ban co the sao chep roi hoi cung duoc ma co gang len nhat ban se lam duoc bai nay va ca nhung bai cau ko hieu moi ngoui dau phai cai gi cung co the biet het dau, co gang len nhe Hai Anh
Điền dấu gì?
\(\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+.....+\frac{1}{90}\)so sánh với 2
Thực hiện phép tính:
a)\(\frac{3}{4}\)x\(\frac{16}{9}\)-\(\frac{7}{5}\):\(\frac{-21}{20}\)
b)\(2\frac{1}{3}\)-\(\frac{1}{3}\)x [\(\frac{-3}{2}\)+(\(\frac{2}{3}\)+0,4x5) ]
c) (20+\(9\frac{1}{4}\)):\(2\frac{1}{4}\)
d) (6-\(2\frac{4}{5}\)x\(3\frac{1}{8}\)-\(1\frac{3}{5}\):\(\frac{1}{4}\)
GIÚP MIK VỚI AI NHANH MIK TICK LUÔN CHO.
ta có \(A=\frac{1}{x^3+y^3}+\frac{4}{xy}=\frac{1}{\left(x+y\right)\left(x^2-xy+y^2\right)}+\frac{4}{xy}=\frac{1}{x^2-xy+y^2}+\frac{1}{xy}+\frac{1}{xy}+\frac{1}{xy}+\frac{1}{xy}\)
áp dụng bất đẳng thức svác sơ ta có
\(\frac{1}{x^2-xy+y^2}+\frac{1}{xy}+\frac{1}{xy}+\frac{1}{xy}\ge\frac{16}{x^2+y^2+2xy}=16\)
mà \(xy\le\frac{\left(x+y\right)^2}{4}=\frac{1}{4}\)
=> \(\frac{1}{xy}\ge4\)
=> \(A\ge20\)
dấu = xảy ra <=> x=y=1/2
e,\(A=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+\left(1-\frac{1}{20}\right)+\left(1-\frac{1}{20}\right)+\left(1-\frac{1}{42}\right)\)
\(\Rightarrow A=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}+1-\frac{1}{42}=4-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
\(\Rightarrow A=4-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)=4-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(\Rightarrow A=4-\left(\frac{1}{1}-\frac{1}{7}\right)=4-\frac{6}{7}=3\frac{1}{7}\)
Tính bằng cách thuận tiện nhất:
\(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{4}{6}+\frac{9}{12}+\frac{16}{20}\)
1.Chứng minh rằng :
\(4\sqrt[4]{\left(a+1\right)\left(b+4\right)\left(c-2\right)\left(d-3\right)}\le a+b+c+d\)với \(a\ge-1;b\ge-4;c\ge2;d>3\)
2. Chứng minh rằng :
\(\frac{a^2}{b^5}+\frac{b^2}{c^5}+\frac{c^2}{d^5}+\frac{d^2}{a^5}\ge\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{1}{d^3}\)với \(a,b,c,d>0\)
\(VT=a+b+\frac{1}{a}+\frac{1}{b}=\left(a+\frac{1}{2a}\right)+\left(b+\frac{1}{2b}\right)+\frac{1}{2a}+\frac{1}{2b}\)
để ý \(1=a^2+b^2\ge2ab\Leftrightarrow ab\le\frac{1}{2}\)
\(\frac{1}{2a}+\frac{1}{2b}\ge2\sqrt{\frac{1}{4ab}}\ge2\sqrt{\frac{1}{2}}\)
\(a+\frac{1}{2a}\ge2\sqrt{\frac{1}{2}}\)
\(b+\frac{1}{2b}\ge2\sqrt{\frac{1}{2}}\)
+ 3 vế thì ta được \(VT\ge6\sqrt{\frac{1}{2}}\) dấu = khi \(\frac{1}{2a}=\frac{1}{2b}....a=\frac{1}{2a}....b=\frac{1}{2b}\)
So sánh A= \(\frac{79^{2018}+1}{79^{2019}+1};B=\frac{79^{2019}-2021}{79^{2020}-2011}\)
Áp dụng bất đẳng thức bu nhi a , ta có
\(\left(a+b+c\right)\left[\frac{a}{\left(ab+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ca+c+1\right)^2}\right]\ge\left(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\right)^2\)
mà bạn dễ dàng chứng minh \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ac+c+1}=1\) với abc=1
=>A(a+b+c)^2>=1
=>\(\frac{a}{\left(ab+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ca+c+1\right)^2}\ge\frac{1}{a+b+c}\left(ĐPCM\right)\)
đấu = xảy ra <=> a=b=c1
