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11 x 34 - (34 + 34 x 6 + 3 x 34)
=11 x 34 - (34 x1 + 34 x 6+34 x 3)
=11 x 34 - 34 x 10
= 34 x 1
= 34
11x34-(34+34x6+3x34)=11x34-[34x(1+6+3)]
=11x34-(34x10)
=11x34-34x 10
=(11-10)x34
=1x34=34
![](https://rs.olm.vn/images/avt/0.png?1311)
34 x 28 - 34 x 11 - 7 x 34
= 34 x (28 - 11 - 7)
= 34 x 10
= 340
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34 x 1 125 -25 x 34- 34 x 100
= 38 250 - 850 - 34 000
= 37 400 - 34 000
= 3 000
~HT~
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a: \(\dfrac{5}{3}\cdot\dfrac{7}{10}=\dfrac{5}{10}\cdot\dfrac{7}{3}=\dfrac{14}{3}>2\)
\(\dfrac{5}{7}:\dfrac{7}{10}=\dfrac{5}{7}\cdot\dfrac{10}{7}=\dfrac{50}{49}< 2\)
=>\(\dfrac{5}{3}\cdot\dfrac{7}{10}>\dfrac{5}{7}:\dfrac{7}{10}\)
b: \(\dfrac{4}{5}\cdot\dfrac{5}{6}=\dfrac{4}{6}=\dfrac{2}{3}\)
\(\dfrac{8}{9}:\dfrac{4}{3}=\dfrac{8}{9}\cdot\dfrac{3}{4}=\dfrac{8}{4}\cdot\dfrac{3}{9}=\dfrac{2}{3}\)
Do đó: \(\dfrac{4}{5}\cdot\dfrac{5}{6}=\dfrac{8}{9}:\dfrac{4}{3}\)
c: \(\dfrac{5}{11}\cdot\dfrac{33}{15}=\dfrac{5}{15}\cdot\dfrac{33}{11}=\dfrac{3}{3}=1\)
\(\dfrac{6}{17}\cdot\dfrac{34}{25}=\dfrac{34}{17}\cdot\dfrac{6}{25}=2\cdot\dfrac{6}{25}=\dfrac{12}{25}< 1\)
=>\(\dfrac{5}{11}\cdot\dfrac{33}{15}>\dfrac{6}{17}\cdot\dfrac{34}{25}\)
d: \(\dfrac{15}{19}\cdot\dfrac{38}{5}=\dfrac{15}{5}\cdot\dfrac{38}{19}=3\cdot2=6\)
\(\dfrac{15}{16}:\dfrac{3}{8}=\dfrac{15}{16}\cdot\dfrac{8}{3}=\dfrac{15}{3}\cdot\dfrac{8}{16}=\dfrac{5}{2}\)<6
=>\(\dfrac{15}{19}\cdot\dfrac{38}{5}>\dfrac{15}{16}:\dfrac{3}{8}\)
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Lời giải:
a. $(132\times 6-66\times 12)\times (132\times 6+66)$
$=(66\times 2\times 6-66\times 12)\times (132\times 6+66)$
$=(66\times 12-66\times 12)\times (132\times 6+66)$
$=0\times (132\times 6+66)=0$
b.
$100-99+98-97+96-95+...+4-3+2$
$=(100-99)+(98-97)+(96-95)+....+(4-3)+2$
$=1+1+1+....+1+2$
Số lần xuất hiện của 1 là: $[(100-3):1+1]:2=49$
Giá trị của tổng cần tính là: $1\times 49+2=51$
![](https://rs.olm.vn/images/avt/0.png?1311)
38 x 99
= 38 x ( 100 - 1 )
= 38 x 100 - 38 x 1
= 3800 - 38
= 3762
38*99=38.(100-1)
=38*100-38*1
=380-38=300