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Bài 1
d, \(x^2+2xy+y^2-2x-2y+1\)
\(\Rightarrow x^2+y^2=1+2xy-2y-2x\)
\(\Rightarrow\left(x+y-1\right)^2\)
Bài 2:
a, \(\left(x+1\right)\left(x+1\right)=\left(x+2\right)\left(x+5\right)\)
\(\Leftrightarrow\left(x+1\right)^2=x^2+5x+2x+10\)
\(\Leftrightarrow x^2+2x+1=x^2=5x+2x+10\)
\(\Leftrightarrow-5x=9\)
\(\Leftrightarrow x=-\frac{9}{5}\)
b,\(\left(x+3\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)
c, \(4x^2-9=0\)
\(\Leftrightarrow4x^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{3}{2}\\\frac{3}{2}\end{matrix}\right.\)
d,\(\left(4x-5\right)^2-\left(3x-4\right)^2=0\)
\(\Leftrightarrow16x^2-40x+25-\left(9x^2-24x+16\right)=0\)
\(\Leftrightarrow16x^2-40x+25-9x^2+24x-16=0\)
\(\Leftrightarrow7x^2-16x+9=0\)
\(\Leftrightarrow x=\frac{-\left(-16\right)\pm\sqrt{\left(-16\right)^2-4.7.9}}{14}\)
\(\Leftrightarrow x=\frac{16\pm\sqrt{256-252}}{14}\)
\(\Leftrightarrow x=\frac{16\pm\sqrt{4}}{14}\)
\(\Leftrightarrow x=\frac{16\pm2}{14}\)
\(\Leftrightarrow x=\left[{}\begin{matrix}\frac{16+2}{14}\\\frac{16-2}{14}\end{matrix}\right.\)
\(\Leftrightarrow x=\left[{}\begin{matrix}\frac{9}{7}\\1\end{matrix}\right.\)
1.a)\(3x-3y+x^2-2xy+y^2\)
\(=3\left(x-y\right)+\left(x-y\right)^2\)
\(=\left(x-y\right)\left(3+x-y\right)\)
d)\(x^2+2xy+y^2-2x-2y+1\)
\(=\left(x+y\right)^2-2\left(x+y\right)+1\)
\(=\left(x+y+1\right)^2\)
2.a)\(\left(x+1\right)\left(x+1\right)=\left(x+2\right)\left(x+5\right)\)
\(\Leftrightarrow\left(x+1\right)^2=x^2+5x+2x+10\)
\(\Leftrightarrow x^2+2x+1-x^2-7x-10=0\)
\(\Leftrightarrow-5x-9=0\)
\(\Leftrightarrow-5x=9\)
\(\Leftrightarrow x=-\frac{9}{5}\). Vậy \(S=\left\{-\frac{9}{5}\right\}\)
b)\(\left(x+3\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\).Vậy \(S=\left\{-3;-5\right\}\)
c)\(4x^2-9=0\)
\(\Leftrightarrow\left(2x+3\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{3}{2}\\x=\frac{3}{2}\end{matrix}\right.\). Vậy \(S=\left\{\pm\frac{3}{2}\right\}\)
d)\(\left(4x-5\right)^2-\left(3x-4\right)^2=0\)
\(\Leftrightarrow\left(4x-5+3x-4\right)\left(4x-5-3x+4\right)=0\)
\(\Leftrightarrow\left(7x-9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}7x-9=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{9}{7}\\x=1\end{matrix}\right.\). Vậy \(S=\left\{1;\frac{9}{7}\right\}\)
3.Ta có:
8x^2-26x+m 2x-3 4x-7 -14x+m m+21
Để \(A\left(x\right)⋮B\left(x\right)\) thì: \(m+21⋮2x-3\)
\(\Rightarrow m+21=0\)
\(\Rightarrow m=-21\)
Vậy...!
Do \(A\left(a;b\right)\in d\Rightarrow3a-2b-1=0\)
\(\Leftrightarrow3a-2b=1\)
\(\Rightarrow1=\left(3a-2b\right)^2\le\left(9+4\right)\left(a^2+b^2\right)\)
\(\Rightarrow a^2+b^2\ge\frac{1}{13}\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}3a-2b=1\\\frac{a}{3}=\frac{b}{-2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{3}{13}\\b=-\frac{2}{13}\end{matrix}\right.\)
Không chắc đâu:v
a) Ta luôn có \(\left(x-1\right)^2+\left(2x-y-3\right)^2+\left(y+z\right)^2\ge0\forall x,y,z\)
Để đẳng thức xảy ra tức là \(\left(x-1\right)^2+\left(2x-y-3\right)^2+\left(y+z\right)^2=0\) (theo đề bài)
Thì \(\left\{{}\begin{matrix}x=1\\y=2x-3=2.1-3=-1\\z=-y=1\end{matrix}\right.\)
Vậy...
b) Ta luôn có \(VT\ge0\) với mọi x, y. Mà theo đề bài \(VT\le0\)
Do vậy \(VT=0\Leftrightarrow\left(2x+3\right)^{1998}+\left(3y-5\right)^{2000}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\frac{3}{2}\\y=\frac{5}{3}\end{matrix}\right.\)
Sửa đề: (C) \(x^2+y^2+2x-4y=0\)
Đường tròn tâm \(I\left(-1;2\right)\) bán kính \(R=\sqrt{5}\)
Do MA; MB là tiếp tuyến \(\Rightarrow MA=MB\)
Mà \(\widehat{AMB}=60^0\Rightarrow\Delta AMB\) đều \(\Rightarrow MA=MB=AB\)
\(\widehat{AIB}=180^0-60^0=120^0\)
\(\Rightarrow AB=\sqrt{IA^2+IB^2-2IA.IB.cos120^0}=\sqrt{15}\)
\(\Rightarrow IM=\sqrt{IA^2+AM^2}=\sqrt{5+15}=2\sqrt{5}\)
Do \(M\in d\Rightarrow M\left(m;m+1\right)\) \(\Rightarrow\overrightarrow{IM}=\left(m+1;m-1\right)\)
\(\Rightarrow\left(m+1\right)^2+\left(m-1\right)^2=20\)
\(\Leftrightarrow2m^2+2=20\Rightarrow m^2=9\Rightarrow\left[{}\begin{matrix}m=3\\m=-3\end{matrix}\right.\)
Có 2 điểm M thỏa mãn: \(\left[{}\begin{matrix}M\left(3;4\right)\\M\left(-3;-2\right)\end{matrix}\right.\)
Bạn ơi (d) này đâu ra vậy?
Chắc là N? Vì M mà sao đằng sau lại là \(NA^2+NB^2\)?
Do N thuộc \(\Delta\) nên tọa độ có dạng \(N\left(6t;4t+2\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AN}=\left(6t-1;4t\right)\\\overrightarrow{BN}=\left(6t+3;4t-3\right)\end{matrix}\right.\)
\(\Rightarrow NA^2+NB^2=\left(6t-1\right)^2+16t^2+\left(6t+3\right)^2+\left(4t-3\right)^2=104t^2+19\ge19\)
Dấu "=" xảy ra khi \(t=0\Rightarrow N\left(0;2\right)\)