Cần các cao nhân giải khác phương pháp SS

Không làm theo cách đánh giá 3(a²b+b²c+c²a)\(\le\)(a+b+c)(a²+b²+c²)=3(a²+b²+c²)

Ai làm được xin cảm ơn trước

#)Giải :

Ta có : \(3\left(a^2+b^2+c^2\right)=\left(a+b+c\right)\left(a^2+b^2+c^2\right)\)

\(=a^3+b^3+c^3+a^2b+b^2c+c^2a+ab^2+bc^2+ca^2\)

Áp dụng BĐT Cauchy :

\(\hept{\begin{cases}a^3+ab^2\ge2a^2b\\b^3+bc^2\ge2b^2c\\c^3+ca^2\ge2c^2a\end{cases}}\)

\(\Rightarrow P\ge a^2+b^2+c^2+\frac{ab+bc+ca}{a^2+b^2+c^2}\)

\(\Rightarrow P\ge a^2+b^2+c^2+\frac{9-\left(a^2+b^2+c^2\right)}{2\left(a^2+b^2+c^2\right)}\)

Đặt \(t=a^2+b^2+c^2\Rightarrow t\ge3\)

\(\Rightarrow P\ge t+\frac{9-t}{2t}=\frac{t}{2}+\frac{9}{2t}+\frac{t}{2}-\frac{1}{2}\ge3+\frac{3}{2}-\frac{1}{2}=4\)

\(\Rightarrow P\ge4\Rightarrow P_{min}=4\)

Dấu ''='' xảy ra khi a = b = c = 1

\(\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{x^4-1}\)

\(\Leftrightarrow\sqrt{x-1}+\sqrt{x^3+x^2+x+1}-1-\sqrt{\left(x-1\right)\left(x^3+x^2+x+1\right)}=0\)

\(\Leftrightarrow\left(\sqrt{x-1}-1\right)+\sqrt{x^3+x^2+x+1}\left(1-\sqrt{x-1}\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-1}-1\right)-\sqrt{x^3+x^2+x+1}\left(\sqrt{x-1}-1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(1-\sqrt{x^3+x^2+x+1}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x-1}-1=0\\1-\sqrt{x^3+x^2+x+1}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x-1}=1\\\sqrt{x^3+x^2+x+1}=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=1\\x\left(x^2+x+1\right)=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=0\left(because:x^2+x+1>0with\forall x\right)\end{cases}}\)

Tiếc quá bạn.Mình vừa giải xong x=2

\(A=x-\sqrt{x}-2=\left(\sqrt{x}\right)^2-2.\sqrt{x}.\frac{1}{2}+\left(\frac{1}{2}\right)^2-2,25=\left(\sqrt{x}-\frac{1}{2}\right)^2-2,25\ge2,25\forall x\ge0\)

ĐK: \(x\ge-2\)

\(B=x-\sqrt{x+2}=x+2-2.\sqrt{x+2}.\frac{1}{2}+\left(\frac{1}{2}\right)^2-2,25=\left(\sqrt{x+2}-\frac{1}{2}\right)^2-2,25\ge-2,25\forall x\ge-2\)........

ĐKXĐ: \(x\ne\pm2y\)

Đặt \(\hept{\begin{cases}\frac{1}{x+2y}=a\\\frac{1}{x-2y}=b\end{cases}\left(a;b\ne0\right)}\)

\(\Rightarrow\hept{\begin{cases}4a-b=1\\20a+3b=1\end{cases}}\Leftrightarrow\hept{\begin{cases}a=\frac{1}{8}\\b=-\frac{1}{2}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x+2y=8\\x-2y=-2\end{cases}\Rightarrow}\hept{\begin{cases}x=3\\y=\frac{5}{2}\end{cases}}\)

Điều kiện xác định x#1; y#3.Đặt: \(\hept{\begin{cases}\frac{1}{x-1}=a\\\frac{1}{y-3}=b\end{cases}}\Rightarrow\hept{\begin{cases}5a+b=10\\a-3b=18\end{cases}}\Rightarrow\hept{\begin{cases}15a+3b=30\\a-3b=18\end{cases}}\)

Cộng theo vế: \(15a+3b+a-3b=48\Rightarrow16a=48\Rightarrow a=3\Rightarrow b=-5\)

\(\Rightarrow\hept{\begin{cases}\frac{1}{x-1}=3\Rightarrow x=\frac{4}{3}\\\frac{1}{y-3}=-5\Rightarrow y=-\frac{14}{5}\end{cases}}\)

\(\hept{\begin{cases}\frac{5}{x-1}+\frac{1}{y-3}=10\\\frac{1}{x-1}-\frac{3}{y-3}=18\end{cases}}\)

Đặt: \(\frac{1}{x-1}=a\left(a>0\right);\frac{1}{y-3}=b\left(b>0\right)\)

Khi đó hpt có dạng:

\(\hept{\begin{cases}5a+b=10\\a-3b=18\end{cases}\Rightarrow\hept{\begin{cases}a=3\\b=-5\end{cases}}\left(Tm\right)}\)

\(\Rightarrow\hept{\begin{cases}\frac{1}{x-1}=3\\\frac{1}{y-3}=-5\end{cases}}\Rightarrow\hept{\begin{cases}3\left(x-1\right)=1\\-5\left(y-3\right)=1\end{cases}}\Rightarrow\hept{\begin{cases}3x-3=1\\-5y+15=1\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{4}{3}\\y=\frac{14}{5}\end{cases}}\)