Tính giá trị biểu thức sau:
A = 1+ 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9
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\(=\dfrac{2}{3}+\dfrac{1}{3}.\left(\dfrac{7}{18}\right):\dfrac{7}{12}\)
\(=\dfrac{2}{3}+\dfrac{7}{54}:\dfrac{7}{12}\)
\(=\dfrac{2}{3}+\dfrac{2}{9}\)
\(=\dfrac{8}{9}\)
A)\(\dfrac{7}{20}-\left(\dfrac{5}{8}-\dfrac{2}{5}\right)\)
\(=\dfrac{7}{20}-\left(\dfrac{25}{40}-\dfrac{16}{40}\right)\)
\(=\dfrac{7}{20}-\dfrac{9}{40}\)
\(=\dfrac{14}{40}-\dfrac{9}{40}=\dfrac{5}{40}=\dfrac{1}{8}\)
B) \(\dfrac{5}{6}+\left(\dfrac{5}{9}-\dfrac{1}{4}\right)\)
\(=\dfrac{5}{6}+\left(\dfrac{20}{36}-\dfrac{9}{36}\right)\)
\(=\dfrac{5}{6}+\dfrac{11}{36}\).
\(=\dfrac{30}{36}+\dfrac{11}{36}=\dfrac{41}{36}\)
C) \(\dfrac{9}{10}-\left(\dfrac{2}{5}-\dfrac{3}{10}\right)+\dfrac{7}{20}\)
\(=\dfrac{9}{10}-\left(\dfrac{4}{10}-\dfrac{3}{10}\right)+\dfrac{7}{20}\)
\(=\dfrac{9}{10}-\dfrac{1}{10}+\dfrac{7}{20}\)
\(=\dfrac{18}{20}-\dfrac{2}{20}+\dfrac{7}{20}=\dfrac{23}{20}\)
a: =7/20-5/8+2/5
=14/40-25/40+16/40
=5/40=1/8
b: =5/6+5/9-1/4
=30/36+20/36-9/36
=41/36
c: =9/10-2/5+3/10+7/20
=12/10-2/5+7/20
=7/20+6/5-2/5
=7/20+4/5
=7/20+16/20
=23/20
1:
a: \(A=2+3\sqrt{x^2+1}>=3\cdot1+2=5\)
Dấu = xảy ra khi x=0
b: \(B=\sqrt{x+8}-7>=-7\)
Dấu = xảy ra khi x=-8
\(=\dfrac{1}{2}\cdot\dfrac{6}{5}\cdot\dfrac{8}{7}+\dfrac{3}{7}\cdot\dfrac{9}{5}=\dfrac{24}{35}+\dfrac{27}{35}=\dfrac{51}{35}\)
Lời giải
0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10
= (1 + 9) + (2 + 8) + (3 + 7) + ( 4 + 6) + 10 + 5
= 10 + 10 + 10 + 10 + 10 + 5
= 55
Vậy đáp án đúng là B
A = (1- 2) \(\times\) ( 4 - 3) \(\times\) (5 - 6) \(\times\) (8 - 7) \(\times\) (9 - 10) \(\times\) (12 - 11) \(\times\)(13 - 14)
A = (-1) \(\times\) 1 \(\times\) (-1) \(\times\) 1 \(\times\) (-1) \(\times\) 1 \(\times\) (-1)
A = 1
A=1+2+3+4+5+6+7+8+9
a=(1+9)+(2+8)+(3+7)+(4+6)+5
A=10+10+10+10+5
A=45
VAY:x=45
A= (1+9) + ( 2+8 ) + ( 3+ 7 )+ (4 + 6) +5
A = 10 + 10 + 10 +10 + 5
A = 40 + 5
A = 45