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- Xét : \(x^2+8x-20\le0\)
\(\Rightarrow-10\le x\le2\)
Mà \(x>0\)
\(\Rightarrow0< x\le2\)
- Xét \(x^2-2\left(m+3\right)x+m^2-2m< 0\)
Có : \(\Delta^,=b^{,2}-ac=\left(m+3\right)^2-\left(m^2-2m\right)\)
\(=m^2+6m+9-m^2+2m=8m+9\)
- Để bất phương trình có nghiệm
\(\Leftrightarrow\Delta>0\)
\(\Leftrightarrow m>-\dfrac{9}{8}\)
=> Bất phương trình có nghiệm \(S=\left(x_1;x_2\right)\)
Mà \(0< x\le2\)
\(\Rightarrow0< x_1< x_2\le2\)
\(TH1:x=2\)
\(\Rightarrow4-4\left(m+3\right)+m^2-2m< 0\)
\(\Rightarrow3-\sqrt{17}< m< 3+\sqrt{17}\)
\(TH2:0< x_1< x_2< 2\)
\(\Rightarrow\left\{{}\begin{matrix}m^2-2m>0\\m^2-6m-8>0\\0< 2\left(m+3\right)< 2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m< 0\\m>2\end{matrix}\right.\\\left[{}\begin{matrix}m>3+\sqrt{17}\\m< 3-\sqrt{17}\end{matrix}\right.\\-3< m< -2\end{matrix}\right.\)
Vậy \(3-\sqrt{7}< m< 3+\sqrt{7}\)
Ban ơi :(( ngay chỗ dấu ngoặc nhọn đầu tiên của TH2 có công thức j k bạn?
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Nếu \(y\le0\Rightarrow\left(y-4\right)^2\ge16>9\left(ktm\right)\Rightarrow y>0\)
Nếu \(x\ge0\Rightarrow\left(x+5\right)^2\ge25>9\left(ktm\right)\Rightarrow x< 0\)
Đặt \(\left\{{}\begin{matrix}-x=a>0\\y=b>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(a-5\right)^2+\left(b-4\right)^2\le9\\3a+b\ge14\end{matrix}\right.\)
Ta có:
\(14^2\le\left(3a+b\right)^2\le\left(3^2+1\right)\left(a^2+b^2\right)\Rightarrow a^2+b^2\ge\dfrac{196}{10}=\dfrac{98}{5}\)
\(P_{min}=\dfrac{98}{5}\) khi \(\left(a;b\right)=\left(\dfrac{21}{5};\dfrac{7}{5}\right)\) hay \(\left(x;y\right)=\left(-\dfrac{21}{5};\dfrac{7}{3}\right)\)
Lại có:
\(\left(a-5\right)^2+\left(b-4\right)^2\le9\Leftrightarrow a^2+b^2\le10a+8b-32\le\sqrt{\left(10^2+8^2\right)\left(a^2+b^2\right)}-32\)
\(\Rightarrow P\le2\sqrt{41}\sqrt{P}-32\Leftrightarrow P-2\sqrt{41}\sqrt{P}+32\le0\)
\(\Rightarrow\left(\sqrt{P}-3-\sqrt{41}\right)\left(\sqrt{P}-3+\sqrt{41}\right)\le0\) (1)
Do \(P\ge\dfrac{98}{5}\Rightarrow\sqrt{P}-3+\sqrt{41}>0\)
Nên (1) tương đương: \(\sqrt{P}-3-\sqrt{41}\le0\Rightarrow P\le50+6\sqrt{41}\)
\(P_{max}=50+6\sqrt{41}\) khi \(\left(a;b\right)=\left(5+\dfrac{15}{\sqrt{41}};4+\dfrac{12}{\sqrt{41}}\right)\)
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Đk:\(y^2-2x-5y+6\ge0\)
Pt (1)\(\Leftrightarrow\left(x^2-1\right)-\left(xy-y\right)+\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)-y\left(x-1\right)+\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2-y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\y=x+2\end{matrix}\right.\)
TH1: Thay x=1 vào pt (2) ta đc: \(3\sqrt{y^2-5y+4}=y+9\)
\(\Leftrightarrow\left\{{}\begin{matrix}y+9\ge0\\9\left(x^2-5y+4\right)=y^2+18y+81\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y\ge-9\\8y^2-63y-45=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}y=\dfrac{63+3\sqrt{601}}{16}\\y=\dfrac{63-3\sqrt{601}}{16}\end{matrix}\right.\) (tm)
TH2: Thay y=x+2 vào pt (2) ta đc:
\(\left(x-1\right)^2+3\sqrt{\left(x+2\right)^2-2x-5\left(x+2\right)+6}=x+2+9\)
\(\Leftrightarrow x^2-3x-10+3\sqrt{x^2-3x}=0\)
Đặt \(t=\sqrt{x^2-3x}\left(t\ge0\right)\)
Pttt: \(t^2-10+3t=0\)\(\Leftrightarrow\left[{}\begin{matrix}t=2\left(tm\right)\\t=-5\left(ktm\right)\end{matrix}\right.\)
\(\Rightarrow2=\sqrt{x^2-3x}\)\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}y=6\\y=1\end{matrix}\right.\) (tm)
Vậy \(\left(x;y\right)=\text{}\left\{\left(1;\dfrac{63+3\sqrt{601}}{16}\right);\left(1;\dfrac{63-3\sqrt{601}}{16}\right),\left(4;6\right),\left(-1;1\right)\right\}\)
Xét pt đầu:
\(\left(x^2+x-2\right)-y\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)-y\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2-y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\y=x+2\end{matrix}\right.\)
- Với \(x=1\) thay xuống pt dưới:
\(3\sqrt{y^2-5y+4}=y+9\) \(\left(y\ge-9\right)\)
\(\Leftrightarrow9\left(y^2-5y+4\right)=y^2+18y+81\)
\(\Leftrightarrow8y^2-63y-45=0\)
\(\Rightarrow y=\dfrac{63\pm3\sqrt{601}}{16}\) (thỏa mãn)
- Với \(y=x+2\) thay xuống pt dưới:
\(\left(x-1\right)^2+3\sqrt{x^2-3x}=x+11\) (ĐKXĐ: ....)
\(\Leftrightarrow x^2-3x+3\sqrt{x^2-3x}-10=0\)
Đặt \(\sqrt{x^2-3x}=t\ge0\)
\(\Rightarrow t^2+3t-10=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-5\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2-3x}=2\Leftrightarrow x^2-3x-4=0\)
\(\Leftrightarrow...\)
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Gọi O là trung điểm IK \(\Rightarrow OI=OK=\dfrac{1}{2}IK\)
\(\left(\overrightarrow{MI}+\overrightarrow{IA}\right)\left(\overrightarrow{MI}+\overrightarrow{IB}\right)+\left(\overrightarrow{MK}+\overrightarrow{KC}\right)\left(\overrightarrow{MK}+\overrightarrow{KD}\right)=\dfrac{1}{2}Ik^2\)
\(\Leftrightarrow MI^2-IA^2+MK^2-KC^2=\dfrac{1}{2}IK^2\)
\(\Leftrightarrow\left(\overrightarrow{MO}+\overrightarrow{OI}\right)^2+\left(\overrightarrow{MO}+\overrightarrow{OK}\right)^2=IA^2+KC^2+\dfrac{1}{2}IK^2\)
\(\Leftrightarrow2MO^2+2OI^2=IA^2+KC^2+\dfrac{1}{2}IK^2\)
\(\Leftrightarrow2MO^2+\dfrac{1}{2}IK^2=IA^2+KC^2+\dfrac{1}{2}IK^2\)
\(\Leftrightarrow MO^2=\dfrac{1}{2}\left(IA^2+KC^2\right)=\dfrac{1}{8}\left(AB^2+CD^2\right)\)
\(\Leftrightarrow MO=\dfrac{1}{2\sqrt{2}}\sqrt{AB^2+CD^2}\)
Tập hợp M là đường tròn tâm O bán kính \(\dfrac{\sqrt{AB^2+CD^2}}{2\sqrt{2}}\)
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Lời giải:
a.
$|x|\leq 2\Leftrightarrow -2\leq x\leq 2$
Tập $A=[-2;2]$
$(x-1)(x-4)< 0\Leftrightarrow 1< x< 4$
Tập $B=(1;4)$
Đến đây bạn có thể dễ dàng biểu diễn nó trên trục số
b.
$A\cap B=[-2;2]\cap (1;4)=(1;2]$
$A\cup B=[-2;2]\cup (1;4)=[-2;4)$
$A\setminus B= [-2;2]\setminus (1;4)=[-2;1]$
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`sin3x sinx+sin(x-π/3) cos (x-π/6)=0`
`<=> 1/2 (cos2x - cos4x) + 1/2(-sin π/6 + sin (2x-π/2)=0`
`<=> cos2x-cos4x-1/2+ sin(2x-π/2)=0`
`<=>cos2x-cos4x-1/2+ sin2x .cos π/2 - cos2x. sinπ/2=0`
`<=> cos2x - cos4x - cos2x = 1/2`
`<=> cos4x = cos(2π)/3`
`<=>` \(\left[{}\begin{matrix}4x=\dfrac{2\text{π}}{3}+k2\text{π}\\4x=\dfrac{-2\text{π}}{3}+k2\text{π}\end{matrix}\right.\)
`<=>` \(\left[{}\begin{matrix}x=\dfrac{\text{π}}{6}+k\dfrac{\text{π}}{2}\\x=-\dfrac{\text{π}}{6}+k\dfrac{\text{π}}{2}\end{matrix}\right.\)
Xét pt hoành độ gđ của đường thẳng và parabol có:
\(\left(m-1\right)x^2+3mx+2m=2x-1\)
\(\Leftrightarrow\left(m-1\right)x^2+x\left(3m-2\right)+2m+1=0\) (1)
Để đt và parabol cắt tại hai điểm pb có hoành độ âm
\(\Leftrightarrow\) Pt (1) có hai nghiệm âm phân biệt
\(\Leftrightarrow\left\{{}\begin{matrix}\Delta>0\\S< 0\\P>0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}m^2-8m+8>0\\\dfrac{2-3m}{m-1}< 0\\\dfrac{2m+1}{m-1}>0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}m\in\left(-\infty;4-2\sqrt{2}\right)\cup\left(4+2\sqrt{2};+\infty\right)\\m\in\left(-\infty;\dfrac{2}{3}\right)\cup\left(1;+\infty\right)\\m\in\left(-\infty;-\dfrac{1}{2}\right)\cup\left(1;+\infty\right)\end{matrix}\right.\)
\(\Rightarrow m\in\left(-\infty;-\dfrac{1}{2}\right)\cup\left(4+2\sqrt{2};+\infty\right)\)
Vậy...