a: ĐKXĐ: x<>-3

b: \(Q=\left(\dfrac{x}{x^2-3x+9}-\dfrac{11}{\left(x+3\right)\left(x^2-3x+9\right)}+\dfrac{1}{x+3}\right)\cdot\dfrac{x+3}{x^2-1}\)

\(=\dfrac{x^2+3x-11+x^2-3x+9}{\left(x+3\right)\left(x^2-3x+9\right)}\cdot\dfrac{x+3}{x^2-1}\)

\(=\dfrac{2x^2-2}{x^2-1}\cdot\dfrac{1}{x^2-3x+9}=\dfrac{2}{x^2-3x+9}\)

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

Đặt \(\frac{x}{-3}=\frac{y}{5}=k\)

\(\Rightarrow x=-3k;y=-5k\left(1\right)\) và \(x.y=-\frac{5}{27}\left(2\right)\)

Thay ( 1 ) và ( 2 ) vào , ta có : \(-3k.5k=-\frac{5}{27}\)

                                                   \(\Rightarrow-15k^2=\frac{-5}{27}\)

                                                   \(\Rightarrow k^2=\frac{1}{81}\)

                                                    \(\Rightarrow k=\pm\frac{1}{9}\)

+ ) Nếu \(k=\frac{1}{9}\) thì \(x=-\frac{1}{3};y=\frac{5}{9}\)

+ ) Nếu \(k=-\frac{1}{9}\) thì \(x=\frac{1}{3};y=-\frac{5}{9}\)

Vậy ..............

\(a,A=\left(x+5\right)^3\)

\(b,B=\left(x-3\right)^3\)

\(c,C=\frac{x^3}{8}+\frac{x^2y}{4}+\frac{xy^2}{6}+\frac{y^3}{27}=\left(\frac{x}{2}+\frac{y}{3}\right)^3\)

Mk nghĩ đề bài phần c fải như trên ,cn đâu bn tự thay số vào nha.

p/s: Nhớ mãi cái hôm thi vio v19 Gặp câu này hong bt làm :((
lg: Đặt biểu thức= A
$<=> A^3 = 9 + 3\sqrt[3]{9-\frac{x}{27}}+A$
$<=> A(A^2- 3\sqrt[3]{9-\frac{x}{27}}) =9 = 1.9 = -1.-9 = -3.-3 = 3.3= -9.-1=9.1$
.... 

\(\Rightarrow\frac{x+1}{2014}+1+\frac{x+2}{2013}+1=\frac{x+3}{2012}+1+\frac{x+4}{2011}+1\)

\(\Rightarrow\frac{x+1+2014}{2014}+\frac{x+2+2013}{2013}=\frac{x+3+2012}{2012}+\frac{x+4+2011}{2011}\)

\(\Rightarrow\frac{x+2015}{2014}+\frac{x+2015}{2013}-\frac{x+2015}{2012}-\frac{x+2015}{2011}=0\)

\(\Rightarrow\left(x+2015\right)\left(\frac{1}{2014}+\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\right)=0\)

Vì \(\frac{1}{2014}+\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\ne0\)

=>x+2015=0

=>x=-2015

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!!!!

a) Theo t/c dãy tỉ số bằng nhau:
 \(\frac{x}{2}=\frac{y}{5}=\frac{x+y}{2+5}=\frac{27}{7}\)

+) \(\frac{x}{2}=\frac{27}{7}\)=> x= (27x2) : 7 =\(\frac{54}{7}\)

+) \(\frac{y}{5}=\frac{27}{7}\)=> y= (27x5) : 7 = \(\frac{135}{7}\)
Vậy x=\(\frac{54}{7}\); y=\(\frac{135}{7}\)

b) Tương tự câu a
\(\frac{x}{3}=\frac{y}{6}=\frac{x+y}{3+6}=\frac{27}{9}=3\)

+) \(\frac{x}{3}=3\)=> x= 3x3 = 9

+) \(\frac{y}{6}=3\)=> y= 3x6 = 18

Vậy x= 9 ; y= 18


 

a, Đặt  : \(\frac{x}{2}=\frac{y}{5}=k\)\(< =>\hept{\begin{cases}x=2k\\y=5k\end{cases}}\)

Ta có : \(x+y=27< =>2k+5k=27< =>7k=27\)

\(< =>k=\frac{27}{7}\)

Suy ra \(x=2k=\frac{54}{7};y=5k=\frac{135}{7}\)