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a) \(\frac{4}{9}x+\frac{2}{5}-\frac{1}{3}x=\frac{2}{9}-\frac{1}{4}x\)
\(\Leftrightarrow\frac{13}{36}x=-\frac{8}{45}\)
\(\Rightarrow x=-\frac{32}{65}\)
b) \(\left(\frac{2}{3}x-\frac{1}{2}\right).\left(-\frac{2}{3}\right)+\frac{1}{5}=-\frac{3}{4}\)
\(\Leftrightarrow-\frac{4}{9}x+\frac{1}{3}+\frac{1}{5}=-\frac{3}{4}\)
\(\Leftrightarrow\frac{4}{9}x=\frac{77}{60}\)
\(\Rightarrow x=\frac{231}{80}\)
a) \(\frac{4}{9}x+\frac{2}{5}-\frac{1}{3}x=\frac{2}{9}-\frac{1}{4}x\)
=> \(\frac{4}{9}x-\frac{1}{3}x+\frac{2}{5}-\frac{2}{9}+\frac{1}{4}x=0\)
=> \(\left(\frac{4}{9}x-\frac{1}{3}x+\frac{1}{4}x\right)+\left(\frac{2}{5}-\frac{2}{9}\right)=0\)
=> \(\frac{13}{36}x+\frac{8}{45}=0\)
=> \(\frac{13}{36}x=-\frac{8}{45}\)
=> \(x=-\frac{32}{65}\)
b) \(\left(\frac{2}{3}x-\frac{1}{2}\right)\cdot\frac{-2}{3}+\frac{1}{5}=\frac{-3}{4}\)
=> \(\left(\frac{2}{3}x-\frac{1}{2}\right)\cdot\frac{-2}{3}=-\frac{19}{20}\)
=> \(\frac{2}{3}x-\frac{1}{2}=\left(-\frac{19}{20}\right):\left(-\frac{2}{3}\right)=\left(-\frac{19}{20}\right)\cdot\left(-\frac{3}{2}\right)=\frac{57}{40}\)
=> \(\frac{2}{3}x=\frac{57}{40}+\frac{1}{2}=\frac{77}{40}\)
=> \(x=\frac{77}{40}:\frac{2}{3}=\frac{77}{40}\cdot\frac{3}{2}=\frac{231}{80}\)
a)\(\left(\frac{4}{5}\right)^{2x+7}=\left(\frac{4}{5}\right)^4\)
=> 2x + 7 = 4
2x = 4 - 7
2x = -3
x = -3 : 2
x = -1,5
Vậy x = -1,5
\(c)\)
\(2x-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-...-\frac{1}{49.50}=\left(7-\frac{1}{50}+x\right)\)
\(\Rightarrow2x-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{49.50}\right)=\left(\frac{350}{50}-\frac{1}{50}+x\right)\)
\(\Rightarrow2x-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\right)=\frac{349}{50}+x\)
\(\Rightarrow2x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)-x=\frac{349}{50}\)
\(\Rightarrow x-\left(1-\frac{1}{50}\right)=\frac{349}{50}\)
\(\Rightarrow x-\frac{49}{50}=\frac{349}{50}\)
\(\Rightarrow x=\frac{349}{50}+\frac{49}{50}\)
\(\Rightarrow x=\frac{199}{25}\)
Vậy \(x=\frac{199}{25}\)
~ Ủng hộ nhé
\(a)2.x-3=x+\frac{1}{2}\)
\(\Rightarrow2x-3-x=\frac{1}{2}\)
\(\Rightarrow x-3=\frac{1}{2}\)
\(\Rightarrow x=\frac{1}{2}+3\)
\(\Rightarrow x=\frac{1}{2}+\frac{6}{2}\)
\(\Rightarrow x=\frac{7}{2}\)
Vậy \(x=\frac{7}{2}\)
\(b)4.x-\left(2.x+1\right)=3-\frac{1}{3}+x\)
\(\Rightarrow4.x-2.x-1=\frac{9}{3}-\frac{1}{3}+x\)
\(\Rightarrow2.x-1=\frac{8}{3}+x\)
\(\Rightarrow2x-1-x=\frac{8}{3}\)
\(\Rightarrow x-1=\frac{8}{3}\)
\(\Rightarrow x=\frac{8}{3}+1\)
\(\Rightarrow x=\frac{8}{3}+\frac{3}{3}\)
\(\Rightarrow x=\frac{11}{3}\)
Vậy \(x=\frac{11}{3}\)
~ Ủng hộ nhé
a) Đặt \(\frac{x}{-2}=\frac{y}{-3}=k\Rightarrow\hept{\begin{cases}x=-2k\\y=-3k\end{cases}}\)
Khi đó 4x - 3y = 9
<=> -8k + 9k = 9
=> k = 9
=> x = -18 ; y = -27
b) Ta có : \(2x=3y\Rightarrow\frac{2x}{6}=\frac{3y}{6}\Rightarrow\frac{x}{2}=\frac{y}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{2}=\frac{y}{3}=\frac{x+y}{2+3}=\frac{10}{5}=2\)
=> x = 4 ; y = 6
c) Đặt \(\frac{x}{3}=\frac{y}{4}=k\Rightarrow\hept{\begin{cases}x=3k\\y=4k\end{cases}}\)
Khi đó (3k)2 + (4k)2 = 100
<=> 9k2 + 16k2 = 100
=> 25k2 = 100
=> k2 = 4
=> k = \(\pm\)2
Khi k = 2 => x = 6 ; y = 8
Khi k = -2 => x = -6 ; y = -8
Vậy các cặp (x;y) thỏa mãn cần tìm là (6;8);(-6;-8)
d) Đặt \(\frac{x}{3}=\frac{y}{4}=k\Rightarrow\hept{\begin{cases}x=3k\\y=4k\end{cases}}\)
Khi đó x3 + y3 = 91
<=> (3k)3 + (4k)3 = 91
=> 27k3 + 64k3 = 91
=> 91k3 = 91
=> k3 = 1
=> k = 1
=> x = 3 ; y = 4
e) Đặt \(\frac{x}{5}=\frac{y}{4}=k\Rightarrow\hept{\begin{cases}x=5k\\y=4k\end{cases}}\)
Khi đó x2y = 100
<=> (5k)2.4k = 100
=> 25k2.4k = 100
=> 100k3 = 100
=> k = 1
=> x = 5 ; y = 4
\(a)\) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)
\(\Leftrightarrow\)\(\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+4}{96}+1\right)=-4+4\)
\(\Leftrightarrow\)\(\frac{x+1+99}{99}+\frac{x+2+98}{98}+\frac{x+3+97}{97}+\frac{x+4+96}{96}=0\)
\(\Leftrightarrow\)\(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)
\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)
Vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\)
Nên \(x+100=0\)
\(\Rightarrow\)\(x=-100\)
Vậy \(x=-100\)
Chúc bạn học tốt ~
\(b)\) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{2008}{2009}\)
\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2008}{2009}\)
\(\Leftrightarrow\)\(1-\frac{1}{x+1}=\frac{2008}{2009}\)
\(\Leftrightarrow\)\(\frac{1}{x+1}=1-\frac{2008}{2009}\)
\(\Leftrightarrow\)\(\frac{1}{x+1}=\frac{1}{2009}\)
\(\Leftrightarrow\)\(x+1=2009\)
\(\Leftrightarrow\)\(x=2009-1\)
\(\Leftrightarrow\)\(x=2008\)
Vậy \(x=2008\)
Chúc bạn học tốt ~
Ta có: \(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{4}\right|\ge0\)
\(\Rightarrow4x\ge0\Rightarrow x\ge0\)
\(\Rightarrow x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{4}=4x\)
\(\Leftrightarrow3x+\frac{13}{12}=4x\)
\(\Leftrightarrow x=\frac{13}{12}\left(tm\right)\)