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a, Ta có
\(\dfrac{111}{37}< x< \dfrac{91}{37}\Leftrightarrow3< x< 7\Leftrightarrow x\in\left\{4,5,6\right\}\)
b,Ta co
\(\dfrac{-84}{14}< 3x< \dfrac{108}{9}\Leftrightarrow3.\dfrac{-28}{14}< 3x< 3.\dfrac{36}{9}\Leftrightarrow\dfrac{-28}{14}< x< \dfrac{36}{9}\Leftrightarrow-2< x< 4\Leftrightarrow x\in\left\{-1,0,1,2,3\right\}\)
\(\dfrac{3}{-2}=\dfrac{-9}{6};\dfrac{-1}{-7}=\dfrac{1}{7}\)
Sắp xếp:
\(\dfrac{-1}{-7};\dfrac{0}{8};\dfrac{-7}{6};\dfrac{3}{-2}\)
\(\dfrac{-5}{11}+\left(\dfrac{-6}{11}+1\right)=\dfrac{-5}{11}+\dfrac{-6}{11}+1=-1+1=0\)
a) +/ \(\dfrac{1}{3}\) và \(\dfrac{-2}{6}\)
Vì\(\dfrac{1}{3}\) > 0 và \(\dfrac{-2}{6}\) < 0 nên \(\dfrac{1}{3}>\dfrac{-2}{6}\)
+/\(\dfrac{-4}{5}\)và \(\dfrac{-20}{25}\)
\(\dfrac{-20}{25}=\dfrac{-20:5}{25:5}=\dfrac{-4}{5}\)
Vì\(\dfrac{-4}{5}\)=\(\dfrac{-4}{5}\)nên \(\dfrac{-4}{5}=\dfrac{-20}{25}\)
+/\(\dfrac{2}{3}\) và \(\dfrac{-5}{-8}\)
\(\dfrac{2}{3}=\dfrac{2.8}{3.8}=\dfrac{16}{24}\) ;\(\dfrac{-5}{-8}=\dfrac{5}{8}=\dfrac{5.3}{8.3}=\dfrac{15}{24}\)
Vì\(\dfrac{16}{24}>\dfrac{15}{24}\) nên \(\dfrac{2}{3}>\dfrac{-5}{-8}\)
Các cặp phân số bằng nhau từ đẳng thức 4.6=12.2 là :
\(\dfrac{4}{12}=\dfrac{2}{6};\dfrac{12}{4}=\dfrac{6}{2};\dfrac{12}{6}=\dfrac{4}{2};\dfrac{6}{12}=\dfrac{2}{4}.\)
a,\(\frac{25}{42}-\frac{20}{63}=\frac{5}{18}\)
b,\(\frac{9}{50}-\frac{13}{75}-\frac{1}{6}\)
=\(\frac{1}{150}-\frac{1}{6}\)
=\(-\frac{4}{25}\)
c,\(\frac{2}{15}+\frac{-2}{65}-\frac{4}{39}\)
=\(\frac{32}{195}-\frac{4}{39}\)
=\(\frac{4}{65}\)
a,=75/126-40/126=35/126=5/18
b,=27/150-26/150-25/150=-4/25
c,=26/195+-6/195-20/195=0
a) 7.28=x.x
=> 196=x2
=> \(\left(\pm14\right)^2=x^2\)
=> x=\(\pm14\)
b) DK: x≠-17
pt<=> 4.(10+2)=6.(17+x)
=> 4.12=17.6+6x
=> 48-102=6x
=>-66=6x
=>x=-11
c) 7.(x+40)=6.(17+x)
=> 7x+280=102+6x
=> 7x-6x=102-280
=> x=-178
Giải:
a) \(\dfrac{7}{x}=\dfrac{x}{28}\)
\(\Leftrightarrow x^2=196\)
\(\Leftrightarrow x=\pm\sqrt{196}=\pm14\)
Vậy ...
b) \(\dfrac{10+2}{17+x}=\dfrac{3}{4}\)
\(\Leftrightarrow40+8=51+3x\)
\(\Leftrightarrow3x=40+8-51=-3\)
\(\Leftrightarrow x=-\dfrac{3}{3}=-1\)
Vậy ...
c) \(\dfrac{40+x}{17+x}=\dfrac{6}{7}\)
\(\Leftrightarrow280+7x=102+6x\)
\(\Leftrightarrow7x-6x=102-280\)
\(\Leftrightarrow x=-178\)
Vậy ...
a)\(\dfrac{-1}{4}\cdot13\dfrac{9}{11}-0,25\cdot6\dfrac{2}{11}\)
\(=\dfrac{-1}{4}\cdot\dfrac{152}{11}-\dfrac{1}{4}\cdot\dfrac{68}{11}\)
=\(\dfrac{1}{4}\cdot\left(\dfrac{-152}{11}-\dfrac{68}{11}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{-220}{11}=-5\)
\(2,25:\dfrac{-6}{25}=\dfrac{9}{4}:\dfrac{-6}{25}\\ =-\dfrac{9}{4}\cdot\dfrac{25}{6}=-\dfrac{3}{4}\cdot\dfrac{25}{2}\\ =-\dfrac{75}{8}\)
\(2.25:\dfrac{-6}{25}\) = \(\dfrac{9}{4}:\dfrac{-6}{25}=\dfrac{9}{4}.\dfrac{25}{-6}=\dfrac{3}{4}.\dfrac{25}{-2}=\dfrac{75}{-8}=\dfrac{-75}{8}\)