Bài 1:

a: \(M=x^2-10x+3\)

\(=x^2-10x+25-22\)

\(=\left(x^2-10x+25\right)-22\)

\(=\left(x-5\right)^2-22>=-22\forall x\)

Dấu '=' xảy ra khi x-5=0

=>x=5

b: \(N=x^2-x+2\)

\(=x^2-x+\dfrac{1}{4}+\dfrac{7}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>=\dfrac{7}{4}\forall x\)

Dấu '=' xảy ra khi x-1/2=0

=>x=1/2

c: \(P=3x^2-12x\)

\(=3\left(x^2-4x\right)\)

\(=3\left(x^2-4x+4-4\right)\)

\(=3\left(x-2\right)^2-12>=-12\forall x\)

Dấu '=' xảy ra khi x-2=0

=>x=2

1) \(\sqrt{4+x}=2-x\) (ĐK: \(x\ge-4\))

\(\Leftrightarrow4+x=\left(2-x\right)^2\)

\(\Leftrightarrow4+x=4-4x+x^2\)

\(\Leftrightarrow x^2-4x-x+4-4=0\)

\(\Leftrightarrow x^2-5x=0\)

\(\Leftrightarrow x\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)

Vậy: \(S=\left\{0;5\right\}\)

2) 

a) ĐKXĐ: \(a>0,a\ne1\)

\(A=\left(\dfrac{a-\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}+1}{a+\sqrt{a}}\right):\dfrac{\sqrt{a}+1}{a}\)

\(A=\left[\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-\dfrac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right]\cdot\dfrac{a}{\sqrt{a}+1}\)

\(A=\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\cdot\dfrac{a}{\sqrt{a}+1}\)

\(A=\dfrac{a-1}{\sqrt{a}}\cdot\dfrac{a}{\sqrt{a}+1}\)

\(A=\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}}\cdot\dfrac{\sqrt{a}\cdot\sqrt{a}}{\sqrt{a}+1}\)

\(A=\sqrt{a}\left(\sqrt{a}-1\right)\)

\(A=a-\sqrt{a}\)

b) Ta có:

\(A=a-\sqrt{a}\)

\(A=\left(\sqrt{a}\right)^2-2\cdot\dfrac{1}{2}\cdot\sqrt{a}+\dfrac{1}{4}-\dfrac{1}{4}\)

\(A=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)

Mà: \(\left(\sqrt{a}-\dfrac{1}{2}\right)^2\ge0\) nên \(A=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

Dấu "=" xảy ra khi:

\(\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}=-\dfrac{1}{4}\)

\(\Leftrightarrow a=\dfrac{1}{4}\)

Vậy: \(A_{min}=-\dfrac{1}{4}\)khi \(a=\dfrac{1}{4}\)