_\(Ú\) \(òa\) \(......!!!!!!~~~\)
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_\(Chúc\) \(mn\) \(btvv...!!!!!~\)
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_\(Tớ\) \(đã\) \(đọc\)\(nội\) \(quy\) \(và\) \(kh\) \(cần\) \(ai\) \(check\) \(vào\) \(đây\) \(......!!!~~\)
_\(Understand....????!!~\)
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_\(11+11=?\)
_\(22+22=?\)
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\(4+5+6+7-1-2-3=\)\(=?\)
\(các\)\(bn\)\(biết\)\(nhóm\)\(nhạc\)\(Kpop\)\(nào\)\(z\)\(?\)
\(62\)\(+\)\(90\)\(-67\)
\(\frac{6}{3}\)\(X\)\(\frac{89}{6}\)
\(78,9\)\(:\)\(59,2\)
\(Ai\)\(\text{đ}ng\)\(on\)\(kb\)\(vs\)\(tui\)\(nha\)
\(1\)\(+\)\(1\)
\(2\)\(+\)\(77\)
\(ai\)\(lamf\)\(gia\)\(sư\)\(cho\)\(mh\)\(không\)
mình kick cho
cho hai phân thức\(\frac{P}{Q}\) và  \(\frac{R}{S}\) chứng tỏ rằng
câu(a) \(\frac{P}{Q}\)=\(\frac{R}{S}\)thì \(\frac{P+Q}{Q}\)=\(\frac{R+S}{S}\)
câu(b) \(\frac{P}{Q}\)=\(\frac{R}{S}\)thì P khác S và R khác S và\(\frac{P}{P-Q}\)=\(\frac{R}{S-R}\)
CHỨNG MINH HỘ MÌNH NHÉ MÌNH LIKE CHO
\(Cho\)\(Xin\)\(Cái\) \(NickBangBang\)\(Vip\)
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.
D= \(\frac{x^2-2x+2000}{x^2}\)= 1-\(\frac{2x}{x^2}\)+\(\frac{2000}{x^2}\)
Đặt t= \(\frac{1}{2}\)=> D= 2000t2-2t+1 = (\(20\sqrt{5}t\))2-2.\(20\sqrt{5}t\).\(\frac{1}{20\sqrt{5}}\)+\(\left(\frac{1}{20\sqrt{5}}\right)^2\)\(-\left(\frac{1}{20\sqrt{5}}\right)^2\)+1
D= (\(20\sqrt{5}t\)-\(\frac{1}{20\sqrt{5}}\)) 2+\(\frac{1999}{2000}\)\(\ge\)\(\frac{1999}{2000}\)
Min D= \(\frac{1999}{2000}\)khi \(20\sqrt{5}t\)\(-\frac{1}{20\sqrt{5}}\)= 0 => t = \(\frac{1}{2000}\)=> \(\frac{1}{x}\)= \(\frac{1}{2000}\)=> x= 2000