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A=\(\frac{5}{7}\)x\(\frac{5}{11}\)-\(\frac{5}{7}\)x\(\frac{2}{11}\)-\(\frac{5}{7}\)x\(\frac{14}{4}\)
A= \(\frac{5}{7}\)x(\(\frac{5}{11}\)-\(\frac{2}{11}\)-\(\frac{14}{4}\))
A=\(\frac{5}{7}\)x(\(\frac{3}{11}\)-\(\frac{14}{4}\))
A=\(\frac{5}{7}\)x\(\frac{-71}{22}\)
A=\(\frac{-355}{154}\)
ta có \(\frac{\frac{4}{5}-\frac{4}{11}+\frac{4}{13}}{\frac{9}{5}-\frac{9}{11}+\frac{9}{13}}+\frac{\frac{5}{7}+\frac{5}{11}-\frac{5}{13}}{\frac{9}{7}+\frac{9}{11}-\frac{9}{13}}\)
\(=\frac{4\left(\frac{1}{5}-\frac{1}{11}+\frac{1}{13}\right)}{9\left(\frac{1}{5}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{5\left(\frac{1}{7}+\frac{1}{11}-\frac{1}{13}\right)}{9\left(\frac{1}{7}+\frac{1}{11}-\frac{1}{13}\right)}\)
\(=\frac{4}{9}+\frac{5}{9}=\frac{9}{9}=1\)
a: =-1/3+1/3=0
b: \(=\dfrac{4}{11}\left(-\dfrac{2}{7}-\dfrac{4}{7}-\dfrac{1}{7}\right)=\dfrac{4}{11}\cdot\left(-1\right)=-\dfrac{4}{11}\)
c: \(=10+\dfrac{5}{9}-3-\dfrac{5}{7}-4-\dfrac{5}{9}=3-\dfrac{5}{7}=\dfrac{16}{7}\)
d: \(=\dfrac{1}{3}+\dfrac{7}{4}-\dfrac{7}{4}+\dfrac{4}{5}=\dfrac{1}{3}+\dfrac{4}{5}=\dfrac{5+12}{15}=\dfrac{17}{15}\)
a: =-1/3+1/3=0
b: =411(−27−47−17)=411⋅(−1)=−411=411(−27−47−17)=411⋅(−1)=−411
c: =10+59−3−57−4−59=3−57=167=10+59−3−57−4−59=3−57=167
d: =13+74−74+45=13+45=5+1215=1715
\(\frac{-7}{9}.\frac{4}{11}+\frac{-7}{9}.\frac{7}{11}+5\frac{7}{9}\)
\(=\frac{-7}{9}.\left(\frac{4}{11}+\frac{7}{11}\right)+5\frac{7}{9}\)
\(=\frac{-7}{9}.1+5\frac{7}{9}\)
\(=\frac{-7}{9}+5\frac{7}{9}=5\)
A= \(11\frac{3}{13}-\left(2\frac{4}{7}+5\frac{3}{13}\right)\)
= \(11\frac{3}{13}-2\frac{4}{7}-5\frac{3}{13}\)
= \(\left(11\frac{3}{13}-5\frac{3}{13}\right)-2\frac{4}{7}\)
= \(6-2\frac{4}{7}\)
= \(3\frac{3}{7}\)
B= \(\left(96\frac{4}{9}+3\frac{7}{11}\right)-4\frac{4}{9}\)
= \(96\frac{4}{9}+3\frac{7}{11}-4\frac{4}{9}\)
= \(\left(96\frac{4}{9}-4\frac{4}{9}\right)+3\frac{7}{11}\)
= \(92+3\frac{7}{11}\)
= \(95\frac{7}{11}\)
C= \(\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}+1\frac{5}{7}\)
= \(\frac{-5}{7}.\left(\frac{2}{11}+\frac{9}{11}\right)+1\frac{5}{7}\)
= \(\frac{-5}{7}.1+1\frac{5}{7}\)
= \(\frac{-5}{7}+1\frac{5}{7}\)
= 1
(ý D bn tự làm ha)
mình vieetts ra thì dài lám
nếu bạn đã chác chăn về kiến thức thì hãy phân tích rồi tìm cách lam
còn ngược lại ban hãy tính máy tính rồi đoán cách làm cũng được
A=(2+3+...+13)-(1+2+...+12)=2+3+...+13-1-2-...-12=(13-1)+(2-2)+(3-3)+...+(12-12)=12
\(\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{9}-\frac{2}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{9}-\frac{4}{11}}\)
\(=\frac{2\cdot\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}{4\cdot\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}\)
\(=\frac{2}{4}=\frac{1}{2}\)
1 .
[ 6 + ( 1/2 ) 3 - /-0,5/ ] : 3/12
= [ 6 + 1/8 - ( -5/10 ) ] . 12/3
= [ 6 + 1/8 + 5/10 ] .4
= [ 6 + 1/8 + 1/2 ] .4
= [ 48/8 + 1/8 + 4/8 ] .4
= 53/8 . 4
= 53 . 1/2
= 53/2
2.
1/3 + 5/4 : (-35/16)
= 1/3 + 5/4 . ( 16/-35 )
= 1/3 + 1 . 4/4 . (-7 )
= 1/3 + 4/28
= 1/3 + 1/7
= 7/21 + 3/21
= 10/21
3 .
11 3/13 - ( 2 4/7 + 5 3/13 )
= 11 3/13 - 2 4/7 - 5 3/13
= { [ ( 11 - 5 ) . ( 3/13 - 3/13 ) } - 2 4/7
= ( 6 . 0 ) - 2 4/7
= 0 - 18/7
= 18/7
4 .
1 3/7 + (-1/3 + 2 4/7 )
= 1 3/7 + ( -1 )/3 + 2 4/7
={ [ ( 1 + 2 ) . ( 3/7 + 4/7 ) } + ( -1 )/3
= ( 3 . 1 ) + ( -1 )/3
= 3 + ( -1 )/3
= 9/3 + ( -1 )/3
= 8/3
5 .
( 6 4/9 + 3 7/11 ) - 4 4/9
= 6 4/9 + 3 7/11 - 4 4/9
= { ( 6 - 4 ) . ( 4/9 - 4/9 ) } + 3 7/11
= ( 2 . 0 ) + 3 7/11
= 0 + 40/11
= 40/11
Nếu đúng thì k cho mình nha !
\(a,15\dfrac{3}{13}-\left(3\dfrac{4}{7}+8\dfrac{3}{13}\right)=15\dfrac{3}{13}-3\dfrac{4}{7}-8\dfrac{3}{13}=\left(15\dfrac{3}{13}-8\dfrac{3}{13}\right)-\dfrac{25}{7}=7-\dfrac{25}{7}=\dfrac{49}{7}-\dfrac{25}{7}=\dfrac{24}{7}\)
\(b,\left(7\dfrac{4}{9}+4\dfrac{7}{11}\right)-3\dfrac{4}{9}=\left(7\dfrac{4}{9}-3\dfrac{4}{9}\right)+4\dfrac{4}{9}=4+\dfrac{40}{9}=\dfrac{36}{9}+\dfrac{40}{9}=\dfrac{76}{9}\)
\(c,\dfrac{-7}{9}.\dfrac{4}{11}+\dfrac{-7}{9}.\dfrac{7}{11}+5\dfrac{7}{9}=\dfrac{-7}{9}\left(\dfrac{4}{11}+\dfrac{7}{11}\right)+\dfrac{52}{9}=\dfrac{-7}{9}.1+\dfrac{52}{9}=\dfrac{-7}{9}+\dfrac{52}{9}=\dfrac{45}{9}=5\)
\(d,50\%.1\dfrac{1}{3}.10.\dfrac{7}{35}.0,75=\dfrac{1}{2}.\dfrac{4}{3}.10.\dfrac{1}{5}.\dfrac{3}{4}=\left(\dfrac{1}{2}.\dfrac{1}{5}.10\right).\left(\dfrac{4}{3}.\dfrac{3}{4}\right)=1.1=1\)
\(e,\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{40.43}=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{40}-\dfrac{1}{43}=1-\dfrac{1}{43}=\dfrac{42}{43}\)
\(\left(7\frac{4}{9}+4\frac{7}{11}\right)-3\frac{4}{9}\)
\(=7\frac{4}{9}+4\frac{7}{11}-3\frac{4}{9}\)
\(=7\frac{4}{9}-3\frac{4}{9}+4\frac{7}{11}\)
\(=4+4\frac{7}{11}=8\frac{7}{11}\)