a) ĐKXĐ: x\(\ge\)-3

PT\(\Leftrightarrow\sqrt{\left(x+7\right)\left(x+3\right)}=3\sqrt{x+3}+2\sqrt{x+7}-6\)

Đặt \(\left(\sqrt{x+3},\sqrt{x+7}\right)=\left(a,b\right)\)                 \(\left(a,b\ge0\right)\)

PT\(\Leftrightarrow ab=3a+2b-6\Leftrightarrow a\left(b-3\right)-2\left(b-3\right)=0\)

\(\Leftrightarrow\left(a-2\right)\left(b-3\right)=0\Leftrightarrow\orbr{\begin{cases}a=2\\b=3\end{cases}}\)(TM ĐK)

TH 1: a=2\(\Leftrightarrow\sqrt{x+3}=2\Leftrightarrow x+3=4\Leftrightarrow x=1\)(tm)

TH 2: b=3\(\Leftrightarrow\sqrt{x+7}=3\Leftrightarrow x+7=9\Leftrightarrow x=2\)(tm)

Vậy tập nghiệm phương trình S={1; 2}

toán lớp mấy đó bạn

Mình giải được rồi nheeee

blabla..

\(\frac{a^3}{\sqrt{b^2+3}}+\frac{a^3}{\sqrt{b^2+3}}+\frac{b^2+3}{8}\ge\frac{3}{2}a^2\)\(\Leftrightarrow\)\(\frac{a^3}{\sqrt{b^2+3}}\ge\frac{3}{4}a^2-\frac{1}{16}b^2-\frac{3}{16}\)

\(P=\Sigma\frac{a^3}{\sqrt{b^2+3}}\ge\frac{3}{4}\left(a^2+b^2+c^2\right)-\frac{1}{16}\left(a^2+b^2+c^2\right)-\frac{9}{16}=\frac{3}{2}\)

Dấu "=" xảy ra khi a=b=c=1 

different way

Ta co:

\(\text{ }P=\Sigma_{cyc}\frac{a^3}{\sqrt{b^2+3}}\ge\Sigma_{cyc}\frac{\left(a^2+b^2+c^2\right)^2}{\Sigma_{cyc}a\sqrt{b^2+3}}\ge\frac{9}{\sqrt{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2+9\right)}}=\frac{3}{2}\)

Dau '=' xay ra khi \(a=b=c=1\)

(-5)+(+2)+I3I+(-4)+I-1I

=(-5)+2+3-4+1

=-3

Hok tốt nhé bạn ^^

#Dũng#