Bấm vào nút kiểm tra nhé

hello

Bài này chưa cần phải nộp nhé

đây nhé:

x/5=5/6+(-19)/30

x/5=25/30+(-19)/30

x/5= 6/30

x/5=1 /5 suy ra x=1

nhớ kik nha bạn ^_^

Ta có: \(\frac{5}{6}+\frac{-19}{30}=\frac{25}{30}+\frac{-19}{30}=\frac{6}{30}\)

Mà \(\frac{x}{5}=x:5\)\(\Rightarrow x:5=\frac{6}{30}\Rightarrow x=\frac{6}{30}\cdot5=\frac{30}{30}=1\)

Lời giải bài 1:

\(\text{a) Trên cùng 1 nửa mặt phẳng bờ chứa tia Ox có}\)\(\widehat{xOy}< \widehat{xOz}\) \(\left(65< 130\right)\)

    \(\Rightarrow\text{ Oy nằm giữa Ox và Oz}\)

b) \(\text{Do Oy nằm giữa Ox và Oz }\)

\(\Rightarrow\widehat{xOy}+\widehat{yOz}=\widehat{xOz}\Rightarrow\widehat{xOz}-\widehat{xOy}=\widehat{yOz}\left(1\right)\)

  mà \(\widehat{xOy}=65^0;\widehat{xOz}=130^0\left(2\right)\)

\(\text{Từ (1) và (2)}\)\(\Rightarrow\)\(\widehat{yOz}=130^0-65^0=65^0\)

\(c.\)

Ta thấy \(\widehat{xOy}=65^0;\widehat{yOz}=65^0\)

\(\Rightarrow\widehat{xOy}=\widehat{yOz}\)

\(\text{d}.\)\(\widehat{yOm}+\widehat{xOy}=180^0\) \(\text{(kề bù)}\)

\(\Rightarrow\widehat{yOm}=\widehat{180^0}-\widehat{xOy}\left(3\right)\)

\(\text{ mà }\)\(\widehat{xOy}=65^0\)

\(\Rightarrow\widehat{yOm}=180^0-65^0=125^0\)

   \(\widehat{xOm}+\widehat{yOm}=180^0\) \(\text{(kề bù)}\)

\(\Rightarrow\widehat{yOm}=180^0-\widehat{xOm}\)

\(\text{mà }\)\(\widehat{xOm}=80^0\)

\(\Rightarrow\widehat{yOm}=100^0\)

B = 1 + 2\(^2\) + 2\(^4\) + ... + 2\(^{2024}\) + 2\(^{2026}\)

2\(^2\) B = 2\(^2\) + 2\(^4\) + ...+ \(2^{2026}\) + 2\(^{2028}\)

4B - B = 2\(^2\) + 2\(^4\) + ...+ \(2^{2026}\) + 2\(^{2028}\) - 1 - 2\(^2\) - 2\(^4\) - ... - 2\(^{2024}\) - 2\(^{2026}\)

3B = (2\(^2\) - 2\(^2\)) + (2\(^4\) - 2\(^4\)) +...+ (2\(^{2026}\) - \(2^{2026}\)) + (2\(^{2028}\) - 1)

3B = 0 + 0 +... + 0 + 2\(^{2028}\) - 1

3B = 2\(^{2028}\) - 1

B = \(\frac{2^{2028}-1}{3}\)

Ta có: \(B=1+2^2+2^4+\cdots+2^{2024}+2^{2026}\)

=>\(4B=2^2+2^4+2^6+\ldots+2^{2026}+2^{2028}\)

=>\(4B-B=2^2+2^4+2^6+\cdots+2^{2026}+2^{2028}-1-2^2-\cdots-2^{2026}\)

=>\(3B=2^{2028}-1\)

=>\(B=\frac{2^{2028}-1}{3}\)