarrow\frac{3}{4}x=\frac{1}{2}\)

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

a) x + 2x + 3x + ... +100x = -213

=>  x . (1 + 2 + 3 +... + 100) = - 213

=> x . 5050 = -213

=> x           = - 213 : 5050

=> x           = -213/5050

b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)

=> \(\frac{1}{2}x-\frac{1}{4}x=\frac{1}{3}-\frac{1}{6}\)

=> \(x.\left(\frac{1}{2}-\frac{1}{4}\right)=\frac{1}{6}\)

=> \(x.\frac{1}{4}=\frac{1}{6}\)

=> \(x=\frac{1}{6}:\frac{1}{4}\)

=> \(x=\frac{2}{3}\)

c) 3(x-2) + 2(x-1) = 10

=> 3x - 6 + 2x - 2 = 10

=> 3x + 2x - 6 - 2 = 10

=> 5x - 8 = 10

=> 5x = 10 + 8

=> 5x = 18

=> x = 18:5

=> x = 3,6

d) \(\frac{x+1}{3}=\frac{x-2}{4}\)

=> \(4\left(x+1\right)=3\left(x-2\right)\)

=>\(4x+4=3x-6\)

=> \(4x-3x=-4-6\)

=> \(x=-10\)

Đặt \(x^2+3=t\) ta có:

\(\frac{6}{t-1}+\frac{12}{t+5}=3-\frac{7}{t}\)

\(\Leftrightarrow\frac{6\left(t+5\right)}{\left(t-1\right)\left(t+5\right)}+\frac{12\left(t-1\right)}{\left(t-1\right)\left(t+5\right)}=\frac{3t-7}{t}\)

\(\Leftrightarrow\frac{18t+18}{t^2+4t-5}=\frac{3t-7}{t}\)

\(\Leftrightarrow3t^3+5t^2-43t+35=18t^2+18t\)

\(\Leftrightarrow3t^3-13t^2-61t+35=0\)

\(\Leftrightarrow\left(t-7\right)\left(3t^2+8t-5\right)=0\)

\(\Leftrightarrow\left[\begin{matrix}t=7\\t=\frac{-8\pm\sqrt{124}}{6}\end{matrix}\right.\)\(\Leftrightarrow\left[\begin{matrix}x^2+3=7\\x^2+3=\frac{-8\pm\sqrt{124}}{6}\end{matrix}\right.\)

\(\Leftrightarrow\left[\begin{matrix}x^2=4\\x^2+3=\frac{-8\pm\sqrt{124}}{6}\left(loai\right)\end{matrix}\right.\)\(\Leftrightarrow x=\pm2\)

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