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}.

\(f\left(x\right)=x^2+4x+\left|x+2\right|-m< 0\) 

\(\Leftrightarrow f\left(x\right)=x^2+4x+4+\left|x+2\right|-4-m< 0\)

\(\Leftrightarrow f\left(x\right)=\left(x+2\right)^2+\left|x+2\right|-4-m< 0\)

\(đặt:\left|x+2\right|=t\ge0\Rightarrow f\left(t\right)=t^2+t-4-m< 0\)

\(có\) \(f\left(x\right)nghiệm\Leftrightarrow f\left(t\right)có\)  \(nghiệm\) \(t\ge0\)

\(f\left(t\right)=t^2+t-4< m\)\(có\) \(nghiệm\) \(t\ge0\)

\(\Leftrightarrow m>minf\left(t\right)\left(trên[0;+\infty\right)\)\(\Leftrightarrow m>-4\)

TL:

= 4.2536277e+13

Ko đúng thì thông cảm cho mình nhé

HT~
@@@

24324243+42536252489952=42536276814195 nhé.

Chúc học tốt!

\(M\in\left(d\right)\Rightarrow M\left(a;a+6\right)\Rightarrow\left\{{}\begin{matrix}MA=\sqrt{\left(a-2\right)^2+\left(a+4\right)^2}=\sqrt{2\left(a+1\right)^2+18}\\MB=\sqrt{\left(a-3\right)^2+\left(a+6\right)^2}=\sqrt{2\left(a+\dfrac{3}{2}\right)^2+\dfrac{81}{2}}=\sqrt{2\left(-\dfrac{3}{2}-a\right)^2+\dfrac{81}{2}}\end{matrix}\right.\)

\(\Rightarrow MA+MB=\sqrt{\sqrt{2}^2\left(a+1\right)^2+18}+\sqrt{\sqrt{2}^2\left(-\dfrac{3}{2}-a\right)^2+\dfrac{81}{2}}\ge\sqrt{\left(\sqrt{2}.a+\sqrt{2}-\dfrac{3}{2}.\sqrt{2}-\sqrt{2}.a\right)^2+\left(\sqrt{18}+\sqrt{\dfrac{81}{2}}\right)^2}=\sqrt{\dfrac{1}{2}+\dfrac{225}{2}}=\sqrt{133}\)

\(dấu"="xayra\Leftrightarrow\dfrac{\sqrt{2}\left(a+1\right)}{\sqrt{18}}=\dfrac{\sqrt{2}\left(-\dfrac{3}{2}-a\right)}{\sqrt{\dfrac{81}{2}}}\Leftrightarrow a=-\dfrac{6}{5}\Rightarrow M\left(-\dfrac{6}{5};\dfrac{24}{5}\right)\)