Vẽ sơ đồ tư duy bài 3 địa lí sách kết nối tri thức
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TD
0
TM
1
NT
3
10 tháng 10 2023
\(\left|5x\right|-3x=2\)
\(\text{⇒}\left|5x\right|=3x+2\)
TH1:
\(5x=3x+2\) \(\left(x\ge0\right)\)
\(\text{⇒}5x-3x=2\)
\(\text{⇒}2x=2\)
\(\text{⇒}x=\dfrac{2}{2}\)
\(\text{⇒}x=1\left(tm\right)\)
TH2:
\(-5x=3x+2\) (x < 0)
\(\text{⇒}-5x-3x=2\)
\(\text{⇒}-8x=2\)
\(\text{⇒}x=-\dfrac{1}{4}\left(tm\right)\)
10 tháng 10 2023
Xét \(x>0\)
\(5x-3x=2\)
\(\left(5-3\right)x=2\)
\(2x=2\)
\(x=1\)
Xét \(x< 0\)
\(\left(-5x\right)-3x=2\)
\(\left(-5-3\right)x=2\)
\(-8x=2\)
\(x=-\dfrac{2}{8}=-\dfrac{1}{4}\)
Vậy: \(\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{4}\end{matrix}\right.\)
TT
2
10 tháng 10 2023
Ta viết lại tổng này thành:
\(P=\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{97.99}\right)+\left(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{98.100}\right)-\dfrac{49}{99}\)
\(P=\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{97.99}\right)+\dfrac{1}{2}\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{98.100}-\dfrac{49}{99}\right)\)
\(P=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)-\dfrac{49}{99}\)
\(P=\dfrac{1}{2}\left(1-\dfrac{1}{99}\right)+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{100}\right)-\dfrac{49}{99}\)
\(P=\dfrac{1}{2}-\dfrac{1}{198}+\dfrac{1}{4}-\dfrac{1}{200}-\dfrac{49}{99}\)
\(P=\dfrac{49}{200}\)
10 tháng 10 2023
Ta có BĐT: \(\left|a-b\right|\ge\left|a\right|-\left|b\right|\)
Áp dụng ta có:
\(A=\left|x-2\right|-\left|x+4\right|\le\left|x-2-x-4\right|=\left|-6\right|=6\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}x-2\ge0\\x+4\ge0\end{matrix}\right.\Leftrightarrow x\ge2\)
Vậy: \(A_{max}=6\Leftrightarrow x\ge2\)
NT
1
10 tháng 10 2023
Theo BĐT: \(\left|a-b\right|\ge\left|a\right|-\left|b\right|\) ta có:
\(B=\left|2x-7\right|-\left|2x-11\right|\le\left|2x-7-2x+11\right|=\left|4\right|=4\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}2x-7\ge0\\2x-11\ge0\end{matrix}\right.\) \(\Leftrightarrow x\ge\dfrac{11}{2}\)
Vậy: \(B_{max}=4\Leftrightarrow x\ge\dfrac{11}{2}\)
TT
0
SV
Sinh Viên NEU
CTVVIP
10 tháng 10 2023
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