a: 

Sửa đề: \(P=\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{4x^2}{x^2-9}\right):\left(\dfrac{5}{3-x}-\dfrac{4x+2}{3x-x^2}\right)\)\(P=\left(\dfrac{-\left(x+3\right)}{x-3}+\dfrac{x-3}{x+3}-\dfrac{4x^2}{\left(x-3\right)\left(x+3\right)}\right):\dfrac{5x-4x-2}{x\left(3-x\right)}\)

\(=\dfrac{-x^2-6x-9+x^2-6x+9-4x^2}{\left(x-3\right)\left(x+3\right)}:\dfrac{x-2}{x\left(3-x\right)}\)

\(=\dfrac{-4x^2-12x}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x\left(3-x\right)}{x-2}\)

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

b: x^2-4x+3=0

=>x=1(nhận) hoặc x=3(loại)

Khi x=1 thì \(P=\dfrac{4\cdot1^2}{1-2}=-4\)

c: P>0

=>x-2>0

=>x>2

d: P nguyên

=>4x^2 chia hết cho x-2

=>4x^2-16+16 chia hết cho x-2

=>x-2 thuộc {1;-1;2;-2;4;-4;8;-8;16;-16}

=>x thuộc {1;4;6;-2;10;-6;18;-14}

Bài 2: 

a: Ta có: \(x^2+4x+7\)

\(=x^2+4x+4+3\)

\(=\left(x+2\right)^2+3\ge3\forall x\)

Dấu '=' xảy ra khi x=-2

\(a,\Rightarrow4x^2-20x-4x^2+3x+4x-3=5\\ \Rightarrow-13x=8\Rightarrow x=-\dfrac{8}{13}\\ b,\Rightarrow3x^2-10x+8-3x^2+27x=-3\\ \Rightarrow17x=-11\Rightarrow x=-\dfrac{11}{17}\\ c,\Rightarrow\left(x+3\right)\left(2-x\right)=0\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\\ d,\Rightarrow2x\left(4x^2-25\right)=0\\ \Rightarrow2x\left(2x-5\right)\left(2x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\\ e,Sửa:\left(4x-3\right)^2-3x\left(3-4x\right)=0\\ \Rightarrow\left(4x-3\right)^2+3x\left(4x-3\right)=0\\ \Rightarrow\left(4x-3\right)\left(7x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)

a.

4x(x-5) - (x-1)(4x-3)-5=0

 4x^2-20x-4x^2+3x+4x+3=0

(4x^2-4x^2)+(-20x+3x+4x)+3=0

 13x+3 = 0

13x=-3

x=-3/13

b,

(3x-4)(x-2)-3x(x-9)+3=0

3x^2-6x-4x+8 - 3x^2+27x+3=0

(3x^2-3x^2)+(-6x-4x+27x)+(8+3)=0

17x+11=0

17x=-11

x=-11/17

c, 2(x+3)-x^2-3x=0

2(x+3) - x(x+3)=0

(x+3)(2-x)=0

TH1: x+3 = 0; x=-3

TH2: 2-x=0;x=2

\(4x\left(3-\dfrac{1}{4}x\right)+\left(x-2\right)\left(x+2\right)=0\)

\(\Rightarrow12x-x^2+x^2-4=0\Rightarrow12x=4\Rightarrow x=\dfrac{1}{3}\)

\(12x-x^2+x^2-2^2=0\)

\(12x-2=0\)

\(12x=2\)

\(x=\dfrac{1}{6}\)

Vậy x=1/6

a) Ta có: \(Q=-x^2-y^2+4x-4y+2=-\left(x^2+y^2-4x+4y-2\right)\)

\(=-\left(x^2-4x+4+y^2+4y+4\right)+10\)

\(=-\left[\left(x-2\right)^2+\left(y+2\right)^2\right]+10\le10\forall x,y\)

Vậy Max_Q=10 khi x=2, y=-2

b) +Ta có: \(A=-x^2-6x+5=-\left(x^2+6x-5\right)=-\left(x^2+6x+9-14\right)\)

\(=-\left(x^2+6x+9\right)+14=-\left(x+3\right)^2+14\le14\forall x\)

Vậy Max_A=14 khi x=-3

+Ta có: \(B=-4x^2-9y^2-4x+6y+3=-\left(4x^2+9y^2+4x-6y-3\right)\)

\(=-\left(4x^2+4x+1+9y^2-6y+1-5\right)\)

\(=-\left[\left(2x+1\right)^2+\left(3y-1\right)^2\right]+5\le5\forall x,y\)

Vậy Max_B=5 khi x=-1/2, y=1/3

c) Ta có: \(P=x^2+y^2-2x+6y+12=x^2-2x+1+y^2+6y+9+2\)

\(=\left(x-1\right)^2+\left(y+3\right)^2+2\ge2\forall x,y\)

Vậy Min_P=2 khi x=1, y=-3

c: Ta có: \(x^3+3x^2+3x-7=0\)

\(\Leftrightarrow x+1=2\)

hay x=1

b: Ta có: \(x\left(x-3\right)-4x+12=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

8: =>6x^2-9x+2x-3-6x^2-12x=16

=>-19x=19

=>x=-1

\(\frac{4x^2+9x+18\sqrt{x}+9}{4x\sqrt{x}+4\sqrt{x}}+\frac{4x\sqrt{x}+4\sqrt{x}}{4x^2+9x+18\sqrt{x}+9}-2=\frac{\left(-4x\sqrt{x}+4x^2+9x+22\sqrt{x}+9\right)^2}{\left(4x^2+9x+18\sqrt{x}+9\right)\left(4x\sqrt{x}+4\sqrt{x}\right)}\ge0\)

Đặt \(M=\frac{4x^2+9x+18\sqrt{x}+9}{4x\sqrt{x}+4x}\left(x>0\right)\Rightarrow M>0\)

Đặt \(y=\sqrt{x}>0\)ta có \(M=\frac{4x^2+9x+18\sqrt{x}+9}{4x\sqrt{x}+4x}=\frac{4y^4+9y^2+18y+9}{4y^3+4y^2}\)\(=\frac{3\left(4y^3+4y^2\right)+\left(4y^2-12y^3-3y^2+18y+9\right)}{4y^3+4y^2}=3+\frac{\left(2y^2-3y-3\right)^2}{4y^3+4y^2}\ge3\)

\(y>0\Rightarrow\hept{\begin{cases}4y^3+4y^2>0\\\left(2y^2-3y-3\right)^2\ge0\end{cases}\Rightarrow\frac{\left(2y-3y-3\right)^2}{4y^3+4y^2}\ge0}\)

Đẳng thức xảy ra \(\Leftrightarrow2y^2-3y-3=0\Leftrightarrow y=\frac{3+\sqrt{33}}{4}\left(y>0\right)\)

\(\Rightarrow x=\left(\frac{3+\sqrt{33}}{4}\right)^2=\frac{21+3\sqrt{33}}{8}\)

Khi đó \(A=M+\frac{1}{M}=\frac{8M}{9}+\left(\frac{M}{9}+\frac{1}{M}\right)\ge\frac{8\cdot3}{9}+2\sqrt{\frac{M}{9}\cdot\frac{1}{M}}=\frac{8}{3}+\frac{2}{3}=\frac{10}{3}\)

Đẳng thức xảy ra \(\Leftrightarrow\hept{\begin{cases}M=3\\\frac{M}{9}=\frac{1}{M}\end{cases}\Leftrightarrow M=3\Leftrightarrow x=\frac{21+3\sqrt{33}}{8}}\)

Vậy \(A_{min}=\frac{10}{3}\Leftrightarrow x=\frac{21+3\sqrt{33}}{8}\)

ta có: \(4x^2+9x+18\sqrt{x}+9=4x^2+9\left(\sqrt{x}+1\right)^2\),\(4x\sqrt{x}+4x=4x\left(\sqrt{x}+1\right)\)
Đặt \(a=x,b=\sqrt{x}+1\)ta có:
\(A=\frac{4a^2+9b^2}{4ab}+\frac{4ab}{4a^2+9b^2}=t+\frac{1}{t},t=\frac{4a^2+9b^2}{4ab}\)
có \(\frac{4a^2+9b^2}{4ab}=t\Rightarrow4a^2-t.4ab+9b^2=0\Leftrightarrow4.\left(\frac{a}{b}\right)^2-4t.\frac{a}{b}+9=0,\)do a khác 0.
Đặt \(\frac{a}{b}=y\Rightarrow4y^2-t.4y+9=0\), \(\Delta=16t^2-36\ge0\Leftrightarrow t\ge\frac{3}{2}\left(t>0\right)\)
xét \(f\left(t\right)=t+\frac{1}{t}\left(t\ge\frac{3}{2}\right)\)
lấy \(\frac{3}{2}< t_1< t_2\)
\(\Rightarrow f\left(t_1\right)-f\left(t_2\right)=\left(t_1-t_2\right)\left(\frac{t_1.t_2-1}{t_1.t_2}\right)< 0\)
suy ra với t càng tăng thì f(t) càng lớn vậy min \(f\left(t\right)=\frac{3}{2}+\frac{2}{3}=\frac{13}{6}\)
các em tự tìm x nhé.

bài này bạn áp dụng BĐT cô si cko 2 số dương là đc.

đáp án: Min A=  2

Phan Quỳnh Anh Cách của bạn không ổn đâu, với lại kết quả bạn chưa đúng ^^