\(\left(1-x\right)^2=2003.\left(x-1\right)\)

\(\left(1-x\right)^2-2003\left(x-1\right)=0\)

\(\left(1-x\right)^2+2003\left(1-x\right)=0\)

\(\left(1-x\right)\left(1-x+2003\right)=0\)

\(\left(1-x\right)\left(2004-x\right)=0\)

\(TH1:1-x=0\)

\(x=1\)

\(TH2:2004-x=0\)

\(x=2004\)

vậy........

\(A=\left(1-\dfrac{1}{4}\right)+\left(\dfrac{1}{4}-\dfrac{1}{9}\right)+\left(\dfrac{1}{9}-\dfrac{1}{16}\right)+...+\left(\dfrac{1}{2401}-\dfrac{1}{2500}\right)\)

\(A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{2401}-\dfrac{1}{2500}\)

\(A=1-\dfrac{1}{2500}=\dfrac{2499}{2500}\)

\(15^8.2^4=\left(15^2\right)^4.2^4=225^4.2^4=\left(225.2\right)^4=450^4\\ 27^5:32^3=\left(3^3\right)^5:\left(2^5\right)^3=3^{15}:2^{15}=\left(\dfrac{3}{2}\right)^{15}\)

\(=12,5:18=\dfrac{216}{5}\)

Lời giải:

$(-2,5).\frac{5}{18}=\frac{-5}{2}.\frac{5}{18}=\frac{-25}{36}$

Lời giải:

Gọi tổng trên là $A$

$A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}$

$2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{18.19.20}$

$=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+....+\frac{20-18}{18.19.20}$

$=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{18.19}-\frac{1}{19.20}$

$=\frac{1}{1.2}-\frac{1}{19.20}=\frac{189}{380}$

$\Rightarrow A=\frac{189}{760}$

ĐÚNG ĐÓ BẠN ƠI

Ai làm được cho nhiều sao đang cần gấp 🌟⭐️

Yêu cầu của đề là gì vậy em.

`#3107`

\(\left(2x-\dfrac{1}{2}\right)^2+\dfrac{3}{7}=\dfrac{19}{8}\\ \Rightarrow\left(2x-\dfrac{1}{2}\right)^2=\dfrac{19}{8}-\dfrac{3}{7}\\ \Rightarrow\left(2x-\dfrac{1}{2}\right)^2=\dfrac{109}{56}\\ \Rightarrow\left(2x-\dfrac{1}{2}\right)^2=\left(\sqrt{\dfrac{109}{56}}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{1}{2}=\sqrt{\dfrac{109}{56}}\\2x-\dfrac{1}{2}=-\sqrt{\dfrac{109}{56}}\end{matrix}\right.\)

Bạn xem lại đề, số lớn quá ;-;.

may ngu nhu cut y dap an la 66

?

Vì Ot là tia phân giác của \(\widehat{xOy}\) nên \(\widehat{xOt}=\widehat{tOy}=\dfrac{\widehat{xOy}}{2}=\dfrac{80^o}{2}=40^o\)