Giải phương trình
( 2sin x - 1)(2sin 2x + 1) = 3 - 4 cos2x
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a)
\(A=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(A=100+99+98+97+...+2+1\)
\(A=\frac{100.101}{2}=5050\)
b)
\(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)
\(B=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)
\(B=\left(2^8-1\right)...\left(2^{64}+1\right)+1\)
\(B=\left(2^{64}-1\right)\left(2^{64}+1\right)+1\)
\(B=2^{128}-1+1=2^{128}\)
c)
\(C=a^2+b^2+c^2+2\left(ab+bc+ca\right)+a^2+b^2+c^2+2ab-2ac-2bc-2a^2-4ab-2b^2\)
\(C=2c^2\)
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Trả lời câu 1: Bà già về đích trước vì con chó có trả biết gì mà chạy cả
Trả lời câu 2: Làm việc chăm chỉ ,đốt .
Trả lời câu 3: Cái roi
Đáp án của mik khác bn ngọc
nhưng câu một và hai bn có ý đúng nhưng câu ba bn chưa hiểu vấn đề
\(20,18\cdot2\frac{2}{5}+201,8\cdot\frac{97}{22}\cdot220\%+0,006\cdot2018\)
\(=2018\cdot\frac{1}{100}\cdot2\frac{2}{5}+2018\cdot\frac{1}{10}\cdot\frac{97}{22}+\frac{6}{1000}+2018\)
\(=2018\cdot\left(\frac{1}{100}\cdot2\frac{2}{5}+\frac{1}{10}\cdot\frac{97}{22}+6\cdot\frac{1}{1000}\right)\)
\(=2018\cdot\left[\frac{1}{1000}\cdot\left(6+100\cdot\frac{97}{22}+10\cdot2\frac{2}{5}\right)\right]\)
\(=\frac{1009}{500}\cdot\left[10\cdot\left(10\cdot\frac{97}{22}+\frac{12}{5}\right)\right]\)
\(=\frac{1009}{50}\cdot45\frac{12}{5}=\frac{239133}{250}\)
lm trên symbolab.com
\(\left(2\sin x-1\right)\left(2\sin2x+1\right)=3-4\cos^2x\)
\(\Leftrightarrow\left(2\sin x-1\right)\left(2\sin2x+1\right)=3-4\left(2-\sin^2x\right)\)
\(\Leftrightarrow\left(2\sin x-1\right)\left(2\sin2x+1\right)=4\sin^2x-1\)
\(\Leftrightarrow\left(2\sin x-1\right)\left(2\sin2x+1\right)=\left(2\sin x-1\right)\left(2\sin x+1\right)\)
\(\Leftrightarrow2\sin2x+1=2\sin x+1\)
\(\Leftrightarrow\sin2x=\sin x\)
\(\Leftrightarrow\sin2x-\sin x=0\)
\(\Leftrightarrow2\cos\frac{3}{2}-\cos\frac{x}{2}=0\)
\(\Leftrightarrow\orbr{\begin{cases}\cos\frac{3}{2}=0\\\cos\frac{x}{2}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{3x}{2}=\frac{\pi}{2}+k2\pi\\\frac{x}{2}=\frac{\pi}{2}+k2\pi\end{cases}\left(k\inℤ\right)}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{\pi}{3}+\frac{2\pi}{3}k\\x=\pi+4k\pi\end{cases}\left(k\inℤ\right)}\)