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\(-2\dfrac{1}{5}=-\dfrac{11}{5}\)
\(\dfrac{110}{-50}=-\dfrac{11}{5}\)
\(\Rightarrow-2\dfrac{1}{5}=\dfrac{110}{-50}\)
Bài 1:
a, Ta có:
\(\dfrac{-8}{15}=-\dfrac{5}{18}+-\dfrac{1}{6}\)
b, Ta có:
\(-\dfrac{8}{15}=\dfrac{11}{15}-\dfrac{19}{15}\)
Bài 2:
a, \(\dfrac{11}{13}-\left(\dfrac{5}{12}-x\right)=-\left(\dfrac{15}{18}-\dfrac{11}{13}\right)\)
\(\Rightarrow\dfrac{11}{13}-\dfrac{5}{12}+x=-\dfrac{15}{18}+\dfrac{11}{13}\)
\(\Rightarrow x=-\dfrac{15}{18}+\dfrac{11}{13}+\dfrac{5}{12}-\dfrac{11}{13}\)
\(\Rightarrow x=-\dfrac{15}{8}+\dfrac{5}{12}=-\dfrac{35}{24}\)
b, \(2x-3=x+\dfrac{1}{2}\)
\(\Rightarrow2x-x=\dfrac{1}{2}+3\Rightarrow x=\dfrac{7}{2}\)
Chúc bạn học tốt!!!
Bài 2:
a: =>11/13-5/42+x=15/18+11/13
=>x-5/42=15/18
=>x=5/6+5/42=35/42+5/42=40/42=20/21
b: 2x-3=x+1/2
=>2x-x=3+1/2
=>x=7/2
a: \(\dfrac{-13}{40}< \dfrac{-12}{40}\)
\(\dfrac{-5}{6}>\dfrac{-91}{104}\)
Câu 1 :
\(\dfrac{-25}{37}\&\dfrac{-20}{31}\)
Ta thấy \(\dfrac{-25}{37}< \dfrac{-20}{37}\)
mà \(\dfrac{-20}{37}< \dfrac{-20}{31}\)
\(\Rightarrow\dfrac{-25}{37}< \dfrac{-20}{31}\)
Câu 2 :
\(\dfrac{2}{3}\&\dfrac{5}{7}\)
\(\dfrac{2}{3}:\dfrac{5}{7}=\dfrac{2}{3}.\dfrac{7}{5}=\dfrac{14}{15}< 1\)
\(\Rightarrow\dfrac{5}{7}>\dfrac{2}{3}\) Câu 3 : \(\dfrac{8}{13}\&\dfrac{5}{7}\)Ta thấy \(\dfrac{8}{13}:\dfrac{5}{7}=\dfrac{8}{13}.\dfrac{7}{5}=\dfrac{56}{65}< 1\)
\(\Rightarrow\dfrac{8}{13}< \dfrac{5}{7}\)
a) \(\dfrac{-5}{9}-\dfrac{-5}{12}=\dfrac{-5}{9}+\dfrac{5}{12}=\dfrac{-20}{36}+\dfrac{15}{36}=-\dfrac{5}{36}\)
b) \(\dfrac{-5}{12}:\dfrac{15}{4}=\dfrac{-5}{12}\times\dfrac{4}{15}=\dfrac{-1}{9}\)
c) \(\dfrac{1}{13}\cdot\dfrac{8}{13}+\dfrac{5}{13}\cdot\dfrac{1}{13}-\dfrac{14}{13}=\dfrac{1}{13}\cdot\left(\dfrac{8}{13}+\dfrac{5}{13}\right)-\dfrac{14}{13}=\dfrac{1}{13}\cdot1-\dfrac{14}{13}=\dfrac{1}{13}-\dfrac{14}{13}=-1\)
\(\dfrac{x}{7}=\dfrac{y}{9}=\dfrac{z}{11}\)
Áp dụng tính chất của dãy tỉ sô bằng nhau ta có :
\(\dfrac{x}{7}=\dfrac{y}{9}=\dfrac{z}{11}=\dfrac{y-x}{9-7}=\dfrac{-1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{7}=\dfrac{-1}{2}\\\dfrac{y}{9}=\dfrac{-1}{2}\\\dfrac{z}{11}=\dfrac{-1}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-7}{2}\\y=\dfrac{-9}{2}\\z=\dfrac{-11}{2}\end{matrix}\right.\)
Ta có : \(\dfrac{x}{7}=\dfrac{y}{9}=\dfrac{z}{11}\)
Áp dụng tính chất dãy tỉ số bằng nhâu , ta có:
\(\dfrac{x}{7}=\dfrac{y}{9}=\dfrac{z}{11}=\dfrac{y-x}{9-7}=-\dfrac{1}{2}\\ \Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}.7=-\dfrac{7}{2}\\y=-\dfrac{1}{2}.9=-\dfrac{9}{2}\\z=-\dfrac{1}{2}.11=-\dfrac{11}{2}\end{matrix}\right.\)
Vậy \(x=-\dfrac{7}{2}\); \(y=-\dfrac{9}{2}\); \(z=-\dfrac{11}{2}\)
a: -3/100=-9/300; -2/3=-200/300
=>-3/100>-2/3
b: -3/5=-9/15
-2/3=-10/15
=>-3/5>-2/3
c: -5/4<-1<-3/8
d: -2/3=-8/12; -3/4=-9/12
=>-2/3>-3/4
e: -267/268>-1
-1>-1347/1343
=>-267/268>-1347/1343
Bài 1:
Ta có: \(3x=2y\)
nên \(\dfrac{x}{2}=\dfrac{y}{3}\)
mà x+y=-15
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{x+y}{2+3}=\dfrac{-15}{5}=-3\)
Do đó:
\(\left\{{}\begin{matrix}\dfrac{x}{2}=-3\\\dfrac{y}{3}=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-6\\y=-9\end{matrix}\right.\)
Vậy: (x,y)=(-6;-9)
Bài 2:
a) Ta có: \(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{5}\)
mà x+y-z=20
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x+y-z}{4+3-5}=\dfrac{20}{2}=10\)
Do đó:
\(\left\{{}\begin{matrix}\dfrac{x}{4}=10\\\dfrac{y}{3}=10\\\dfrac{z}{5}=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=40\\y=30\\z=50\end{matrix}\right.\)
Vậy: (x,y,z)=(40;30;50)
(Sửa \(cn-bm\rightarrow cn-dm\))
Ta có :
\(\left\{{}\begin{matrix}ad-bc=1\\cn-dm=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}ad=1+bc\\cn=1+dm\end{matrix}\right.\)
\(\dfrac{x}{y}=\dfrac{a}{b}.\dfrac{d}{c}=\dfrac{ad}{bc}=\dfrac{1+bc}{bc}=1+\dfrac{1}{bc}>1\left(bc>0\right)\)
\(\Rightarrow x=\dfrac{a}{b}>y=\dfrac{c}{d}\left(2\right)\)
\(\dfrac{y}{z}=\dfrac{c}{d}.\dfrac{n}{m}=\dfrac{cn}{dm}=\dfrac{1+dm}{dm}=1+\dfrac{1}{dm}>1\left(dc>0\right)\)
\(\Rightarrow y=\dfrac{c}{d}>z=\dfrac{m}{n}\left(2\right)\)
\(\left(1\right);\left(2\right)\Rightarrow x>y>z\)
