A= \(|\sqrt{x^2}+\sqrt{1}-9|+|\sqrt{x^2}+\sqrt{1}-12|\)

A=\(|x+1-9|+|x+1-12|\)

A=\(|x-8|+|x-11|\)

TH1: x<0

=> A= (-x)-8 + (-x) -11

A=(-x-x)-(8+11)

A=-2x-19

TH2:x>0

=> A=x-8+x-11

A=(x+x)-(8+11)

A=2x-19

Tương tự x=0 sau đấy cậu KL nhé, phần sau mình lười

Áp dụng BĐT \(\left|x\right|+\left|y\right|\ge\left|x+y\right|\):

\(\left|\sqrt{x^2+1}-9\right|+\left|\sqrt{x^2+1}-12\right|\)\(=\left|\sqrt{x^2+1}-9\right|+\left|12-\sqrt{x^2+1}\right|\)

\(\ge\left|\left(\sqrt{x^2+1}-9\right)+\left(12-\sqrt{x^2+1}\right)\right|=3\)

Vậy \(A_{min}=3\Leftrightarrow\left(\sqrt{x^2+1}-9\right)\left(12-\sqrt{x^2+1}\right)\ge0\)

\(TH1:\hept{\begin{cases}\sqrt{x^2+1}-9\ge0\\12-\sqrt{x^2+1}\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x^2+1\ge81\\x^2+1\le144\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x^2\ge80\\x^2\le143\end{cases}}\Leftrightarrow\orbr{\begin{cases}\sqrt{80}\le x\le\sqrt{143}\\-\sqrt{80}\ge x\ge-\sqrt{143}\end{cases}}\)

\(TH2:\hept{\begin{cases}\sqrt{x^2+1}-9\le0\\12-\sqrt{x^2+1}\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x^2+1\le81\\x^2+1\ge144\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x^2\le80\\x^2\ge143\end{cases}}\left(L\right)\)

\(A=\left|2021-x\right|+\dfrac{1}{2}\left|4040-2x\right|\)

\(A=\left|2021-x\right|+\left|2020-x\right|\)

\(A=\left|2021-x\right|+\left|x-2020\right|\ge\left|2021-x+x-2020\right|=1\)

\(A_{min}=1\) khi \(2020\le x\le2021\)

Ta có: \(M=\sqrt{\left(1993-x\right)^2}+\sqrt{\left(1994-x\right)^2}>0\)

ĐKXĐ: \(\sqrt{\left(1993-x\right)^2}\ge0,\sqrt{\left(1994-x\right)^2}\ge0\forall x\inℝ\)

\(M=|1993-x|+|1994-x|\)

Ta có: GTNN của \(\sqrt{\left(1993-x\right)^2}=0\left(\sqrt{\left(1993-x\right)^2}\ge0\right)\)

GTNN của \(\sqrt{\left(1994-x\right)^2}=0\left(\sqrt{\left(1994-x\right)^2}\ge0\right)\)

=> GTNN của \(M=|1993-1994|hay|1994-1993|=1\)

Ta có: M = \(\sqrt{\left(1993-x\right)^2}+\sqrt{\left(1994-x\right)^2}\)

\(\Leftrightarrow\)M = \(\left|1993-x\right|+\left|1994-x\right|\)

              = \(\left|x-1993\right|+\left|1994-x\right|\)

              \(\ge\left|x-1993+1994-x\right|\)\(=\left|1\right|\)= 1

\(\Rightarrow M\ge1\)

Dấu "=" xảy ra khi: \(\left(x-1993\right)\left(1994-x\right)\ge0\)

                              \(\Leftrightarrow\hept{\begin{cases}x-1993\ge0\\1994-x\ge0\end{cases}}\)

                                \(\Leftrightarrow\hept{\begin{cases}x\ge1993\\x\le1994\end{cases}}\)

                                  \(\Leftrightarrow1993\le x\le1994\)

Vậy: min M = 1  \(\Leftrightarrow1993\le x\le1994\)

nhờ ng` khácđi nhé giờ bận rùi

\(A=x\left(x+1\right)\left(x+2\right)\left(x+3\right)+\sqrt{13}\)

\(=x\left(x+3\right)\left(x+1\right)\left(x+2\right)+\sqrt{13}\)

\(=\left(x^2+3x\right)\left(x^2+3x+2\right)+\sqrt{13}\)

Đặt \(t=x^2+3x+1\)

\(=>A=\left(t-1\right)\left(t+1\right)+\sqrt{13}\)

\(=t^2-1+\sqrt{13}\)

\(=\left(x^2+3x+1\right)^2-1+\sqrt{13}\ge-1+\sqrt{13}\)

Vậy ....................

Câu B thôi buồn ngủ qué ko làm được , xin lỗi bạn .

máy lag + mệt = nản, vô đây tham khảo HERE

ta có :\(a^2-ab+b^2=\left(a+b\right)^2-3ab\ge\left(a+b\right)^2-\dfrac{3}{4}\left(a+b\right)^2=\dfrac{1}{4}\left(a+b\right)^2\)(theo BĐT AM-GM)

\(\Rightarrow P\ge\sum\dfrac{a+b}{2\sqrt{ab+1}}\)

ÁP dụng BĐT AM-GM:

\(\dfrac{a+b}{2\sqrt{ab+1}}+\dfrac{b+c}{2\sqrt{bc+1}}+\dfrac{c+a}{2\sqrt{ca+1}}\ge3\sqrt[3]{\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{8\sqrt{\left(ab+1\right)\left(bc+1\right)\left(ca+1\right)}}}=\dfrac{3}{2}.\dfrac{1}{\sqrt[3]{\sqrt{\left(ab+1\right)\left(bc+1\right)\left(ca+1\right)}}}\)

Mà \(\sqrt[3]{\left(ab+1\right)\left(bc+1\right)\left(ca+1\right)}\le\dfrac{1}{3}\left(ab+bc+ca+3\right)\)

\(\Rightarrow P\ge\dfrac{3\sqrt{3}}{2\sqrt{\left(ab+bc+ca+3\right)}}\)(*)

ta liên tưởng đến BĐT phụ:\(\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+xz\right)\)

Cm: phân tích :\(VT=xy\left(x+y\right)+yz\left(y+z\right)+zx\left(x+z\right)+2xyz\)

\(=xy\left(x+y\right)+yz\left(y+z\right)+xz\left(z+x\right)+3xyz-xyz\)

\(=\left(x+y+z\right)\left(xy+yz+xz\right)-xyz\)

mà \(\left(x+y+z\right)\left(xy+yz+xz\right)\ge3\sqrt[3]{xyz}.3\sqrt[3]{x^2y^2z^2}=9xyz\)

nên \(\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge\left(x+y+z\right)\left(xy+yz+xz\right)-\dfrac{1}{9}\left(x+y+z\right)\left(xy+yz+xz\right)=\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)\)

Áp dụng:

\(1=\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\dfrac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\)

mặt khác,theo AM-GM,dễ dàng chứng minh được \(a+b+c\ge\dfrac{3}{2}\)

nên \(1\ge\dfrac{8}{9}.\dfrac{3}{2}\left(ab+bc+ca\right)\Leftrightarrow ab+bc+ca\le\dfrac{3}{4}\)

từ (*)\(\Rightarrow P\ge\dfrac{3\sqrt{3}}{2\sqrt{\dfrac{3}{4}+3}}=\dfrac{3}{\sqrt{5}}\)

Dấu = xảy ra khi \(a=b=c=\dfrac{1}{2}\)

sửa đề: \(\left(\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{2}{\sqrt{2}+\sqrt{3}}\right).\sqrt{21-12\sqrt{3}}\)

\(\left(\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{2}{\sqrt{2}+\sqrt{3}}\right).\sqrt{21-12\sqrt{3}}\\ =\left(\dfrac{\sqrt{2}+\sqrt{3}+2\sqrt{2}-2\sqrt{3}}{2-3}\right).\sqrt{3}.\sqrt{7-4\sqrt{3}}\\ =\dfrac{3\sqrt{2}-\sqrt{3}}{-1}.\sqrt{3}.\sqrt{\left(2-\sqrt{3}\right)^2}\\ =\left(-3\sqrt{2}+\sqrt{3}\right).\sqrt{3}.\left(2-\sqrt{3}\right)\\ =\left(\sqrt{3}-3\sqrt{2}\right)\left(2\sqrt{3}-3\right)\\ =6-3\sqrt{3}-6\sqrt{6}+9\sqrt{2}\)

tới đó thì ko bt còn hay hết