Ta có: \(x^2-3x+5=\left(x^2-2.x.\frac{3}{2}+\frac{9}{4}\right)+\frac{11}{4}\)

                                    \(=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}=0\)

Vì \(\left(x-\frac{3}{2}\right)^2\ge0\forall x\)\(\Rightarrow\)\(\left(x-\frac{3}{2}\right)^2+\frac{11}{4}>0\forall x\)

mà \(\left(x-\frac{3}{2}\right)^2+\frac{11}{4}=0\)\(\Rightarrow\)Phương trình trên vô nghiệm

Vậy.......

Ta có : \(x^2-3x+5=0\)

\(\Leftrightarrow\left(x^2-3x+\frac{9}{4}\right)+\frac{11}{4}=0\)

\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2+\frac{11}{4}=0\)

\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2=-\frac{11}{4}\)(Vô lí , do \(\left(x-\frac{3}{2}\right)^2\ge0\forall x\))

Vậy PT vô nghiệm

<=>x-45/55 -1 + x-47/53  -1=x-55/45 -1 + x-53/47-1

<=>x-100/55 + x-100/53 = x-100/45 + x-100/47

<=>(x-100)(1/55 + 1/53 - 1/45 - 1/47 )=0

vi (1/55 + 1/53 - 1/45 - 1/47 ) luon khac 0 nen x-100=0 <=>x=100

x³ - 7x² = 3x² - 12x

<=> x³ - 10x² + 12x = 0

<=> x(x² - 10x + 12) = 0

<=> x(x² - 10x + 25 - 13) = 0

<=> x[(x - 5)² - 13] = 0

<=> \(x\left(x-5+\sqrt{13}\right)\left(x-5-\sqrt{13}\right)=0\)

<=> x = 0 hoặc x = \(5-\sqrt{13}\)hoặc x = \(5+\sqrt{13}\)

Vậy \(x\in\left\{0;5-\sqrt{13};5+\sqrt{13}\right\}\)là nghiệm phương trình

haha Felix kìa :))

pt <=> \(\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)

<=> \(\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x+2009}{2003}=0\)

<=> \(\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)

Vì \(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\ne0\)

=> x + 2009 = 0 <=> x = -2009

Vậy phương trình có nghiệm x = -2009

Ta có : \(\frac{1}{x}+\frac{1}{y}=\frac{1}{4}\)(x;y \(\ne\)0)

<=> \(\frac{x+y}{xy}=\frac{1}{4}\)

<=> 4(x + y) = xy

<=> xy - 4x - 4y =0

<=> x(y - 4) - 4y + 16 = 16

<=> x(y - 4) - 4(y - 4) = 16

<=> (x - 4)(y - 4) = 16

Ta có 16 = 1.16 = 4.4 = (-4).(-4) = (-1).(-16)

Lập bảng xét các trường hợp

Vây các cặp (x;y) thỏa mãn là (5;20) ; (20;5) ; (8;8)

Áp dụng BĐT Bunyakovsky dạng phân thức ta có:
\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{a^2}{ab+ac}+\frac{b^2}{bc+ba}+\frac{c^2}{ca+cb}\)

\(\ge\frac{\left(a+b+c\right)^2}{ab+bc+bc+ca+ca+ab}=\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\)

Mà \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\)

\(\Rightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{\left(a+b+c\right)^2}{\frac{2\left(a+b+c\right)^2}{3}}=\frac{3}{2}\)

Dấu "=" xảy ra khi: a = b = c

Ta có x² + 3y² = 4xy

=> x² - 4xy + 3y² = 0

=> x² - xy - 3xy + 3y² = 0

<=> x(x - y) - 3y(x - y) = 0

<=> (x - 3y)(x - y) = 0

<=> \(\orbr{\begin{cases}x-y=0\\x-3y=0\end{cases}}\)

Ta có x - y > 0 (vì x > y > 0) => x - y = 0 loại

Ta có : x - 3y = 3x - 3y - 2y = 3(x - y) - 2y \(\le\) 0 (vì x - y > 0 ; y > 0) 

=> x - 3y = 0 tm

Khi đó x = 3y

Với x = 3y => A = \(\frac{2x+5y}{x-2y}=\frac{2.3y+5y}{3y-2y}=\frac{11y}{y}=11\)

a) Ta có: \(\left(8x+7\right)^2\left(4x+3\right)\left(x+1\right)=\frac{7}{2}\)

\(\Leftrightarrow\left(8x+7\right)^2\cdot2\left(4x+3\right)\cdot8\left(x+1\right)=16\cdot\frac{7}{2}\)

\(\Leftrightarrow\left(8x+7\right)^2\left(8x+6\right)\left(8x+8\right)=56\)

Đặt \(8x+7=a\) khi đó:

\(a^2\left(a-1\right)\left(a+1\right)=56\)

\(\Leftrightarrow a^2\left(a^2-1\right)=56\)

\(\Leftrightarrow a^4-a^2-56=0\)

\(\Leftrightarrow\left(a^2-8\right)\left(a^2+7\right)=0\)

\(\Leftrightarrow a^2-8=0\Leftrightarrow\left(8x+7\right)^2-8=0\)

\(\Leftrightarrow\left(8x+7\right)^2=8\Leftrightarrow\orbr{\begin{cases}8x+7=2\sqrt{2}\\8x+7=-2\sqrt{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}8x=2\sqrt{2}-7\\8x=-2\sqrt{2}-7\end{cases}}\Rightarrow x=\frac{\pm2\sqrt{2}-7}{8}\)

b) Ta có: \(x^2+5y^2-4xy+10x-22y+\left|x+y+z\right|+26=0\)

\(\Leftrightarrow\left(x^2-4xy+4y^2\right)+\left(10x-20y\right)+25+y^2-2y+1+\left|x+y+z\right|=0\)

\(\Leftrightarrow\left(x-2y\right)^2+10\left(x-2y\right)+25+\left(y-1\right)^2+\left|x+y+z\right|=0\)

\(\Leftrightarrow\left(x-2y+5\right)^2+\left(y-1\right)^2+\left|x+y+z\right|=0\)

\(\Rightarrow\hept{\begin{cases}\left(x-2y+5\right)^2=0\\\left(y-1\right)^2=0\\\left|x+y+z\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\\z=2\end{cases}}\)

Vậy x = -3 , y = 1 , z = 2