53 x 39 + 49 x 39 - 53 x 21 - 47 x 21

= 39 x ( 53 + 49 ) - 21 x ( 53 - 47 )

= 39 x 102 - 21 x 6

= 3978 - 136

= 3842

53 x 39 + 49 x 39 - 53 x 21 - 47 x 21

= 39 x ( 53 + 49 ) - 21 x ( 53 - 47 )

= 39 x 102 - 21 x 6

= 3978 - 136

= 3842

bàng 3842

\(=\dfrac{5}{21}+\dfrac{16}{21}-\left(\dfrac{19}{23}+\dfrac{4}{23}\right)+\dfrac{1}{2}=\dfrac{1}{2}\)

\(\dfrac{x+35}{65}+\dfrac{x+39}{61}=\dfrac{x+43}{57}+\dfrac{x+47}{53}\)

\(\Leftrightarrow\dfrac{x+35}{65}+1+\dfrac{x+39}{61}+1=\dfrac{x+43}{57}+1+\dfrac{x+47}{53}+1\)

\(\Leftrightarrow\dfrac{x+100}{65}+\dfrac{x+100}{61}-\dfrac{x+100}{57}-\dfrac{x+100}{53}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{65}+\dfrac{1}{61}-\dfrac{1}{57}-\dfrac{1}{53}\ne0\right)=0\Leftrightarrow x=-100\)

Ta có:

\(\dfrac{x+35}{65}+\dfrac{x+39}{61}=\dfrac{x+43}{57}+\dfrac{x+47}{53}\\ \Rightarrow\left(\dfrac{x+35}{65}+1\right)+\left(\dfrac{x+39}{61}+1\right)=\left(\dfrac{x+43}{57}+1\right)+\left(\dfrac{x+47}{53}+1\right)\\ \Rightarrow\dfrac{x+100}{53}+\dfrac{x+100}{61}=\dfrac{x+100}{57}+\dfrac{x+100}{53}\\ \Rightarrow\left(x+100\right)\left(\dfrac{1}{65}+\dfrac{1}{61}-\dfrac{1}{57}-\dfrac{1}{53}\right)=0\)

Ta thấy:

\(\dfrac{1}{65}< \dfrac{1}{57}\\ \dfrac{1}{61}< \dfrac{1}{53}\\ \Rightarrow\left(\dfrac{1}{65}+\dfrac{1}{62}\right)-\left(\dfrac{1}{57}+\dfrac{1}{53}\right)< 0\)

Hay \(\dfrac{1}{65}+\dfrac{1}{62}-\dfrac{1}{57}-\dfrac{1}{53}\ne0\)

\(\Rightarrow x+100=0\\ \Rightarrow x=0-100\\ \Rightarrow x=-100\)

 Vậy \(x=-100\)

Ta có: \(\dfrac{x+35}{65}+\dfrac{x+39}{61}=\dfrac{x+43}{57}+\dfrac{x+47}{53}\)

\(\Leftrightarrow\dfrac{x+100}{65}+\dfrac{x+100}{61}-\dfrac{x+100}{57}-\dfrac{x+100}{53}=0\)

\(\Leftrightarrow x+100=0\)

hay x=-100

=>x-1/2=3+1/14-5-1/3-4/7+11/21

=>x-1/2=-97/42

=>x-21/42=-97/42

=>x=-97/42+21/42=-76/42=-38/21

-(-21+43+7) - (11-53-17)
= -(22+7) - (-42 -17) 
= (-29) - (-59)
= +( 59-29)
=+30 ( hay bn ghi 30 cx đc nha) 
bài này dễ mà

30

~HT~

giúp mình giải thích câu tục ngữ : thuốc đắng giã tật

B1

Số nhóm biểu thức nhò là

(2001+3)/2+1=1003(số)
mà giá trị mỗi biểu thức là 1

=> 1+1*1003=1004

\(a,\dfrac{53}{57}=\dfrac{53.10}{57.10}=\dfrac{530}{570}=1-\dfrac{40}{570}\\ \dfrac{531}{571}=1-\dfrac{40}{571}\\ Vì:\dfrac{40}{570}>\dfrac{40}{571}\Rightarrow1-\dfrac{40}{570}< 1-\dfrac{40}{571}\\ \Leftrightarrow\dfrac{53}{57}< \dfrac{531}{571}\)

\(b,\dfrac{25}{26}=\dfrac{25.1010}{26.1010}=\dfrac{25250}{26260}=1-\dfrac{1010}{26260}\\ \dfrac{25251}{26261}=1-\dfrac{1010}{26261}\\ Vì:\dfrac{1010}{26260}>\dfrac{1010}{26261}\Rightarrow1-\dfrac{1010}{26260}< \dfrac{1010}{26261}\\ \Leftrightarrow\dfrac{25}{26}< \dfrac{25251}{26261}\)

\(\dfrac{2}{x}\) + \(\dfrac{1}{y}\) = \(\dfrac{1}{6}\) (\(x;y\) \(\in\) N*)

\(\dfrac{2}{x}\)         = \(\dfrac{1}{6}\) - \(\dfrac{1}{y}\)

\(\dfrac{2}{x}\)         = \(\dfrac{y-6}{6y}\)

\(x\)          = 2: \(\dfrac{y-6}{6y}\)

\(x\)         = \(\dfrac{12y}{y-6}\)

Vì \(x\); y \(\in\) N* nên 12\(y\) ⋮ y - 6 ( và y > 6)

12y ⋮ y - 6 ⇔ 12y - 72 + 72 ⋮ y - 6 ⇔ 12.(y-6) + 72 ⋮ y - 6 ⇔ 72⋮ y - 6  72 = 2³.3² 

Ư(72) = { 1; 2; 3; 4; 6; 8; 9; 12; 18; 24; 36; 72}

Lập bảng ta có:

Theo bảng trên ta có các cặp số tự nhên \(x\); y thỏa mãn đề bài lần lượt là:

(\(x\);y) =(84;7); (48;8); (36;9); (30;10);(34;12); (21;14); (20;15);(18;18);

(16;24); (15; 30); (14;42);(13;78)

\(\dfrac{2}{x}+\dfrac{1}{y}=\dfrac{1}{6}\left(x;y\inℕ^∗\right)\)

\(\Leftrightarrow\dfrac{2y+x}{xy}=\dfrac{1}{6}\)

\(\Leftrightarrow6\left(2y+x\right)=xy\)

\(\Leftrightarrow12y+6x=xy\)

\(\Leftrightarrow12y-xy+6x=0\)

\(\Leftrightarrow y\left(12-x\right)+6x-72+72=0\)

\(\Leftrightarrow-y\left(x-12\right)+6\left(x-12\right)=-72\)

\(\Leftrightarrow\left(x-12\right)\left(6-y\right)=-72\)

\(\Leftrightarrow\left(x-12\right);\left(6-y\right)\in\left\{-1;1;-2;2;-3;3;-4;4;-8;8;-9;9;-18;18;-24;24;-36;36;-72;72\right\}\)

Lập bảng sẽ ra \(\left(x;y\inℕ^∗\right)\) cần tìm...