b xem hinh31 có BÉ //CD và AD vuông AC chứng minh rằng
BE<CE
CE<CD
BE<CD
giúp mình với mai lôp rôi
b xem hinh31 có BÉ //CD và AD vuông AC chứng minh rằng
BE<CE
CE<CD
BE<CD
giúp mình với mai lôp rôi
(x+2)(x-3)=5
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 nha
Tính
A=\(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
Sửa đề
\(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{\dfrac{1}{4}-\dfrac{1}{6}-\dfrac{1}{64}-\dfrac{1}{256}}+\dfrac{5}{8}\)
\(=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}\cdot\dfrac{3\left(\dfrac{1}{4}-\dfrac{1}{6}-\dfrac{1}{64}-\dfrac{1}{256}\right)}{\dfrac{1}{4}-\dfrac{1}{6}-\dfrac{1}{64}-\dfrac{1}{256}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}\cdot3+\dfrac{5}{8}=\dfrac{3}{2}+\dfrac{5}{8}=\dfrac{17}{8}\)
A= \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
=> \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2.(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13})}.\dfrac{3.(\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}-\dfrac{1}{256})}{\dfrac{4}{4}-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
=> \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2.(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13})}.\dfrac{3.(\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}-\dfrac{1}{256})}{4.(\dfrac{1}{4})-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
=> \(\dfrac{1}{2}.\dfrac{3.(\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{4^3}-\dfrac{1}{16^2})}{4.(\dfrac{1}{4})-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
=> \(\dfrac{1}{2}.\dfrac{3.(-\dfrac{1}{4^2}-\dfrac{1}{16^2})}{4-\dfrac{1}{4^3}}+\dfrac{5}{8}\)
=> \(\dfrac{1}{2}.\dfrac{3.(-\dfrac{1}{16^2})}{4.-\dfrac{1}{4^2}}+\dfrac{5}{8}\)
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
1)Tính
A=\(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{16}\left(1+2+3+...+16\right)\)
\(A=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{16}\left(1+..+16\right)\\ =1+\dfrac{1}{2}.\dfrac{\left(1+2\right).2}{2}+\dfrac{1}{3}.\dfrac{\left(1+3\right).3}{2}+....+\dfrac{1}{16}.\dfrac{\left(16+1\right).16}{2}\\ =\dfrac{2}{2}+\dfrac{\left(1+2\right)}{2}+...+\dfrac{16+1}{2}\\ =\dfrac{2+...+17}{2}\\ =\dfrac{152}{2}=76\)
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
Tính nhanh
\(\dfrac{1}{2}.\dfrac{1}{-3}+\dfrac{1}{-3}.\dfrac{1}{4}+\dfrac{1}{4}.\dfrac{1}{-5}+\dfrac{1}{-5}.\dfrac{1}{6}\)
\(A=\dfrac{81^4.3^{10}.27^5.3^{12}}{3^{18}.9^3.243^2}\)
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) \(A=\dfrac{81^4.3^{10}.27^5.3^{12}}{3^{18}.9^3.243^2}\)
\(=\dfrac{9^8.9^8.9^{13}.9^{10}}{9^{16}.9^3.9^3}\)
\(=\dfrac{9^{39}}{9^{22}}\)
\(=9^{17}\)
\(A=\dfrac{81^4\cdot3^{10}\cdot27^5\cdot3^{12}}{3^{18}\cdot9^3\cdot243^2}=\dfrac{3^{16}\cdot3^{10}\cdot3^{15}\cdot3^{12}}{3^{18}\cdot3^6\cdot3^{10}}=\dfrac{3^{53}}{3^{34}}=3^{19}\)
Vậy A = 319
Ngân Hà làm đúng phần a) nhưng làm sai phần b) nên mk chỉ làm phần b) thôi
Tính nhanh
\(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{1}{3}-0,25+0,2}{1\dfrac{1}{6}-0,875+0,7}+\dfrac{6}{7}\)
\(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{1}{3}-0,25+0,2}{1\dfrac{1}{6}-0,875+0,7}+\dfrac{6}{7}\)
\(=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2\cdot\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}\cdot\dfrac{\dfrac{2}{6}-\dfrac{2}{8}+\dfrac{2}{10}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}+\dfrac{6}{7}\)
\(=\dfrac{1}{2}\cdot\dfrac{2\cdot\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{10}\right)}{7\cdot\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{10}\right)}+\dfrac{6}{7}\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{7}+\dfrac{6}{7}\)
\(=\dfrac{1}{7}+\dfrac{6}{7}=\dfrac{7}{7}=1\)
\(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}.\dfrac{\dfrac{1}{3}-0,25+0,2}{1\dfrac{1}{6}+0,875+0,7}+\dfrac{6}{7}.\)
\(=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}.\dfrac{\dfrac{2}{6}-\dfrac{2}{8}+\dfrac{2}{10}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}+\dfrac{6}{7}.\)
\(=\dfrac{1}{2}.\dfrac{2\left(\dfrac{1}{6}+\dfrac{1}{8}+\dfrac{1}{10}\right)}{7\left(\dfrac{1}{6}+\dfrac{1}{8}+\dfrac{1}{10}\right)}+\dfrac{6}{7}.\)
\(=\dfrac{1}{2}.\dfrac{2}{7}+\dfrac{6}{7}.\)
\(=\dfrac{1}{7}+\dfrac{6}{7}=\dfrac{7}{7}=1.\)
Vậy.....
bài 1 tìm x, y biết:
a,\(\dfrac{x}{2}=\dfrac{y}{5}\)và x+y=-14
b,\(\dfrac{x}{7}=\dfrac{y}{5}\)và x-y=8
Bài 2: Tìm x, y, z biết:
a,\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}\)và x+y+z=56
b,\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{8}\)và 2x+y-z=12
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 ..
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
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:
\(\dfrac{2x}{6}=\dfrac{y}{5}=\dfrac{z}{8}=\dfrac{2x+y-z}{6+5-8}=\dfrac{12}{3}=4\)
\(\Rightarrow\dfrac{2x}{6}=4\Rightarrow2x=4.6=24\Rightarrow x=24:2=12\)
\(\Rightarrow\dfrac{y}{5}=4\Rightarrow y=4.5=20\)
\(\Rightarrow\dfrac{z}{8}=4\Rightarrow z=4.8=32\)
\(\text{Vậy }\left\{{}\begin{matrix}x=12\\y=20\\z=32\end{matrix}\right.\)
bài 1: Tìm x biết:
a,\(x+\dfrac{1}{4}=\dfrac{3}{4}\)
b,\(\dfrac{3}{4}-\dfrac{2}{5}x=\dfrac{29}{60}\)
c,\(2x-\dfrac{1}{3}=\dfrac{-5}{6}\)
d,\(2-x:\dfrac{3}{4}=\dfrac{1}{2}\)
e,\(\dfrac{11}{12}-\dfrac{2}{3}|x|=\dfrac{3}{8}\)
f,\(|2x-1|=5\)
lưu ý: Có câu có 2 trường hợp
Các bn giúp mk với
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
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
thực hiện phép tính:
a,\(\left(\dfrac{2}{3}\right)^3-\left(\dfrac{3}{4}\right)^2.\left(-1\right)^5\)
b,\(12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)
c,\(\dfrac{7}{22}:\dfrac{3}{11}+\dfrac{7}{22}:\dfrac{4}{11}\)
d,\(\dfrac{12}{35}.\left(\dfrac{7}{4}+\dfrac{13}{4}\right)-\dfrac{1}{3}\)
Giúp mk với
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) \(\dfrac{12}{35}.\left(\dfrac{7}{4}+\dfrac{13}{4}\right)-\dfrac{1}{3}\)
\(=\dfrac{12}{35}.5-\dfrac{1}{3}\)
\(=\dfrac{12}{7}-\dfrac{1}{3}\)
\(=\dfrac{36}{21}-\dfrac{7}{21}\)
\(=\dfrac{29}{21}\)
Thực hiện phép tính
a,\(\left(2-\dfrac{3}{4}\right)^2:\dfrac{11}{16}\)
b,\(2^3.\dfrac{7}{20}+\dfrac{7}{10}\)
c,\(\sqrt{3^2+4^2-\sqrt{1^3+2^3+3^3}}\)
d,\(21^3:\left(-7\right)^3\)
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, \(21^3:\left(-7\right)^3=\left[21:\left(-7\right)\right]^3=\left(-3\right)^3=-27\)
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) \(21^3:\left(-7\right)^3=\left[21:\left(-7\right)\right]^3=\left(-3\right)^3=-27\)