Giải:

a) Có: \(0,\left(37\right)=0,373737373737...\)

\(0,\left(62\right)=0,626262626262...\)

\(\Leftrightarrow0,\left(37\right)+0,\left(62\right)=0,99999999999...\)

Mà \(0,9999999999999...\simeq1\)

Hay \(0,\left(9\right)=1\)

Vậy \(0,\left(37\right)+0,\left(62\right)=1\).

b) \(0,\left(33\right).3=0,99999...=0,\left(9\right)=1\)

Vậy \(0,\left(33\right).3=1\).

Chúc bạn học tốt!!!

\(a)0,\left(37\right)=0,37373737....\)

\(0,\left(62\right)=0,62626262....\)\(\Leftrightarrow0,\left(37\right)+0,\left(62\right)=0,99999999....\)

Mà \(0,99999999....\simeq1\)

hoặc \(0,\left(9\right)\simeq1\)

\(\Rightarrow0,\left(37\right)+\left(0,62\right)=1\)

\(b)0,\left(33\right).3=1\)

\(\Leftrightarrow0,99999999....=0,\left(9\right)\simeq1\)

\(\Rightarrow0,\left(33\right).3=1\)

Chúc bạn học tốt!

không có câu hỏi sao trả lời

Ta có hình vẽ:

x x' O y y' \(\widehat{xOy}+\widehat{yOx'}+\widehat{x'Oy'}=297^o\)

\(\widehat{xOy}\) và \(\widehat{x'Oy'}\) đối đỉnh \(\Rightarrow\widehat{xOy}=\widehat{x'Oy'}\)

\(\widehat{x'Oy}\) và \(\widehat{x'Oy'}\) kề bù nên:

\(\widehat{x'Oy'}+\widehat{x'Oy}=180^o\)

\(\Rightarrow\widehat{xOy}+180^0=297^o\)

\(\Rightarrow\widehat{xOy}=117^o\)

\(\widehat{xOy}=\widehat{x'Oy'}=117^o\)

\(\Rightarrow\widehat{x'Oy}=297^o-117^o-177^o=3^o\)

\(\widehat{x'Oy}\) đối đỉnh với \(\widehat{xOy'}\) nên

\(\widehat{x'Oy}=\widehat{xOy'}=3^o\)

Vậy...

Ta có :

\(\left(x-10\right)^{x+1}-\left(x-10\right)^{x+11}=0\)

\(\Leftrightarrow\left(x-10\right)^{x+1}\left[1-\left(x-10\right)^{10}\right]=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-10=0\\1-\left(x-10\right)^{10}=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=10\\\left[\begin{array}{nghiempt}x-10=1\\x-10=-1\end{array}\right.\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=10\\\left[\begin{array}{nghiempt}x=11\\x=9\end{array}\right.\end{array}\right.\)

Vậy x = 10 ; x = 11 ; x =  9

\(\left(x-10\right)^{x+1}-\left(x-10\right)^{x+11}=0\)

\(\Rightarrow\left(x-10\right)^{x+1}.\left[1-\left(x-10\right)^{10}\right]=0\)

\(\Rightarrow\left(x-10\right)^{x+1}=0\) hoặc \(1-\left(x-10\right)^{10}=0\)

+) \(\left(x-10\right)^{x+1}=0\)

\(\Rightarrow x-10=0\)

\(\Rightarrow x=10\)

+) \(1-\left(x-10\right)^{10}=0\)

\(\Rightarrow\left(x-10\right)^{10}=1\)

\(\Rightarrow x-10=\pm1\)

+ \(x-10=1\Rightarrow x=11\)

+ \(x-10=-1\Rightarrow x=9\)

Vậy \(x\in\left\{10;11;9\right\}\)

\(3x^2y^4\)-\(5xy^3\)-\(\dfrac{3}{2}x^2y^4\)+\(3xy^3\)+\(2xy^3\)+1=1,5\(x^2y^4\)+1>0

thank you!!!!!!

2ⁿ⁺³ + 2ⁿ⁺² - 2ⁿ⁺¹ + 2ⁿ = 2ⁿ.2³ + 2ⁿ.2² - 2ⁿ.2 + 2ⁿ

= 2ⁿ.(2³ + 2² - 2 + 1)

= 2ⁿ.11

P= \(x^{2y5}-3y^3+3x^3-x^3y-2015\)

P +Q =0 => P = -Q = x²y⁵ - 3y³ + 3x³ - x³y -2015

ủa sao ngộ z ?

bn dợi mk lát nhé

mk ko chép đề mà tách luôn nha

M = x²x² + x²x² + x²y² + x²y² + x²y² + y²y² + y²

= ( x²x² + x²y² ) + ( x²x² + x²y² ) + ( x²y² + y²y² ) + y²

= x²( x² + y² ) + x²( x² + y² ) + y²( x² + y² ) + y²

= ( x² + y² ) (x² + x² + y² ) + y²

= 1( x² + 1) + y²

= x² + y² +1 = 2

thanks bn

\(A=\dfrac{4^2}{1.3}+\dfrac{4^2}{3.5}+\dfrac{4^2}{5.8}+...+\dfrac{4^2}{45.47}.\dfrac{1-3-5-...-49}{8}\)

\(A=4\left(\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.8}+...+\dfrac{4}{45.47}\right).\dfrac{1-3-5-...-49}{8}\)\(A=4\left[2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{45}-\dfrac{1}{47}\right)\right].\dfrac{1-3-5-...-49}{8}\)\(A=8\left(1-\dfrac{1}{47}\right).\dfrac{1-3-5-...-49}{8}\)

\(A=8\left(1-\dfrac{1}{47}\right).\dfrac{-623}{8}\)

\(A=\dfrac{368}{47}.\dfrac{-623}{8}=\dfrac{-28658}{47}\)