a) Vì \(\sqrt{x-5}\) ≥0

⇒ \(\sqrt{x-5}+7\) ≥ 7

Min A=7⇔x-5=0

             ⇔x=5

b) Vì \(\sqrt{3x-5}\) ≥0

⇒ 8-\(\sqrt{3x-5}\) ≤8

Max=8⇔3x-5\(=\)0

           ⇔\(x=\dfrac{5}{3}\)

a: \(=\sqrt{x-3-2\sqrt{x-3}+3}\)

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

Dấu = xảy ra khi x-3=1

=>x=4

\(A=\dfrac{x^2-4x+1}{x^2}=\dfrac{1}{x^2}-\dfrac{4}{x}+1=\left(\dfrac{1}{x^2}-\dfrac{4}{x}+4\right)-3=\left(\dfrac{1}{x}-2\right)^2-3\ge-3\)

\(A_{min}=-3\) khi \(x=\dfrac{1}{2}\)

2. \(P=\dfrac{x^2}{y+z}+\dfrac{y^2}{x+z}+\dfrac{z^2}{x+y}\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}\) (BĐT Cauchy-Schwarz) 

\(=\dfrac{1}{2}\)

\(\Rightarrow P_{min}=\dfrac{1}{2}\) khi \(\dfrac{x}{y+z}=\dfrac{y}{z+x}=\dfrac{z}{x+y}\Rightarrow x=y=z=\dfrac{1}{3}\)

1, đặt \(x^2+x=t\)

=>\(A=t\left(t-4\right)=t^2-4t=t^2-4t+4-4\)

\(=>A=\left(t-2\right)^2-4\ge-4\) dấu"=' xảy ra\(t=2\)

\(=>x^2+x=2< =>x^2+x-2=0\)

\(< =>x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{9}{4}=0\)

\(< =>\left(x+\dfrac{1}{2}\right)^2-\left(\dfrac{3}{2}\right)^2=0< =>\left(x-1\right)\left(x+2\right)=0\)

\(=>\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\) vậy Amin=-4<=>\(\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

B2

\(=>P=\dfrac{x^2}{y+z}+\dfrac{y+z}{4}+\dfrac{y^2}{x+z}+\dfrac{x+z}{4}+\dfrac{z^2}{x+y}+\dfrac{x+y}{4}\)

\(-\left(\dfrac{y+z+x+z+x+y}{4}\right)\)

áp dụng BDT AM-GM

\(=>\dfrac{x^2}{y+z}+\dfrac{y+z}{4}\ge2\sqrt{\dfrac{x^2}{4}}=x^{ }\left(1\right)\)

\(\)tương tự \(=>\dfrac{y^2}{x+z}+\dfrac{x+z}{4}\ge y\left(2\right)\)

\(=>\dfrac{z^2}{x+y}+\dfrac{x+y}{4}\ge z\left(3\right)\)

(1)(2)(3) \(=>P\ge x+y+z-\dfrac{1}{2}.x+y+z=1-\dfrac{1}{2}=\dfrac{1}{2}\)

dấu"=" xảy ra<=>x=y=z=1/3

\(A=\left|x-201\right|+\left|x-204\right|=\left|x-201\right|+\left|204-x\right|\ge\left|x-201+204-x\right|=\left|3\right|=3\)

\(minA=3\Leftrightarrow\left(x-201\right)\left(204-x\right)\ge0\Leftrightarrow204\ge x\ge201\)

a. ĐKXĐ: \(x\ge-1\)

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

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

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

b.

\(f\left(x\right)=\dfrac{x-1}{2}+\dfrac{2}{x-1}+\dfrac{1}{2}\ge2\sqrt{\dfrac{2\left(x-1\right)}{2\left(x-1\right)}}+\dfrac{1}{2}=\dfrac{5}{2}\)

c.

\(y=\dfrac{x-2018+1}{\sqrt{x-2018}}=\sqrt{x-2018}+\dfrac{1}{\sqrt{x-2018}}\ge2\sqrt{\dfrac{\sqrt{x-2018}}{\sqrt{x-2018}}}=2\)

a) có nghĩa khi \(x-1\ne0\Rightarrow x\ne1\)

b)\(f\left(7\right)=\frac{7+2}{7-1}=\frac{9}{6}\)

c)\(f\left(x\right)=\frac{x+2}{x-1}=\frac{1}{4}\Leftrightarrow x+2=4x-4\)

\(\Leftrightarrow-3x=-6\Leftrightarrow x=2\)

e)\(f\left(x\right)>1\Rightarrow\frac{x+2}{x-1}-1>0\)

\(\Rightarrow\frac{3}{x-1}>0\) thấy 3>0 nên x-1>0 =>x>1

Bài 2:

a)\(P=9-2\left|x-3\right|\)

Thấy: \(\left|x-3\right|\ge0\)\(\Rightarrow2\left|x-3\right|\ge0\)

\(\Rightarrow-2\left|x-3\right|\le0\)

\(\Rightarrow9-2\left|x-3\right|\le9\)

Khi x=3

b)Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:

\(Q=\left|x-2\right|+\left|x-8\right|\)

\(=\left|x-2\right|+\left|8-x\right|\)

\(\ge\left|x-2+8-x\right|=6\)

Khi \(2\le x\le8\)

bài 1 bựa nớ ta làm rồi chj nựa