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a: \(A=\dfrac{-7}{28}\cdot\dfrac{15}{25}=\dfrac{-1}{4}\cdot\dfrac{3}{5}=\dfrac{-3}{20}\)
b: \(B=\dfrac{-5\cdot7}{14\cdot\left(-3\right)}=\dfrac{35}{42}=\dfrac{5}{6}\)
c: \(C=\dfrac{-1}{5}-\dfrac{1}{5}\cdot\dfrac{3}{5}=\dfrac{-1}{5}-\dfrac{3}{25}=\dfrac{-8}{25}\)
d: \(D=\dfrac{-3}{4}-\dfrac{1}{4}=-1\)
e: \(E=\dfrac{-4}{5}\left(1-\dfrac{15}{16}\right)=\dfrac{-4}{5}\cdot\dfrac{1}{16}=\dfrac{-1}{20}\)
f: \(F=\dfrac{6-7}{4}\cdot\dfrac{4+12}{22}=\dfrac{-1}{4}\cdot\dfrac{8}{11}=\dfrac{-2}{11}\)
Ta có: \(\dfrac{x+1}{17}+\dfrac{x+2}{16}=\dfrac{x+3}{15}+\dfrac{x+4}{14}\)
\(\Rightarrow\dfrac{x+1}{17}+1+\dfrac{x+2}{16}+1=\dfrac{x+3}{15}+1+\dfrac{x+4}{14}+1\)
\(\Rightarrow\dfrac{x+18}{17}+\dfrac{x+18}{16}=\dfrac{x+18}{15}+\dfrac{x+18}{14}\)
\(\Rightarrow\dfrac{x+18}{17}+\dfrac{x+18}{16}-\dfrac{x+18}{15}-\dfrac{x+18}{14}=0\)
\(\Rightarrow\left(x+18\right).\left(\dfrac{1}{17}+\dfrac{1}{16}-\dfrac{1}{15}-\dfrac{1}{14}\right)=0\) (1)
Mà \(\dfrac{1}{17}+\dfrac{1}{16}-\dfrac{1}{15}-\dfrac{1}{14}\ne0\) (2)
Từ (1) và (2) => x+18=0 => x=-18
Vậy x=-18
1) \(\left(-1\dfrac{1}{5}+x\right):\left(-3\dfrac{3}{5}\right)=\dfrac{-7}{4}+\dfrac{1}{4}:\dfrac{1}{8}\)
\(\Leftrightarrow\left(-1\dfrac{1}{5}+x\right):\left(-3\dfrac{3}{5}\right)=\dfrac{-7}{4}+2\)
\(\Leftrightarrow\left(-1\dfrac{1}{5}+x\right):\left(-3\dfrac{3}{5}\right)=\dfrac{1}{4}\)
\(\Leftrightarrow-1\dfrac{1}{5}+x=\dfrac{1}{4}.\left(-3\dfrac{3}{5}\right)\)
\(\Leftrightarrow-1\dfrac{1}{5}+x=\dfrac{1}{4}.\left(-\dfrac{18}{5}\right)\)
\(\Leftrightarrow-1\dfrac{1}{5}+x=-\dfrac{9}{10}\)
\(\Leftrightarrow x=\left(-\dfrac{9}{10}\right)-\left(-1\dfrac{1}{5}\right)\)
\(\Leftrightarrow x=\dfrac{3}{10}\)
\(\left(1+\dfrac{x+1}{17}\right)+\left(1+\dfrac{x+2}{16}\right)=\left(1+\dfrac{x+3}{15}\right)+\left(1+\dfrac{x+4}{14}\right)\)
\(\dfrac{x+18}{17}+\dfrac{x+18}{16}=\dfrac{x+18}{15}+\dfrac{x+18}{14}\)
\(\dfrac{x+18}{17}+\dfrac{x+18}{16}-\dfrac{x+18}{15}-\dfrac{x+18}{14}=0\)
\(\left(x+18\right)\left(\dfrac{1}{17}+\dfrac{1}{16}-\dfrac{1}{15}-\dfrac{1}{14}\right)=0\)
Vì : \(\dfrac{1}{17}< \dfrac{1}{15};\dfrac{1}{16}< \dfrac{1}{14}\Rightarrow\dfrac{1}{17}+\dfrac{1}{16}-\dfrac{1}{15}-\dfrac{1}{14}< 0\)
\(\Rightarrow x+18=0\Rightarrow x=0-18=-18\)
Từ đầu đến đến dòng thứ tư thì mình đồng ý, nhưng mình nghĩ không nhất thiết phải so sánh, chỉ cần làm tiếp như sau:
\(\Leftrightarrow\left(x+18\right)\left(\dfrac{1}{17}+\dfrac{1}{16}-\dfrac{1}{15}-\dfrac{1}{14}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+18=0\\\dfrac{1}{17}+\dfrac{1}{16}-\dfrac{1}{15}-\dfrac{1}{14}=0\end{matrix}\right.\)
rồi tính tiếp là OK rồi. Dù gì thì cũng xin cảm ơn nha ;)
\(a.-8:\left(4\dfrac{1}{5}x+\dfrac{3}{10}\right)=4\dfrac{4}{9}\)
\(4\dfrac{1}{5}x+\dfrac{3}{10}=\left(-8\right):4\dfrac{4}{9}\)
\(4\dfrac{1}{5}x+\dfrac{3}{10}=\dfrac{-9}{5}\)
\(4\dfrac{1}{5}x=\dfrac{-9}{5}-\dfrac{3}{10}\)
\(4\dfrac{1}{5}x=\dfrac{-21}{10}\)
\(x=\dfrac{-21}{10}:\dfrac{21}{5}\)
\(x=\dfrac{-1}{2}\)
Vay \(x=\dfrac{-1}{2}\).
\(b.4\dfrac{2}{3}-\left(\dfrac{3}{5}:x\right)=-20\%\)
\(\dfrac{14}{3}-\left(\dfrac{3}{5}:x\right)=\dfrac{-1}{5}\)
\(\dfrac{3}{5}:x=\dfrac{14}{3}-\dfrac{-1}{5}\)
\(\dfrac{3}{5}:x=\dfrac{73}{15}\)
\(x=\dfrac{3}{5}:\dfrac{73}{15}\)
\(x=\dfrac{9}{73}\)
Vay \(x=\dfrac{9}{73}\).
Câu c; d; e tương tự nhé.
\(\dfrac{x-1}{10}+\dfrac{x-2}{11}+\dfrac{x-3}{12}=\dfrac{x-4}{13}+\dfrac{x-5}{14}+\dfrac{x-6}{15}\)
Dựa vào t/c dãy tỉ số = nhau ta có:
\(\dfrac{x-1+x-2+x-3}{10+11+12}=\dfrac{x-4+x-5+x-6}{13+14+15}\)
\(\dfrac{3x-6}{33}=\dfrac{3x-15}{42}\)
\(42\left(3x-6\right)=33\left(3x-15\right)\)
\(126x-252=99x-495\)
\(126-99x=594-252\)
\(27x=342\)
\(x=\dfrac{38}{3}\)
a ) \(5\left(x^2\right)+7x+2\)
\(\Leftrightarrow5x^2+7x+2=0\)
\(\Leftrightarrow5x^2+5x+2x+2=0\)
\(\Leftrightarrow\left(5x+2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{5}\\x=-1\end{matrix}\right.\)
Vậy .............
b ) \(\dfrac{x+1}{17}+\dfrac{x+2}{16}=\dfrac{x+3}{15}+\dfrac{x+4}{14}\)
\(\Leftrightarrow\dfrac{x+1}{17}+1+\dfrac{x+2}{16}+1=\dfrac{x+3}{15}+1+\dfrac{x+4}{14}+1\)
\(\Leftrightarrow\dfrac{x+18}{17}+\dfrac{x+18}{16}=\dfrac{x+18}{15}+\dfrac{x+18}{14}\)
\(\Leftrightarrow\dfrac{x+18}{17}+\dfrac{x+18}{16}-\dfrac{x+18}{15}-\dfrac{x+18}{14}=0\)
\(\Leftrightarrow\left(x+18\right)\left(\dfrac{1}{17}+\dfrac{1}{16}-\dfrac{1}{15}-\dfrac{1}{14}\right)=0\)
Vì \(\left(\dfrac{1}{17}+\dfrac{1}{16}-\dfrac{1}{15}-\dfrac{1}{14}\right)\ne0\)
Ta có : \(x+18=0\Leftrightarrow x=-18\)
Vậy ......
c ) \(\dfrac{x-1}{x-3}=\dfrac{x-4}{x-7}\)
\(\Leftrightarrow\left(x-1\right)\left(x-7\right)=\left(x-3\right)\left(x-4\right)\)
\(\Leftrightarrow x^2-7x-x+7=x^2-4x-3x+12\)
\(\Leftrightarrow-x=5\)
\(\Leftrightarrow x=-5\)
Vậy ..
a, 15 - 2 | x + \(\dfrac{1}{3}\)| = \(\dfrac{4}{7}\)
=>2| x + \(\dfrac{1}{3}\)| = 15 - \(\dfrac{4}{7}\)
=> 2 | x + \(\dfrac{1}{3}\)| =\(\dfrac{101}{7}\)
=> | x + \(\dfrac{1}{3}\)| = \(\dfrac{101}{14}\)
=> x+ \(\dfrac{1}{3}\)= \(\dfrac{101}{14}\) hoặc x +\(\dfrac{1}{3}\)= - \(\dfrac{101}{14}\)
+) x+\(\dfrac{1}{3}\)= \(\dfrac{101}{14}\)=> x = \(\dfrac{302}{3}\)
+) x + \(\dfrac{1}{3}\)= - \(\dfrac{101}{14}\)=> x =-\(\dfrac{317}{42}\)
Vậy ...........
b, \(\dfrac{13}{11}\).\(\dfrac{22}{26}\)-x2=\(\dfrac{7}{16}\)
=>1 -x2 = \(\dfrac{7}{16}\)
=> x2= \(\dfrac{9}{16}\)
=> x= \(\dfrac{3}{4}\)hoặc x = -\(\dfrac{3}{4}\)
Vậy ..................
\(a,15-2\left|x+\dfrac{1}{3}\right|=\dfrac{4}{7}\)
\(2\left|x+\dfrac{1}{3}\right|=15-\dfrac{4}{7}\)
\(2\left|x+\dfrac{1}{3}\right|=\dfrac{101}{7}\)
\(\left|x+\dfrac{1}{3}\right|=\dfrac{101}{7}:2\)
\(\left|x+\dfrac{1}{3}\right|=\dfrac{101}{14}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{101}{14}\) hoặc \(x+\dfrac{1}{3}=-\dfrac{101}{14}\)
\(x=\dfrac{101}{14}-\dfrac{1}{3}\) \(x=-\dfrac{101}{14}-\dfrac{1}{3}\)
\(x=\dfrac{302}{3}\) \(x=-\dfrac{317}{42}\)
\(1\dfrac{3}{5}+0,25\cdot\dfrac{16}{15}-\dfrac{13}{15}\\ =\dfrac{8}{5}+\dfrac{1}{4}\cdot\dfrac{16}{15}-\dfrac{13}{15}\\ =\dfrac{8}{5}+\dfrac{4}{15}-\dfrac{13}{15}\\ =\dfrac{8}{5}-\dfrac{3}{5}\\ =1\)
\(1\dfrac{3}{5}\)+0,25 x \(\dfrac{16}{15}-\dfrac{13}{15}\)
=\(\dfrac{8}{5}+\dfrac{1}{4}\) x \(\dfrac{1}{5}\)
=\(\dfrac{8}{5}+\dfrac{1}{20}\)
=\(\dfrac{32}{20}+\dfrac{1}{20}\)
=\(\dfrac{33}{20}\)
`(x-3)/13+(x-3)/14=(x-3)/15+(x-3)/16`
`=>(x-3).(1/13+1/14-1/15-1/16)=0`
`=>x-3=0`
`=>x=3`
\(\dfrac{x-3}{13}+\dfrac{x-3}{14}=\dfrac{x-3}{15}+\dfrac{x-3}{16}\)
\(\Leftrightarrow\dfrac{x-3}{13}+\dfrac{x-3}{14}-\dfrac{x-3}{15}-\dfrac{x-3}{16}=0\)
\(\Leftrightarrow\left(x-3\right).\left(\dfrac{1}{13}+\dfrac{1}{14}-\dfrac{1}{15}-\dfrac{1}{16}\right)=0\)
\(\Leftrightarrow x-3=0\)( Vì \(\dfrac{1}{13}+\dfrac{1}{14}-\dfrac{1}{15}-\dfrac{1}{16}>0\))
\(\Leftrightarrow x-3=0\Rightarrow x=3\)