Tim x biet
a) \(\left(x+\frac{1}{4}-\frac{1}{3}\right):\) \(\left(2+\frac{1}{6}-\frac{1}{4}\right)=\frac{7}{46}\)
b) \(2\frac{1}{2}x+x=2\frac{2}{15}\)
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1.
\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
\(MC:12\)
Quy đồng :
\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)
\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)
\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)
\(\Leftrightarrow6x+9-3x=-4-9+16\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=\frac{-3}{7}\)
2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
\(MC:20\)
Quy đồng :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
\(\frac{7}{8}.(\frac{2}{12}+\frac{4}{10})\)
\(\Rightarrow\frac{7}{8}.(\frac{10+24}{60})\)
\(\Rightarrow\frac{7}{8}.\frac{34}{60}=\frac{238}{480}\)
bt2
\(2.x-\frac{5}{4}=\frac{20}{15}\)
\(\Leftrightarrow2x=\frac{20}{15}+\frac{5}{4}\)
\(\Leftrightarrow2x=\frac{80+75}{60}\)
\(\Leftrightarrow2x=2,5\)
\(\Leftrightarrow x=1,25\)
.7/8.(1/6+2/5)=7/8.17/30=119/240
3/2-5/6:1/4+\(\sqrt{4}\)=3/2-10/3+2=1/6
2x=20/15+5/4
2x=31/12
x=31/12:2
x=31/24
ko bt nha thông cảm
1,\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\left(7-\frac{1}{6}\right)+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\frac{41}{6}+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{41}{14}+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{137}{42}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{137}{42}-\frac{1}{2}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{58}{21}\)
\(\left(x-\frac{9}{4}\right)=\frac{5}{2}:\frac{2}{9}\)
\(\left(x-\frac{9}{4}\right)=\frac{45}{4}\)
\(x=\frac{45}{4}+\frac{9}{4}\)
\(x=\frac{27}{2}\)
cau a dau nhi cuoi cung k phai j dau nha ! mk an lom !
\(a,\)\(\left|x+5\right|=\frac{1}{7}-\left|\frac{4}{3}-\frac{1}{6}\right|\)
\(\Leftrightarrow\left|x+5\right|=\frac{1}{7}-\frac{7}{6}\)
\(\Leftrightarrow\left|x+5\right|=\frac{-43}{42}\)
ta có |x+5| \(\ge\)0 \(\forall x\)
Mà \(-\frac{43}{42}< 0\)nên ko có giá trị x thoả mãn
b,
\(\left|x+\frac{2}{3}\right|=\frac{1}{2}-\left(\frac{1}{4}+\frac{2}{3}\right)\)
\(\Leftrightarrow\left|x+\frac{2}{3}\right|=\frac{11}{12}\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{2}{3}=\frac{11}{12}\forall x\ge-\frac{2}{3}\\-x-\frac{2}{3}=\frac{11}{12}\forall< -\frac{2}{3}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=-\frac{19}{12}\end{cases}}\)(thoả mãn đk)
\(\left(x+\frac{1}{4}-\frac{1}{3}\right)\div\left(2+\frac{1}{6}-\frac{1}{4}\right)=\frac{7}{46}\)
\(\Leftrightarrow\left(x+\frac{1}{4}-\frac{1}{6}\right)\div\frac{23}{12}=\frac{7}{46}\)
\(\Leftrightarrow x+\frac{1}{4}-\frac{1}{6}=\frac{7}{46}\times\frac{23}{12}\)
\(\Leftrightarrow x+\frac{1}{4}-\frac{1}{6}=\frac{7}{24}\)
\(\Leftrightarrow x+\frac{1}{4}=\frac{7}{24}+\frac{1}{6}\)
\(\Leftrightarrow x+\frac{1}{4}=\frac{11}{24}\)
\(\Leftrightarrow x=\frac{11}{24}-\frac{1}{4}\)
\(\Leftrightarrow x=\frac{5}{24}\)
Học tốt !
\(\left(x+\frac{1}{4}-\frac{1}{3}\right):\left(2+\frac{1}{6}-\frac{1}{4}\right)=\frac{7}{46}\)
\(\left(x+\frac{1}{4}-\frac{1}{3}\right):\) \(\frac{23}{12}\) \(=\frac{7}{46}\)
\(x+\frac{1}{4}-\frac{1}{3}\) \(=\frac{7}{45}\times\frac{23}{12}\)
\(x+\frac{1}{4}-\frac{1}{3}\) \(=\frac{161}{540}\)
\(x+\frac{1}{4}\) \(=\frac{161}{450}+\frac{1}{3}\)
\(x+\frac{1}{4}\) \(=\frac{311}{450}\)
\(x\) \(=\frac{311}{450}-\frac{1}{4}\)
\(x\) \(=\frac{397}{900}\)
\(\left(x+\frac{1}{4}-\frac{1}{3}\right):\left(2+\frac{1}{6}-\frac{1}{4}\right)=\frac{7}{46}\)
\(\Leftrightarrow\left(x-\frac{1}{12}\right):\frac{23}{12}=\frac{7}{46}\)
\(\Leftrightarrow x-\frac{1}{12}=\frac{7}{24}\)
\(\Leftrightarrow x=\frac{3}{8}\)