có 3x5+84875:4343=?
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a, \(\dfrac{15\times14-1}{13\times15+14}\)
= \(\dfrac{15\times\left(13+1\right)-1}{13\times15+14}\)
= \(\dfrac{13\times15+15-1}{13\times15+14}\)
= \(\dfrac{13\times15+14}{13\times15+14}\)
= 1
c, \(\dfrac{1+2+3+4+5+6+7+8+9}{10}\)
= \(\dfrac{\left(9+1\right)\times\left\{\left(9-1\right):1+1\right\}:2}{10}\)
= \(\dfrac{10\times9:2}{10}\)
= \(\dfrac{9}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
bài 2
a] = 3 x \(\frac{4343}{7171}\)= \(\frac{17372}{7171}\)= \(\frac{172}{71}\)
b] = \(\frac{1}{33}\)x \(\frac{44}{7}\)= \(\frac{1}{3}\)x \(\frac{4}{7}\)=\(\frac{4}{21}\)
bài 1
a] y là 9
b] <=> 64y + 36y = 700 - 75 - 225
<=> 100y = 400
<=> y = 4
trên lớp cô sửa rồi nên mình giải luôn:
1) Tìm y
a) y3 + 3y = 12 x 11
y3 + 3y = 132
y x 10 + 3 + 3 x 10 + y = 132
( y x 10 + y ) + ( 3 x 10 + 3 ) = 132
11 x y + 33 = 132
11 x y = 132 - 33
11 x y = 99
y = 99 : 11
y = 9
b) 64 x y + 225 = 700 - 75 - 36 x y
64 x y + 225 = 625 - 36 x y
64 x y + 36 x y = 625 -225
64 x y + 36 x y = 400
( 64 + 36 ) x y = 400
100 x y = 400
y = 400 : 100
y = 4
2) Tính
a) \(\frac{4343}{7171}+\frac{4343}{7171}+\frac{4343}{7171}+\frac{4343}{7171}\)
\(=\frac{4343}{7171}\times4\)
\(=\frac{43}{71}\times4\)
\(=\frac{172}{71}\)
b) A = \(\frac{1}{33}\times\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)
Ta có:
\(\frac{3333}{2020}=\frac{3333:101}{2020:101}=\frac{33}{20}\)
\(\frac{333333}{303030}=\frac{333333:10101}{303030:10101}=\frac{33}{30}\)
\(\frac{33333333}{42424242}=\frac{33333333:1010101}{42424242:1010101}=\frac{33}{42}\)
A = \(\frac{1}{33}\times\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
A = \(\frac{1}{33}\times33\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
A = 1 x \(\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
A = 1 x \(\left(\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}\right)\)
A = 1 x \(\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
A = 1 x \(\left(\frac{1}{3}-\frac{1}{7}\right)\)
A = 1 x \(\left(\frac{7}{21}-\frac{3}{21}\right)\)
A = 1 x \(\frac{4}{21}\)
A = \(\frac{4}{21}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{121212}{232323}=\dfrac{12x10101}{23x10101}=\dfrac{12}{23}=\dfrac{12.54}{23.54}=\dfrac{648}{1242}\)
\(\dfrac{4343}{5454}=\dfrac{43x101}{54x101}=\dfrac{43}{54}=\dfrac{43.23}{23.54}=\dfrac{989}{1242}\)
mà \(\dfrac{989}{1242}>\dfrac{648}{1242}\)
Nên \(\dfrac{121212}{232323}< \dfrac{4343}{5454}\)
\(\Rightarrow-\dfrac{121212}{232323}>-\dfrac{4343}{5454}\)
\(\Rightarrow\dfrac{-121212}{232323}>\dfrac{4343}{-5454}\)
Đính chính, có thể so sánh cách 2 nhanh hơn.
...\(\dfrac{12}{23}< \dfrac{12+31}{23+31}=\dfrac{43}{54}\)
\(\Rightarrow\dfrac{121212}{232323}< \dfrac{4343}{5454}\)
\(...\Rightarrow\dfrac{-121212}{232323}>\dfrac{4343}{-5454}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(43^{43}-17^{17}\)
\(=43^{40}.43^3-17^{16}.17\)
\(=\overline{.....1}.\overline{.....7}-\overline{.....1}.7\)
\(=\overline{.....7}-\overline{.....7}\)
\(=\overline{.....0⋮}10\)
\(\Rightarrow dpcm\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=15x\left(\dfrac{21}{43}+\dfrac{22}{43}\right)=15x1=15\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(-8686\right)^5:\left(4343\right)^5\)
\(=\left(\frac{-8686}{4343}\right)^5\)
\(=\left(\frac{-2}{1}\right)^5\)
\(=\left(-2\right)^5=-32\)
![](https://rs.olm.vn/images/avt/0.png?1311)
3737 x 43 - 4343 x 37 / 2 + 4 + 6 + ... +100
= 3737 x 43 - 4343 x 37 / 2 + 4 + 6+ ....+ 100
= 0/ 2 + 4+ 6 + ...+ 100
= 0
het rui
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Em xem lại đề bài nhé, lớp 3 chưa học số thập phân đâu em?
ok