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Với mọi x ta có :
+) \(\left|x+\dfrac{1}{1.3}\right|\ge0; \)
+) \(\left|x+\dfrac{1}{3.5}\right|\ge0;\)
.....................................
+) \(\left|x+\dfrac{1}{97.99}\right|\ge0\)
\(\Leftrightarrow\left|x+\dfrac{1}{1.3}\right|+\left|x+\dfrac{1}{3.5}\right|+.......+\left|x+\dfrac{1}{97.99}\right|\ge0\)
\(\Leftrightarrow50x\ge0\)
\(\Leftrightarrow x\ge0\)
Khi \(x\ge0\) ta được :
+) \(\left|x+\dfrac{1}{1.3}\right|=x+\dfrac{1}{1.3}\)
+) \(\left|x+\dfrac{1}{3.5}\right|=x+\dfrac{1}{3.5}\)
.............................................
+) \(\left|x+\dfrac{1}{97.99}\right|=x+\dfrac{1}{97.99}\)
\(\Leftrightarrow\left(x+\dfrac{1}{1.3}\right)+\left(x+\dfrac{1}{3.5}\right)+......+\left(x+\dfrac{1}{97.99}\right)=50x\)
\(\Leftrightarrow49x+\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+....+\dfrac{1}{97.99}\right)=50x\)
\(\Leftrightarrow x=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{97}-\dfrac{1}{99}\)
\(\Leftrightarrow x=\dfrac{16}{99}\)
Vậy...
a, \(\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2x-1}-\dfrac{1}{2x+1}\right)=\dfrac{49}{99}\)
\(\Leftrightarrow\dfrac{1}{2}.\left(1-\dfrac{1}{2x+1}\right)=\dfrac{49}{99}\)
\(\Leftrightarrow\dfrac{2x+1-1}{2x+1}=\dfrac{98}{99}\)
\(\Leftrightarrow98\left(2x+1\right)=99.2x\)
\(\Leftrightarrow2x=98\Rightarrow x=49\)
b: Đặt \(A=1-3+3^2-3^3+...+\left(-3\right)^x\)
\(=\left(-3\right)^0+\left(-3\right)^1+\left(-3\right)^2+...+\left(-3\right)^x\)
\(\Leftrightarrow-3A=\left(-3\right)^1+\left(-3\right)^2+...+\left(-3\right)^{x+1}\)
\(\Leftrightarrow-3A-A=\left(-3\right)^1+\left(-3\right)^2+...+\left(-3\right)^{x+1}-...-1\)
\(\Leftrightarrow-4A=\left(-3\right)^{x+1}-1\)
\(\Leftrightarrow A=\dfrac{\left(-3\right)^{x+1}-1}{-4}=\dfrac{-\left(-3\right)^{x+1}+1}{4}\)
\(\Leftrightarrow\dfrac{-\left(-3\right)^{x+1}+1}{4}=\dfrac{3^{2012}-1}{2}\)
\(\Leftrightarrow-\left(-3\right)^{x+1}+1=2\cdot3^{2012}-2\)
\(\Leftrightarrow-\left(-3\right)^{x+1}=2\cdot3^{2012}-3\)
\(\Leftrightarrow-\left(-3\right)^{x+1}=3\left(2\cdot3^{2011}-1\right)\)
\(\Leftrightarrow-\left(-3\right)^x=2\cdot3^{2011}-1\)
=>x=2010
a) \(\left|3x-\dfrac{1}{2}\right|+\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|=0\)
Do \(\left|3x-\dfrac{1}{2}\right|,\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|\ge0\forall x,y\)
\(\Rightarrow\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\\dfrac{1}{4}y+\dfrac{3}{5}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-\dfrac{12}{5}\end{matrix}\right.\)
b) \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|+\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\le0\)
Do \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|,\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\ge0\forall x,y\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{9}=0\\\dfrac{5}{7}y-\dfrac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{27}\\y=\dfrac{7}{10}\end{matrix}\right.\)
Mấy câu này dễ mà,động não lên chứ bạn:v
Link______________Link
h) \(\left|x-1\right|+\left|x-3\right|=\left|x-1\right|+\left|3-x\right|\)
\(\ge\left|x-1+3-x\right|=2\)
\(\Rightarrow x+1>2\Leftrightarrow x>1\)
Vậy: \(\left\{{}\begin{matrix}x>1\\x\in R\end{matrix}\right.\)
Câu b xét khoảng tương tự với cái link t đưa thôi
hơi bức xúc rồi đó
tau chỉ muốn kiểm tra lại thôi
A = \(\dfrac{-101}{200}>\dfrac{-100}{200}=\dfrac{-1}{2}\)
Nhận thấy A có 99 hạng tử mà mỗi hạng tử chứa dấu âm nên viết gọn\(A=-\dfrac{3}{4}.\dfrac{8}{9}.....\dfrac{9999}{10000}=-\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}....\dfrac{99.101}{100^2}=-\dfrac{\left(1.2...99\right).\left(3.4...101\right)}{\left(2.3..100\right).\left(2.3...100\right)}=-\dfrac{101}{2.100}=-\dfrac{101}{200}< -\dfrac{1}{2}\)
b) Vì \(\left|x+\dfrac{1}{1.3}\right| \ge0;\left|x+\dfrac{1}{3.5}\right|\ge0;...;\left|x+\dfrac{1}{97.99}\right|\ge0\)
\(\Rightarrow50x\ge0\Rightarrow x\ge0\)
Khi đó: \(\left|x+\dfrac{1}{1.3}\right|=x+\dfrac{1}{1.3};\left|x+\dfrac{1}{3.5}\right|=x+\dfrac{1}{3.5};...;\left|x+\dfrac{1}{97.99}\right|=x+\dfrac{1}{97.99}\left(1\right)\)
Thay (1) vào đề bài:
\(x+\dfrac{1}{1.3}+x+\dfrac{1}{3.5}+...+x+\dfrac{1}{97.99}=50x\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\right)=50x\)
\(\Rightarrow49x+\left[\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\right]=50x\)
\(\Rightarrow49x+\dfrac{16}{99}=50x\)
\(\Rightarrow x=\dfrac{16}{99}\)
Vậy \(x=\dfrac{16}{99}.\)
thank bn nhìu nhìu