Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
há há.. bài này mà lớp 8 hã?
\(50^2+48^2+...+4^2+2^2-49^2-47^2-...-1^2\)
\(=50^2-49^2+48^2-47^2+...+2^2-1^2\)
\(=\left(50+49\right)\left(50-49\right)+\left(48+47\right)\left(48-47\right)+...\left(2+1\right)\left(2-1\right)\)
\(=99+95+...+3\)
\(=\frac{\left(99+3\right)\left(99-3\right):4+1}{2}\)
\(=1275\)
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....
=(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
\(B=\left(50^2+48^2+46^2+...+4^2+2^2\right)-\left(49^2+47^2+45^2+...+3^2+1^2\right)\)
\(B=50^2+48^2+46^2+...+4^2+2^2-49^2-47^2-...-3^2-1^2\)
\(B=\left(50^2-49^2\right)+\left(48^2-47^2\right)+...+\left(4^2-3^2\right)+\left(2^2-1^2\right)\)
\(B=\left(50-49\right)\left(50+49\right)+\left(48-47\right)\left(48+47\right)+...+\left(4-3\right)\left(4+3\right)+\left(2-1\right)\left(2+1\right)\)
\(B=50+49+48+47+...+4+3+2+1\)
\(B=1+2+3+...+48+49+50\)
\(B=\dfrac{50-1+1}{2}.\left(1+50\right)\)
\(B=25.51\)
\(B=1275\)
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)
(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
= \(\frac{\left(50+1\right).50}{2}=\frac{51.50}{2}=1275\)