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ĐKXĐ:\(x\ne0,x\ne-12\)
\(\dfrac{270}{x}-\dfrac{270}{x+12}=\dfrac{7}{10}\\ \Leftrightarrow270\left(\dfrac{1}{x}-\dfrac{1}{x+12}\right)=\dfrac{7}{10}\\ \Leftrightarrow\dfrac{x+12-x}{x\left(x+12\right)}=\dfrac{7}{2700}\\ \Leftrightarrow\dfrac{12}{x^2+12x}=\dfrac{7}{2700}\\ \Leftrightarrow7x^2+84x-32400=0\)
Xem lại đề
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xin lỗi nhé mình mới có lớp 6 à nên ko bít
tha lỗi cho mình nhé!
\(\left(x^2+22x-120\right)\left(x^2+33x+270\right)-2x^2\)
\(=x^4+55x^3+876x^2+1980x-32400-2x^2\)
\(=x^4+55x^3+874x^2+1980x-32400\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x^2+22x+120\right)\left(x^2+33x+270\right)-2x^2\)
\(=\left(x+10\right)\left(x+12\right)\left(x+18\right)\left(x+15\right)-2x^2\)
\(=\left(x^2+28x+180\right)\left(x^2+27x+280\right)-2x^2\)
\(=\left(x^2+180\right)^2+55x\left(x^2+180\right)+754x^2\)
\(=\left(x^2+29x+180\right)\left(x^2+26x+180\right)\)
\(=\left(x+9\right)\left(x+20\right)\left(x^2+26x+180\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x+2\right)\left(x+3\right)\left(17x^2-17x+8\right)=\left(x+2\right)\left(x+3\right)\left(x^2-17x+33\right)\\ \Leftrightarrow\left(x+2\right)\left(x+3\right)\left(17x^2-17x+8\right)-\left(x+2\right)\left(x+3\right)\left(x^2-17x+33\right)=0\\\Leftrightarrow \left(x+2\right)\left(x+3\right)\left(16x^2-25\right)=0\\\Leftrightarrow \left(x+2\right)\left(x+3\right)\left(4x-5\right)\left(4x+5\right)=0\\\Leftrightarrow \left[{}\begin{matrix}x+2=0\\x+3=0\\4x-5=0\\4x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-3\\x=\frac{5}{4}\\x=-\frac{5}{4}\end{matrix}\right.\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{-2;-3;-\frac{5}{4};\frac{5}{4}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x^3 -9x^2 +27x -27 -(x^3 -27) +6(x^2 +2x+1) +3x^2 =-33
x^3 -9x^2 +27x -27 -x^3 +27 +6x^2 + 12x+ 6 +3x^2 =-33
39x+6=-33
39x=-39
x=-1
Vậy x=-1