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\(\frac{5}{6}x+\frac{1}{2}-\frac{1}{3}x=0.75x-\frac{7}{8}\)
\(\frac{5}{6}x-\frac{1}{3}x-\frac{3}{4}x=-\frac{7}{8}-\frac{1}{2}\) ( 3/4x là 0,75x nha)
\(x\times\left(\frac{10}{12}-\frac{4}{12}-\frac{9}{12}\right)=-\frac{7}{8}-\frac{4}{8}\)
\(x\times\left(-\frac{3}{12}\right)=-\frac{11}{8}\Rightarrow x=\frac{11}{8}\div\left(-\frac{3}{12}\right)=-\frac{11}{2}\)
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, \(\left|x-3,5\right|+\left|x-\frac{1}{3}\right|=0\)
\(\hept{\begin{cases}x-3,5\ge0\forall x\\x-\frac{1}{3}\ge0\forall x\end{cases}\Rightarrow\left|x-3,5\right|+\left|x-\frac{1}{3}\right|\ge0\forall x}\)
Dấu ''='' xảy ra <=> \(x-3,5=0\Leftrightarrow x=3,5\)
\(x-\frac{1}{3}=0\Leftrightarrow x=\frac{1}{3}\)
b, \(\left|x\right|+x=\frac{1}{3}\Leftrightarrow\left|x\right|=\frac{1}{3}-x\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}-x\\x=-\frac{1}{3}+x\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=\frac{1}{3}\\0\ne-\frac{1}{3}\end{cases}\Leftrightarrow}x=\frac{1}{6}}\)
c, \(\left|x-2\right|=x\Leftrightarrow\orbr{\begin{cases}x-2=x\\x-2=-x\end{cases}\Leftrightarrow\orbr{\begin{cases}-2\ne0\\x=1\end{cases}}}\)
d, tương tự c
Sửa ý a) của bạn @akirafake
a) \(\left|x-3,5\right|+\left|x-1,3\right|=0\)
Ta có : \(\left|x-3,5\right|+\left|x-1,3\right|=\left|-\left(x-3,5\right)\right|+\left|x-1,3\right|=\left|3,5-x\right|+\left|x-1,3\right|\)
Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)ta có :
\(\left|3,5-x\right|+\left|x-1,5\right|\ge\left|3,5-x+x-1,5\right|=\left|2\right|=2\)
mà \(\left|x-3,5\right|+\left|x-1,3\right|=0\)( vô lí )
Vậy không có giá trị của x thỏa mãn
b) \(\left|x\right|+x=\frac{1}{3}\)
=> \(\left|x\right|=\frac{1}{3}-x\)
=> \(\orbr{\begin{cases}x=\frac{1}{3}-x\\x=x-\frac{1}{3}\end{cases}\Rightarrow}\orbr{\begin{cases}2x=\frac{1}{3}\\0x=-\frac{1}{3}\end{cases}\Rightarrow}2x=\frac{1}{3}\Rightarrow x=\frac{1}{6}\)
c) \(\left|x\right|-x=\frac{3}{4}\)
=> \(\left|x\right|=\frac{3}{4}+x\)
=> \(\orbr{\begin{cases}x=\frac{3}{4}+x\\x=-x-\frac{3}{4}\end{cases}\Rightarrow}\orbr{\begin{cases}0x=\frac{3}{4}\\2x=-\frac{3}{4}\end{cases}}\Rightarrow2x=-\frac{3}{4}\Rightarrow x=-\frac{3}{8}\)
d) \(\left|x-2\right|=x\)
=> \(\orbr{\begin{cases}x-2=x\\x-2=-x\end{cases}}\Rightarrow\orbr{\begin{cases}0x=2\\2x=2\end{cases}}\Rightarrow2x=2\Rightarrow x=1\)
e) \(\left|x+2\right|=x\)
=> \(\orbr{\begin{cases}x+2=x\\x+2=-x\end{cases}}\Rightarrow\orbr{\begin{cases}0x=-2\\2x=-2\end{cases}}\Rightarrow2x=-2\Rightarrow x=-1\)
Thế x = -1 ta được :
\(\left|-1+2\right|=-1\)( vô lí )
=> Không có giá trị của x thỏa mãn
đặt bt=k
x=2k+1;y=3k+2;z=4k+3
2x+3y-z=4k+2+9k+6-4k-3=9k+5=50
k=5
x=11;y=17;z=23
ÁP dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)<=> \(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x-2+3y-6-z+3}{4+9-4}=\frac{50-5}{9}=\frac{45}{9}=5\)
=> \(\hept{\begin{cases}\frac{x-1}{2}=5\\\frac{y-2}{3}=5\\\frac{z-3}{4}=5\end{cases}}\) <=> \(\hept{\begin{cases}x=11\\y=17\\z=23\end{cases}}\)
\(\frac{5}{x}\)+ \(\frac{y}{4}\) = \(\frac{1}{8}\)
\(\frac{5}{x}\)+ \(\frac{2y}{8}\) = \(\frac{1}{8}\)
\(\frac{5}{x}\) = \(\frac{1-2y}{8}\)
Tiếp đó quy đồng rùi tự làm nốt hem
\(\frac{5}{x}+\frac{y}{4}=\frac{1}{8}\)
\(\Leftrightarrow\frac{5}{x}=\frac{1}{8}-\frac{2y}{8}\)
\(\Leftrightarrow\frac{5}{x}=\frac{1-2y}{8}\)
\(\Leftrightarrow40=x\left(1-2y\right)\)
Xét Ư thì có hơi dài nhỉ ?
\(\Leftrightarrow3^{x-1}\left(1+5\right)=162\)
\(\Leftrightarrow3^{x-1}.6=162\)
\(\Leftrightarrow3^{x-1}=27\)
\(\Leftrightarrow x-1=3\)
\(\Leftrightarrow x=3+1=4\)
\(3^{x-1}+5.3^{x-1}=162\)
\(3^{x-1}.\left(1+5\right)=162\)
\(3^{x-1}.6=162\)
\(3^{x-1}=27\)
\(3^{x-1}=3^3\)
\(\Leftrightarrow x-1=3\)
\(x=4\)
Vậy x = 4