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

Bài làm:

c) \(\left(x-2\right)\left(x+3\right)>0\)

Ta xét 2 trường hợp sau:

+ Nếu \(\hept{\begin{cases}x-2>0\\x+3>0\end{cases}\Rightarrow}\hept{\begin{cases}x>2\\x>-3\end{cases}\Rightarrow}x>2\)

+ Nếu \(\hept{\begin{cases}x-2< 0\\x+3< 0\end{cases}}\Rightarrow\hept{\begin{cases}x< 2\\x< -3\end{cases}}\Rightarrow x< -3\)

Vậy \(\orbr{\begin{cases}x>2\\x< -3\end{cases}}\)

d) \(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)

\(\Leftrightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)

\(\Leftrightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)

\(\Leftrightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)

\(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\)

\(\Leftrightarrow3x=\frac{1}{9}\)

\(\Leftrightarrow x=\frac{1}{27}\)

Vậy \(x=\frac{1}{27}\)

Học tốt!!!!

\(\left(3-x\right)^3=-\dfrac{27}{64}\)

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

\(=>3-x=\dfrac{-3}{4}\)

\(x=3-\dfrac{-3}{4}=\dfrac{12}{4}+\dfrac{3}{4}\)

\(x=\dfrac{15}{4}\)

________

\(\left(x-5\right)^3=\dfrac{1}{-27}\)

\(\left(x-5\right)^3=\left(\dfrac{-1}{3}\right)^3\)

\(=>x-5=\dfrac{-1}{3}\)

\(x=\dfrac{-1}{3}+5=\dfrac{-1}{3}+\dfrac{15}{3}\)

\(x=\dfrac{14}{3}\)

_____________

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{27}{8}\)

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

\(=>x-\dfrac{1}{2}=\dfrac{3}{2}\)

\(x=\dfrac{3}{2}+\dfrac{1}{2}\)

\(x=2\)

________

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

\(\left(2x-1\right)^2=\left(\dfrac{1}{2}\right)^2\)           hoặc              \(\left(2x-1\right)^2=\left(\dfrac{-1}{2}\right)^2\)

\(=>2x-1=\dfrac{1}{2}\)                                       \(2x-1=\dfrac{-1}{2}\)

\(2x=\dfrac{1}{2}+1=\dfrac{1}{2}+\dfrac{2}{2}\)                               \(2x=\dfrac{-1}{2}+1=\dfrac{-1}{2}+\dfrac{2}{2}\)

\(2x=\dfrac{3}{2}\)                                                     \(2x=\dfrac{1}{2}\)

\(x=\dfrac{3}{2}:2=\dfrac{3}{2}.\dfrac{1}{2}\)                                     \(x=\dfrac{1}{2}:2=\dfrac{1}{2}.\dfrac{1}{2}\)

\(x=\dfrac{3}{4}\)                                                       \(x=\dfrac{1}{4}\)

____________

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

\(\left(2-3x\right)^2=\left(\dfrac{3}{2}\right)^2\)                hoặc                  \(\left(2-3x\right)^2=\left(\dfrac{-3}{2}\right)^2\)

\(=>2-3x=\dfrac{3}{2}\)                                               \(2-3x=\dfrac{-3}{2}\)

\(3x=2-\dfrac{3}{2}=\dfrac{4}{2}-\dfrac{3}{2}\)                                      \(3x=2-\dfrac{-3}{2}=\dfrac{4}{2}+\dfrac{3}{2}\)

\(3x=\dfrac{1}{2}\)                                                            \(3x=\dfrac{7}{2}\)

\(x=\dfrac{1}{2}.\dfrac{1}{3}\)                                                          \(x=\dfrac{7}{2}.\dfrac{1}{3}\)

\(x=\dfrac{1}{6}\)                                                               \(x=\dfrac{7}{6}\)

______________

\(\left(1-\dfrac{2}{3}\right)^2=\dfrac{4}{9}\) -> Kiểm tra đề câu này

(3-x)^{3_{=(-\(\dfrac{3}{4}\))}3}

^{3-x=-\(\dfrac{3}{4}\)}

  ^{x=3-(-\(\dfrac{3}{4}\))}

  ^{x=\(\dfrac{15}{4}\)}

a)(x+3)³-x(3x+1)²+(2x+1)(4x²-2x+1-3x²)=54

\(\Rightarrow\)x³+9x²+27x+27-x(9x²+6x+1)+(2x+1)(x²-2x+1)=54

\(\Rightarrow\)x³+9x²+27x+27-9x³-6x²-x+2x³-4x²+2x+x²-2x+1=54

\(\Rightarrow\)-6x³+26x+28=54

\(\Rightarrow\)-6x³+26x=54-28

\(\Rightarrow\)-6x³+26x=26

\(\Rightarrow\)-6x³+26x-26=0

\(\Rightarrow\)-2(3x³+13x+14)

Bài 2: 

a: \(B=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{6}{3\left(x-2\right)}+\dfrac{1}{x-2}\right):\left(\dfrac{x^2-4+16-x^2}{x+2}\right)\)

\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)

\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)

\(=\dfrac{x-x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{12}=\dfrac{-1}{6\left(x-2\right)}\)

b: Thay x=1/2 vào B, ta được:

\(B=\dfrac{-1}{6\cdot\left(\dfrac{1}{2}-2\right)}=\dfrac{-1}{6\cdot\dfrac{-3}{2}}=\dfrac{1}{9}\)

Thay x=-1/2 vào B, ta được:

\(B=\dfrac{-1}{6\cdot\left(-\dfrac{1}{2}-2\right)}=-\dfrac{1}{15}\)

c: Để B=2 thì \(\dfrac{-1}{6\left(x-2\right)}=2\)

=>6(x-2)=-1/2

=>x-2=-1/12

hay x=23/12

  \(\left(\frac{x^2+3x}{x^3+3x^2+9x+27}\right)\): \(\left(\frac{1}{x-3}-\frac{6x}{x^3-3x^2+9x-27}\right)\)

=\(\left[\frac{x\left(x+3\right)}{x^2\left(x+3\right)+9\left(x+3\right)}\right]\):\(\left[\frac{1}{x-3}-\frac{6x}{x^2\left(x-3\right)+9\left(x-3\right)}\right]\)

=\(\left[\frac{x\left(x-3\right)}{\left(x^2+9\right)\left(x-3\right)}\right]\):\(\left[\frac{1}{x-3}-\frac{6x}{\left(x^2+9\right)\left(x-3\right)}\right]\)

=\(\frac{x}{x^2+9}\):\(\left[\frac{x^2+9}{\left(x-3\right)\left(x^2+9\right)}-\frac{6x}{\left(x-3\right)\left(x^2+9\right)}\right]\)

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

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

=\(\frac{x}{x^2+9}\).\(\frac{x^2+9}{x-3}\)

=\(\frac{x}{x-3}\)

= \(\left[\frac{x.\left(x+3\right)}{\left(x+3\right).\left(x^2+9\right)}+\frac{3}{x+9}\right]:\left[\frac{1}{x-3}-\frac{6x}{\left(x-3\right)\left(x^2+9\right)}\right]\) ]

 \(=\frac{x+3}{x^2-9}.\frac{\left(x-3\right).\left(x^2+9\right)}{x^2+9-6x}\)

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

= \(\frac{x+3}{x-3}\)

k mik nhé. Plssss~

a ) \(3x\left(x-1\right)-x\left(3x-2\right)=5\)

\(\Leftrightarrow3x^2-3x-3x^2+2x=5\)

\(\Leftrightarrow-x=5\)

\(\Leftrightarrow x=-5\)

Vậy phương trình có nghiệm x = - 5 .

a, \(3x\left(x-1\right)-x\left(3x-2\right)=5\)

\(\Rightarrow3x^2-3x-\left(3x^2-2x\right)=5\)

\(\Rightarrow3x^2-3x-3x^2+2x=5\)

\(\Rightarrow5x=5\Rightarrow x=1\)

Câu b,c làm tương tự! Cứ tách ra là làm được à!

b ) \(8x\left(2x+1\right)-4x\left(2x-3\right)=-40\)

\(\Leftrightarrow16x^2+8x-8x^2+12x=-40\)

\(\Leftrightarrow20x=-40\)

\(\Leftrightarrow x=-2\)

Vậy phương trình có nghiệm x = - 2