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a.17⋅(29−(−111)]+29⋅(−17)17⋅(29−(−111))+29⋅(−17)
=17⋅(29+111)−29⋅17=17⋅(29+111)−29⋅17
=17⋅(29+111−29)=17⋅(29+111−29)
=17⋅(29−29+111)=17⋅(29−29+111)
=17⋅111=17⋅111
=2997=2997
b.19⋅43+(−20)⋅43−(−40)19⋅43+(−20)⋅43−(−40)
=19⋅43−20⋅43+40=19⋅43−20⋅43+40
=43(19−20)+40=43(19−20)+40
=43⋅(−1)+40=43⋅(−1)+40
=−43+40=−43+40
=−3=−3
[ chúc bạn học tốt ]
a) Ta có: \(17\cdot\left[29-\left(-111\right)\right]+29\cdot\left(-17\right)\)
\(=17\cdot\left[29+111-29\right]\)
\(=17\cdot111=1887\)
b) Ta có: \(19\cdot43+\left(-20\right)\cdot43-\left(-40\right)\)
\(=43\cdot\left(19-20\right)+40\)
\(=-43+40=-3\)
\(a,\left(4+32+6\right)+\left(10-32-2\right)\\ =\left(10+32\right)+\left(10-32-2\right)\\ =10+32+10-32-2\\ =20-2\\ =18\\ b,300:4+300:6-25\\ =300:\left(4+6\right)-25\\ =300:10-25\\ =30-25\\ =5\\ c,17.\left[29-\left(-111\right)\right]+29.\left(-17\right)\\ =17.\left(29+111\right)+29.\left(-1\right).17\\ =17.\left(29+111-29\right)\\ =17.111=1887\\ d,19.43+\left(-20\right).43-\left(-40\right)\\ =19.43+\left(-20\right).43+40\\ =43.\left(19-20\right)+40\\ =-43+40\\ =-3\)
17[21-(-111)]+17[29-121]=17.132+17.(-92)=17.[132+(-92)]=17.40=680
\(D=\left(\dfrac{3}{111}+\dfrac{29}{17}-\dfrac{15}{59}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
\(D=\left(\dfrac{3}{111}+\dfrac{29}{17}-\dfrac{15}{59}\right).\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)\)
\(D=\left(\dfrac{3}{111}+\dfrac{29}{17}-\dfrac{15}{59}\right)\cdot\left(\dfrac{3-2-1}{6}\right)\)
\(D=\left(\dfrac{3}{111}+\dfrac{29}{17}-\dfrac{15}{59}\right)\cdot\dfrac{0}{6}\)
\(D=\left(\dfrac{3}{111}+\dfrac{29}{17}-\dfrac{15}{59}\right)\cdot0=0\)
\(a,0,5x-\frac{2}{3}x=\frac{7}{12}\Rightarrow\frac{1}{2}x-\frac{2}{3}x=\frac{7}{12}\)
\(\Rightarrow x\left(\frac{1}{2}-\frac{2}{3}\right)=\frac{7}{12}\Rightarrow x\cdot\left(\frac{3}{6}-\frac{4}{6}\right)=\frac{7}{12}\)
\(\Rightarrow x\cdot\left(-1\right)=\frac{7}{12}\Rightarrow x=\frac{7}{12}:\left(-1\right)=\frac{7}{-12}\)
\(c,\frac{\left(x-5\right)}{12}\cdot\frac{9}{29}=\frac{-6}{29}\Rightarrow\frac{\left(x-5\right)}{12}=\frac{-6}{29}:\frac{9}{26}\)
\(\frac{\Rightarrow\left(x-5\right)}{12}=\frac{-6}{9}=\frac{-2}{3}\Rightarrow x-5=-\frac{2}{3}\cdot12\)
\(\Rightarrow x-5=\frac{-24}{3}=-8\Rightarrow x=-8+5=-3\)
\(a,0,5x-\frac{2}{3}x=\frac{7}{12}\)
\(\Rightarrow\frac{1}{2}x-\frac{2}{3}x=\frac{7}{12}\)
\(\Rightarrow-\frac{1}{6}x=\frac{7}{12}\)
\(\Rightarrow x=-\frac{7}{2}\)
\(c,\frac{x-5}{12}\cdot\frac{9}{29}=-\frac{6}{29}\)
\(\Rightarrow\frac{x-5}{12}=-\frac{2}{3}\)
\(\Rightarrow x-5=12.\left(-\frac{2}{3}\right)\)
\(\Rightarrow x-5=-8\)
\(\Rightarrow x=-3\)
\(\dfrac{17}{251}\).\(\dfrac{13}{29}\)+\(\dfrac{17}{251}\).\(\dfrac{16}{29}\)=\(\dfrac{17}{251}\).(\(\dfrac{13}{29}\)+\(\dfrac{16}{29}\))=\(\dfrac{17}{251}\).1=\(\dfrac{17}{251}\)
Áp dụng tính chất phân phối của phép nhân đối với phép cộng, ta có :
\(\dfrac{17}{251}\cdot\left(\dfrac{13}{29}+\dfrac{16}{29}\right)=\dfrac{17}{251}\cdot\dfrac{29}{29}\)
\(=\dfrac{17}{251}\cdot1\)
\(=\dfrac{17}{251}\)
Chúc bạn học tốt !
Bằng 1887
bằng 1887