Giúp tôi giải toán:
Tính bằng cách thuận tiện nhất
15 x \(\frac{2121}{4343}\)+ 15 x \(\frac{222222}{434343}\)
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15×2121/4343+15×222222/434343
= 15 x (21x101)/(101x43) + 15x(22x 10101)/(43x10101)
= 15 x 21/43 + 15 x 22/43
= 15x(21/43 + 22/43)
= 15x1 = 15
\(=15x\left(\dfrac{21}{43}+\dfrac{22}{43}\right)=15x1=15\)
\(1,\\ a,=\dfrac{7}{19}\times\left(\dfrac{1}{3}+\dfrac{2}{3}\right)=\dfrac{7}{19}\times1=\dfrac{7}{19}\\ b,=\dfrac{2}{5}+\left(\dfrac{3}{4}+\dfrac{1}{4}\right)=\dfrac{2}{5}+1=\dfrac{7}{5}\\ 2,\\ a,=15\times\left(\dfrac{2121}{4343}+\dfrac{222222}{434343}\right)\\ =15\times\left(\dfrac{2121:101}{4343:101}+\dfrac{222222:10101}{434343:10101}\right)\\ =15\times\left(\dfrac{21}{43}+\dfrac{22}{43}\right)=15\times\dfrac{43}{43}=15\times1=15\)
\(3,\)
Cạnh \(AC=\) chu vi ABC \(-AB-BC=\dfrac{4}{5}-\dfrac{1}{5}-\dfrac{1}{4}=\dfrac{3}{5}-\dfrac{1}{4}=\dfrac{7}{20}\left(m\right)\)
Vì \(\dfrac{7}{20}>\dfrac{5}{20}>\dfrac{4}{20}\Rightarrow\dfrac{7}{20}>\dfrac{1}{4}>\dfrac{1}{5}\) nên \(AC>BC>AB\)
`a)2018xx(1/2+1212/2424)`
`=2018xx(1/2+1/2)`
`=2018xx2/2=2018`
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`b)15xx2121/4343+15xx222222/434343`
`=15xx(2121/4343+222222/434343)`
`=15xx(21/43+22/43)`
`=15xx43/43=15`
b. 16×25+44×100/ 29×96+142×48
=400+4×11×100/29×96+71×2×48
=400+400×11/29×96+71×96
=400×(1+11)/96×(29+71)
=400×12/96×100=1/2
\(a,=15\left(\dfrac{2121}{4343}+\dfrac{222222}{434343}\right)=15\left(\dfrac{21}{43}+\dfrac{22}{43}\right)=15\cdot1=15\)
\(15x\frac{2121}{4343}+15x\frac{222222}{434343}\)
\(=15x\frac{21}{43}+15x\frac{22}{43}\)
\(=15x\left(\frac{21}{43}+\frac{22}{43}\right)\)
\(=15x1\)
\(=15\)
= 15 x 21/43 + 15x 22/43
= 15 x (21/43 + 22/43)
= 15 x 1 =15