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a) \(2\left(3x-1\right)-\left(5+3x\right)=3\left(2x-1\right)\)
\(\Leftrightarrow6x-2-5-3x=6x-3\)
\(\Leftrightarrow6x-3x-6x=-3+2+5\)
\(\Leftrightarrow-3x=4\)
\(\Leftrightarrow x=-\frac{4}{3}\)
b) \(3\left(x-\frac{1}{2}\right)+4\left(\frac{x}{3}-\frac{1}{3}\right)=\frac{x}{4}\)
\(\Leftrightarrow3x-\frac{3}{2}+\frac{4}{3}x-\frac{4}{3}=\frac{x}{4}\)
\(\Leftrightarrow3x+\frac{4}{3}x-\frac{x}{4}=\frac{3}{2}+\frac{4}{3}\)
\(\Leftrightarrow\frac{49}{12}x=\frac{17}{6}\)
\(\Leftrightarrow x=\frac{34}{49}\)
c) \(\frac{1}{5}\left(x-\frac{1}{3}\right)-4\left(\frac{x}{5}-\frac{1}{2}\right)=x\)
\(\Leftrightarrow\frac{1}{5}x-\frac{1}{15}-\frac{4}{5}x+2=x\)
\(\Leftrightarrow\frac{1}{5}x-\frac{4}{5}x-x=\frac{1}{15}-2\)
\(\Leftrightarrow-\frac{8}{5}x=-\frac{29}{15}\)
\(\Leftrightarrow x=\frac{29}{24}\)
a, Ta có : \(\frac{x+1}{2}+\frac{x-2}{4}=1-\frac{2\left(x-1\right)}{3}\)
=> \(\frac{6\left(x+1\right)}{12}+\frac{3\left(x-2\right)}{12}=\frac{12}{12}-\frac{8\left(x-1\right)}{12}\)
=> \(6\left(x+1\right)+3\left(x-2\right)=12-8\left(x-1\right)\)
=> \(6x+6+3x-6=12-8x+8\)
=> \(17x=20\)
=> \(x=\frac{20}{17}\)
b, Ta có : \(\frac{5x-1}{6}+x=\frac{6-x}{4}\)
=> \(\frac{5x-1+6x}{6}=\frac{6-x}{4}\)
=> \(4\left(11x-1\right)=6\left(6-x\right)\)
=> \(44x-4-36+6x=0\)
=> \(\)\(50x=40\)
=> \(x=\frac{4}{5}\)
c, Ta có : \(\frac{5\left(1-2x\right)}{3}+\frac{x}{2}=\frac{3\left(x-5\right)}{4}-2\)
=> \(\frac{20\left(1-2x\right)}{12}+\frac{6x}{12}=\frac{9\left(x-5\right)}{12}-\frac{24}{12}\)
=> \(20\left(1-2x\right)+6x=9\left(x-5\right)-24\)
=> \(20-40x+6x-9x+45+24=0\)
=> \(43x=89\)
=> \(x=\frac{89}{43}\)
\(\left(x-3\right)^3-2\left(x-1\right)=x\left(x-2\right)^2-5x^2\)
\(\Leftrightarrow x^3-9x^2+27x-27-2x+2=x^3-4x^2+4x-5x^2\)
\(\Leftrightarrow27x-2x-4x-27+2=0\)
\(\Leftrightarrow21x=25\)
\(\Leftrightarrow x=\frac{25}{21}\)
Hết ý tưởng,phá tung ra,sai chỗ nào tự sửa nhé !
\(\frac{\left(x+1\right)^2}{3}+\frac{\left(x+2\right)\left(x-3\right)}{2}=\frac{\left(5x-1\right)\left(x-4\right)}{6}+\frac{28}{3}\)
\(\Leftrightarrow\frac{2\left(x+1\right)^2+3\left(x+2\right)\left(x-3\right)-\left(5x-1\right)\left(x-4\right)}{6}=\frac{28}{3}\)
\(\Leftrightarrow\frac{2x^2+4x+2+3x^2-3x-18-5x^2-21x+4}{6}=\frac{28}{3}\)
\(\Leftrightarrow\frac{\left(4x-3x-21x\right)+\left(2-18+4\right)}{6}=\frac{56}{6}\)
\(\Leftrightarrow-20x-12=56\)
\(\Leftrightarrow-20x=68\)
\(\Leftrightarrow x=-\frac{17}{5}\)
Tự check lại nhá
a)
\(\frac{7}{x-5}-2=\frac{3}{5-x}\\ \Leftrightarrow\frac{-7}{5-x}-2-\frac{3}{5-x}=0\\ \Leftrightarrow\frac{-7}{5-x}-\frac{10-2x}{5-x}-\frac{3}{5-x}=0\\ \Leftrightarrow\frac{-7-10+2x-3}{5-x}=0\\ \Leftrightarrow\frac{2x-20}{5-x}=0\\ \Rightarrow2x-20=0\\ \Rightarrow x=10\)
b)
\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\cdot\left(x-2\right)}\\ \Leftrightarrow\frac{2}{x+1}-\frac{1}{x-2}-\frac{3x-11}{\left(x+1\right)\cdot\left(x-2\right)}=0\\ \Leftrightarrow\frac{2x-4}{\left(x+1\right)\cdot\left(x-2\right)}-\frac{x+1}{\left(x+1\right)\cdot\left(x-2\right)}-\frac{3x-11}{\left(x+1\right)\cdot\left(x-2\right)}=0\\ \Leftrightarrow\frac{2x-4-x-1-3x+11}{\left(x+1\right)\cdot\left(x-2\right)}=0\\ \Leftrightarrow\frac{6-2x}{\left(x+1\right)\cdot\left(x-2\right)}=0\\ \Rightarrow6-2x=0\\ \Rightarrow x=3\)
c)
\(\frac{1}{x}-\frac{x+2}{x-2}=\frac{2}{x\cdot\left(2-x\right)}\\ \Leftrightarrow\frac{1}{x}-\frac{x-2}{2-x}-\frac{2}{x\cdot\left(2-x\right)}=0\\ \Leftrightarrow\frac{2-x}{x\cdot\left(2-x\right)}-\frac{x^2-2x}{x\cdot\left(2-x\right)}-\frac{2}{x\cdot\left(2-x\right)}=0\\ \Leftrightarrow\frac{2-x-x^2+2x-2}{x\cdot\left(2-x\right)}=0\\ \Leftrightarrow\frac{x-x^2}{x\cdot\left(2-x\right)}=0\\ \Rightarrow x-x^2=0\\ \Rightarrow x\cdot\left(1-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\1-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
mới lớp 6 à!!!!