Đk: x \(\ne\)0; x \(\ne\)\(\pm\)3

Ta có: A = \(\left(\frac{1}{3}+\frac{3}{x^2-3x}\right):\left(\frac{x^2}{27-3x^2}+\frac{1}{x+3}\right)\)

A = \(\frac{x^2-3x+9}{3x\left(x-3\right)}:\frac{x^2+3\left(3-x\right)}{3\left(x+3\right)\left(3-x\right)}\)

A = \(\frac{x^2-3x+9}{3x\left(x-3\right)}\cdot\frac{3\left(3-x\right)\left(x+3\right)}{x^2-3x+9}\)

A = \(\frac{-\left(x+3\right)}{x}\)

Để A < -1 <=> \(-\frac{\left(x+3\right)}{x}< -1\) <=> \(\frac{-x-3}{x}+1< 0\)

<=> \(\frac{-x-3+x}{x}< 0\) <=> \(-\frac{3}{x}< 0\) 

Do -3 <0 => x> 0

Vậy Để A < -1 <=> x > 0 và x khác 3

a) Ta có : \(AC=\frac{3}{8}.CE\)

\(\Leftrightarrow AE-CE=\frac{3}{8}.CE\)

\(\Leftrightarrow\frac{AE-CE}{CE}=\frac{3}{8}\)

\(\Leftrightarrow8AE-8CE=3CE\)

\(\Leftrightarrow8AE=11CE\)

\(\Leftrightarrow\frac{AE}{CE}=\frac{11}{8}\)

mà \(\frac{AD}{BD}=\frac{11}{8}\)

\(\Rightarrow\frac{AE}{CE}=\frac{AD}{BD}\)

\(\Rightarrow BC//DE\)( định lý Ta lét đảo )

b) Xét \(\Delta DAE\)có BC // DE : theo hệ quả của định lý Ta lét ta có :

\(\frac{AD}{BD}=\frac{DE}{BC}\)mà \(\frac{AD}{BD}=\frac{11}{8}\)

\(\Rightarrow\frac{11}{8}=\frac{DE}{3}\)

\(\Rightarrow DE=\frac{3.11}{8}\)

\(\Rightarrow DE=\frac{33}{8}\left(cm\right)\)

ta có phương trình tương đương

\(4x^2-10x-6=0\)

\(\Leftrightarrow4x^2-2.2.\frac{5}{2}x+\frac{25}{4}=\frac{49}{4}\)

\(\Leftrightarrow\left(2x-\frac{5}{2}\right)^2=\left(\frac{7}{2}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}2x-\frac{5}{2}=\frac{7}{2}\\2x-\frac{5}{2}=-\frac{7}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-\frac{1}{2}\end{cases}}}\)

vậy phương trình có hai nghiệm như trên

-2x² + 5x + 3 = 0

<=> -2x² + 6x - x + 3 = 0

<=> -2x( x - 3 ) - ( x - 3 ) = 0

<=> ( x - 3 )( -2x - 1 ) = 0

<=> x - 3 = 0 hoặc -2x - 1 = 0

<=> x = 3 hoặc x = -1/2

Vậy tập nghiệm của phương trình là : S = { 3 ; -1/2 }

\(\frac{7-x}{2}+\frac{2}{3}\left(x-7\right)\left(x-3\right)=0\)

=> \(\left(x-7\right)\left(-\frac{1}{2}+\frac{2}{3}\left(x-3\right)\right)=0\)

=> \(\left(x-7\right)\left(-\frac{1}{2}+\frac{2}{3}x-2\right)=0\)

=> \(\left(x-7\right)\left(\frac{2}{3}x-\frac{5}{2}\right)=0\)

=> \(\orbr{\begin{cases}x-7=0\\\frac{2}{3}x-\frac{5}{2}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\x=\frac{15}{4}\end{cases}}\)

( 3x - 5 )( x + 2 ) = x² - 5x 

<=> 3x² + 6x - 5x - 10 - x² + 5x = 0

<=> 2x² + 6x - 10 = 0

Δ = b² - 4ac = 6² - 4.2.(-10) = 36 + 80 = 116

Δ > 0 nên phương trình có hai nghiệm phân biệt :

\(x_1=\frac{-b+\sqrt{\text{Δ}}}{2a}=\frac{-6+\sqrt{116}}{4}=\frac{-3+\sqrt{29}}{2}\)

\(x_2=\frac{-b-\sqrt{\text{Δ}}}{2a}=\frac{-6-\sqrt{116}}{4}=\frac{-3-\sqrt{29}}{2}\)

Vậy tập nghiệm của phương trình là : \(S=\left\{\frac{-3\pm\sqrt{29}}{2}\right\}\)

\(\left(3x-5\right)\left(x+2\right)=x^2-5x\)

\(\Leftrightarrow3x^2+6x-5x-10=x^2-5x\)

\(\Leftrightarrow3x^2-x^2+x+5x-10=0\)

\(\Leftrightarrow2x^2+6x-10=0\)

\(\Leftrightarrow2\left(x^2+3x-5\right)=0\Leftrightarrow x^2+3x+5=0\)giải delta ta được : 

\(x=\frac{-3\pm\sqrt{29}}{2}\)

ta có 

\(PT\Leftrightarrow\left(x-1\right)\left(2x+3\right)+2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x+3+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)

( x - 1 )( 2x + 3 ) + 2x = 2

<=> ( x - 1 )( 2x + 3 ) + 2x - 2 = 0

<=> ( x - 1 )( 2x + 3 ) + 2( x - 1 ) = 0

<=> ( x - 1 )( 2x + 3 + 2 ) = 0

<=> ( x - 1 )( 2x + 5 ) = 0

<=> x - 1 = 0 hoặc 2x + 5 = 0

<=> x = 1 hoặc x = -5/2

Vậy tập nghiệm của phương trình là : S = { 1 ; -5/2 }

1) a) \(\frac{x}{x+1}+\frac{x^3-2x^2}{x^3+1}=\frac{x}{x+1}+\frac{x^3-2x^2}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\frac{x\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{x^3-2x^2}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{x^3-x^2+x+x^3-2x^2}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\frac{2x^3-3x^2+x}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{x\left(x-1\right)\left(2x-1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)

b) \(\frac{x+1}{2x-2}+\frac{3}{x^2-1}+\frac{x+3}{2x+2}=\frac{x+1}{2\left(x-1\right)}+\frac{3}{\left(x-1\right)\left(x+1\right)}+\frac{x+3}{2\left(x+1\right)}\)

\(=\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}+\frac{6}{2\left(x-1\right)\left(x+1\right)}+\frac{\left(x+3\right)\left(x-1\right)}{2\left(x+1\right)\left(x-1\right)}\)

\(=\frac{\left(x+1\right)^2+6+\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}=\frac{x^2+2x+1+6+x^2+2x-3}{2\left(x-1\right)\left(x+1\right)}\)

\(=\frac{2x^2+4x+2}{2\left(x-1\right)\left(x+1\right)}=\frac{2\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}=\frac{x+1}{x-1}\)

2) Ta có A = \(\left(\frac{x^2+y^2}{x^2-y^2}-1\right).\frac{x-y}{4y}=\frac{2y^2}{x^2-y^2}.\frac{x-y}{4y}=\frac{2y^2\left(x-y\right)}{\left(x-y\right)\left(x+y\right).4y}=\frac{y}{2\left(x+y\right)}\)

Thay x = 14 ; y = -15 vào biểu thức ta được 

\(A=\frac{y}{2\left(x+y\right)}=\frac{-15}{2\left(14-15\right)}=\frac{-15}{-2}=7,5\)

Tìm GTNN??

Ta có: \(A=\frac{2x^2+2x+7}{x^2+x+1}=\frac{2\left(x^2+x+1\right)+5}{x^2+x+1}=2+\frac{5}{x^2+x+1}\)

(Vì \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\) )

\(\Rightarrow A=2+\frac{5}{x^2+x+1}\le2+\frac{5}{\frac{3}{4}}=\frac{26}{3}\)

Dấu "=" xảy ra khi: x = -1/2