bài giải:
đặt biểu thức bằng A
=> A= \(\dfrac{1}{5}+\left(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}\right)+\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\right)\)
ta thấy:\(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< 3.\dfrac{1}{13}\)
\(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< 3.\dfrac{1}{61}\)
=> A<\(\dfrac{1}{5}+\dfrac{3}{13}+\dfrac{3}{61}\)<\(\dfrac{1}{2}\)
=> đpcm.
cho A =\(\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{16}+\dfrac{1}{17}+\dfrac{1}{18}+\dfrac{1}{19}+\dfrac{1}{20}\)SO SÁNH A VỚI \(\dfrac{1}{2}\)
Cho A= \(\dfrac{1}{1.102}+\dfrac{1}{2.103}+.....+\dfrac{1}{299.400}\)
Chứng minh rằng: A=\(\dfrac{1}{101}\left[\left(1+\dfrac{1}{2}+...+\dfrac{1}{101}\right)-\left(\dfrac{1}{300}+\dfrac{1}{301}+...+\dfrac{1}{400}\right)\right]\)
Help me please.....
S=\(\dfrac{1}{2}+\dfrac{1}{14}+\dfrac{1}{35}+\dfrac{1}{65}+\dfrac{1}{104}+\dfrac{1}{152}\)
Tìm x, y, z
\(\dfrac{x+y+2}{z}=\dfrac{y+z+1}{x}=\dfrac{z+x-3}{y}=\dfrac{1}{x+y+z}\)
Áp dụng tích chất của dãy tỉ số bằng nhau, ta có
\(\dfrac{x+y+2}{z}=\dfrac{y+z+1}{x}=\dfrac{z+x-3}{y}\\ =\dfrac{x+y+2+y+z+1+z+x-3}{z+x+y}=\dfrac{2\left(x+y+z\right)+\left(1+2-3\right)}{z+x+y}=2\\ Vì\dfrac{x+y+2}{z}=\dfrac{y+z+1}{x}=\dfrac{z+x-3}{y}=\dfrac{1}{x+y+z}\\ =>2=\dfrac{1}{x+y+z}=>2\left(x+y+z\right)=1=>x+y+z=\dfrac{1}{2}\\ =>\dfrac{x+y+2}{z}=2=>x+y+2=2z\\ \dfrac{y+z+1}{x}=2=>y+z+1=2x\\ \dfrac{z+x-3}{y}=2=>z+x-3=2y\\ \dfrac{1}{x+y+z}=2=>x+y+z=\dfrac{1}{2}\)
+) x+y+z = \(\dfrac{1}{2}=>y+z=\dfrac{1}{2}-x=>\dfrac{1}{2}-x+1=2x=>3x=\dfrac{3}{2}=>x=\dfrac{1}{2}\)
+)\(x+y+z=\dfrac{1}{2}=>x+y=\dfrac{1}{2}-z=>\dfrac{1}{2}-z+2=2z=>3z=\dfrac{5}{2}=>z=\dfrac{5}{6}\)
\(=>x+y+z=\dfrac{1}{2}+\dfrac{5}{6}+y=\dfrac{1}{2}=>\dfrac{4}{3}+y=\dfrac{1}{2}=>y=\dfrac{-5}{6}\)
Vậy \(x=\dfrac{1}{2}\\ y=\dfrac{-5}{6}\\ z=\dfrac{5}{6}\)
Ê mấy bọn 7B Nguyễn Lương Bằng ơi bài 2 Toán chiều làm thế này đúng chưa! Góp ý nha!
tính hợp li A=\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}{\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)-\dfrac{1}{2}\cdot\dfrac{1}{3}\cdot\dfrac{1}{4}}\)
Rút gọn:
A=\(\dfrac{\sqrt{a}+2}{\sqrt{a+3}}-\dfrac{5}{a+\sqrt{a}-6}+\dfrac{1}{2-\sqrt{a}}\)
B=\((1-\dfrac{\sqrt{a}}{\sqrt{a}+1})\div(\dfrac{\sqrt{a}+3}{\sqrt{a}-2}+\dfrac{\sqrt{a}+2}{3-\sqrt{a}}+\dfrac{\sqrt{a}+2}{a-5\sqrt{a}+6)}\)
C=\((\dfrac{\sqrt{a}-1}{3\sqrt{a}-1}-\dfrac{1}{3\sqrt{a}+1}+\dfrac{8\sqrt{a}}{9a-1})\div(1-\dfrac{3\sqrt{a}-2}{3\sqrt{a}+1)}\)
Giups mình cái mình đag cần gấp
Chứng minh rằng : \(\dfrac{1}{a}\) - \(\dfrac{1}{a+n}=\dfrac{n}{a\left(a+n\right)}\)
M.n giúp em vs ạ, một bài thôi cũng được, rất cần luôn!!!
1.Giải hệ phương trình:
\(\left\{{}\begin{matrix}x^4+3=4y\\x^4+3=4x\end{matrix}\right.\)
2. Viết tính chất đặc trưng cho các phân tử của tập hợp sau:
a) \(A=\left\{\dfrac{1}{2},\dfrac{1}{6},\dfrac{1}{12},\dfrac{1}{20},\dfrac{1}{30}\right\}\)
b) \(B=\left\{\dfrac{2}{3},\dfrac{3}{8},\dfrac{4}{15},\dfrac{5}{24},\dfrac{6}{35}\right\}\)
3. Tìm m để phương trình \(\left|x^2-1\right|=m^4-m^2+1\) có 4 nghiệm phân biệt.
So sánh
\(\dfrac{1}{2}\)- \(\dfrac{1}{4}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{16}\)+\(\dfrac{1}{32}\)-\(\dfrac{1}{64}\)<\(\dfrac{1}{3}\)