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a: \(\Leftrightarrow x=\dfrac{1}{9}+\dfrac{5}{7}=\dfrac{52}{63}\)
b: \(\Leftrightarrow x=\dfrac{1}{10}+\dfrac{1}{15}=\dfrac{1}{6}\)
c: \(\Leftrightarrow x=\dfrac{-3}{7}-\dfrac{4}{5}+\dfrac{2}{3}=-\dfrac{59}{105}\)
d: \(\Leftrightarrow x=\dfrac{-2}{15}+\dfrac{3}{10}=\dfrac{1}{6}\)
a: =-5/6-3/7=-35/42-18/42=-53/42
b: =2/5-4/9=18/45-20/45=-2/45
c: =-24/35
d: =2/3x-5/4=-10/12=-5/6
\(\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\\ =\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{1}{7\times9}+\dfrac{1}{9\times11}+\dfrac{1}{11\times13}\\ =\left(\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+\dfrac{2}{9\times11}+\dfrac{2}{11\times13}\right).\dfrac{1}{2}\\ =\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\right).\dfrac{1}{2}\\ =\left(\dfrac{1}{3}-\dfrac{1}{13}\right).\dfrac{1}{2}\\ =\dfrac{10}{39}.\dfrac{1}{2}\\ =\dfrac{5}{39}\)
115+135+163+199+1143=13×5+15×7+17×9+19×11+111×13=(23×5+25×7+27×9+29×11+211×13).12=(13−15+15−17+17−19+19−111+111−113).12=(13−113).12=1039.12=539
a) .
Ta có :
\(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+....+\dfrac{1}{200^2}< \dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{199.200}\)
Gọi \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{199.200}\left(là.B\right)\)
ta có:
\(B=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{199}-\dfrac{1}{200}\)
\(\dfrac{1}{2}-\dfrac{1}{200}< \dfrac{1}{2}\) (1 )
Từ (1) ta suy ra được :
\(A=\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+....+\dfrac{1}{200^2}< \dfrac{1}{2}\left(đ.p.c.m\right)\)
b ) .
Ta có :
\(B=2\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{200^2}\right)\)\(>2\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{199.200}\right)\)
gọi vế thứ (2) ở B là C
Ta có :
\(C=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{199}-\dfrac{1}{200}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{200}\right)=\dfrac{99}{100}>\dfrac{9}{20}\) (1)
Từ (1) ta suy ra được :
\(B=\dfrac{2}{2^2}+\dfrac{2}{3^2}+\dfrac{2}{4^2}+.....+\dfrac{2}{200^2}>\dfrac{9}{20}\)
1:
1: \(A=\dfrac{3}{14}+\dfrac{2}{7}-\dfrac{3}{7}=\dfrac{3}{14}-\dfrac{1}{7}=\dfrac{3}{14}-\dfrac{2}{14}=\dfrac{1}{14}\)
2: \(\Leftrightarrow-\dfrac{2}{3}\cdot x=-\dfrac{2}{3}\)
=>x=1
3:
a: Cùng phía với điểm I: C,K,B
b: Tia đối của tia CK là CI hoặc CA
c: AC=2*1=2cm
=>CB=8-2=6cm
KB=6/2=3cm
a, Nửa chu vi là \(36:2=18\left(m\right)\)
Chiều dài là \(18:\left(2+1\right)\times2=12\left(m\right)\)
Chiều rộng là \(18-12=6\left(m\right)\)
Diện tích phòng là \(12\times6=72\left(m^2\right)\)
b, Đổi \(72\left(m^2\right)=720000\left(cm^2\right)\)
Diện tích 1 viên gạch là \(30\times30=900\left(cm^2\right)\)
Cần sd \(720000:900=800\left(viên.gạch\right)\) để lát đầy sàn phòng
a)\(=1,25\cdot35,1-35,1\cdot\dfrac{1}{4}=35,1\cdot\left(1,25-\dfrac{1}{4}\right)\)
\(=35,1\cdot\left(\dfrac{5}{4}-\dfrac{1}{4}\right)=35,1\)
b)\(=-62-\dfrac{9}{5}+3,1\cdot\left(27-5,2\right)\)
\(=-62-\dfrac{9}{5}+3,1\cdot\dfrac{109}{5}=\dfrac{189}{50}=3,78\)
c)\(=11,44+12,56-1+64=87\)
d)\(=2,5\cdot\left(-1,4+14,4\right)+2,1=2,5\cdot13+2,1=34,6\)
e)\(=11,44+12,56-30,05+9=2,95\)