Bài 3: Nhân, chia số hữu tỉ

Hải Nam Trần
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Nguyễn Hạnh Ngân
22 tháng 7 2018 lúc 8:37

vi (x+2)(x-3)=5

suy ra(x+2),(x-3)={1,5,-1,-5}

x+2=1 va x-3=5

x=-1 x=8

mấy trường hợp kia tương tự nhahihi

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Cô bé áo xanh
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Luân Đào
1 tháng 1 2018 lúc 18:10

Sửa đề

\(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{\dfrac{1}{4}-\dfrac{1}{6}-\dfrac{1}{64}-\dfrac{1}{256}}+\dfrac{5}{8}\)

\(=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}\cdot\dfrac{3\left(\dfrac{1}{4}-\dfrac{1}{6}-\dfrac{1}{64}-\dfrac{1}{256}\right)}{\dfrac{1}{4}-\dfrac{1}{6}-\dfrac{1}{64}-\dfrac{1}{256}}+\dfrac{5}{8}\)

\(=\dfrac{1}{2}\cdot3+\dfrac{5}{8}=\dfrac{3}{2}+\dfrac{5}{8}=\dfrac{17}{8}\)

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Ái Nữ
1 tháng 1 2018 lúc 18:11

A= \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

=> \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2.(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13})}.\dfrac{3.(\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}-\dfrac{1}{256})}{\dfrac{4}{4}-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

=> \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2.(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13})}.\dfrac{3.(\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}-\dfrac{1}{256})}{4.(\dfrac{1}{4})-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

=> \(\dfrac{1}{2}.\dfrac{3.(\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{4^3}-\dfrac{1}{16^2})}{4.(\dfrac{1}{4})-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

=> \(\dfrac{1}{2}.\dfrac{3.(-\dfrac{1}{4^2}-\dfrac{1}{16^2})}{4-\dfrac{1}{4^3}}+\dfrac{5}{8}\)

=> \(\dfrac{1}{2}.\dfrac{3.(-\dfrac{1}{16^2})}{4.-\dfrac{1}{4^2}}+\dfrac{5}{8}\)

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Ái Nữ
1 tháng 1 2018 lúc 18:14

A= \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

=> \(\dfrac{1}{2}.\dfrac{3.(\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}-\dfrac{1}{256})}{4.\dfrac{1}{4}-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

=> \(\dfrac{1}{2}.\dfrac{3.(\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}-\dfrac{1}{4^4})}{4.\dfrac{1}{4}-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

=> \(\dfrac{1}{2}.\dfrac{3.\dfrac{1}{4^3}}{4.}+\dfrac{5}{8}\)

Cách này cũng được và gọn hơn

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Cô bé áo xanh
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Ngô Tấn Đạt
1 tháng 1 2018 lúc 19:18

\(A=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{16}\left(1+..+16\right)\\ =1+\dfrac{1}{2}.\dfrac{\left(1+2\right).2}{2}+\dfrac{1}{3}.\dfrac{\left(1+3\right).3}{2}+....+\dfrac{1}{16}.\dfrac{\left(16+1\right).16}{2}\\ =\dfrac{2}{2}+\dfrac{\left(1+2\right)}{2}+...+\dfrac{16+1}{2}\\ =\dfrac{2+...+17}{2}\\ =\dfrac{152}{2}=76\)

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 Mashiro Shiina
2 tháng 1 2018 lúc 0:21

Ủng hộ 1 cách khác :v

Xét thừa số tổng quát: \(\dfrac{1}{n}\left(1+2+3+...+n\right)=\dfrac{1+2+3+...+n}{n}=\dfrac{\left[\left(n-1\right):1+1\right]:2\left(n+1\right)}{n}=\dfrac{\dfrac{n}{2}\left(n+1\right)}{n}=\dfrac{\dfrac{n^2}{2}+\dfrac{n}{2}}{n}=\dfrac{n\left(\dfrac{n}{2}+\dfrac{1}{2}\right)}{n}=\dfrac{n+1}{2}\)

Hay:

\(A=\dfrac{2}{2}+\dfrac{2+1}{2}+\dfrac{3+1}{2}+...+\dfrac{16+1}{2}\)

\(A=\dfrac{2+3+4+...+17}{2}\)

Đơn giản r :v

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Nguyễn Thị Trà
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Phạm Ngân Hà
17 tháng 12 2017 lúc 20:19

a) \(\dfrac{1}{2}.\dfrac{1}{-3}+\dfrac{1}{-3}.\dfrac{1}{4}+\dfrac{1}{4}.\dfrac{1}{-5}+\dfrac{1}{-5}.\dfrac{1}{6}\)

\(=\dfrac{1}{-3}\left(\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{1}{-5}\left(\dfrac{1}{4}+\dfrac{1}{6}\right)\)

\(=\dfrac{1}{-3}.\dfrac{3}{4}+\dfrac{1}{-5}.\dfrac{5}{12}\)

\(=\left(-\dfrac{1}{4}\right)+\left(-\dfrac{1}{12}\right)\)

\(=-\dfrac{1}{3}\)

b) \(A=\dfrac{81^4.3^{10}.27^5.3^{12}}{3^{18}.9^3.243^2}\)

\(=\dfrac{9^8.9^8.9^{13}.9^{10}}{9^{16}.9^3.9^3}\)

\(=\dfrac{9^{39}}{9^{22}}\)

\(=9^{17}\)

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Nguyễn Thị Thu
20 tháng 12 2017 lúc 21:38

\(A=\dfrac{81^4\cdot3^{10}\cdot27^5\cdot3^{12}}{3^{18}\cdot9^3\cdot243^2}=\dfrac{3^{16}\cdot3^{10}\cdot3^{15}\cdot3^{12}}{3^{18}\cdot3^6\cdot3^{10}}=\dfrac{3^{53}}{3^{34}}=3^{19}\)

Vậy A = 319

Ngân Hà làm đúng phần a) nhưng làm sai phần b) nên mk chỉ làm phần b) thôi

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Aria Nguyễn
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Giang Thủy Tiên
25 tháng 12 2017 lúc 21:15

\(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{1}{3}-0,25+0,2}{1\dfrac{1}{6}-0,875+0,7}+\dfrac{6}{7}\)

\(=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2\cdot\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}\cdot\dfrac{\dfrac{2}{6}-\dfrac{2}{8}+\dfrac{2}{10}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}+\dfrac{6}{7}\)

\(=\dfrac{1}{2}\cdot\dfrac{2\cdot\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{10}\right)}{7\cdot\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{10}\right)}+\dfrac{6}{7}\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{7}+\dfrac{6}{7}\)

\(=\dfrac{1}{7}+\dfrac{6}{7}=\dfrac{7}{7}=1\)

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Nam Nguyễn
25 tháng 12 2017 lúc 21:16

\(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}.\dfrac{\dfrac{1}{3}-0,25+0,2}{1\dfrac{1}{6}+0,875+0,7}+\dfrac{6}{7}.\)

\(=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}.\dfrac{\dfrac{2}{6}-\dfrac{2}{8}+\dfrac{2}{10}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}+\dfrac{6}{7}.\)

\(=\dfrac{1}{2}.\dfrac{2\left(\dfrac{1}{6}+\dfrac{1}{8}+\dfrac{1}{10}\right)}{7\left(\dfrac{1}{6}+\dfrac{1}{8}+\dfrac{1}{10}\right)}+\dfrac{6}{7}.\)

\(=\dfrac{1}{2}.\dfrac{2}{7}+\dfrac{6}{7}.\)

\(=\dfrac{1}{7}+\dfrac{6}{7}=\dfrac{7}{7}=1.\)

Vậy.....

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Vũ Phương Anh
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Nguyễn Thanh Hằng
21 tháng 12 2017 lúc 20:49

1,a/ Theo t,c dãy tỉ số bằng nhau ta có :

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{-14}{7}=2\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=-2\\\dfrac{y}{5}=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=-10\end{matrix}\right.\)

Vậy ...

b, Theo t,c dãy tỉ số bằng nhau ta có :

\(\dfrac{x}{7}=\dfrac{y}{5}=\dfrac{x-y}{7-5}=\dfrac{8}{2}=4\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{7}=4\\\dfrac{y}{5}=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=28\\y=20\end{matrix}\right.\)

Vậy ...

2/a, Theo t,c dãy tỉ số bằng nhau ta có :

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{x+y+z}{2+5+7}=\dfrac{56}{14}=4\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=4\\\dfrac{y}{5}=4\\\dfrac{z}{7}=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=10\\z=28\end{matrix}\right.\)

Vậy ...

b/ \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{8}\)

\(\Leftrightarrow\dfrac{2x}{6}=\dfrac{y}{5}=\dfrac{z}{8}\)

Theo t,c dãy tỉ số bằng nhau ta có :

\(\dfrac{2x}{6}=\dfrac{y}{5}=\dfrac{z}{8}=\dfrac{2x+y-z}{6+5-8}=\dfrac{12}{3}=4\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2x}{6}=4\\\dfrac{y}{5}=4\\\dfrac{z}{8}=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=24\\y=20\\z=32\end{matrix}\right.\)

Vậy ..

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Thiên Phong
21 tháng 12 2017 lúc 20:55

Bài Giải:

Bài 1:

a) Theo đề bài, ta có:

\(\dfrac{x}{2}=\dfrac{y}{5}\)và x+y=-4

Áp dụng tính chất của dãy tỉ số bằng nhau

Ta có: \(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{-14}{7}=-2\)

Suy ra: x = 2 . (-2) =-4

y = 5 . (-2) =-10

Vậy: x = -4 và y = -10

Mấy câu sau cậu cứ dựa vào bài trên để giải nhé!

Tick cho Phong nhé:>

Yêu nhiều>3

#Phong_419

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Nguyễn Thị Bích Thủy
21 tháng 12 2017 lúc 20:59

Bài 1
a) \(\dfrac{x}{2}=\dfrac{y}{5}\)
\(\text{Mà }x+y=-14\)
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{-14}{7}=-2\)
\(\Rightarrow\dfrac{x}{2}=-2\Rightarrow x=2.\left(-2\right)=-4\)
\(\Rightarrow\dfrac{y}{5}=-2\Rightarrow y=5.\left(-2\right)=-10\)
\(\text{Vậy }\left\{{}\begin{matrix}x=-4\\y=-10\end{matrix}\right.\)
b) \(\dfrac{x}{7}=\dfrac{y}{5}\)
\(\text{Mà }x-y=8\)
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{7}=\dfrac{y}{5}=\dfrac{x-y}{7-5}=\dfrac{8}{2}=4\)
\(\Rightarrow\dfrac{x}{7}=4\Rightarrow x=4.7=28\)
\(\Rightarrow\dfrac{y}{5}=4\Rightarrow y=4.5=20\)
\(\text{Vậy }\left\{{}\begin{matrix}x=28\\y=20\end{matrix}\right.\)
Bài 2:
a) \(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}\)
\(\text{Mà }x+y+z=56\)
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{x+y+z}{2+5+7}=\dfrac{56}{14}=4\)
\(\Rightarrow\dfrac{x}{2}=4\Rightarrow x=4.2=8\)
\(\Rightarrow\dfrac{y}{5}=4\Rightarrow y=4.5=20\)
\(\Rightarrow\dfrac{z}{7}=4\Rightarrow z=4.7=28\)
\(\text{Vậy }\left\{{}\begin{matrix}x=8\\y=20\\z=28\end{matrix}\right.\)
b) \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{8}\)
\(\Rightarrow\dfrac{2x}{6}=\dfrac{y}{5}=\dfrac{z}{8}\)
\(\text{Mà }2x+y-z=12\)
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\dfrac{2x}{6}=\dfrac{y}{5}=\dfrac{z}{8}=\dfrac{2x+y-z}{6+5-8}=\dfrac{12}{3}=4\)
\(\Rightarrow\dfrac{2x}{6}=4\Rightarrow2x=4.6=24\Rightarrow x=24:2=12\)
\(\Rightarrow\dfrac{y}{5}=4\Rightarrow y=4.5=20\)
\(\Rightarrow\dfrac{z}{8}=4\Rightarrow z=4.8=32\)
\(\text{Vậy }\left\{{}\begin{matrix}x=12\\y=20\\z=32\end{matrix}\right.\)

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Vũ Phương Anh
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Đạt Trần Tiến
21 tháng 12 2017 lúc 21:01

a,\(x+\frac{1}{4}=\frac{3}{4} \)

<=>x=\(\frac{3}{4}-\frac{1}{4} \)

<=>\(x=\frac{1}{2} \)

b,\(\frac{3}{4}-\frac{2}{5}x=\frac{29}{60} \)

<=>\(\frac{2}{5}x =\frac{3}{4}-\frac{29}{60} \)

<=>\(\frac{2}{5}x=\frac{4}{15} \)

<=x=\(\frac{2}{3} \)

c,\(2x-\frac{1}{3}=\frac{-5}{6} \)

2x=\(\frac{-5}{6}+\frac{1}{3} \)

2x=\(\frac{-1}{2} \)

x=\(\frac{-1}{4} \)

d,2-\(\frac{3}{4x}=\frac{1}{2} \)

<=>\(\frac{3}{4x}=2-\frac{1}{2} \)

<=>\(\frac{3}{4}x=\frac{3}{2} \)

<=>x=2

e,\(\frac{11}{12}- \frac{2}{3}|x| =\frac{3}{8} \)

<=>\(\frac{2}{3}|x|=\frac{13}{24} \)

<=>\(|x|=\frac{13}{16} \)

<=>x=\(\pm\frac{13}{16} \)

f,\(|2x-1|=5\)

<=>2x-1=5 hoặc 2x-1=-5

<=> 2x=6 2x=-4

<=> x=3 x=-2

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Thiên Phong
21 tháng 12 2017 lúc 21:22

Giải

a) \(x+\dfrac{1}{4}=\dfrac{3}{4}\)

=> \(x=\dfrac{3}{4}-\dfrac{1}{4}\)=>\(x=\dfrac{1}{2}\)

b)\(\dfrac{3}{4}-\dfrac{2}{5}x=\dfrac{29}{60}\)

=>\(\dfrac{2}{5}x=\dfrac{3}{4}-\dfrac{29}{60}\)=>\(\dfrac{2}{5}x=\dfrac{4}{15}\)

=>\(x=\dfrac{4}{15}:\dfrac{2}{5}=\dfrac{2}{3}\)

c)\(2x-\dfrac{1}{3}=\dfrac{-5}{6}\)=>\(2x=\dfrac{-5}{6}+\dfrac{1}{3}\)=\(\dfrac{-1}{2}\)

=>\(x=\dfrac{-1}{2}:2=\dfrac{-1}{4}\)

d)\(2-x:\dfrac{3}{4}=\dfrac{1}{2}\)=>\(x:\dfrac{3}{4}=2-\dfrac{1}{2}\)=\(\dfrac{3}{2}\)

=>\(x=\dfrac{3}{2}.\dfrac{3}{4}=\dfrac{9}{8}\)

e)\(\dfrac{11}{12}-\dfrac{2}{3}.\left|x\right|=\dfrac{3}{8}\)=>\(\dfrac{2}{3}.\left|x\right|=\dfrac{11}{12}-\dfrac{3}{8}\)=\(\dfrac{13}{24}\)

=>\(\left|x\right|=\dfrac{13}{24}:\dfrac{2}{3}=\dfrac{13}{16}\)

Vậy: \(x=\dfrac{13}{16}\)hoặc\(x=\dfrac{-13}{16}\)

f)\(\left|2x-1\right|=5\)

*2x-1=5 =>2x=5+1=6 =>x=6:2=3

*2x-1=-5 =>2x=(-5)+1=-4 =>x=-4:2=-2

Vậy: x=3 hoặc x=-2

Tick cho Phong nhé:>

Yêu nhiều>3

#Phong_419

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Vũ Phương Anh
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Phạm Ngân Hà
22 tháng 12 2017 lúc 19:39

a) \(\left(\dfrac{2}{3}\right)^3-\left(\dfrac{3}{4}\right)^2.\left(-1\right)^5\)

\(=\dfrac{8}{27}-\dfrac{9}{16}.\left(-1\right)\)

\(=\dfrac{8}{27}-\left(-\dfrac{9}{16}\right)\)

\(=\dfrac{8}{27}+\dfrac{9}{16}\)

\(=\dfrac{128}{432}+\dfrac{243}{432}\)

\(=\dfrac{371}{432}\)

b) \(12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)

\(=12:\left(\dfrac{9}{12}-\dfrac{10}{12}\right)^2\)

\(=12:\left(\dfrac{-1}{12}\right)^2\)

\(=12:\dfrac{1}{144}\)

\(=12.144\)

\(=1728\)

c) \(\dfrac{7}{22}:\dfrac{3}{11}+\dfrac{7}{22}:\dfrac{4}{11}\)

\(=\dfrac{7}{22}:\left(\dfrac{3}{11}+\dfrac{4}{11}\right)\)

\(=\dfrac{7}{22}:\dfrac{7}{11}\)

\(=\dfrac{7}{22}.\dfrac{11}{7}\)

\(=\dfrac{1}{2}\)

d) \(\dfrac{12}{35}.\left(\dfrac{7}{4}+\dfrac{13}{4}\right)-\dfrac{1}{3}\)

\(=\dfrac{12}{35}.5-\dfrac{1}{3}\)

\(=\dfrac{12}{7}-\dfrac{1}{3}\)

\(=\dfrac{36}{21}-\dfrac{7}{21}\)

\(=\dfrac{29}{21}\)

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Vũ Phương Anh
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Trương Tú Nhi
20 tháng 12 2017 lúc 20:35

a,\(\left(2-\dfrac{3}{4}\right)^2:\dfrac{11}{6}=\left(\dfrac{5}{4}\right)^2:\dfrac{11}{6}=\dfrac{75}{88}\)

b,\(2^3.\dfrac{7}{20}+\dfrac{7}{10}=8.\dfrac{7}{20}+\dfrac{7}{10}=\dfrac{14}{5}+\dfrac{7}{10}=\dfrac{7}{2}\)

c, \(\sqrt{3^2+4^2-\sqrt{1^3+2^3+3^3}=}\sqrt{9+16-6}\)

\(=\sqrt{19}\)

d, \(21^3:\left(-7\right)^3=\left[21:\left(-7\right)\right]^3=\left(-3\right)^3=-27\)

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Hoàng Phương Linh
20 tháng 12 2017 lúc 20:59

a) \(\left(2-\dfrac{3}{4}\right)^2\): \(\dfrac{11}{16}\) = \(\left(\dfrac{8}{4}-\dfrac{3}{4}\right)^2:\dfrac{11}{16}\)

= \(\left(\dfrac{5}{4}\right)^2:\dfrac{11}{16}\)

= \(\dfrac{25}{16}:\dfrac{11}{16}=\dfrac{25}{16}.\dfrac{16}{11}=\dfrac{25}{11}\)

b) \(2^3.\dfrac{7}{20}+\dfrac{7}{10}\) = \(8.\dfrac{7}{20}+\dfrac{7}{10}=\dfrac{14}{5}+\dfrac{7}{10}\)

= \(\dfrac{28}{10}+\dfrac{7}{10}=\dfrac{7}{2}=3,5\)

c) \(\sqrt{3^2+4^2-\sqrt{1^3+2^3+3^3}}\)

= \(\sqrt{9+16-\sqrt{1+8+27}}\)

= \(\sqrt{9+16-\sqrt{36}}=\sqrt{9+16-6}\)

= \(\sqrt{19}\)

d) \(21^3:\left(-7\right)^3=\left[21:\left(-7\right)\right]^3=\left(-3\right)^3=-27\)

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