Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
=(100+99)(100-99)+(98+97)(98-97)+....+(2+1)(2-1)
=199+195+....+3
dãy số trên có số số hạng là :
(199-3):4+1=50 (số hạng)
tổng dãy số trên là :
(199+3)50/2=5050
vậy 100^2-99^2+98^2-97^2+...+2^2-1^2=5050
a) \(\frac{5-x}{4x^2-8x}\) + \(\frac{7}{8x}\) = \(\frac{x-1}{2x\left(x-2\right)}\) +\(\frac{1}{8x-16}\) ĐKXĐ : x #0, x#2, x#-2
<=> \(\frac{5-x}{4x\left(x-2\right)}\) + \(\frac{7}{8x}=\frac{x-1}{2x\left(x-2\right)}\) + \(\frac{1}{8\left(x-2\right)}\)
<=> \(\frac{2\left(5-x\right)}{8x\left(x-2\right)}+\frac{7\left(x-2\right)}{8x\left(x-2\right)}=\frac{4\left(x-1\right)}{8x\left(x-2\right)}+\frac{x}{8x\left(x-2\right)}\)
=> 10 - 2x + 7x - 14 = 4x - 4 + x
<=>-2x + 7x - 4x + x = -4 - 10 + 14
<=>x=-14
a/ A = 1002 - 992 + 982 -...+22 - 12
= (1002 - 992) + (982 - 972) +...+ (22 - 12)
= 199 + 195 + 191 + ... + 1
= (\(\frac{199-1}{4}+1\))(\(\frac{199+1}{2}\)) = 5050
b/ Y chang câu a luôn nha
c/ \(C=\frac{780^2-220^2}{125^2+150.125+75^2}=\frac{\left(780-220\right)\left(780+220\right)}{\left(125+75\right)^2}\)
\(=\frac{560.1000}{200^2}=14\)
(502+482+...+22) - (492+472+...+12)
= (502-492) + (482-472) + ... + (22-12)
= (50+49)(50-49) + (48+47)(48-47) + ... + (2+1)(2-1)
= 50+49+48+47+...+1
= \(\frac{\left(50+1\right).50}{2}=\frac{51.50}{2}=1275\)
\(R=\frac{43^2-11^2}{\left(36,5\right)^2-\left(27,5\right)^2}\)
\(=\frac{\left(43-11\right)\left(43+11\right)}{\left(36,5-27,5\right)\left(36,5+27,5\right)}\)
\(=\frac{32.54}{9.64}\)
\(=\frac{6}{2}=3\)
Bạn viết sai đề bài rồi
\(S=\frac{97^3+83^3}{180}-97.83\)
\(=\frac{\left(97+83\right)\left(97^2-97.83+83^2\right)}{180}-97.83\)
\(=97^2-97.83+83-97.83\)
\(=\left(97-83\right)^2=14^2=196\)
Trả lời:
\(R=\frac{43^2-11^2}{36,5^2-27,5^2}\)
\(R=\frac{\left(43-11\right).\left(43+11\right)}{\left(36,5-27,5\right).\left(36,5+27,5\right)}\)
\(R=\frac{32.54}{9.64}\)
\(R=3\)
Đề bài sai bạn nhé
\(S=\frac{97^3+83^3}{180}-97.83\)
\(S=\frac{\left(97+83\right).\left(97^2-97.23+83^2\right)}{180}-97.83\)
\(S=97^2-97.83+83^2-97.83\)
\(S=97^2-2.97.83+83^2\)
\(S=\left(97-83\right)^2\)
\(S=14^2\)
\(S=196\)
b)\(\dfrac{x+14}{86}+\dfrac{x+15}{85}+\dfrac{x+16}{84}+\dfrac{x+17}{83}+\dfrac{x+116}{4}=0\)
\(\Leftrightarrow\dfrac{x+14}{86}+1+\dfrac{x+15}{85}+1+\dfrac{x+16}{84}+1+\dfrac{x+17}{83}+1+\dfrac{x+116}{4}-4=0\)
\(\Leftrightarrow\dfrac{x+100}{86}+\dfrac{x+100}{85}+\dfrac{x+100}{84}+\dfrac{x+100}{83}+\dfrac{x+100}{4}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{86}+\dfrac{1}{85}+\dfrac{1}{84}+\dfrac{1}{83}+\dfrac{1}{4}\right)=0\)
\(\Leftrightarrow x+100=0\).Do \(\dfrac{1}{86}+\dfrac{1}{85}+\dfrac{1}{84}+\dfrac{1}{83}+\dfrac{1}{4}\ne0\)
\(\Leftrightarrow x=-100\)
c)\(\dfrac{1}{\left(x^2+5\right)\left(x^2+4\right)}+\dfrac{1}{\left(x^2+4\right)\left(x^2+3\right)}+\dfrac{1}{\left(x^2+3\right)\left(x^2+2\right)}+\dfrac{1}{\left(x^2+2\right)\left(x^2+1\right)}=-1\)
\(\Leftrightarrow\dfrac{1}{\left(x^2+1\right)\left(x^2+2\right)}+\dfrac{1}{\left(x^2+2\right)\left(x^2+3\right)}+...+\dfrac{1}{\left(x^2+4\right)\left(x^2+5\right)}=-1\)
\(\Leftrightarrow\dfrac{1}{x^2+1}-\dfrac{1}{x^2+2}+\dfrac{1}{x^2+2}-\dfrac{1}{x^2+3}+...+\dfrac{1}{x^2+4}-\dfrac{1}{x^2+5}=-1\)
\(\Leftrightarrow\dfrac{1}{x^2+1}-\dfrac{1}{x^2+5}=-1\)\(\Leftrightarrow\dfrac{4}{x^4+6x^2+5}=-1\)
\(\Leftrightarrow\dfrac{x^4+6x^2+9}{x^4+6x^2+5}=0\Leftrightarrow x^4+6x^2+9=0\)
\(\Leftrightarrow\left(x^2+3\right)^2>0\forall x\) (vô nghiệm)
a) \(A=\frac{97^3+83^3}{180}-97\cdot83\)
\(A=\frac{\left(97+83\right)\left(97^2-97\cdot83+83^2\right)}{180}-97\cdot83\)
\(A=\frac{180\cdot\left(97^2-97\cdot83+83^2\right)}{180}-97\cdot83\)
\(A=97^2-97\cdot83+83^2-97\cdot83\)
\(A=9409-2\cdot8051+6889\)
\(A=196\)
b) \(B=\left(50^2+48^2+...+2^2\right)-\left(49^2+47^2+...+1^2\right)\)
\(B=50^2+48^2+...+2^2-49^2-47^2-...-1^2\)
\(B=\left(50^2-49^2\right)+\left(48^2-47^2\right)+...+\left(2^2-1^2\right)\)
\(B=\left(50+49\right)\left(50-49\right)+\left(48+47\right)\left(48-47\right)+...+\left(2+1\right)\left(2-1\right)\)
\(B=50+49+48+47+...+2+1\)
Số số hạng là : \(\left(50-1\right):1+1=50\)( số )
Tổng B là : \(\left(50+1\right)\cdot50:2=1275\)
Vậy....