\(1,ĐK:x\ge2\\ PT\Leftrightarrow\sqrt{3x-6}+x-2-\left(\sqrt{2x-3}-1\right)=0\\ \Leftrightarrow\dfrac{3\left(x-2\right)}{\sqrt{3x-6}}+\left(x-2\right)-\dfrac{2\left(x-2\right)}{\sqrt{2x-3}+1}=0\\ \Leftrightarrow\left(x-2\right)\left(\dfrac{3}{\sqrt{3x-6}}-\dfrac{2}{\sqrt{2x-3}+1}+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\\dfrac{3}{\sqrt{3x-6}}-\dfrac{2}{\sqrt{2x-3}+1}+1=0\left(1\right)\end{matrix}\right.\)
Với \(x>2\Leftrightarrow-\dfrac{2}{\sqrt{2x-3}+1}>-\dfrac{2}{1+1}=-1\left(3x-6\ne0\right)\)
\(\Leftrightarrow\left(1\right)>0-1+1=0\left(vn\right)\)
Vậy \(x=2\)
\(2,ĐK:x\ge-1\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\\\sqrt{x^2-x+1}=b\end{matrix}\right.\left(a,b\ge0\right)\Leftrightarrow a^2+b^2=x^2+2\)
\(PT\Leftrightarrow2a^2+2b^2-5ab=0\\ \Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=2b\\b=2a\end{matrix}\right.\)
Với \(a=2b\Leftrightarrow x+1=4x^2-4x+4\left(vn\right)\)
Với \(b=2a\Leftrightarrow4x+4=x^2-x+1\Leftrightarrow x^2-5x-3=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5+\sqrt{37}}{2}\left(tm\right)\\x=\dfrac{5-\sqrt{37}}{2}\left(tm\right)\end{matrix}\right.\)
Vậy ...
\(3,ĐK:x\ge-1\\ PT\Leftrightarrow3\left(x^2-x+1\right)-2\left(x+1\right)=5\sqrt{x^3+1}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\\\sqrt{x^2-x+1}=b\end{matrix}\right.\left(a,b\ge0\right)\)
\(PT\Leftrightarrow3b^2-2a^2=5ab\\ \Leftrightarrow2a^2+5ab-3b^2=0\\ \Leftrightarrow\left[{}\begin{matrix}a=2b\\a=-3b\left(vn\right)\end{matrix}\right.\Leftrightarrow a=2b\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5+\sqrt{37}}{2}\\x=\dfrac{5-\sqrt{37}}{2}\end{matrix}\right.\left(\text{giống bài 2}\right)\)
\(4,\)
Đặt \(\left(x;y;z\right)=\left(a^3;b^3;c^3\right)\Leftrightarrow abc=1\)
\(\Leftrightarrow VT=\dfrac{1}{a^3+b^3+1}+\dfrac{1}{b^3+c^3+1}+\dfrac{1}{c^3+a^3+1}\)
Cần cm \(a^3+b^3\ge ab\left(a+b\right)\Leftrightarrow\left(a+b\right)\left(a-b\right)^2\ge0\left(\text{luôn đúng}\right)\)
\(\Leftrightarrow VT\le\sum\dfrac{abc}{ab\left(a+b\right)+abc}=\sum\dfrac{abc}{ab\left(a+b+c\right)}=\sum\dfrac{c}{a+b+c}=1\)
Dấu \("="\Leftrightarrow a=b=c=1\Leftrightarrow x=y=z=1\)
\(5,\\ a,b,c\in\left[0;2\right]\Leftrightarrow\left(2-a\right)\left(2-b\right)\left(2-c\right)\le0\\ \Leftrightarrow2\left(ab+bc+ca\right)-4\left(a+b+c\right)-abc+8\le0\\ \Leftrightarrow2\left(ab+bc+ca\right)-abc\le4\left(a+b+c=3\right)\\ \Leftrightarrow2\left(ab+bc+ca\right)\le4+abc=4\left(a,b,c\ge0\right)\\ \Leftrightarrow\left(a+b+c\right)^2-a^2-b^2-c^2\le4\\ \Leftrightarrow9-\left(a^2+b^2+c^2\right)\le4\\ \Leftrightarrow a^2+b^2+c^2\ge5\)
Dấu \("="\Leftrightarrow a=2;b=1;c=0\) và các hoán vị
\(6,ĐK:-\sqrt{5}\le x\le\sqrt{5}\)
\(A^2=\left(2x+\sqrt{5-x^2}\right)^2\le\left(2^2+1^2\right)\left(x^2+5-x^2\right)=5\cdot5=25\\ \Leftrightarrow A\le5\)
Dấu \("="\Leftrightarrow\dfrac{x}{2}=\sqrt{5-x^2}\Leftrightarrow x^2=20-4x^2\Leftrightarrow x=2\left(x\ge0\right)\)
\(A=2x+\sqrt{5-x^2}\ge-2\sqrt{5}\)
Dấu \("="\Leftrightarrow x=-\sqrt{5}\)
\(7,M=\sum\dfrac{a^2}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{a+b+b+c+c+a}=\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}=5\)
Dấu \("="\Leftrightarrow a=b=c=\dfrac{10}{3}\)
