\(\frac{2}{3}\)+ \(\frac{2}{5}\): \(\frac{5}{4}\)
ghi cach lam cu the
5/2+ 5/4 + 5/8 + 5/16 + 5 /32 + 5/64
ghi cach lam cu the
5/2 + 5/4 + 5/8 + 5/16 + 5/32 + 5/64
= 5(1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64)
Đặt A = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64
=> 2A = 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32
=> 2A - A = 1 - 1/64
=> A = 1 - 1/64
Do đó : 5(1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64) = 5(1 - 1/64) = 5 . 63/64 = 315/64
\(2A=5+\frac{5}{2}+\frac{5}{4}+\frac{5}{8}+\frac{5}{16}+\frac{5}{32}.\)
\(A=2A-A=5-\frac{5}{64}=\frac{315}{64}\)
Tinh
D= \(\frac{\frac{1}{3}+\frac{1}{17}-\frac{1}{13}}{\frac{2}{3}+\frac{2}{17}-\frac{2}{13}}.\frac{\frac{3}{4}-\frac{3}{16}+\frac{3}{64}-\frac{3}{256}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)
Trinh bay cach lam nhe
Ko hieu vao DOC THEM
1/2 . ( 1/3- 1/11)=.....
ghi cach lam cu the
\(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(=\frac{1}{2}\times\frac{8}{33}\)
\(=\frac{4}{33}\)
*Cách làm làm trong ngoặc tròn trước rồi nhân :v
\(\frac{1}{2}\)x(\(\frac{1}{3}\)-\(\frac{1}{11}\))
=\(\frac{8}{66}\)
\(\frac{2}{7}-x=-\frac{3}{4}\)
ket qua va cach lam
\(\frac{2}{7}-x=\frac{-3}{4}\)
\(x=\frac{2}{7}-\frac{-3}{4}\)
\(x=\frac{8}{28}+\frac{21}{28}=\frac{29}{28}\)
vậy \(x=\frac{29}{28}\)
chúc bn học tốt!
= 2/7- -3/4
= 8/28 + 21/28
= 29/28
vậy x =29/28
~Study Well~
tk mk nha
2/7-x=-3/4
x=2/7-(-3/4)
x=2/7+3/4
x=8/28+21/28
X=29/28
Tìm x biết \(\frac{x+24}{1996}+\frac{x+25}{1995}+\frac{x+26}{1994}+\frac{x+27}{1993}+\frac{x+2036}{4}=0\)
xin ghi ro cach lam
\(\frac{x+24}{1996}+\frac{x+25}{1995}+\frac{x+26}{1994}+\frac{x+27}{1993}+\frac{x+2036}{4}=0\)
\(\Leftrightarrow\frac{x+24}{1996}+1+\frac{x+25}{1995}+1+\frac{x+26}{1994}+\frac{x+27}{1993}+1+\frac{x+2036}{4}-4==0\)
\(\Leftrightarrow\frac{x+2020}{1996}+\frac{x+2020}{1995}+\frac{x+2020}{1994}+\frac{x+2020}{1993}+\frac{x+2020}{4}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{1996}+\frac{1}{1995}+\frac{1}{1994}+\frac{1}{1993}+\frac{1}{4}\right)=0\)
<=> x+2020=0 \(\left(\frac{1}{1996}+\frac{1}{1955}+\frac{1}{1994}+\frac{1}{1993}+\frac{1}{4}\right)=0\)
<=> x=-2020
Mình sửa lại dòng thứ 2 từ dưới lên
\(\Leftrightarrow x+2020=0\left(\frac{1}{1996}+\frac{1}{1995}+\frac{1}{1994}+\frac{1}{1993}+\frac{1}{4}\ne0\right)\)
<=> x=-2020
Vậy x=-2020
\(\frac{3}{4}x\left(\frac{8}{3}-\frac{4}{5}\right)+\frac{3}{4}-\left(\frac{4}{5}-\frac{2}{3}\right)\)
Giup minh lam bai nay voi
Mình cần giúp. Ai đó có thể giúp mình đc ko zậy
\(\frac{3}{4}+\frac{3}{16}+\frac{3}{16} \)
lam hai cach gium minh nha 1 cach cung dc :)
=\(\frac{12}{16}+\frac{3}{16}+\frac{3}{16}\)
=\(\frac{18}{16}=\frac{9}{8}\)
\(c1:\frac{3}{4}+\frac{3}{16}+\frac{3}{16}=\frac{15}{16}+\frac{3}{16}=\frac{9}{8}.\)
\(c2:\frac{3}{4}+\frac{3}{16}+\frac{3}{16}=\frac{3}{4}+\frac{6}{16}=\frac{9}{8}\)
3/4 + ( 3/16+3/16) = 3/4 + 6/16 = 18/16
bai: Thuc hien phep tinh bang cach hop ly nhat:
a) \((\frac{1}{5}-\frac{1}{4})^2\)
b) \((\frac{5}{3}+\frac{1}{2}):\frac{-13}{5}+1\frac{5}{6}\)
c) \((\frac{3}{17})^4.(\frac{-17}{6})^4\)
a, \(\left(\frac{1}{5}-\frac{1}{4}\right)^2=\left(\frac{-1}{20}\right)^2=\frac{1}{400}\)
b, \(\left(\frac{5}{3}+\frac{1}{2}\right):\frac{-13}{5}+1\frac{5}{6}\)
=> \(\frac{13}{6}:\frac{-13}{5}+\frac{11}{6}\)
=> \(\frac{13}{6}.\frac{5}{-13}+\frac{11}{6}\)
=> \(\frac{-5}{6}+\frac{11}{6}=\frac{6}{6}=1\)
c, \(\left(\frac{3}{17}\right)^4.\left(\frac{-17}{6}\right)^4\)
=> \(\left(\frac{3}{17}.\frac{-17}{6}\right)^4=\left(\frac{-1}{2}\right)^4=\frac{1}{16}\)
a) \(\left(\frac{1}{5}-\frac{1}{4}\right)^2=\left(-\frac{1}{20}\right)^2=\frac{1}{400}.\)
b) \(\left(\frac{5}{3}+\frac{1}{2}\right):-\frac{13}{5}+1\frac{5}{6}\)
= \(\frac{13}{6}:\left(-\frac{13}{5}\right)+\frac{11}{6}\)
= \(\left(-\frac{5}{6}\right)+\frac{11}{6}\)
= \(1.\)
c) \(\left(\frac{3}{17}\right)^4.\left(-\frac{17}{6}\right)^4\)
= \(\left[\frac{3}{17}.\left(-\frac{17}{6}\right)\right]^4\)
= \(\left(-\frac{1}{2}\right)^4=\frac{1}{16}.\)
Chúc bạn học tốt!
help me
chung minh :\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{149}+\frac{1}{150}>\frac{1}{3}\)
nho ghi ca cach lam nua nhe
Đặt A = \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{150}\)(50 số hạng)
=> A > \(\frac{1}{150}+\frac{1}{150}+\frac{1}{150}+...+\frac{1}{150}\)(50 số hạng)
=> A > \(\frac{1}{150}.50\)
=> A > \(\frac{1}{3}\)
=> \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{150}\) > \(\frac{1}{3}\)(Đpcm)
từ \(\frac{1}{101}\)đến \(\frac{1}{150}\)có 50 phân số.
có :\(\frac{1}{101}\)lớn hơn \(\frac{1}{150}\)
\(\frac{1}{102}\)lớn hơn \(\frac{1}{150}\)........cứ như vậy cho đến \(\frac{1}{149}\)lớn hơn \(\frac{1}{150}\).suy ra tổng 50 phân số đã cho lớn hơn 50 nhân vơi \(\frac{1}{150}\)=\(\frac{1}{3}\)