\(\dfrac{x-1}{x+5}=\dfrac{6}{7}\)
a,\(\dfrac{1}{7}\text{x}\dfrac{2}{7}+\dfrac{1}{7}\text{x}\dfrac{5}{7}+\dfrac{6}{7}\) b,\(\dfrac{6}{11}\text{x}\dfrac{4}{9}+\dfrac{6}{11}\text{x}\dfrac{7}{9}-\dfrac{6}{11}\text{x}\dfrac{2}{9}\)
c, \(\dfrac{4}{25}\text{x}\dfrac{5}{8}\text{x}\dfrac{25}{4}\text{x}24\)
`a)1/7xx2/7+1/7xx5/7+6/7`
`=1/7xx(2/7+5/7)+6/7`
`=1/7xx1+6/7`
`=1/7+6/7=1`
`b)6/11xx4/9+6/11xx7/9-6/11xx2/9`
`=6/11xx(4/9+7/9-2/9)`
`=6/11xx9/9`
`=6/11`
Sorry nãy ghi thiếu.
`c)4/25xx5/8xx25/4xx24`
`=(4xx5xx25xx24)/(25xx8xx4)`
`=(4xx5xx24)/(4xx8)`
`=(5xx24)/8`
`=5xx3=15`
a, \(\dfrac{1}{7}.\dfrac{2}{7}+\dfrac{1}{7}.\dfrac{5}{7}+\dfrac{6}{7}\)
\(=\dfrac{1}{7}.\left(\dfrac{2}{7}+\dfrac{5}{7}\right)+\dfrac{6}{7}\)
\(=\dfrac{1}{7}.1+\dfrac{6}{7}\)
\(=\dfrac{1}{7}+\dfrac{6}{7}=1\)
b, \(\dfrac{6}{11}.\dfrac{4}{9}+\dfrac{6}{11}.\dfrac{7}{9}-\dfrac{6}{11}.\dfrac{2}{9}\)
\(=\dfrac{6}{11}.\left(\dfrac{4}{9}+\dfrac{7}{9}-\dfrac{2}{9}\right)\)
\(=\dfrac{6}{11}.1=\dfrac{6}{11}\)
c, \(\dfrac{4}{25}.\dfrac{5}{8}.\dfrac{25}{4}.24\)
\(=\left(\dfrac{4}{25}.\dfrac{25}{4}\right).\left(\dfrac{5}{8}.24\right)\)
\(=1.15=15\)
Tìm x:
a) \(\dfrac{-3}{7}\).x=\(\dfrac{3}{56}\).\(\dfrac{28}{9}\)
b) x-\(\dfrac{3}{16}\)=\(\dfrac{7}{15}\):\(\dfrac{3}{5}\)
c) \(\dfrac{2}{5}\)+\(\dfrac{1}{5}\).x=\(\dfrac{5}{6}\)
d) \(\dfrac{3}{4}\)x-\(\dfrac{2}{5}\)x=\(\dfrac{3}{7}\).\(\dfrac{1}{6}\)+\(\dfrac{5}{7}\).\(\dfrac{1}{6}\)
*Lưu ý: Trình bày chi tiết kết quả.
a)\(x=\left(\dfrac{3}{56}\cdot\dfrac{28}{9}\right):\dfrac{-3}{7}=\dfrac{1}{6}:\dfrac{-3}{7}=-\dfrac{7}{18}\)
b)\(x=\left(\dfrac{7}{15}\cdot\dfrac{5}{3}\right)+\dfrac{3}{16}=\dfrac{7}{9}+\dfrac{3}{16}=\dfrac{139}{144}\)
c)\(x=\left(\dfrac{5}{6}-\dfrac{2}{5}\right).5=\dfrac{13}{6}\)
d)\(=>x\left(\dfrac{3}{4}-\dfrac{2}{5}\right)=\dfrac{1}{6}\cdot\left(\dfrac{3}{7}+\dfrac{5}{7}\right)\)
\(x\cdot\dfrac{7}{20}=\dfrac{4}{21}=>x=\dfrac{4}{21}\cdot\dfrac{20}{7}=\dfrac{80}{147}\)
\(\dfrac{5}{7}-x=\dfrac{9}{21}\)
\(-x-\dfrac{1}{3}=\dfrac{2}{6}\)
\(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\dfrac{5}{7}-x=\dfrac{9}{21}\\ \Rightarrow\dfrac{5}{7}-x=\dfrac{3}{7}\\ \Rightarrow x=\dfrac{5}{7}-\dfrac{3}{7}\\ \Rightarrow x=\dfrac{2}{7}\\ b,-x-\dfrac{1}{3}=\dfrac{2}{6}\\ \Rightarrow-x=\dfrac{2}{6}+\dfrac{1}{3}\\ \Rightarrow-x=\dfrac{2}{3}\\ \Rightarrow x=-\dfrac{2}{3}\\ \dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\\ \Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\\ \Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\\ \Rightarrow x=-\dfrac{13}{12}\)
Tính theo cách thuận tiện :
\(\dfrac{13}{7}+\dfrac{5}{6}+\dfrac{2}{7}+\dfrac{7}{6}\) = ?
\(\dfrac{1}{2}x\dfrac{5}{6}+\dfrac{1}{2}x\dfrac{11}{6}=?\)
\(\dfrac{13}{7}+\dfrac{5}{6}+\dfrac{2}{7}+\dfrac{7}{6}=\dfrac{15}{7}+\dfrac{12}{6}=\dfrac{29}{7}\)
\(\dfrac{1}{2}\times\dfrac{5}{6}+\dfrac{1}{2}\times\dfrac{11}{6}=\dfrac{1}{2}\times\left(\dfrac{5}{6}+\dfrac{11}{6}\right)=\dfrac{1}{2}\times\dfrac{16}{6}=\dfrac{4}{3}\)
a,\(\dfrac{2}{3}\)x\(\dfrac{5}{2}\):\(\dfrac{9}{5}\)
b,\(\dfrac{1}{3}\)x\(\dfrac{1}{4}\)+\(\dfrac{5}{6}\)
c,\(\dfrac{1}{2}\)-\(\dfrac{7}{8}\):\(\dfrac{7}{4}\)
d,\(\dfrac{6}{5}\)-\(\dfrac{4}{5}\)x\(\dfrac{3}{2}\)
Tìm x, biết:
a) \(x:\dfrac{6}{13}=\dfrac{13}{7}\) b) \(\dfrac{4}{7}.x-\dfrac{2}{3}=\dfrac{1}{5}\) c) \(\left(\dfrac{3}{10}-x\right):\dfrac{2}{5}=\dfrac{3}{5}\) d) \(\dfrac{2}{3}.x-\dfrac{7}{6}=\dfrac{5}{2}\)
a) \(x:\dfrac{6}{13}=\dfrac{13}{7}\\ \Rightarrow x=\dfrac{13}{7}.\dfrac{6}{13}\\ \Rightarrow x=\dfrac{6}{7}\)
b) \(\dfrac{4}{7}.x-\dfrac{2}{3}=\dfrac{1}{5}\\ \Rightarrow\dfrac{4}{7}.x=\dfrac{13}{15}\\ \Rightarrow x=\dfrac{91}{60}\)
c) \(\left(\dfrac{3}{10}-x\right):\dfrac{2}{5}=\dfrac{3}{5}\\ \Rightarrow\dfrac{3}{10}-x=\dfrac{6}{25}\\ \Rightarrow x=\dfrac{3}{50}\)
d) \(\dfrac{2}{3}x-\dfrac{7}{6}=\dfrac{5}{2}\\ \Rightarrow\dfrac{2}{3}x=\dfrac{11}{3}\\ \Rightarrow x=\dfrac{11}{2}\)
\(a,\)\(x:\dfrac{6}{13}=\dfrac{13}{7}\)
\(x=\dfrac{13}{7}.\dfrac{6}{13}\)
\(x=\dfrac{6}{7}\)
b,\(\dfrac{4}{7}.x-\dfrac{2}{3}=\dfrac{1}{5}\)
\(\dfrac{4}{7}.x=\dfrac{1}{5}+\dfrac{2}{3}\)
\(\dfrac{4}{7}.x=\dfrac{3}{15}+\dfrac{10}{15}\)
\(\dfrac{4}{7}.x=\dfrac{13}{15}\)
\(x=\dfrac{13}{15}:\dfrac{4}{7}\)
\(x=\dfrac{13}{15}.\dfrac{7}{4}\)
\(x=\dfrac{91}{60}\)
a: Ta có: \(x:\dfrac{6}{13}=\dfrac{13}{7}\)
\(\Leftrightarrow x=\dfrac{13}{7}\cdot\dfrac{6}{13}\)
hay \(x=\dfrac{6}{7}\)
b: Ta có: \(\dfrac{4}{7}x-\dfrac{2}{3}=\dfrac{1}{5}\)
\(\Leftrightarrow x\cdot\dfrac{4}{7}=\dfrac{13}{15}\)
hay \(x=\dfrac{91}{60}\)
Tìm x:
a) (2x - 3)(6 - 2x) = 0
b) \(5\dfrac{4}{7}:x=13\)
c) 2x - \(\dfrac{3}{7}\) = \(6\dfrac{2}{7}\)
d) \(\dfrac{x}{5}\) + \(\dfrac{1}{2}\) = \(\dfrac{6}{10}\)
e) \(\dfrac{x+3}{15}=\dfrac{1}{3}\)
f) \(\dfrac{x-12}{4}=\dfrac{1}{2}\)
g) \(2\dfrac{1}{4}\).\(\left(x-7\dfrac{1}{3}\right)=1,5\)
h) \(\left(4,5-2x\right).1\dfrac{4}{7}=\dfrac{11}{14}\)
i) \(\dfrac{2}{3}\left(x-25\%\right)=\dfrac{1}{6}\)
k) \(\dfrac{3}{2}x-1\dfrac{1}{2}=x-\dfrac{3}{4}\)
a) (2x - 3)(6 - 2x) = 0
=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)
b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)
c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)
d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)
e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)
f)\(\dfrac{x-12}{4}=\dfrac{1}{2}=\dfrac{x-12}{4}=\dfrac{2}{4}\)
⇒\(x-12=2\)
\(x=2+12\)
x = 14
g)2\(\dfrac{1}{4}.\left(x-7\dfrac{1}{3}\right)=1,5\)
\(\dfrac{9}{4}.\left(x-\dfrac{22}{3}\right)=1,5\)
\(\left(x-\dfrac{22}{3}\right)=\dfrac{3}{2}:\dfrac{9}{4}\)
\(x-\dfrac{22}{3}=\dfrac{2}{3}\)
\(x=\dfrac{2}{3}+\dfrac{22}{3}\)
\(x=8\)
Tính:
a) (6 : \(\dfrac{3}{5}\) \(1\dfrac{1}{6}\) x \(\dfrac{6}{7}\) ) : ( \(4\dfrac{1}{5}\) x \(\dfrac{10}{11}\) + \(5\dfrac{2}{11}\) )
b) (\(1-\dfrac{1}{2}\)) x (\(1-\dfrac{1}{3}\)) x (\(1-\dfrac{1}{4}\)) x ..... x (\(1-\dfrac{1}{2003}\)) x (\(1-\dfrac{1}{2007}\))
\(\dfrac{15}{14}\) : \(\dfrac{10}{21}\) x \(\dfrac{1}{5}\)
\(5\) x \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\)
\(7\) : \(\dfrac{1}{5}\) - \(\dfrac{1}{5}\)
\(6\) + \(\dfrac{1}{5}\) : \(2\)
\(8\) - \(\dfrac{1}{5}\) x \(7\)
\(\dfrac{15}{14}\): \(\dfrac{10}{21}\) \(\times\) \(\dfrac{1}{5}\) = \(\dfrac{15}{14}\) \(\times\) \(\dfrac{21}{10}\) \(\times\) \(\dfrac{1}{5}\) = \(\dfrac{5\times3\times7\times3}{7\times2\times10\times5}\) = \(\dfrac{9}{20}\)
5 \(\times\) \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) = 1 + \(\dfrac{1}{5}\) = \(\dfrac{6}{5}\)
7 : \(\dfrac{1}{5}\) - \(\dfrac{1}{5}\) = 35 - \(\dfrac{1}{5}\) = \(\dfrac{174}{5}\)
6 + \(\dfrac{1}{5}\): 2 = 6 + \(\dfrac{1}{10}\) = \(\dfrac{61}{10}\)
8 - \(\dfrac{1}{5}\) \(\times\) 7 = 8 - \(\dfrac{7}{5}\) = \(\dfrac{33}{5}\)
\(\dfrac{15}{14}\) : \(\dfrac{10}{21}\) x \(\dfrac{1}{5}\) = \(\dfrac{15}{14}\) x \(\dfrac{21}{10}\) x \(\dfrac{1}{5}\) = \(\dfrac{9}{4}\) x \(\dfrac{1}{5}\) = \(\dfrac{9}{20}\)
5 x \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) = \(\dfrac{5}{1}\) x \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) = 1 x \(\dfrac{1}{5}\) = \(\dfrac{1}{5}\)
7 : \(\dfrac{1}{5}-\dfrac{1}{5}\) = \(\dfrac{7}{1}\) x \(\dfrac{5}{1}-\dfrac{1}{5}\) = \(\dfrac{35}{1}\) - \(\dfrac{1}{5}\) = \(\dfrac{175}{5}\) - \(\dfrac{1}{5}\) = \(\dfrac{174}{5}\)
6 + \(\dfrac{1}{5}\) : 2 = \(\dfrac{6}{1}\) + \(\dfrac{1}{5}\) x \(\dfrac{1}{2}\) = \(\dfrac{6}{1}+\dfrac{1}{10}\) = \(\dfrac{60}{10}\) + \(\dfrac{1}{10}\) = \(\dfrac{61}{10}\)
8 - \(\dfrac{1}{5}\) x 7 = \(\dfrac{8}{1}\) - \(\dfrac{1}{5}\) x \(\dfrac{7}{1}\) = \(\dfrac{8}{1}-\dfrac{7}{5}\) = \(\dfrac{40}{5}\) - \(\dfrac{7}{5}\) = \(\dfrac{33}{5}\)
Sai Báo Lại Mình Nha!
ツvõ•тнùʏ• ᴅươɴԍ⁀ɪdoʟ
Sai dấu cuối câu 5*1/5+1/5 nhé!
k) 8 - \(\dfrac{x-2}{2}\) = \(\dfrac{x}{4}\)
m) \(\dfrac{3x+2}{2}\) - \(\dfrac{3x+1}{6}\) = 2x + \(\dfrac{5}{3}\)
n) \(\dfrac{x+1}{7}\)+ \(\dfrac{x+2}{6}\) = \(\dfrac{x+3}{5}\) + \(\dfrac{x+4}{4}\)
o) \(\dfrac{x+5}{6}\) + \(\dfrac{x+6}{5}\) = x + 9
\(\begin{array}{l} n) \Leftrightarrow \dfrac{{x + 1}}{7} + 1 + \dfrac{{x + 2}}{6} + 1 = \dfrac{{x + 3}}{5} + 1 + \dfrac{{x + 4}}{4} + 1\\ \Leftrightarrow \dfrac{{x + 8}}{7} + \dfrac{{x + 8}}{6} - \dfrac{{x + 8}}{5} - \dfrac{{x + 8}}{4} = 0\\ \Leftrightarrow \left( {x + 8} \right)\underbrace {\left( {\dfrac{1}{7} + \dfrac{1}{8} - \dfrac{1}{5} - \dfrac{1}{6}} \right)}_{ < 0} = 0\\ \Leftrightarrow x + 8 = 0\\ \Leftrightarrow x = - 8 \end{array}\)
k/
\(8-\dfrac{x-2}{3}=\dfrac{x}{4}\)
\(\Leftrightarrow\dfrac{96}{12}-\dfrac{4\left(x-2\right)}{12}=\dfrac{3x}{12}\)
\(\Leftrightarrow96-4x+8=3x\)
\(\Leftrightarrow96-4x+8-3x=0\)
\(\Leftrightarrow104-7x=0\)
\(\Leftrightarrow7x=104\)
\(\Leftrightarrow x=104:7\)
\(\Leftrightarrow x=\dfrac{104}{7}\)
Vậy tập nghiệm của phương trình là \(S=\left\{\dfrac{104}{7}\right\}\)
m/
\(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)
\(\Leftrightarrow9x+6-3x-1-12x-10=0\)
\(\Leftrightarrow-6x-5=0\)
\(\Leftrightarrow-6x=5\)
\(\Leftrightarrow x=-\dfrac{5}{6}\)
Vậy tập nghiệm của phương trình là \(S=\left\{-\dfrac{5}{6}\right\}\)
k) Ta có: \(8-\dfrac{x-2}{2}=\dfrac{x}{4}\)
\(\Leftrightarrow\dfrac{32}{4}-\dfrac{2\left(x-2\right)}{4}=\dfrac{x}{4}\)
\(\Leftrightarrow32-2x+4-x=0\)
\(\Leftrightarrow28-x=0\)
hay x=28
Vậy: S={28}
m) Ta có: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)
\(\Leftrightarrow9x+6-3x-1=12x+10\)
\(\Leftrightarrow6x+5-12x-10=0\)
\(\Leftrightarrow-6x=5\)
hay \(x=-\dfrac{5}{6}\)
Vậy: \(S=\left\{-\dfrac{5}{6}\right\}\)
n) Ta có: \(\dfrac{x+1}{7}+\dfrac{x+2}{6}=\dfrac{x+3}{5}+\dfrac{x+4}{4}\)
\(\Leftrightarrow\dfrac{x+1}{7}+1+\dfrac{x+2}{6}+1=\dfrac{x+3}{5}+1+\dfrac{x+4}{4}+1\)
\(\Leftrightarrow\dfrac{x+8}{7}+\dfrac{x+8}{6}=\dfrac{x+8}{5}+\dfrac{x+8}{4}\)
\(\Leftrightarrow\dfrac{x+8}{7}+\dfrac{x+8}{6}-\dfrac{x+8}{5}-\dfrac{x+8}{4}=0\)
\(\Leftrightarrow\left(x+8\right)\left(\dfrac{1}{7}+\dfrac{1}{6}-\dfrac{1}{5}-\dfrac{1}{4}\right)=0\)
mà \(\dfrac{1}{7}+\dfrac{1}{6}-\dfrac{1}{5}-\dfrac{1}{4}\ne0\)
nên x+8=0
hay x=-8
Vậy: S={-8}