Tính bằng cách thuận tiện 5/6 x 4/11 + 5/6 x 7/11
1.Tính bằng cách thuận tiện nhất :
(6/7 - 2/7) : 4/9 x 3 =
7/11 x 5/6 + 4/11 x 5/6 =
1,
a,=4/7:4/9.3
=9/7.3
=27/7
b,=(7/11+4/11).5/6
=1.5/6
=5/6
`(6/7-2/7):4/9xx3=4/7xx9/4xx3=[4xx9xx3]/[7xx4]=27/7`
`7/11xx5/6+4/11xx5/6=(7/11+4/11)xx5/6=11/11xx5/6=5/6`
Tính bằng cách thuận tiện:
9/5 + 4/7 + 6/5 + 3/7 = .........................................................
1/2 x 45/33 x 1/9 x 11/6 = ....................................................
9/5 + 4/7 + 6/5 + 3/7
= ( 9/5 + 6/5 ) + ( 4/7 + 3/7)
= 3 + 1
= 4
9/5 + 4/7 + 6/5 + 3/7 = (9/5 + 6/5) + (4/7 + 3/7)
= 15/5 + 7/7
= 3 + 1
= 4
1/2 x 45/33 x 1/9 x 11/6 = (1/2 x 1/9) x (45/33 x 11/6)
= 1/18 x 45/18
= 5/2
1/2 x 45/33 x 1/9 x 11/6
= (1/2 x 1/9) x (45/33 x 11/6)
= 1/18 x 45/18
= 5/2
Tính bằng cách thuận tiện nhất:
a) 16, 2 x 3, 7 + 5, 7 x 16, 2 - 7, 8 x 4, 8 - 4, 6 x 7, 8 : 11, 2 + 12, 3 + 13, 4 - 12, 6 - 11, 5 - 10, 4
tính bằng cách thuận tiện nhất:
a,7/4 x 11/21 + 11/21 x 5/4
b,23/14 x 6/13 -9/14 x 6/13
a, = \(\frac{11}{21}\)x \(\left(\frac{7}{4}+\frac{5}{4}\right)\)
= \(\frac{11}{21}\)x 3
= \(\frac{33}{21}\)
b) = \(\left(\frac{23}{14}-\frac{9}{14}\right)\)x \(\frac{6}{13}\)
= 1 x \(\frac{6}{13}\)
= \(\frac{6}{13}\)
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Ai xong kết bạn nhé
tính bằng cách thuận tiện
6 x 9 x 49/ 81 x 7 x 6
2/5 + 11/9 + 3/5 + 7/9
2/5+11/9+3/5+7/9= (2/5+3/5)+(11/9+7/9) = 1+2 = 3
Bài 6: Tính bằng cách thuận tiện nhất.
3 x 6/11 + 6/11 x 5 + 2 x 6/11
\(3\times\frac{6}{11}+\frac{6}{11}\times5+2\times\frac{6}{11}\text{ }\)
\(=\frac{6}{11}\times\left(3+5+2\right)\)
\(=\frac{6}{11}\times10\)
\(=\frac{60}{11}\)
tính bằng cách thuận tiện
6 x 9 x 49/ 81 x 7 x 6
2/5 + 11/9 + 3/5 + 7/9
Bài 3. Một hình bình hành có chiều cao là 5/7 m. Độ dài đáy gấp hai lần chiều cao, cạnh bên là 6/7m. Tính chu vi của hình bình hành.
Bài 4. Tính bằng cách thuận tiện nhất
a) 7/4 x 11/21 + 11/21 x 5/4 =
b) 23/14 x 6/13 - 9/14 x 6/13 =
Bài 4:
a: \(=\dfrac{11}{21}\left(\dfrac{7}{4}+\dfrac{5}{4}\right)=\dfrac{11}{21}\cdot3=\dfrac{11}{7}\)
b: \(=\dfrac{6}{13}\left(\dfrac{23}{14}-\dfrac{9}{14}\right)=\dfrac{6}{13}\)
Bài 3. Một hình bình hành có chiều cao là 5/7 m. Độ dài đáy gấp hai lần chiều cao, cạnh bên là 6/7m. Tính chu vi của hình bình hành.
Bài 4. Tính bằng cách thuận tiện nhất
a) 7/4 x 11/21 + 11/21 x 5/4 =
b) 23/14 x 6/13 - 9/14 x 6/13 =
Bài 4:
a: =11/21(7/4+5/4)=11/21x3=11/7
b: =6/13(23/14-9/14)=6/13