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Bài 1:
60= 22.3.5 ; 88 = 23.11
ƯCLN(60;88)= 22 = 4
ƯC(60;88)=Ư(4)={1;2;4}
Bài 2:
24= 23.3 ; 30=2.3.5 ; 40 = 23.5
BCNN(24;30;40)=23.3.5= 120
BC(24;30;40)=B(120)={0;120;240;360;...}
a) \(24=2^3.3\)
\(60=2^2.3.5\)
\(UCLN\left(a;b\right)=UCLN\left(24;60\right)=2^2.3=6\)
\(BCNN\left(a;b\right)=BCNN\left(24;60\right)=2^3.3.5=120\)
\(a.b=UCLN\left(a;b\right).BCNN\left(a;b\right)\)
\(\Rightarrow a.b=6.120=720\)
mà \(\dfrac{a}{b}=\dfrac{24}{60}\Rightarrow\dfrac{a}{24}=\dfrac{b}{60}=\dfrac{720}{24.60}=\dfrac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}a=24.\dfrac{1}{2}=12\\b=60.\dfrac{1}{2}=30\end{matrix}\right.\)
Vậy Phân số cần tìm là \(\dfrac{12}{30}\)
b) \(\left\{{}\begin{matrix}14=2.7\\21=3.7\end{matrix}\right.\)
\(\Rightarrow UCLN\left(a;b\right)=UCLN\left(14;21\right)=7\)
\(a.b=UCLN\left(14;21\right).BCNN\left(14;21\right)\)
\(\Rightarrow a.b=7.3456=24192\)
\(\dfrac{a}{b}=\dfrac{14}{21}\Rightarrow\dfrac{a}{14}=\dfrac{b}{21}=\dfrac{a.b}{14.21}=\dfrac{24192}{294}=\dfrac{576}{7}\)
\(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{576}{7}.14=1152\\b=\dfrac{576}{7}.21=1728\end{matrix}\right.\)
Vậy phân số cần tìm là \(\dfrac{1152}{1728}\)
\(24^{54}.54^{24}.2^{10}\\ =8^{54}.3^{54}.27^{54}.2^{54}.2^{10}\)
\(=2^{162}.3^{54}.3^{72}.2^{54}.2^{10}\\
=2^{226}.3^{126}\\
=2^{3.63+37}.3^{2.63}\\
=8^{63}.9^{63}.2^{37}\\
=72^{63}.2^{37}\)
Dễ thấy \(72^{63}.2^{37}⋮̸72^{63}\)
Mk phải đi học quá nhiều , gần như không có ngày nghỉ nên mình muốn bỏ
Mk không muốn cận nữa vì mỗi lần lên Học 24 là mình không biết gì hết cứ cắm đầu vào kiếm điểm .
Mk cần thời gian thư giãn và đỡ cận
a) goi hai so la a ; b va a >b
vi UCLN(a,b)=18=>a=18k ; b=18q (trong do UCLN (k,q)=1 va k>q)
=>a+b=162
18k+18q =162
18(k+q)=162
k+q=9
ta co bang sau | |||||||||||||||||||||||
vay ........... | |||||||||||||||||||||||
21453
52542000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 | 542454550212.100000000000000000000000000000000000000000000000000000000000000000000000000000 |
Đường chéo là cạnh huyền.
402+302=2500=502
=> Đường chéo màn hình là 50 inch.
=> Ti vi thuộc loại 50inch.
(Ti vi này cũng chưa quá to đâu nhỉ??)
Ta có: Theo định lí Pitago: AB2 + AC2 =BC2
=> 402+302 =BC2
=> 2500 =BC2
=> BC=50(inch)Vậy tivi đó thuộc loại 50 inch
ƯCLN(24;40)
24=2^3 x 3
40=2^3 x 5
=> ƯCLN(24;40)=2^3=8
k cho mik nha!!!
là 8 nhếii
k mk đy