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sai đề rồi nha...bạn thay dấu suy ra thành dấu tương đương giùm mik..mik bị nhầm
\(\frac{x+5}{65}+\frac{x+10}{60}=\frac{x+15}{35}+\frac{x+20}{50}\)
\(\Rightarrow\frac{x+5}{65}+\frac{x+10}{60}-\frac{x+15}{55}-\frac{x+20}{50}+2-2=0\)
\(\Rightarrow\left(\frac{x+5}{65}+1\right)+\left(\frac{x+10}{60}+1\right)-\left(\frac{x+15}{55}+1\right)-\left(\frac{x+20}{50}+1\right)=0\\ \)
\(\Rightarrow\left(\frac{x+5}{65}+\frac{65}{65}\right)+\left(\frac{x+10}{60}+\frac{60}{60}\right)-\left(\frac{x+15}{55}+\frac{55}{55}\right)-\left(\frac{x+20}{50}+\frac{50}{50}\right)=0\)
\(\Rightarrow\frac{x+70}{65}+\frac{x+70}{60}-\frac{x+70}{55}-\frac{x+70}{50}=0\)
\(\Rightarrow\left(x+70\right)\left(\frac{1}{65}+\frac{1}{60}-\frac{1}{55}-\frac{1}{50}\right)=0\)
\(\Rightarrow x+70=0\left(\frac{1}{65}+\frac{1}{60}-\frac{1}{55}-\frac{1}{50}\nè0\right)\)
\(\Leftrightarrow x=-70\)
học tốt...............nhớ k cho mik nha
(3x-2)(2x-1)=(2-3x)(x+3)
(3x-2)(2x-1)-(2-3x)(x+3)=0
(3x-2)(2x-1)+(3x-2)(x+3)=0
(3x-2)(2x-1+x+3)=0
(3x-2)(3x+2)=0
\(\orbr{\begin{cases}3x-2=0\\3x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=2\\3x=-2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{-2}{3}\end{cases}}}\)
Vậy........
(=) 6x2 - 3x - 4x - 2 = 2x + 6 - 3x2 -9x
(=) 6x2 +3x2 - 7x + 7x + 2 -6 = 0
(=) 9x2 - 4 = 0
(=) 9x2 = 4
(=) x2 = \(\frac{9}{4}\)
(=) x = +- \(\frac{3}{2}\)
\(\frac{9x^2}{11y^2}:\frac{3x}{2y}:\frac{6x}{11y}=\frac{9x^2}{11y^2}.\frac{2y}{3x}.\frac{11y}{6x}=\frac{18y^3x^2}{66x^2y^2}=\frac{3y}{11}\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(=> a=k\)x\(b\)
\(c=k\)x\(d\)
Rồi thay vào sẽ làm ra
CHÚC BẠN HOC
\( a)5\left( {x - 3} \right) - 4 = 2\left( {x - 1} \right) + 7\\ \Leftrightarrow 5x - 15 - 4 = 2x - 2 + 7\\ \Leftrightarrow 5x - 19 = 2x + 5\\ \Leftrightarrow 5x - 2x = 5 + 19\\ \Leftrightarrow 3x = 24\\ \Leftrightarrow x = 8\\ b)\dfrac{{8x - 3}}{4} - \dfrac{{3x - 2}}{2} = \dfrac{{2x - 1}}{2} + \dfrac{{x + 3}}{4}\\ \Leftrightarrow 8x - 3 - \left( {3x - 2} \right).2 = \left( {2x - 1} \right).2 + x + 3\\ \Leftrightarrow 8x - 3 - 6x + 4 = 4x - 2 + x + 3\\ \Leftrightarrow 2x + 1 = 5x + 1\\ \Leftrightarrow 2x - 5x = 0\\ \Leftrightarrow - 3x = 0\\ \Leftrightarrow x = 0 \)
\( c)\dfrac{{2\left( {x + 5} \right)}}{3} + \dfrac{{x + 12}}{2} - \dfrac{{5\left( {x - 2} \right)}}{6} = \dfrac{x}{3} + 11\\ \Leftrightarrow 4\left( {x + 5} \right) + 3\left( {x + 12} \right) - \left[ {5\left( {x - 2} \right)} \right] = 2x + 66\\ \Leftrightarrow 4x + 20 + 3x + 36 - 5x + 10 = 2x + 66\\ \Leftrightarrow 2x + 66 = 2x + 66\\ \Leftrightarrow 0x = 0\left( {VSN} \right)\\ \Leftrightarrow x = 0 \)
\(d)\dfrac{x-10}{1994}+\dfrac{x-8}{1996}+\dfrac{x-6}{1998}+\dfrac{x-4}{2000}+\dfrac{x-2}{2002}=\dfrac{x-2002}{2}+\dfrac{x-2000}{4}+\dfrac{x-1998}{6}+\dfrac{x-1996}{8}+\dfrac{x-1994}{10}\\ \Leftrightarrow \dfrac{x-10}{1994}-1+\dfrac{x-8}{1996}-1+\dfrac{x-6}{1998}-1+\dfrac{x-4}{2000}-1+\dfrac{x-2}{2002}-1=\dfrac{x-2002}{2}-1+\dfrac{x-2000}{4}-1+\dfrac{x-1998}{6}-1+\dfrac{x-1996}{8}-1+\dfrac{x-1994}{10}-1\\ \Leftrightarrow \dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}+\dfrac{x-2004}{2000}\dfrac{x-2004}{2002}=\dfrac{x-2004}{2}+\dfrac{x-2004}{4}+\dfrac{x-2004}{6}+\dfrac{x-2004}{8}+\dfrac{x-2004}{10}\\ \Leftrightarrow \dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}+\dfrac{x-2004}{2000}\dfrac{x-2004}{2002}-\dfrac{x-2004}{2}-\dfrac{x-2004}{4}-\dfrac{x-2004}{6}-\dfrac{x-2004}{8}-\dfrac{x-2004}{10}=0\\ \Leftrightarrow \left(x-2004\right)\left(\dfrac{1}{1994}+\dfrac{1}{1996}+\dfrac{1}{1998}+\dfrac{1}{2000}+\dfrac{1}{2002}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{10}=0\right)\\ \Leftrightarrow x-2004=0\\ \Leftrightarrow x=2004\)
Yêu cầu:tìm x