sai roi 0 co 1/30

\(\frac{1}{12}+\frac{1}{20}+\frac{1}{42}+.....+\frac{1}{132}\)

\(=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{11.12}\)

\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+......+\frac{1}{11}-\frac{1}{12}\)

\(=\frac{1}{3}-\frac{1}{12}\)

\(=\frac{1}{4}\)

\(S=\frac{1}{12}+\frac{1}{20}+\frac{1}{42}+....+\frac{1}{132}\)

\(=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{11.12}\)

\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{12}\)

\(=\frac{1}{3}-\frac{1}{12}=\frac{4-1}{12}=\frac{3}{12}=\frac{1}{4}\)

\(\frac{1}{12}\)+ \(\frac{1}{20}\)+ \(\frac{1}{42}\)+ .... + \(\frac{1}{132}\)

= \(\frac{1}{3.4}\)+ \(\frac{1}{4.5}\)+ \(\frac{1}{5.6}\)+ ... + \(\frac{1}{11.12}\)

= \(\frac{1}{3}\)- \(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{5}\)+ ... + \(\frac{1}{11}\)- \(\frac{1}{12}\)

= \(\frac{1}{3}\)- \(\frac{1}{12}\)

= \(\frac{1}{4}\)

\(\frac{1}{12}+\frac{1}{20}+\frac{1}{42}+.....+\frac{1}{132}\)

\(=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+......+\frac{1}{11.12}\)

\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{11}-\frac{1}{12}\)

\(=\frac{1}{3}-\frac{1}{12}\)

\(=\frac{1}{4}\)

cám ơn 

1.\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{4}{23}-\frac{4}{27}\)

\(=\frac{1}{3}-\frac{1}{27}=\frac{9}{27}-\frac{1}{27}=\frac{8}{27}\)

2. Đặt \(A=\frac{3}{14}+\frac{3}{84}+\frac{3}{204}+\frac{3}{374}+\frac{3}{594}+\frac{3}{864}\)

\(\Rightarrow A=\frac{3}{2.7}+\frac{3}{7.12}+...+\frac{3}{27.32}\)

\(\Rightarrow5A=3.\left(\frac{5}{2.7}+\frac{5}{7.12}+...+\frac{5}{27.32}\right)\)

\(\Rightarrow5A=3.\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+...+\frac{1}{27}-\frac{1}{32}\right)\)

\(\Rightarrow5A=3.\left(\frac{1}{2}-\frac{1}{32}\right)\)

\(\Rightarrow5A=3.\frac{15}{32}=\frac{45}{32}\Rightarrow A=\frac{45}{32}:5=\frac{9}{32}\)

3. Đặt \(S=\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{340}\)

\(\Rightarrow3S=\frac{3}{10}+\frac{3}{40}+...+\frac{3}{340}\)

\(\Rightarrow3S=\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{17.20}\)

\(\Rightarrow3S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\)

\(\Rightarrow3S=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\Rightarrow S=\frac{9}{20}:3=\frac{3}{20}\)

Câu 1:

\(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+\frac{4}{15.19}+\frac{4}{19.23}+\frac{4}{23.27}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}+\frac{1}{23}-\frac{1}{27}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{27}\)

\(=\frac{8}{27}\)

Câu 1:

\(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{23\cdot27}\)

Áp dụng tính chất \(\frac{b}{a\left[a+b\right]}=\frac{1}{a}-\frac{1}{a+b}\), ta có:

\(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{23\cdot27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{23}-\frac{1}{27}\)

\(=\frac{1}{3}-\frac{1}{27}=\frac{9}{27}-\frac{1}{27}=\frac{8}{27}\)

Câu 2 tương tự nhưng phải phân

\(\frac{3}{14}+\frac{3}{84}+\frac{3}{204}+...+\frac{3}{864}=\frac{3}{2\cdot7}+\frac{3}{7\cdot12}+\frac{3}{12\cdot17}+...+\frac{3}{27\cdot32}\)

Cái này áp dụng công thức \(\frac{a}{b\left[b+c\right]}=\frac{a}{c}\left[\frac{1}{b}-\frac{1}{b+c}\right]\), ta có:

\(\frac{3}{2\cdot7}+\frac{3}{7\cdot12}+\frac{3}{12\cdot17}+...+\frac{3}{27\cdot32}=\frac{3}{5}\left[\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}+...+\frac{1}{27}-\frac{1}{32}\right]\)

\(=\frac{3}{5}\left[\frac{1}{2}-\frac{1}{32}\right]=\frac{3}{5}\cdot\frac{15}{32}=\frac{9}{32}\)

Câu 3:

tương tự quy laautj mẫu là 2.5; 5.8 ....

Câu 4: qL mẫu là 1.7; 7.13; ....

Câu 5: \(=\left[1-\frac{1}{2}\right]+\left[1-\frac{1}{6}\right]+...+\left[1-\frac{1}{110}\right]\)

\(=\left[1-\frac{1}{1\cdot2}\right]+\left[1-\frac{1}{2.3}\right]+\left[1-\frac{1}{3\cdot4}\right]+...+\left[1-\frac{1}{10.11}\right]\)

\(=10-\frac{9}{10}=\frac{91}{10}\)

x ở đâu thế

198 - 42 : 6 x y + 15 = 200

198 - 7 x y               = 200 - 15

198 - 7 x y               = 185

7 x y                       = 198 - 185

7 x y                       = 13

y                             = 13 : 7

y                             = 13/7

Vậy y = 13/7

<=> 198 - 42 : 6y = 185

<=> 42 : 6y = 13

<=> 6y = 42/13

<=> y = 7/13

198 - 42 : 6 x y +15 = 200 

42 : 6 x y = 198 - ( 200 - 15 ) = 13

42 : 6 = 7 

7 x y = 13 

y = 13 :7 

y = 13/7

(135-35).(-37)+37.(-42-58)

= 135-35.0.(-42)-58

=100.0.100

=0

(vì số nào nhân với 0 vẫn bằng 0)

CHÚC BẠN HỌC GIỎI.NĂM MỚI VUI VẺ , HẠNH PHÚC

TK MÌNH NHÉ

(135 - 35) . (-37) + 37 . (-42 - 58)

= 100 . (37 - 37) . (-100)

= 100 . 0 . (-100)

= 0 - (-100)

= 0

số 0 chia hay nhân với số nào cũng bằng 0 nha

( 135 - 35 ) . (-37) + 37 . (-42 - 58 )

= 100 . 0 - (-100)

= 0

( vì số nào nhân 0 vẫn chính bằng 0 )

k mk nha

Mk cảm ơn bạn nhiều 

Ta có : 

\(2x3y=42\)

\(\Rightarrow6xy=42\)

\(\Rightarrow xy=42:6\)

\(\Rightarrow xy=7\)

Do \(x;y\inℤ\)

\(\Rightarrow x;y\inƯ\left(7\right)\)

\(\Rightarrow x;y\in\left\{\pm1;\pm7\right\}\)

Ta có bảng sau : 

Vậy \(\left(x,y\right)\in\left\{\left(1,7\right);\left(7,1\right);\left(-1,-7\right);\left(-7,-1\right)\right\}\)

~ Ủng hộ nhé 

a. 3^6 : 3^2 + 2^3 . 2^ 2 = 3^ (6-3) + 2^(3+2) = 3^3 + 2^5= 27 +  32 =59

b. ( 39. 42 - 37.42) : 42 = [ 42. ( 39-37) ] :42 = 39 -37 = 2

a) 59

b) 2

4 nhân 5 mũ 2 -3 nhân(24-9)

= 1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7 

= 1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7

=1-1/7=6/7

=1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7

=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7

=1-1/7

=6/7

\(E=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)

\(E=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)

\(E=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)

\(E=1-\frac{1}{7}\)

\(E=\frac{6}{7}\)

Vậy \(E=\frac{6}{7}\)

Chúc bạn học tốt ~