((27*11-(x+12)*8)*3-79):=100
DẤU * LÀ DẤU NHÂN
GIÚP MÌNH NHA
{ [ 27 x 11 - ( x + 12 ) x 8 ] x 3 - 79 } : 5 =
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( (27*11-(X+12)*8)*3-79):5=100
{ [27 * 11 - ( x + 12 ) * 8 ] * 3 - 79 } : 5 = 100
{ [ 27 * 11 - ( x + 12 ) * 8 ] * 3 - 79 } : 5 = 100
[ 297 - ( x + 12 ) * 8 ] * 3 - 79 = 100 * 5
[ 297 - ( x + 12 ) * 8 ] * 3 - 79 = 500
[ 297 - ( x + 12 ) * 8 ] * 3 = 500 + 79
[ 297 - ( x + 12 ) * 8 ] * 3 = 579
297 - ( x + 12 ) * 8 = 579 : 3
297 - ( x + 12 ) * 8 = 193
( x + 12 ) * 8 = 297 - 193
( x + 12 ) * 8 = 104
x + 12 = 104 : 8
x + 12 = 13
x = 13 - 12
x = 1.^^
( (27*11-(X+12)*8)*3-79):5=100
{[27 nhân 11-(x+12)nhân 8]nhân 3 -79}
(297-(x+12)*8)*3-79=297-79-(x+12)*8*3
=218-(x+12)*24
=218-24x+288
=218-288+24x
=-70+24x
TÌM X:
{[27*11-(x+12)*8]*3-79}:5=100
TÌM X:
{[27*11-(x+12)*8]*3-79}:5=100
TÌM X:
{[27*11-(X+12)*8]*3-79}:5=100
\(\left\{\left[27\times11-\left(x+12\right)\times8\right]\times3-79\right\}\div5=100\)
\(\left[27\times11-\left(x+12\right)\times8\right]\times3-79=100\times5\)
\(\left[27\times11-\left(x+12\right)\times8\right]\times3-79=500\)
\(\left[27\times11-\left(x+12\right)\times8\right]\times3=500+79\)
\(\left[27\times11-\left(x+12\right)\times8\right]\times3=579\)
\(27\times11-\left(x+12\right)\times8=579:3\)
\(27\times11-\left(x+12\right)\times8=193\)
\(297-\left(x+12\right)\times8=193\)
\(\left(x+12\right)\times8=297-193\)
\(\left(x+12\right)\times8=104\)
\(x+12=104:8\)
\(x+12=13\)
\(x=13-12\)
\(x=1\)
\(k\)\(dung\)\(cho\)\(minh\)\(nhe!\)
TÍNH NHANH
a) 1/10*11+1/11*12+1/12*13+1/13*14+........+1/78*79
b) 8/7*9+8/9*11+8/11*13+8/13*15+.......+8/133*135
c) 12/8*11+12/11*14+12/14*17+.........+12/503*506
d) 1/4*7+1/7*10+1/10*13+1/13*16+..........+1/391*394
e) 4/5*8+4/8*11+4/11*14+4/14*17+.........+4/602*605
g) 1+1/3+1/6+1/10+1/15+........+1/802
DẤU NÀY * LÀ DẤU NHÂN
CÁC BẠN GIẢI GIÚP MÌNH VỚI Ạ
MÌNH CHỈ CÒN KHOẢNG 15 PHÚT NỮA THÔI CẦU XIN CÁC BẠN ĐÓ
a) = 1/10 - 1/11 + 1/11 -1/12 + 1/12 - 1/13 +1/13 1/14 +...+ 1/78 - 1/79
= 1/10 - 1/79
= máy tính ok
mấy câu khác bn làm tương tự là đc nhưng nhớ nhanh thêm khoảng cách giữa các mẫu nha
a)\(\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{78.79}=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{78}-\frac{1}{79}=\frac{1}{10}-\frac{1}{79}=\frac{69}{790}\)
b) \(\frac{8}{7.9}+\frac{8}{9.11}+...+\frac{8}{133.135}=4\left(\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{133.135}\right)\)
\(=4\left(\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{133}-\frac{1}{135}\right)=4\left(\frac{1}{7}-\frac{1}{135}\right)=4.\frac{128}{945}=\frac{456}{945}\)
c) \(\frac{12}{8.11}+\frac{12}{11.14}+...+\frac{12}{503.506}=4\left(\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{503.506}\right)\)
\(=4\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{503}-\frac{1}{506}\right)=4\left(\frac{1}{8}-\frac{1}{506}\right)=\frac{249}{506}\)
d) \(\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{391.394}=\frac{1}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{391.394}\right)\)
\(=\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{391}-\frac{1}{394}\right)=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{394}\right)=\frac{1}{3}.\frac{195}{788}=\frac{65}{788}\)
e) \(\frac{4}{5.8}+\frac{4}{8.11}+...+\frac{4}{602.605}=\frac{4}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{602.605}\right)\)
\(=\frac{4}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{602}-\frac{1}{605}\right)=\frac{4}{3}\left(\frac{1}{5}-\frac{1}{605}\right)=\frac{4}{3}.\frac{24}{121}=\frac{32}{121}\)
g) Sửa đề\(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{820}=2\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{1640}\right)=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{40.41}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{40}-\frac{1}{41}\right)=2\left(1-\frac{1}{41}\right)=2.\frac{40}{41}=\frac{80}{41}\)
Bài làm:
a) \(\frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{78.79}\)
\(=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{78}-\frac{1}{79}\)
\(=\frac{1}{10}-\frac{1}{79}=\frac{69}{790}\)
b) \(\frac{8}{7.9}+\frac{8}{9.11}+...+\frac{8}{133.135}\)
\(=4\left(\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{133.135}\right)\)
\(=4\left(\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{133}-\frac{1}{135}\right)\)
\(=4\left(\frac{1}{7}-\frac{1}{135}\right)\)
\(=4.\frac{128}{945}=\frac{512}{945}\)
c) \(\frac{12}{8.11}+\frac{12}{11.14}+...+\frac{12}{503.506}\)
\(=4\left(\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{503.506}\right)\)
\(=4\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{503}-\frac{1}{506}\right)\)
\(=4\left(\frac{1}{8}-\frac{1}{506}\right)\)
\(=4.\frac{249}{2024}=\frac{249}{506}\)
d) \(\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{391.394}\)
\(=\frac{1}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{391.394}\right)\)
\(=\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{391}-\frac{1}{394}\right)\)
\(=\frac{1}{3}\left(\frac{1}{4}-\frac{1}{394}\right)\)
\(=\frac{1}{3}.\frac{195}{788}=\frac{65}{788}\)
e) \(\frac{4}{5.8}+\frac{4}{8.11}+...+\frac{4}{602.605}\)
\(=\frac{4}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{602.605}\right)\)
\(=\frac{4}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{602}-\frac{1}{605}\right)\)
\(=\frac{4}{3}\left(\frac{1}{5}-\frac{1}{605}\right)\)
\(=\frac{4}{3}.\frac{24}{121}=\frac{32}{121}\)
g) Phải sửa \(\frac{1}{802}\) thành \(\frac{1}{820}\) nhé
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{820}\)
\(=1+\frac{1}{1.3}+\frac{1}{2.3}+\frac{1}{2.5}+...+\frac{1}{41.20}\)
\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{40.41}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{40}-\frac{1}{41}\right)\)
\(=2\left(1-\frac{1}{41}\right)\)
\(=2.\frac{40}{41}=\frac{80}{41}\)