Với \(x\ge-\frac{1}{2}\)

2f(x) = \(2\sqrt{\left(2x+1\right)\left(x+2\right)}+4\sqrt{x+3}-4x\)

\(=-\left(2x+1\right)+2\sqrt{\left(2x+1\right)\left(x+2\right)}-\left(x+2\right)-\left(x+3\right)+4\sqrt{x+3}-4+10\)

\(=-\left(\sqrt{2x+1}-\sqrt{x+2}\right)^2-\left(\sqrt{x+3}-2\right)^2+10\le10\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}2x+1=x+2\\x+3=4\end{cases}}\Leftrightarrow x=1\)

=> min 2f(x) = 10 tại x = 1

=> min f(x) = 5 tại x = 1

A= cos² 20^o + cos² 30^o + cos² 40^o + sin² 40^o +sin² 30^o +sin²20^o

A = (cos² 20^o +sin²20^o) + ( cos² 30^o+sin² 30^o) +(cos² 40^o + sin² 40^o)

A= 1+1+1=3

\(\frac{1}{R_{td}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\) (Thay số rồi tính)

\(U_{R_1}=U_{R_2}=U_{R_3}=I_{R_1}.R_1=I_{R_2}.R_2=I_{R_3}.R_3\)

\(\Rightarrow2.I_{R_1}=4.I_{R_2}=6.0,6=3,6\) Từ đây tính được I ở hai nhánh còng lại

I mạch chính = tổng các I mạch nhánh

Đặt S = \(\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ac+a^2}\)

\(S=\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2}\\ =\sqrt{a^2+2ab+b^2-3ab}+\sqrt{b^2+2bc+c^2-3bc}+\sqrt{c^2+2ca+a^2-3ca}\\ =\sqrt{\left(a+b\right)^2-\dfrac{3}{4}\cdot4ab}+\sqrt{\left(b+c\right)^2-\dfrac{3}{4}\cdot4bc}+\sqrt{\left(c+a\right)^2-\dfrac{3}{4}\cdot4ca}\)

Áp dụng BĐT cô - si ta có :

\(\Rightarrow S=\sqrt{\left(a+b\right)^2-\dfrac{3}{4}\cdot4ab}+\sqrt{\left(b+c\right)^2-\dfrac{3}{4}\cdot4bc}+\sqrt{\left(c+a\right)^2-\dfrac{3}{4}\cdot4ca}\\ \ge\sqrt{\left(a+b\right)^2-\dfrac{3}{4}\cdot\left(a+b\right)^2}+\sqrt{\left(b+c\right)^2-\dfrac{3}{4}\left(b+c\right)^2}+\sqrt{\left(c+a\right)^2-\dfrac{3}{4}\left(c+a\right)^2}\\ =\sqrt{\dfrac{1}{4}\left(a+b\right)^2}+\sqrt{\dfrac{1}{4}\left(b+c\right)^2}+\sqrt{\dfrac{1}{4}\left(c+a\right)^2}\\ =\dfrac{1}{2}\left(a+b\right)+\dfrac{1}{2}\left(b+c\right)+\dfrac{1}{2}\left(c+a\right)\\ =\dfrac{1}{2}\left(a+b+b+c+c+a\right)\\ =a+b+c\\ =2019\)

Dấu " = " xảy ra khi \(\hept{\begin{cases}a=b=c\\a+b+c=2019\end{cases}\Rightarrow\hept{\begin{cases}a=673\\b=673\\c=673\end{cases}}}\)

Vậy Min S = 2019 <=> a=b=c = 673

a) x = 16 (tm) => A = \(\frac{\sqrt{16}-2}{\sqrt{16}+1}=\frac{4-2}{4+1}=\frac{2}{5}\)

b) B = \(\left(\frac{1}{\sqrt{x}+5}-\frac{x+2\sqrt{x}-5}{25-x}\right):\frac{\sqrt{x}+2}{\sqrt{x}-5}\)

B = \(\frac{\sqrt{x}-5+x+2\sqrt{x}-5}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\cdot\frac{\sqrt{x}-5}{\sqrt{x}+2}\)

B = \(\frac{x+3\sqrt{x}-10}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)

B = \(\frac{x+5\sqrt{x}-2\sqrt{x}-10}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)

B = \(\frac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}-2}{\sqrt{x}+2}\)

c) P = \(\frac{B}{A}=\frac{\sqrt{x}-2}{\sqrt{x}+2}:\frac{\sqrt{x}-2}{\sqrt{x}+1}=\frac{\sqrt{x}+1}{\sqrt{x}+2}\)

=> \(P\left(\sqrt{x}+2\right)\ge x+6\sqrt{x}-13\)

<=> \(\frac{\sqrt{x}+1}{\sqrt{x}+2}.\left(\sqrt{x}+2\right)-x-6\sqrt{x}+13\ge0\)

<=> \(-x-6\sqrt{x}+13+\sqrt{x}+1\ge0\)

<=> \(-x-5\sqrt{x}+14\ge0\)

<=> \(x+5\sqrt{x}-14\le0\)

<=> \(x+7\sqrt{x}-2\sqrt{x}-14\le0\)

<=> \(\left(\sqrt{x}+7\right)\left(\sqrt{x}-2\right)\le0\)

Do \(\sqrt{x}+7>0\) với mọi x => \(\sqrt{x}-2\le0\)

<=> \(\sqrt{x}\le2\) <=> \(x\le4\)

Kết hợp với Đk: x \(\ge\)0; x \(\ne\)4; x \(\ne\)25

và x thuộc Z => x = {0; 1; 2; 3}

d) M = \(3P\cdot\frac{\sqrt{x}+2}{x+\sqrt{x}+4}\) <=>M = \(3\cdot\frac{\sqrt{x}+1}{\sqrt{x}+2}\cdot\frac{\sqrt{x}+2}{x+\sqrt{x}+4}\)

M = \(\frac{3\sqrt{x}+3}{x+\sqrt{x}+4}=\frac{x+\sqrt{x}+4-x+2\sqrt{x}-1}{\left(x+\sqrt{x}+\frac{1}{4}\right)+\frac{15}{4}}=1-\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{15}{4}}\le1\)(Do \(\left(\sqrt{x}-1\right)^2\ge0\) và \(\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{15}{4}>0\))

Dấu "=" xảy ra <=> \(\sqrt{x}-1=0\) <=> \(x=1\)

Vậy MaxM = 1 khi x = 1

Ta có: \(A=\left(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\left(1-\frac{3-\sqrt{x}}{\sqrt{x}+1}\right)\)     (   ĐKXĐ: \(x>0,\)\(x\ne0,\)\(x\ne1\))

    \(\Leftrightarrow A=\left(\frac{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right).\left(x-\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}+1\right)}\right):\left(\frac{\sqrt{x}+1-3+\sqrt{x}}{\sqrt{x}+1}\right)\)

    \(\Leftrightarrow A=\left(\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}+1}\right)\)

    \(\Leftrightarrow A=\left(\frac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\right).\left(\frac{\sqrt{x}+1}{2.\left(\sqrt{x}-1\right)}\right)\)

    \(\Leftrightarrow A=\left(\frac{2\sqrt{x}}{\sqrt{x}}\right).\left(\frac{\sqrt{x}+1}{2.\left(\sqrt{x}-1\right)}\right)\)

    \(\Leftrightarrow A=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

Để \(A\ge\frac{3}{2}\)\(\Rightarrow\)\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\ge\frac{3}{2}\)

Ta có: \(\frac{\sqrt{x}+1}{\sqrt{x}-1}\ge\frac{3}{2}\)

    \(\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{3}{2}\ge0\)

    \(\Leftrightarrow\frac{2\sqrt{x}+2-3\sqrt{x}+3}{2.\left(\sqrt{x}-1\right)}\ge0\)

    \(\Leftrightarrow\frac{5-\sqrt{x}}{2.\left(\sqrt{x}-1\right)}\ge0\)

+ TH₁: \(\hept{\begin{cases}5-\sqrt{x}\ge0\\2\sqrt{x}-2\ge0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}\sqrt{x}\le5\\\sqrt{x}\ge1\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x\le25\\x\ge1\end{cases}}\)\(\Rightarrow\)\(1\le x\le25\)\(\left(TM\right)\)

+ TH₂: \(\hept{\begin{cases}5-\sqrt{x}\le0\\2\sqrt{x}-2\le0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}\sqrt{x}\ge5\\\sqrt{x}\le1\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x\ge25\\x\le1\end{cases}}\)\(\left(L\right)\)

            \(\Rightarrow\)\(1\le x\le25.\)Kết hợp ĐKXĐ: \(x\ne1\)

                         \(\Rightarrow\)\(1< x\le25\)

Vậy để \(A\ge\frac{3}{2}\)\(\Leftrightarrow\)\(1< x\le25\)

\(\left(\sqrt{b}-\sqrt{c}\right)^2\ge0\Leftrightarrow b-2\sqrt{bc}+c\ge0\Leftrightarrow b+c\ge2\sqrt{bc}\) dấu "="xảy ra khi b=c

\(\left(a+2b\right)\left(a+2c\right)=a^2+2a\left(b+c\right)+4bc\ge a^2+4a\sqrt{bc}+4bc=\left(a+2\sqrt{bc}\right)^2\)

\(\Rightarrow\sqrt{\left(a+2b\right)\left(a+2c\right)}\ge a+2\sqrt{bc}\)

tương tự ta có \(\hept{\begin{cases}\sqrt{\left(b+2c\right)\left(b+2c\right)}\ge b+2\sqrt{bc}\\\sqrt{\left(c+2a\right)\left(a+2b\right)}\ge c+2\sqrt{ab}\end{cases}}\)

dấu "=" xảy ra khi a=b=c

\(\Rightarrow A=\sqrt{\left(a+2b\right)\left(a+2c\right)}+\sqrt{\left(b+2a\right)\left(b+2c\right)}+\sqrt{\left(c+2a\right)\left(c+2b\right)}\)\(\ge a+b+c+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ac}\)

hay \(A\ge\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=\left(\sqrt{3}\right)^2=3\)

dấu "="xảy ra khi \(\hept{\begin{cases}a=b=c\\\sqrt{a}+\sqrt{b}+\sqrt{c}=3\end{cases}\Leftrightarrow a=b=c=\frac{\sqrt{3}}{3}}\)

\(M=\left(2\sqrt{a}+3\sqrt{b}-4\sqrt{c}\right)^2=\left(2\sqrt{a}+3\sqrt{a}-4\sqrt{a}\right)^2=\left(\sqrt{a}\right)^2=\frac{\sqrt{3}}{3}\)

bổ sung thêm điều kiện x,y là số thực

với x>=1; y>=1 từ giả thiết ta có \(x\sqrt{x}-y\sqrt{y}=\sqrt{y-1}-\sqrt{x-1}\left(1\right)\)

nếu x=y=1 thì S=6 (*)

nếu x,y không đồng thời bằng 1 thì \(\sqrt{y-1}+\sqrt{x-1}>0\)vì vậy

(1) \(\Leftrightarrow x\sqrt{x}-y\sqrt{y}=\frac{\left(y-1\right)-\left(x-1\right)}{\sqrt{y-1}+\sqrt{x-1}}\)

\(\Leftrightarrow\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x+1}+\sqrt{y+1}}\right)=0\left(2\right)\)

vì x>=1; y>=1 nên từ (2) => x=y

vì vậy S=2x²-8x+12=2(x-2)²+4>=4 (**) với mọi x

dấu "=" xảy ra khi x=2

vậy minS=4 <=> x=y=2