Gọi tâm đường tròn là \(I\left(a;b\right)\Rightarrow IA=IB=d\left(I;Ox\right)=b\)

\(\left\{{}\begin{matrix}\overrightarrow{AI}=\left(a-1;b-1\right)\\\overrightarrow{BI}=\left(a-1;b-4\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}AI^2=\left(a-1\right)^2+\left(b-1\right)^2\\BI^2=\left(a-1\right)^2+\left(b-4\right)^2\end{matrix}\right.\)

\(AI^2=BI^2\Rightarrow\left(b-1\right)^2=\left(b-4\right)^2\)

\(\Rightarrow-2b+1=-8b+16\Rightarrow b=\dfrac{5}{2}\)

Lại có:

\(IA=b\Rightarrow IA^2=b^2\Rightarrow\left(a-1\right)^2+\left(\dfrac{5}{2}-1\right)^2=\left(\dfrac{5}{2}\right)^2\)

\(\Rightarrow\left(a-1\right)^2=4\Rightarrow\left[{}\begin{matrix}a=3\\a=-1\end{matrix}\right.\)

Có 2 đường tròn thỏa mãn: \(\left[{}\begin{matrix}\left(x-3\right)^2+\left(y-\dfrac{5}{2}\right)^2=\dfrac{25}{4}\\\left(x+1\right)^2+\left(y-\dfrac{5}{2}\right)^2=\dfrac{25}{4}\end{matrix}\right.\)

có làm thì mới có ăn ko làm mà đòi có ăn thì ăn đồng bằng ăn cát

có làm thì mới có ăn ko làm mà đòi có ăn thì ăn đồng bằng ăn cát

ĐKXĐ: \(x\ge4\)

\(\sqrt{\left(x-1\right)\left(x-4\right)}-\sqrt{x-4}+3-3\sqrt{x-1}\le0\)

\(\Leftrightarrow\sqrt{x-4}\left(\sqrt{x-1}-1\right)-3\left(\sqrt{x-1}-1\right)\le0\)

\(\Leftrightarrow\left(\sqrt{x-4}-3\right)\left(\sqrt{x-1}-1\right)\le0\)

Do \(x\ge4\Rightarrow\sqrt{x-1}-1\ge\sqrt{3}-1>0\) nên BPT tương đương:

\(\sqrt{x-4}-3\le0\)

\(\Leftrightarrow\sqrt{x-4}\le3\)

\(\Leftrightarrow x\le13\)

Kết hợp điều kiện ban đầu ta được: \(4\le x\le13\)

11.

\(T=sin\left(\dfrac{7\pi}{2}\right)+2cos\left(\dfrac{\pi}{3}-\dfrac{7\pi}{3}\right).sin\left(\dfrac{\pi}{6}-\dfrac{7\pi}{6}\right)=-1\)

12.

Do \(-1\le sinx\le1\)

\(\Rightarrow P=\left(sinx+1\right)\left(sinx+2\right)+2\ge2\Rightarrow m=2\)

\(P=sin^2x+6sinx-7+14=\left(sinx-1\right)\left(sinx+7\right)+14\le14\Rightarrow M=14\)

\(\Rightarrow M+m=16\)

Hình như căn thức cuối cùng phải là \(\sqrt{\dfrac{c}{a+b}}\) chứ nhỉ?

\(\sqrt{\dfrac{b+c}{a}.1}\le\dfrac{\dfrac{b+c}{a}+1}{2}=\dfrac{a+b+c}{2a}\Rightarrow\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\)

\(tương\) \(tự\Rightarrow\sqrt{\dfrac{b}{c+a}}\ge\dfrac{2b}{a+b+c};\sqrt{\dfrac{c}{a+b}}\ge\dfrac{2c}{a+b+c}\)

\(\Rightarrow VP\ge\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)

\(dấu"="\Leftrightarrow\sqrt{\dfrac{b+c}{a}}=\sqrt{\dfrac{c+a}{b}}=\sqrt{\dfrac{a+b}{c}}=1\Leftrightarrow\left\{{}\begin{matrix}b+c=a\\c+a=b\\a+b=c\end{matrix}\right.\)

\(\Leftrightarrow a+b+c=2\left(a+b+c\right)\)\(vô\) \(lí\) \(do:a,b,c>0\)

\(\Rightarrow VP>2\)

\(VT< \dfrac{a+c}{a+b+c}+\dfrac{a+b}{a+b+c}+\dfrac{c+b}{a+b+c}=2\Rightarrow VT< 2\)

\(\Rightarrow VT< VP\left(đpcm\right)\)

= 0 =))

bằng 87326894586419483726264927837475758689798085740293746563739203857567389725869916490572496217946 bn nhé

Bài 2.

Ta có \(A=\left\{x\in R,3x+2\le14\right\}=\left\{x\in R,x\le4\right\}\) = (\(-\infty\);4]

Để \(A\cap B=\varnothing\Leftrightarrow4< 3m+2\Leftrightarrow m>\dfrac{2}{3}\)

Bài 3.

a) TXĐ \(D=R\backslash\left\{-2\right\}\)

b) ĐK: \(12-3x\ge0\Leftrightarrow x\le4\). Vậy TXĐ D=(\(-\infty\);4].

c) ĐK: \(x-4>0\Leftrightarrow x>4\). Vậy TXĐ \(D=\left(4;+\infty\right)\).

d) ĐK: \(\left\{{}\begin{matrix}x-1\ne0\\3-x>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x< 3\end{matrix}\right.\)

Vậy TXĐ \(D=\left(-\infty;3\right)\backslash\left\{1\right\}\).

e) ĐK: \(\left\{{}\begin{matrix}5-x\ge0\\x^2-3x-10\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le5\\x\ne-2\\x\ne5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 5\\x\ne-2\end{matrix}\right.\)

Vậy TXĐ \(D=\left(-\infty;5\right)\backslash\left\{-2\right\}\).

f) ĐK: \(\left\{{}\begin{matrix}2x+1\ge0\\4-3x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{1}{2}\\x\le\dfrac{4}{3}\end{matrix}\right.\Leftrightarrow-\dfrac{1}{2}\le x\le\dfrac{4}{3}\)

Vậy TXĐ \(D=\left[-\dfrac{1}{2};\dfrac{4}{3}\right]\).

g) ĐK: \(\left\{{}\begin{matrix}2x-5\ge0\\x^2-4x-5\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\x\ne-1\\x\ne5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\x\ne5\end{matrix}\right.\)

Vậy TXĐ D=[\(\dfrac{5}{2};+\infty\))\{5}.

h) ĐK: \(\left\{{}\begin{matrix}-x+4\ge0\\x^2-x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le4\\x\ne0\\x\ne1\end{matrix}\right.\)

Vậy TXĐ \(D=\)(\(-\infty;4\)]\{0;1}.