x.(x+1)=90
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Ta có các công thức cơ bản sau: \(cos\left(90^0+x\right)=-sinx;sin\left(90^0-x\right)=cosx\)
\(cot\left(90^0-x\right)=tanx;tan\left(90^0+x\right)=-cotx\)
Thay vào bài toán:
\(\dfrac{1-\left(-sinx\right)^2}{1-cos^2x}-tanx.\left(-cotx\right)=\dfrac{1-sin^2x}{1-cos^2x}+tanx.cotx\)
\(=\dfrac{cos^2x}{sin^2x}+1=\dfrac{cos^2x+sin^2x}{sin^2x}=\dfrac{1}{sin^2x}\)
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5x.5x+1.5x+2<100.................00:224
Có 24 số 0
53x.51.52<1024:2224
53x.53<524
53x<524:53
53x<521
=>3x=21
x=21:3
x=7\(\in\)N
Vậy x=7
Chúc bn học tốt
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Ta có \(\left(x-\frac{1}{2}\right)+\left(x-\frac{1}{6}\right)+\left(x-\frac{1}{12}\right)+...+\left(x-\frac{1}{90}\right)=1\)
\(\Rightarrow\left(x+x+x+...+x\right)-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)=1\)
\(\Rightarrow9x-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{9.10}\right)=1\)
\(\Rightarrow9x-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)=1\)
\(\Rightarrow9x-\left(1-\frac{1}{10}\right)=1\)
\(\Rightarrow9x-\frac{9}{10}=1\)
\(\Rightarrow9x=\frac{19}{10}\)
\(\Rightarrow x=\frac{19}{10}\)
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1: 90<x<180
=>cosx<0
=>\(cosx=-\sqrt{1-\left(\dfrac{4}{5}\right)^2}=-\dfrac{3}{5}\)
\(sin2x=2\cdot sinx\cdot cosx=2\cdot\dfrac{4}{5}\cdot\dfrac{-3}{5}=\dfrac{-24}{25}\)
\(cos2x=2\cdot cos^2x-1=2\cdot\dfrac{9}{25}-1=-\dfrac{7}{25}\)
\(tan2x=\dfrac{-24}{25}:\dfrac{-7}{25}=\dfrac{24}{7}\)
2: 0<x<90
=>cosx>0
=>\(cosx=\sqrt{1-\left(\dfrac{1}{2}\right)^2}=\dfrac{\sqrt{3}}{2}\)
\(cos2x=2\cdot cos^2x-1=2\cdot\dfrac{3}{4}-1=\dfrac{6}{4}-1=\dfrac{2}{4}=\dfrac{1}{2}\)
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`(x/(x+1))^2+(x/(x-1))^2=90(x ne -1,1)`
`<=>x^2/(x+1)^2+x^2/(x-1)^2=90`
`<=>x^2(x-1)^2+x^2(x-1)^2=90(x^2-1)^2`
`<=>x^2(2x^2+2)=90(x^4-2x^2+1)`
`<=>2x^4+2x^2=90x^4-180x^2+90`
`<=>88x^4-182x^2+90=0`
`<=>88x^4-110x^2-72x^2+90=0`
`<=>22x^2(4x^2-5)-18(4x^2-5)=0`
`<=>(4x^2-5)(22x^2-18)=0`
`<=>(4x^2-5)(11x^2-9)=0`
`<=>` $\left[ \begin{array}{l}4x^2=5\\11x^2=9\end{array} \right.$
`<=>` $\left[ \begin{array}{l}x=\sqrt{\dfrac{5}{4}}\\x=-\sqrt{\dfrac{5}{4}}\\x=\sqrt{\dfrac{9}{11}}\\x=-\sqrt{\dfrac{9}{11}}\end{array} \right.$
Vậy `S={\sqrt{9/11},-\sqrt{9/11},\sqrt{5/4},-\sqrt{5/4}}`
\(\left(\dfrac{x}{x+1}\right)^2+\left(\dfrac{x}{x-1}\right)^2=90\)
\(\Leftrightarrow\dfrac{x^2}{\left(x+1\right)^2}+\dfrac{x^2}{\left(x-1\right)^2}=90\)
\(\Leftrightarrow\dfrac{x^2\left(x-1\right)^2}{\left(x+1\right)^2\left(x-1\right)^2}+\dfrac{x^2\left(x+1\right)^2}{\left(x+1\right)^2\left(x-1\right)^2}=90\)
\(\Leftrightarrow\dfrac{x^2\left(x-1\right)^2+x^2\left(x+1\right)^2-90\left(x-1\right)^2\left(x+1\right)^2}{\left(x+1\right)^2\left(x-1\right)^2}=0\)
\(\Rightarrow x^2\left(x^2-2x+1\right)+x^2\left(x^2+2x+1\right)-90\left(x^2-1\right)^2=0\)
\(\Leftrightarrow x^4-2x^3+x^2+x^4+2x^3+x^2-90x^4+90x^2-90=0\)
\(\Leftrightarrow-88x^4+92x^2-90=0\)
\(x\times\left(x+1\right)=90\)
\(x\times\left(x+1\right)-90=0\)
\(x^2+x-90=0\)
\(x^2+10x-9x-90=0\)
\(x\left(x+10\right)-9\left(x+10\right)=0\)
\(\left(x-9\right)\left(x+10\right)=0\)
\(x=9;x=-10\)
Vậy: \(x=9;x=-10\)
\(x\cdot\left(x+1\right)=90\\ x\cdot\left(x+1\right)=3\cdot3\cdot2\cdot5\\ x\cdot\left(x+1\right)=9\cdot10\)
vì x và x+1 là 2 số nguyên liên tiếp
=> x = 9