\(\dfrac{1}{3\cdot63}\)+\(\dfrac{1}{4\cdot64}\)+...+\(\dfrac{1}{1802\cdot1862}\)
\(\dfrac{1}{3\cdot1803}\)+\(\dfrac{1}{4\cdot1804}\)+...+\(\dfrac{1}{62\cdot1862}\)
tính nhanh : A, \(\dfrac{7,2:2\cdot28,6+1,43\cdot2\cdot64}{13+5+7+...+49-339}\)
tìm x : B, 2+( \(\dfrac{3}{2}\)+ x - \(\dfrac{5}{8}\) ) : \(\dfrac{13}{4}\)=6
a: Sửa đề; \(A=\dfrac{7.2:2\cdot28.6+1.43\cdot2\cdot64}{1+3+5+7+...+49-339}\)
\(=\dfrac{3.6\cdot28.6+2.86\cdot64}{1+3+5+...+49-339}\)
\(=\dfrac{2.86\left(64+36\right)}{25^2-339}=\dfrac{286}{286}=1\)
b: =>2(x+7/8)=6*13/4=78/4=39/2
=>x+7/8=39/4
=>x=71/8
Rút gọn rồi so sánh
A=\(\dfrac{8056}{2012\cdot16-1982}\)
B=\(\dfrac{1\cdot2\cdot6+2\cdot4\cdot12+4\cdot8\cdot24+7\cdot14\cdot42}{1\cdot6\cdot9+2\cdot12\cdot18+4\cdot24\cdot36+7\cdot42\cdot63}\)
So sánh A và B
\(A=\frac{8056}{2012.16-1982}\)= \(\frac{2014.4}{2012.15+2012-1982}\)=\(\frac{2014.4}{2012.15+30}\)=\(\frac{2014.4}{2012.15+2.15}\)=\(\frac{2014.4}{15.\left(2012+2\right)}=\frac{2014.4}{15.2014}=\frac{4}{15}\)
B = \(\frac{1.2.6+2.4.12+4.8.24+7.14.42}{1.6.9+2.12.18+4.24.36+7.42.63}\)
= \(\frac{1.2.3.2+2.2.2.12+4.4.2.24+7.7.2.42}{1.2.3.9+2.12.2.9+4.24.4.9+7.42.7.9}\)
= \(\frac{2\left(1.2.3+2.2.12+4.4.24+7.7.42\right)}{9\left(1.2.3+2.2.12+4.4.24+7.7.42\right)}\)
= \(\frac{2}{9}\)
Ta có: \(\frac{4}{15}=\frac{4.3}{15.3}=\frac{12}{45};\frac{2}{9}=\frac{2.5}{9.5}=\frac{10}{45}\)
Vì \(\frac{12}{45}>\frac{10}{45}\Rightarrow\frac{4}{15}>\frac{2}{9}\Rightarrow A>B\)
Vậy A > B
Tính :
\(B=\dfrac{45}{12\cdot21}+\dfrac{45}{21\cdot31}-\dfrac{40}{24\cdot34}-\dfrac{40}{34\cdot44}-\dfrac{40}{44\cdot54}-\dfrac{40}{54\cdot64}\)
ta có B = \(\dfrac{45}{12.21}+\dfrac{45}{21.30}-\left(\dfrac{40}{24.34}+...+\dfrac{40}{54.64}\right)\)
\(=5\left(\dfrac{9}{12.21}+\dfrac{9}{21.30}\right)-4\left(\dfrac{10}{24.34}+...+\dfrac{10}{54.64}\right)\)
\(=5\left(\dfrac{1}{12}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{30}\right)-4\left(\dfrac{1}{24}-\dfrac{1}{34}+\dfrac{1}{34}-\dfrac{1}{44}+...+\dfrac{1}{54}-\dfrac{1}{64}\right)\)
\(=5\left(\dfrac{1}{12}-\dfrac{1}{30}\right)-4\left(\dfrac{1}{24}-\dfrac{1}{64}\right)\)
\(=5.\dfrac{1}{20}-4.\dfrac{5}{192}\)
\(=5.\dfrac{1}{20}-\dfrac{4}{192}.5\)
\(=5\left(\dfrac{1}{20}-\dfrac{4}{192}\right)=5.\dfrac{7}{240}=\dfrac{7}{48}\)
Rút gọn rồi so sánh
A=\(\dfrac{8056}{2012\cdot16-1982}\)
B=\(\dfrac{1\cdot2\cdot6+2\cdot4\cdot12+4\cdot8\cdot24+7\cdot14\cdot42}{1\cdot6\cdot9+2\cdot12\cdot18+4\cdot24\cdot36+7\cdot42\cdot63}\)
So sánh A và B
\(A=\dfrac{8056}{2012.16-1982}\)
\(A=\dfrac{8056}{32192-1982}\)
\(A=\dfrac{8056}{30210}=\dfrac{12}{45}\)
\(B=\dfrac{1.2.6+2.4.12+4.8.24+7.14.42}{1.6.9+2.12.18+4.24.36+7.42.63}\)
\(B=\dfrac{12+96+768+4116}{54+432+3456+18522}\)
\(B=\dfrac{4992}{22464}=\dfrac{10}{45}\)
Vậy: \(\dfrac{12}{45}>\dfrac{10}{45}\Rightarrow A>B\)
Tính:
a/\(\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}}{1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}:\dfrac{3+\dfrac{3}{2}+\dfrac{3}{3}+\dfrac{3}{4}}{2-\dfrac{2}{2}+\dfrac{2}{3}-\dfrac{2}{4}}\)
b/\(\dfrac{1+\dfrac{1}{4}+\dfrac{1}{1+\dfrac{1}{4}}}{1-\dfrac{1}{4}-\dfrac{1}{1-\dfrac{1}{4}}}\)
c/\(\dfrac{\dfrac{2}{5}-\dfrac{7}{5}}{\dfrac{2}{5}-\dfrac{\dfrac{3}{4}}{\dfrac{3}{4}.\dfrac{3}{7}-1}}-\dfrac{1}{\dfrac{3}{7}\left(\dfrac{3}{4}.\dfrac{3}{7}.\dfrac{2}{5}-\dfrac{2}{5}-\dfrac{3}{4}\right)}\)
d/\(\left(\dfrac{\dfrac{4}{3}}{2+\dfrac{4}{3}}+\dfrac{2-\dfrac{4}{3}}{\dfrac{4}{3}}\right).\left(\dfrac{\dfrac{2}{3}}{4+\dfrac{2}{3}}-\dfrac{4-\dfrac{2}{3}}{\dfrac{2}{3}}\right)\)
Giúp mik với các bạn ơi 1 bài thôi cug đc.
a
= { 1*( 1+1/2+1/3+1/4) } / { 1 * ( 1-1/2 +1/3-1/4)} : { 3*(1+1/2+1/3+1/4)} / { 2*( 1-1/2 +1/3-1/4)}
Sau đó bn tự tính ra nhé cứ tính nhu bình thường sẽ ra.
Mà mình thấy máy câu này yêu cầu tính chứ có bảo tính theo cách hợp lí đâu? Vì thế bn cứ lấy máy tính tính như bình thường là được .
chứng minh rằng
a , \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+...+\dfrac{1}{512}-\dfrac{1}{1024}\) < \(\dfrac{1}{3}\)
b , \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) < \(\dfrac{3}{16}\)
b) B=\(\dfrac{\dfrac{1}{22}+\dfrac{1}{13}-0,5}{\dfrac{3}{13}-\dfrac{3}{2}+\dfrac{3}{22}}.\dfrac{\dfrac{3}{4}-0,375+\dfrac{3}{16}-\dfrac{3}{32}}{1-\dfrac{1}{2}+\dfrac{1}{4}-0,875}+\dfrac{3}{4}\)
Ta có: \(B=\dfrac{\dfrac{1}{22}-\dfrac{1}{2}+\dfrac{1}{13}}{\dfrac{3}{22}-\dfrac{3}{2}+\dfrac{3}{13}}\cdot\dfrac{\dfrac{3}{4}-0.375+\dfrac{3}{16}-\dfrac{3}{32}}{1-\dfrac{1}{2}+\dfrac{1}{4}-0.875}+\dfrac{3}{4}\)
\(=\dfrac{1}{3}\cdot\dfrac{-15}{4}+\dfrac{3}{4}\)
\(=\dfrac{-5}{4}+\dfrac{3}{4}=\dfrac{-1}{2}\)
1/ \(\dfrac{5x+1}{8}-\dfrac{x-2}{4}=\dfrac{1}{2}\)
2/ \(\dfrac{x+3}{4}+\dfrac{1-3x}{3}=\dfrac{-x+1}{18}\)
3/ \(\dfrac{x+2}{4}-\dfrac{5x}{6}=\dfrac{1-x}{3}\)
4/ \(\dfrac{x-3}{2}-\dfrac{x+1}{10}=\dfrac{x-2}{5}\)
5/ \(\dfrac{4x+1}{4}-\dfrac{9x-5}{12}+\dfrac{x-2}{3}=0\)
1: Ta có: \(\dfrac{5x+1}{8}-\dfrac{x-2}{4}=\dfrac{1}{2}\)
\(\Leftrightarrow5x+1-2\left(x-2\right)=4\)
\(\Leftrightarrow5x+1-2x+4=4\)
\(\Leftrightarrow3x=-1\)
hay \(x=-\dfrac{1}{3}\)
2: Ta có: \(\dfrac{x+3}{4}+\dfrac{1-3x}{3}=\dfrac{-x+1}{18}\)
\(\Leftrightarrow9x+27+12-36x=-2x+2\)
\(\Leftrightarrow-27x+2x=2-39\)
hay \(x=\dfrac{37}{25}\)
3: Ta có: \(\dfrac{x+2}{4}-\dfrac{5x}{6}=\dfrac{1-x}{3}\)
\(\Leftrightarrow3x+6-10x=4-4x\)
\(\Leftrightarrow-7x+4x=4-6=-2\)
hay \(x=\dfrac{2}{3}\)
4: Ta có: \(\dfrac{x-3}{2}-\dfrac{x+1}{10}=\dfrac{x-2}{5}\)
\(\Leftrightarrow5x-15-x-1=2x-4\)
\(\Leftrightarrow4x-2x=-4+16=12\)
hay x=6
5: Ta có: \(\dfrac{4x+1}{4}-\dfrac{9x-5}{12}+\dfrac{x-2}{3}=0\)
\(\Leftrightarrow12x+3-9x+5+4x-8=0\)
\(\Leftrightarrow7x=0\)
hay x=0