x^2.(x^2 + 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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`(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}}`
\(\left(\dfrac{x}{x+1}\right)^2+\left(\dfrac{x}{x-1}\right)^2=90\)
\(\Leftrightarrow\dfrac{x^2}{\left(x+1\right)^2}+\dfrac{x^2}{\left(x-1\right)^2}=90\)
\(\Leftrightarrow\dfrac{x^2\left(x-1\right)^2}{\left(x+1\right)^2\left(x-1\right)^2}+\dfrac{x^2\left(x+1\right)^2}{\left(x+1\right)^2\left(x-1\right)^2}=90\)
\(\Leftrightarrow\dfrac{x^2\left(x-1\right)^2+x^2\left(x+1\right)^2-90\left(x-1\right)^2\left(x+1\right)^2}{\left(x+1\right)^2\left(x-1\right)^2}=0\)
\(\Rightarrow x^2\left(x^2-2x+1\right)+x^2\left(x^2+2x+1\right)-90\left(x^2-1\right)^2=0\)
\(\Leftrightarrow x^4-2x^3+x^2+x^4+2x^3+x^2-90x^4+90x^2-90=0\)
\(\Leftrightarrow-88x^4+92x^2-90=0\)
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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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a) quá dài
b)<=>x^2+2x+1=90
=>x^2+2x-89=0
áp dụng denta
=>2^2-(-4(1.89))=360
\(\Rightarrow x_{1,2}=\frac{-b+-\sqrt{\Delta}}{2a}=\frac{-2+-\sqrt{360}}{2}\)
=>x=\(+-3\sqrt{10}-1\)
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\(\dfrac{5\left|x+1\right|}{2}=\dfrac{90}{\left|x+1\right|}\left(x\ne-1\right)\\ \Rightarrow5\left|x+1\right|^2=90\cdot2=180\\ \Rightarrow\left|x+1\right|^2=36\\ \Rightarrow\left|x+1\right|=6\left(\left|x+1\right|>0\right)\\ \Rightarrow\left[{}\begin{matrix}x+1=6\\x+1=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=-7\end{matrix}\right.\)
\(x^4+x^2-90=0\)
Đặt x^2 = t ( t >= 0 )
\(t^2+t-90=0\Leftrightarrow t=9;t=-10\left(loại\right)\)
Theo cách đặt \(x^2=9\Leftrightarrow x=3;x=-3\)
hảo hán là gì ?