timf x biet 1+2+3+...+ x =55
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Timf GTNN,GTLN cua \(A=x^2+y^2\)
biet rang: \(x^2\left(x^2+2y^2-3\right)+\left(y^2-2\right)^2=1\)
\(x^4+2x^2y^2+y^4-3x^2-4y^2+4=1\)
\(\Leftrightarrow\left(x^2+y^2\right)^2-4\left(x^2+y^2\right)+4=1-x^2\)
\(\Leftrightarrow\left(x^2+y^2-2\right)^2=1-x^2\)
Do \(1-x^2\le1\) \(\forall x\)
\(\Rightarrow-1\le x^2+y^2-2\le1\)
\(\Rightarrow1\le x^2+y^2\le3\)
\(A_{min}=1\) khi \(\left\{{}\begin{matrix}x=0\\y=\pm1\end{matrix}\right.\)
\(A_{max}=3\) khi \(\left\{{}\begin{matrix}x=0\\y=\pm\sqrt{3}\end{matrix}\right.\)
timf x biet x^13=27.x^10
x^13=27.x^10
x^13:x^10=27
x^3=27
vì 3^3=27
=> x=3
vậy x = 3
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x13 = 27 . x10
=> x13 : x10 = 27
=> x3 = 27
=> x = 3
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x13=27.x10
=> x13:x10=27
=>x3=27
=>x3=33
=>x=3
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timf x,y thuoc Z biet \(xy^2+2xy-4y+x=0\)
tim x,biet: 1+2+3+5+...+x=55
Timf x,y,z Biet
a)\(\left(x^2-1\right)^2+\left(x-y+3\right)^2=0\)
b)\(\frac{2x-3y}{2}=\frac{4y-2Z}{3}=\frac{3Z-4x}{4}và3x+2y+Z=17\)
Tim x,y biet
a)(2x +1)(3y-2)=55
b)(x+1)(xy-1)=3
timf x-210=-1+-2+-3+....+(-x+1)+(-x) vay x=
timf x,y (x-1)^2+(y+3)^2=0
\(\left(x-1\right)^2+\left(y+3\right)^2=0\left(1\right)\)
Ta thấy \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0,\forall x\\\left(y+3\right)^2\ge0,\forall y\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+3\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)
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=> \(\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+3\right)^2=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\left(x+1\right)^2=0^2\\\left(y+3\right)^2=0^2\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x+1=0\\y+3=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-1\\y=-3\end{matrix}\right.\)
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Tim x biet: 1+2+3+...+x=55
Ai tra loi nhanh nhat mik k cho
