1.

7/15:(1/2-9/10xX)-1/6=13/6

=>7/15:(1/2-9/10xX)=13/6+1/6

=>7/16:(1/2-9/10xX)=7/3

=>1/2-9/10xX=7/16:7/3

=>1/2-9/10xX=3/16

=>9/10xX=1/2-3/16

=>9/10xX=5/16

=>X=5/16:9/10

=>X=25/72

vì x+20 chia hết cho 10 mà 0<x<300 nên x= 80;180 hoặc 280

ta thấy số 80+20 chia hết cho 10 

80+1 chia hết cho 9 và 

Câu 1:

Ta có: \(\dfrac{x-4}{y-3}=\dfrac{4}{3}\)

=> \(3.\left(x-4\right)=4.\left(y-3\right)\)

=>\(3x-12=4y-12\)

=>\(3x=4y\) (1)

Ta có: \(x-y=5\)

=> \(y=y+5\) Thay vào (1) ta có:

\(3.\left(y+5\right)=4.\)y

=>\(3y+15=4y\)

=> \(15=4y-3y\)

=> 15 = y

=> y =15

ta có: x = y +5

=> x = 15 +5

=> x =20

Câu 2:

\(B=\dfrac{10}{56}+\dfrac{10}{140}+\dfrac{10}{260}+...+\dfrac{10}{1400}\)

\(B=\dfrac{5}{28}+\dfrac{6}{70}+\dfrac{5}{130}+...+\dfrac{5}{700}\)

\(B=\dfrac{5}{4.7}+\dfrac{5}{7.10}+\dfrac{5}{10.13}+...+\dfrac{5}{25.28}\)

\(B=5,\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)

\(3B=5.\left(\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{3}{25.28}\right)\)

\(3B=5.\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)

\(3B=5.\left(\dfrac{1}{4}-\dfrac{1}{28}\right)\)

\(3B=5.\dfrac{3}{14}\)

\(B=\dfrac{15}{14}:3=\dfrac{5}{14}\)

Câu 3:

38 - (|x+10|+13) = \(\left(-6\right)^{20}:\left(9^9.4^{10}\right)\)

=> \(38-\left(\left|x+10\right|+13\right)=\left(2.3\right)_{ }^{20}:\)\(\left[\left(3^2\right)^9.\left(2^2\right)^4\right]\)

=>\(38-\left(\left|x+10\right|+13\right)=2^{20}.3^{20}:\left(3^{18}.2^{20}\right)\)

=> \(38-\left(\left|x+10\right|+13\right)=\dfrac{3^{20}.2^{20}}{3^{18}.2^{20}}\)

=> \(38-\left(\left|x+10\right|+13\right)=9\)

=> |x +10| + 13 = 38 -9

=> |x+10| +13 = 29

=> |x+10| = 29 -13

=> |x+10| = 16

\(\Rightarrow\left[{}\begin{matrix}x+10=16\\x+10=-16\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=6\\x=-26\end{matrix}\right.\)

Câu 2:

\(B=\dfrac{10}{56}+\dfrac{10}{140}+\dfrac{10}{260}+....+\dfrac{10}{1400}\)
\(\Rightarrow B=\dfrac{20}{112}+\dfrac{20}{280}+\dfrac{20}{520}+.....+\dfrac{20}{2800}\)
\(\Rightarrow B=20\left(\dfrac{1}{112}+\dfrac{1}{280}+\dfrac{1}{520}+...+\dfrac{1}{2800}\right)\)
\(\Rightarrow B=20\left(\dfrac{1}{8.14}+\dfrac{1}{14.20}+\dfrac{1}{20.26}+...+\dfrac{1}{50.56}\right)\)
\(\Rightarrow B=\dfrac{20}{6}\left(\dfrac{6}{8.14}+\dfrac{6}{14.20}+\dfrac{6}{20.26}+...+\dfrac{6}{50.56}\right)\)
\(\Rightarrow B=\dfrac{20}{6}\left(\dfrac{1}{8}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{26}+...+\dfrac{1}{50}-\dfrac{1}{56}\right)\)
\(\Rightarrow B=\dfrac{20}{6}\left(\dfrac{1}{8}-\dfrac{1}{56}\right)\)
\(\Rightarrow B=\dfrac{20}{6}\left(\dfrac{7}{56}-\dfrac{1}{56}\right)\)
\(\Rightarrow B=\dfrac{20.6}{6.56}\)
\(\Rightarrow B=\dfrac{20}{56}\)
\(\Rightarrow B=\dfrac{5}{14}\)

a: \(x⋮10\)

b: \(x⋮5\)

x=80 vì 80 đạt đủ mọi tiêu chẩn

x=80

x + 20 \(⋮\) 10 

\(\Rightarrow\) CSTC của x là 0

Mà x \(⋮\) 8

\(\Rightarrow\) x \(\in\) {0; 80; 160; 240; 320; 400; ...}

Mà x < 300

\(\Rightarrow\) x \(\in\) {0; 80; 160; 240}

Mà x + 1 \(⋮\) 9

\(\Rightarrow\) x + 1 = 81

\(\Rightarrow\) x = 80

20 - 19 + 18 - 17 + 16 - 15 + 14 - 13 + 12 - 11 + 10 - 9 + 8 - 7 + 6 - 5 + 4 - 3 + 2 - 1

= ( 20 - 19 ) + ( 18 - 17 ) + ( 16 - 15 ) + ( 14 - 13 ) + ( 12 - 11 ) + ( 10 - 9 ) + ( 8 - 7 ) + ( 6 - 5 ) + ( 4 - 3 ) + ( 2 - 1 )

= 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1

= 1 . 10

= 10

20 - 19 + 18 - 17 + 16 - 15 + 14 - 13 + 12 - 11 + 10 - 9 + 8 - 7 + 6 - 5 + 4 - 3 + 2 - 1  

=  1      +    1      +     1      +    1      +     1    +    1      +  1    +   1    +   1    +  1

= 10

20 - 19 + 18 - 17 + 16 - 15 + 14 - 13 + 12 - 11 + 10 - 9 + 8 - 7 + 6 - 5 + 4 - 3  + 2 - 1 

=    1 +   1 + 1 + 1 + 1 + 1 + 1 + 1 +1+ 1

= 10

4*X +50=74

4*X=74-50=24

4*X=24/4=6

6 - 8 + 10 - 12 + ... - x = - 200

( 6 - 8 ) +( 10 - 12 ) + ... + ( x - 2 - x ) = - 200

( - 2 ) + ( - 2 ) + ... + ( - 2 ) = - 200 ( Có ( x - 6 ) : 2 + 1 = x : 2 - 3 + 1 = x : 2 - 2 số hạng - 2 )

( - 2 ) x x : 2 - 2 = 200

( - 2 ) x x : 2 = 202

x = - 202