2410806        

Ta có: 443 =443

5442 =2 x 3 x 907

Do đó:BCNN(443,5442) = 443 x 2 x 3 x 907 =2410806

toàn ra mấy bài kiểu này

đau đầu chết

5442 + 565 = 6007

hok tốt

# Puka#

6007

có nha

5442+565=6007

hok tot nha

bye

\(\left[195-\left(3x-27\right)\right]\cdot39=4212\)

\(\Rightarrow\left(195-3x+27\right)\cdot39=4212\)

\(\Rightarrow\left(-3x+222\right)\cdot39=4212\)

\(\Rightarrow-3x+222=\dfrac{4212}{39}\)

\(\Rightarrow-3x+222=108\)

\(\Rightarrow3x=222-108\)

\(\Rightarrow3x=114\)

\(\Rightarrow x=\dfrac{114}{3}\)

\(\Rightarrow x=38\)

=>[195-(3x-27)]=4212:39=108

=>3x-27=195-108=87

=>3x=27+87=114

=>x=38

\(95-\left(3x-27\right).39=4212\)

\(95-\left(3x-27\right)=108\)

\(3x-27=-13\)

\(3x=14\)

\(x=\dfrac{14}{3}\)

\(95-\left(3x-27\right).39=4212\)

\(95-\left(3x-27\right)=4212:39\)

\(95-\left(3x-27\right)=108\)

\(3x-27=95-108\)

\(3x-27=-13\)

\(3x=-13+27\)

\(3x=14\)

\(x=14:3\)

\(x=\dfrac{14}{3}\)

Vậy \(x=\dfrac{14}{3}\)

\(\left[195-\left(3x-27\right)\right]\cdot39=4212\\ \Rightarrow195-\left(3x-27\right)=4212:39\\ \Rightarrow195-\left(3x-27\right)=108\\ \Rightarrow3x-27=195-108\\ \Rightarrow3x-27=87\\ \Rightarrow3x=87+27\\ \Rightarrow3x=114\\ \Rightarrow x=114:3\\ \Rightarrow x=38\)

\(\left[195-\left(3x-27\right)\right]\cdot39=4212\\ \Leftrightarrow195-\left(3x-27\right)=108\\ \Leftrightarrow3x-27=87\\ \Leftrightarrow3x=114\\ \Leftrightarrow x=38\)

Vậy x = 38.

[195−(3x−27)]⋅39=4212

⇒195−(3x−27)=4212:39

⇒195−(3x−27)=108

⇒3x−27=195−108

⇒3x−27=87⇒3x=87+27✿

⇒3x=114⇒x=114:3⇒x=38

[195-(15x-27).39=4212

195-(15x-27)=4212 : 39

195-(15x-27)=108

15x-27=195-108

15x-27=87

15x=87+27

15x=114

x=114:15

x=38/5

x=7,6

k đúng cho m nha bạn !

Đề thiếu bạn ạ.

thank you quang anh nguyễn

\(a,91-5.\left(5+x\right)=61\\ \Rightarrow=5.\left(5+x\right)=30\\ \Rightarrow5+x=6\\ \Rightarrow x=1.\\ b,\left[195-\left(3x-27\right)\right].39=4212\\ \Rightarrow195-\left(3x-27\right)=108\\ \Rightarrow3x-27=87\\ \Rightarrow3x=114\\ \Rightarrow x=38.\)

\(91-5.\left(5+x\right)=61\)

\(\Rightarrow5.\left(5+x\right)=91-61\)

\(\Rightarrow5.\left(5+x\right)=30\)

\(\Rightarrow5+x=\dfrac{30}{5}=6\)

\(\Rightarrow x=6-5=1\)

\(\left[195-\left(3x-27\right)\right].39=4212\)

\(\Rightarrow195-3x+27=\dfrac{4212}{39}=108\)

\(\Rightarrow222-3x=108\)

\(\Rightarrow3x=222-108=114\)

\(\Rightarrow x=\dfrac{114}{3}=38\)

$91-5.\left(5+x\right)=61$

$\Rightarrow 5.\left(5+x\right)=91-61$

$\Rightarrow 5.\left(5+x\right)=30$

⇒5+�=305=6⇒5+x=530​=6

$\Rightarrow x=6-5=1$

$\left[195-\left(3x-27\right)\right].39=4212$

⇒195−3�+27=421239=108⇒195−3x+27=
​4212/39=108

$\Rightarrow 222-3x=108$

$\Rightarrow 3x=222-108=114$

⇒�=1143=38⇒x=114/3
​=38

\(\frac{2106}{7320}=\frac{351}{1220}\)

\(\frac{4212}{14604}=\frac{351}{1217}\)

\(\frac{6318}{21960}=\frac{351}{1220}\)

=> \(\frac{351}{1217}>\frac{351}{1220}\);\(\frac{351}{1217}>\frac{351}{1220}\)

Vậy : \(\frac{2106}{7320}< \frac{6318}{21960}\);\(\frac{4212}{14604}>\frac{6318}{21960}\)

Ta có:

\(\frac{2106}{7320}=\frac{2106:3}{7320:3}=\frac{720}{2440}\)   (1)

\(\frac{4212}{14604}=\frac{4212:6}{14604:6}=\frac{702}{2434}\)

\(\frac{6218}{21960}=\frac{6218:9}{21960:9}=\frac{702}{2440}\)    (2)

Từ (1) và (2)=>\(\frac{2106}{7320}=\frac{6318}{21960}\)