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Mai Trung Hải Phong
Giới thiệu về bản thân
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Đề có sai không ạ?
\(M=-x^2-6x+1\)
\(M=-\left(x^2+6x-1\right)\)
\(M=-\left(x^2+6x+9-10\right)\)
\(M=-\left[\left(x+3\right)^2-10\right]\)
\(M=-\left(x+3\right)^2+10\)
\(\Rightarrow M\le10\)
\(\Rightarrow M_{max}=10\Leftrightarrow x=-3\)
Sửa:
\(\left|x\right|=\left|y\right|\) và \(x>0;y< 0\)
\(\Rightarrow y=-x\)
\(\Rightarrow2x+\left(-x\right)=x\)
Vậy \(2x+y=x\)
\(\left|x\right|=\left|y\right|\) và \(x>0;y< 0\)
\(\Rightarrow y=-x\)
\(\Rightarrow2x\pm x=x\)
Vậy \(2x+y=x\)
Ta có:
\(a^2+a+1=\left(a^2+2.a.\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(a+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall a\)
\(\Rightarrow\)PT đã cho vô nghiệm
Vậy không có giá trị \(a\) thỏa mãn \(P=a^{2014}+\dfrac{1}{a^{2014}}\)
11
Ta có:
\(\left|x-\dfrac{1}{3}\right|+\left|x-\dfrac{1}{15}\right|+...+\left|x-\dfrac{1}{399}\right|\ge0\forall x\)
\(\Rightarrow-11x\ge0\forall x\Rightarrow x\le0\)
\(\Rightarrow x-\dfrac{1}{3};x-\dfrac{1}{15};...;x-\dfrac{1}{399}< 0\)
\(\Rightarrow x-\dfrac{1}{3}+x-\dfrac{1}{15}+...+x-\dfrac{1}{399}=11x\)
\(\Rightarrow x+x+...+x-\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{19.21}\right)=11x\)
Vì số lượng \(x\) ở vế trái bằng số lượng số hạng là phân số
\(\Rightarrow\) Số lượng \(x\) ở vế trái là:\(\left(19-1\right):2+1=10\left(số\right)\)
\(\Rightarrow10x-\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)=11x\)
\(\Rightarrow-x=1-\dfrac{1}{21}\)
\(\Rightarrow x=-\dfrac{20}{21}\)
\(Ư\left(240\right)=\left\{\text{1;2,3,4,5,6,8,10,24,30,40,48,60,80,120,240}\right\}\)
\(3^{x+1}=27\)
\(\Rightarrow3^{x+1}=3^3\)
\(\Rightarrow x+1=3\)
\(\Rightarrow x=3-1\)
\(\Rightarrow x=2\)
Xem lại đề!