Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)
\(\Leftrightarrow\frac{12\left(x-3\right)}{12}-\frac{2\left(x-3\right)\left(2x-5\right)}{12}=\frac{3\left(x-3\right)\left(3-x\right)}{12}\)
\(\Leftrightarrow12\left(x-3\right)-2\left(x-3\right)\left(2x-5\right)=3\left(x-3\right)\left(3-x\right)\)
\(\Leftrightarrow12\left(x-3\right)-2\left(x-3\right)\left(2x-5\right)-3\left(x-3\right)\left(3-x\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(13-x\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x-3=0\\13-x=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=3\\x=13\end{cases}}}\)
Vậy tập nghiệm của phương trình trên là:\(S=\left\{3;13\right\}\)
#hoktot<3#
\(\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)
\(\frac{12\left(x-3\right)}{12}-\frac{\left(x-3\right)\left(2x-5\right)2}{12}=\frac{\left(x-3\right)\left(3-x\right)3}{12}\)
Khử mẫu : \(12\left(x-3\right)-\left(x-3\right)\left(2x-5\right)2=\left(x-3\right)\left(3-x\right)3\)
\(34x-66-4x^2=18x-3x^2-27\)
\(34x-66-4x^2-18x+3x^2+27=0\)
\(16x-39-x^2=0\)
Phân tích nốt nhé !
\(\Leftrightarrow\frac{12\left(x-3\right)-2\left[\left(x-3\right)\left(2x-5\right)\right]}{12}=\frac{3\left[\left(x-3\right)\left(3-x\right)\right]}{12}\)
\(\Leftrightarrow12x-36-2\left(2x^2-5x-6x+15\right)=3\left(3x-x^2-9+3x\right)\)
\(\Leftrightarrow12x-36-4x^2+10x+12x-30=9x-3x^2-27+9x\)
\(\Leftrightarrow-x^2+34x-39=0\)
\(\Leftrightarrow-\left(x-17\right)^2-250=0\)
\(\Leftrightarrow-\left(x-17\right)^2=250\left(voly\right)\)
vay \(S=\varnothing\)
a) \(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4}=\frac{x-4}{5}+\frac{x-5}{6}\)
\(\left(\frac{x-1}{2}+1\right)+\left(\frac{x-2}{3}+3\right)+\left(\frac{x-3}{4}+1\right)=\left(\frac{x-4}{5}+1\right)+\left(\frac{x-5}{6}+1\right)\)
\(\frac{x-1}{2}+\frac{x-1}{3}+\frac{x-1}{4}=\frac{x-1}{5}+\frac{x-1}{6}\)
\(\left(x-1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)\)=0
\(x-1=0\)
\(x=1\)
\(\frac{25x-655}{95}-\frac{5\left(x-12\right)}{209}=\frac{89-3x-\frac{2\left(x-18\right)}{5}}{11}\)
\(< =>\frac{5x-131}{19}=\frac{1631-52x-\frac{38x-684}{5}}{209}\)
\(< =>\left(5x-131\right)209=\left(1631-52x-\frac{38x-684}{5}\right)19\)
\(< =>55x-1441=1631-52x-\frac{38x-684}{5}\)
\(< =>3072-107x=\frac{38x-684}{5}\)
\(< =>\left(3072-107x\right)5=38x-684\)
\(< =>15360-535x-38x-684=0\)
\(< =>14676=573x< =>x=\frac{14676}{573}=\frac{4892}{191}\)
nghệm xấu thế
\(\frac{8\left(x+22\right)}{45}-\frac{7x+149+\frac{6\left(x+12\right)}{5}}{9}=\frac{x+35+\frac{2\left(x+50\right)}{9}}{5}\)
\(< =>\frac{8x+176}{45}-\frac{41x+817}{45}=\frac{11x+415}{45}\)
\(< =>993-33x-11x-415=0\)
\(< =>578=44x< =>x=\frac{289}{22}\)
