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\(=15x\left(\dfrac{21}{43}+\dfrac{22}{43}\right)=15x1=15\)
\(a,=15\left(\dfrac{2121}{4343}+\dfrac{222222}{434343}\right)=15\left(\dfrac{21}{43}+\dfrac{22}{43}\right)=15\cdot1=15\)
\(1,\\ a,=\dfrac{7}{19}\times\left(\dfrac{1}{3}+\dfrac{2}{3}\right)=\dfrac{7}{19}\times1=\dfrac{7}{19}\\ b,=\dfrac{2}{5}+\left(\dfrac{3}{4}+\dfrac{1}{4}\right)=\dfrac{2}{5}+1=\dfrac{7}{5}\\ 2,\\ a,=15\times\left(\dfrac{2121}{4343}+\dfrac{222222}{434343}\right)\\ =15\times\left(\dfrac{2121:101}{4343:101}+\dfrac{222222:10101}{434343:10101}\right)\\ =15\times\left(\dfrac{21}{43}+\dfrac{22}{43}\right)=15\times\dfrac{43}{43}=15\times1=15\)
\(3,\)
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 \(AC>BC>AB\)
Bài 1
a; \(\dfrac{7}{19}\) x \(\dfrac{1}{3}\) + \(\dfrac{7}{19}\) x \(\dfrac{2}{3}\)
= \(\dfrac{7}{19}\) x (\(\dfrac{1}{3}+\dfrac{2}{3}\))
= \(\dfrac{7}{19}\) x 1
= \(\dfrac{7}{19}\)
b; 15 x \(\dfrac{2121}{4343}\) + 15 x \(\dfrac{212121}{434343}\)
= 15 x \(\dfrac{21}{43}\) + 15 x \(\dfrac{21}{43}\)
= 15 x \(\dfrac{21}{43}\) x (1 + 1)
= 15 x \(\dfrac{21}{43}\) x 2
= (15 x 2) x \(\dfrac{21}{43}\)
= 30 x \(\dfrac{21}{43}\)
= \(\dfrac{630}{43}\)
\(=15.\left(\frac{21.10101}{43.10101}-\frac{33.10101}{35.10101}\right)\)
\(=15.\left(\frac{21}{43}-\frac{33}{35}\right)\)
\(=15.\frac{-684}{1505}=\frac{-2052}{301}\)
1/2x3+1/3x4+1/4x5+1/5x6
= 1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6
=1/2-1/6
=1/3
15:x= 0.85+0.35
15:x= 1,2
x= 15:1,2
x= 12,5
nha bn, mk nghĩ vậy
Câu 2: Bằng 50/101
(Đặt biểu thức đó là A. Nhân 2 vào 2A = 2/1x3+2/3x5+2/5x7+...+2/99x101
=>2A= 1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101
=1-1/101=100/101
=>A=100/101:2=50/101)
=15 nhé
=15 nhé bạn