giúp em phần 4 với ạ
Giúp em câu Phần 4 với ạ
giúp em bài 1-4 phần Tự Luận với ạ
II/ Bài tập tham khảo:
Bài 4:
\(A=sin^21^0+sin^22^0+sin^23^0+...+sin^288^0+sin^289^0\)
\(A=\left(sin^21^0+sin^289^0\right)+\left(sin^22^0+sin^288^0\right)+...+\left(sin^244^0+sin^246^0\right)+sin^245^0\)
\(A=\left(sin^21^0+cos^21^0\right)+\left(sin^22^0+cos^22^0\right)+...+\left(sin^244^0+cos^244^0\right)+\left(\frac{\sqrt{2}}{2}\right)^2\)
\(A=1+1+...+1+1\)(45 số hạng tất cả)
(vì \(\sin^2\alpha+\cos^2\alpha=1\)và \(\left(\frac{\sqrt{2}}{2}\right)^2=1\)
A = 45
1 phần 5 : 1 phần 10 - 1 phần 3 ( 6 phần 5 - 9 phần 4 )2
Giúp em với ạ
\(\frac{1}{5}\div\frac{1}{10}-\frac{1}{3}\cdot\left(\frac{6}{5}-\frac{9}{4}\right)^2=2-\frac{1}{3}\cdot-\frac{21}{20}^2\)
\(=2-\frac{1}{3}\cdot\frac{441}{400}\)
\(=2-\frac{147}{400}=\frac{800}{400}-\frac{147}{400}=\frac{653}{400}\)
Chúc chị/anh học tốt.
\(\frac{1}{5}:\frac{1}{10}-\frac{1}{3}\left(\frac{6}{5}-\frac{9}{4}\right)^2\)
\(=2-\frac{1}{3}\left[\frac{\left(-21\right)}{20}\right]^2\)
\(=2-\frac{1}{3}\cdot\frac{441}{400}\)( số âm. số âm = số dương )
\(=2-\frac{147}{400}\)
\(=\frac{800-147}{400}\)
\(=\frac{653}{400}=1\frac{253}{400}=1,6325\)
\(\frac{1}{5}:\frac{1}{10}-\frac{1}{3}\left(\frac{6}{5}-\frac{9}{4}\right)^2\)
\(=\frac{1}{5}.10-\frac{1}{3}\left(\frac{24}{20}-\frac{45}{20}\right)^2\)
\(=2-\frac{1}{3}\left(\frac{-21}{20}\right)^2\)
\(=2-\frac{1}{3}.\frac{441}{400}\)
\(=2-\frac{147}{400}\)
\(=\frac{800}{400}-\frac{147}{400}\)
\(=\frac{653}{400}\)
Giúp bài tập này cho em với ạ
4 - 5 phần 7
\(4-\frac{5}{7}\)
\(=\frac{4}{1}-\frac{5}{7}\)
\(=\frac{28-5}{7}\)
\(=\frac{23}{7}\)
Chuẩn 100%
ai vẽ giúp em cái sơ đồ tư duy về phần địa lí dân cư với ạ. Bài gồm bài 1-4 luôn ạ
giúp em với 2 phần 3 x với 4 phần 5 x với 25 thì làm thế nào ạ, bài này làm bằng 2 cách ạ
tìm phần số x/9 (x thuộc z) sao cho:
x/9<4/7<x+1/9
GIÚP MÌNH VỚI Ạ. MÌNH CẦN GẤP. CẢM ƠN CÁC BẠN NHIỀU!
CÔ NGUYỄN THỊ THƯƠNG HOÀI GIÚP EM VỚI Ạ
\(\dfrac{x}{9}< \dfrac{4}{7}< \dfrac{x+1}{9}\)
=>\(\dfrac{7x}{63}< \dfrac{36}{63}< \dfrac{7x+7}{63}\)
\(\Rightarrow7x< 36< 7x+7\)
\(\Rightarrow x< \dfrac{36}{7}< x+1\)
\(\Rightarrow x< 5\dfrac{1}{7}< x+1\)
\(\Rightarrow x=5\)
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tìm phần số x/9 (x thuộc z) sao cho:
x/9<4/7<x+1/9
GIÚP MÌNH VỚI Ạ. MÌNH CẦN GẤP. CẢM ƠN CÁC BẠN NHIỀU!
CÔ NGUYỄN THỊ THƯƠNG HOÀI GIÚP EM VỚI Ạ
\(\dfrac{x}{9}\) < \(\dfrac{4}{7}\) < \(x\) + \(\dfrac{1}{9}\)
\(\dfrac{7x}{63}\) < \(\dfrac{36}{63}\) < \(\dfrac{63x}{63}\) + \(\dfrac{7}{63}\)
7\(x\) < 36 < 63\(x\) + 7
⇒\(\left\{{}\begin{matrix}7x< 36\\63x+7>36\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\63x>36-7\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\63x>29\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\x>\dfrac{29}{63}\end{matrix}\right.\)
\(\dfrac{29}{63}\)< \(x\) < \(\dfrac{36}{7}\) vì \(x\in\) Z nên \(x\in\) { 1; 2; 3; 4; 5}
⇒ \(\dfrac{x}{9}\) = \(\dfrac{1}{9}\); \(\dfrac{2}{9}\); \(\dfrac{3}{9}\); \(\dfrac{4}{9}\);\(\dfrac{5}{9}\)
giúp em phần trắc nghiệm với ạ ảnh em để ở dưới ạ