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1)\(\dfrac{-5}{2}:\dfrac{1}{4}\) = \(\dfrac{-5}{2}\) x \(\dfrac{4}{1}\) = \(\dfrac{-20}{2}\)
\(\dfrac{298}{719}:\left(\dfrac{1}{4}+\dfrac{1}{12}-\dfrac{1}{3}\right)-\dfrac{2011}{2012}\)
\(=\dfrac{298}{719}.0-\dfrac{2011}{2012}\)
\(=0-\dfrac{2011}{2012}\)
\(=-\dfrac{2011}{2012}\)
\(=\dfrac{298}{719}.0-\dfrac{2011}{2012}\)
\(=0-\dfrac{2011}{2012}\)
\(=-\dfrac{2011}{2012}\)
\(\frac{27.\left(18+103-120\right)}{33.\left(15+12\right)}\)=\(\frac{27.1}{33.27}\)=\(\frac{1}{33}\)
mik dang ban moi giai duoc mot bai ha, sorry
\(\frac{27\cdot18+27\cdot103-27\cdot120}{15\cdot33+12\cdot33}=\frac{27\left(18+103-120\right)}{33\left(15+12\right)}\)\(=\frac{27\cdot1}{33\cdot27}=\frac{1}{33}\)
\(\Leftrightarrow x-\frac{1}{3}=\left(+-\right)\frac{5}{6}\)
nếu \(x-\frac{1}{3}=\frac{5}{6}\) nếu \(x-\frac{1}{3}=-\frac{5}{6}\)
\(\Leftrightarrow x=\frac{5}{6}+\frac{1}{3}\) \(\Leftrightarrow x=-\frac{5}{6}+\frac{1}{3}\)
\(\Leftrightarrow x=\frac{7}{6}\) \(\Leftrightarrow x=-\frac{1}{2}\)
Vậy ....
nhớ k mk nha bạn , mk nhanh nhất
thanks
\(\frac{27.18+27.103-120.27}{15.33+33.12}\)
\(=\)\(\frac{27.\left(18+103-120\right)}{33.\left(15+12\right)}\)
\(=\)\(\frac{27.1}{33.27}\)
\(=\)\(\frac{1}{33}\)
1/ \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}\)
\(B< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{7.8}\)
\(B< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}\)
\(B< \dfrac{1}{1}-\dfrac{1}{8}< 1\)
\(B< 1\)
2/ \(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{20}\right)\)
\(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{19}{20}\)
\(B=\dfrac{1\times2\times3\times...\times19}{2\times3\times4\times...\times20}\)
\(B=\dfrac{1}{20}\)
3/ \(A=\dfrac{7}{4}\cdot\left(\dfrac{3333}{1212}+\dfrac{3333}{2020}+\dfrac{3333}{3030}+\dfrac{3333}{4242}\right)\)
\(A=\dfrac{7}{4}\cdot\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)
\(A=\dfrac{7}{4}\cdot\left(\dfrac{33}{3.4}+\dfrac{33}{4.5}+\dfrac{33}{5.6}+\dfrac{33}{6.7}\right)\)
\(A=\dfrac{7}{4}.33.\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)
\(A=\dfrac{231}{4}.\left(\dfrac{1}{3}-\dfrac{1}{7}\right)\)
\(A=\dfrac{231}{4}\cdot\dfrac{4}{21}\)
\(A=11\)
4/ A phải là \(\dfrac{2011+2012}{2012+2013}\)
Ta có : \(B=\dfrac{2011}{2012}+\dfrac{2012}{2013}>\dfrac{2011}{2013}+\dfrac{2012}{2013}=\dfrac{2011+2012}{2013}>\dfrac{2011+2012}{2012+2013}=A\)
\(\Rightarrow B>A\)
Áp dụng tính chất phân phối, rồi tính giá trị biểu thức.
Chẳng hạn,
Với , thì
ĐS. ; C = 0.
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Bài 1)
Ta có:
A = \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}\)
A < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)
A < \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)
A < \(1-\dfrac{1}{8}\) = \(\dfrac{7}{8}\) < 1
Vậy A < 1
Bài 2)
Ta thấy:
\(\dfrac{2011}{2012+2013}< \dfrac{2011}{2012};\dfrac{2012}{2012+2013}< \dfrac{2012}{2013}\)
\(\Rightarrow\) \(\dfrac{2011}{2012+2013}+\dfrac{2012}{2012+2013}< \dfrac{2011}{2012}+\dfrac{2012}{2013}\)
\(\Rightarrow\) \(\dfrac{2011+2012}{2012+2013}< \dfrac{2011}{2012}+\dfrac{2012}{2013}\)
\(\Rightarrow\) A < B
Bài 3)
Ta có:
B = \(\left(1-\dfrac{1}{1}\right)\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)......\left(1-\dfrac{1}{20}\right)\)
= \(0.\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)......\left(1-\dfrac{1}{20}\right)\)
= 0
Bài 3)
Ta có:
A = \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.....+\dfrac{1}{2^{2012}}\)
\(\Rightarrow\) 2A = \(2\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.....+\dfrac{1}{2^{2012}}\right)\)
\(\Rightarrow\) 2A = \(2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{2011}}\)
\(\Rightarrow\) 2A - A = \(\left(2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{2011}}\right)\)-\(\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.....+\dfrac{1}{2^{2012}}\right)\)
\(\Rightarrow\) A = 2 - \(\dfrac{1}{2^{2012}}\) = \(\dfrac{2^{2013}-1}{2^{2012}}\)
Bài 5)
\(\pi\) + 5 \(⋮\) \(\pi\) - 2
\(\Leftrightarrow\) \(\pi\) - 2 + 7 \(⋮\) \(\pi\) - 2
\(\Leftrightarrow\) 7 \(⋮\) \(\pi\) - 2 (vì \(\pi\) - 2 \(⋮\) \(\pi\) - 2)
\(\Leftrightarrow\) \(\pi\) - 2 \(\in\) Ư(7)
\(\Leftrightarrow\) \(\pi\) - 2 \(\in\) \(\left\{\pm1;\pm7\right\}\)
\(\Leftrightarrow\) \(\pi\) \(\in\) \(\left\{1;3;-5;9\right\}\)
a) \(\dfrac{298}{719}:\left(\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{3}\right)-\dfrac{2011}{2012}=\dfrac{298}{719}:\left(\dfrac{3}{12}+\dfrac{1}{12}+\dfrac{4}{12}\right)-\dfrac{2011}{2012}=\dfrac{298}{719}:\left(\dfrac{3+1+4}{12}\right)-\dfrac{2011}{2012}=\dfrac{298}{719}:\dfrac{2}{3}-\dfrac{2011}{2012}=\dfrac{298}{719}\cdot\dfrac{3}{2}-\dfrac{2011}{2012}=\dfrac{149.3}{719.1}-\dfrac{2011}{2012}=\dfrac{447}{719}-\dfrac{2011}{2012}=\dfrac{889364}{1446628}-\dfrac{1445909}{1446628}=\dfrac{889364-1445909}{1446628}=-\dfrac{556545}{1446628}.\)b)\(\dfrac{27\cdot18+27+103-120\cdot27}{15\cdot33+33\cdot12}=\dfrac{27\left(18+103-120\right)}{33\left(15+12\right)}=\dfrac{27\cdot1}{33\cdot27}=\dfrac{1\cdot1}{33\cdot1}=\dfrac{1}{33}\)