2cos2x+tanx=4/5
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\(\dfrac{1}{\sqrt{\dfrac{5}{7}}+\sqrt{\dfrac{5}{13}}+1}+\dfrac{1}{\sqrt{\dfrac{7}{13}}+\sqrt{\dfrac{7}{5}}+1}+\dfrac{1}{\sqrt{1\dfrac{6}{7}}+\sqrt{2\dfrac{3}{5}}+1}\\ =\dfrac{1}{\dfrac{\sqrt{5}}{\sqrt{7}}+\dfrac{\sqrt{5}}{\sqrt{13}}+\dfrac{\sqrt{5}}{\sqrt{5}}}+\dfrac{1}{\dfrac{\sqrt{7}}{\sqrt{13}}+\dfrac{\sqrt{7}}{\sqrt{5}}+\dfrac{\sqrt{7}}{\sqrt{7}}}+\dfrac{1}{\dfrac{\sqrt{13}}{\sqrt{7}}+\dfrac{\sqrt{13}}{\sqrt{5}}+\dfrac{\sqrt{13}}{\sqrt{13}}}\\ =\left(\dfrac{1}{\sqrt{5}}+\dfrac{1}{\sqrt{7}}+\dfrac{1}{\sqrt{13}}\right)\cdot\dfrac{1}{\dfrac{1}{\sqrt{5}}+\dfrac{1}{\sqrt{7}}+\dfrac{1}{\sqrt{13}}}\\ =1\)
a, \(\left(-17\right)+5+8+17+\left(-3\right)\)
\(=\left(-17+17\right)+\left[5+\left(-3\right)\right]+8\)
\(=0+8+8=8+8=16\)
b, \(\left(5^{19}:5^{17}+3\right):7=\left(5^2+3\right):7\)
\(=\left(25+3\right):7=28:7=4\)
c, \(|-8|+\left(-5\right)+9+\left(-7\right)+|-4|\)
\(=8-5+9-7+4=3+2+4=5+4=9\)
ý d mk ko biết nha.
thông cảm cho mk nha.
k mk nha.
#mon
a.
Đặt \(cos2x=t\Rightarrow t\in\left[-1;1\right]\)
Xét hàm \(y=f\left(t\right)=2t^2+2t-4\) trên \(\left[-1;1\right]\)
\(-\dfrac{b}{2a}=-\dfrac{1}{2}\in\left[-1;1\right]\)
\(f\left(-1\right)=-4\) ; \(f\left(-\dfrac{1}{2}\right)=-\dfrac{9}{2}\) ; \(f\left(1\right)=0\)
\(\Rightarrow y_{min}=-\dfrac{9}{2}\) khi \(t=-\dfrac{1}{2}\) hay \(cos2x=-\dfrac{1}{2}\)
\(y_{max}=0\) khi \(cos2x=1\)
b. Đặt \(tanx=t\Rightarrow t\in\left[-1;\sqrt{3}\right]\)
Xét hàm \(f\left(t\right)=t^2-2\sqrt{3}t-1\) trên \(\left[-1;\sqrt{3}\right]\)
\(-\dfrac{b}{2a}=\sqrt{3}\in\left[-1;\sqrt{3}\right]\)
\(f\left(-1\right)=2\sqrt{3}\) ; \(f\left(\sqrt{3}\right)=-4\)
\(y_{min}=-4\) khi \(x=\dfrac{\pi}{3}\) ; \(y_{max}=2\sqrt{3}\) khi \(x=-\dfrac{\pi}{4}\)
Ta có \(E=\frac{5n-4}{2n+5}\)
\(\Rightarrow2E=\frac{10n-8}{2n+5}=\frac{5\left(2n+5\right)-33}{2n+5}=5-\frac{33}{2n+5}\)
Để E nguyên => 2E nguyên => 5-\(\frac{33}{2n+5}\)nguyên
=> \(\frac{33}{2n+5}\)nguyên
=> \(33⋮2n+5\)
\(\Rightarrow2n+5=Ư_{\left(33\right)}=\left\{-33;-1;1;33\right\}\)
Ta có bảng
2n+5 | -33 | -1 | 1 | 33 |
2n | -38 | -6 | -4 | 28 |
n | -19 | -3 | -2 | 14 |
Vậy n={-19;-3;-2;14}