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!!!!

a) \(\frac{1}{9}=\frac{x}{27}\)

\(\Rightarrow x=\frac{1}{9}\cdot27\)

\(\Rightarrow x=3\)

b) \(\frac{4}{x}=\frac{8}{6}\)

\(\Rightarrow x=4:\frac{8}{6}\)

\(\Rightarrow x=3\)

c) \(\frac{x}{3}-\frac{1}{2}=\frac{1}{5}\)

\(\Rightarrow\frac{x}{3}=\frac{1}{5}+\frac{1}{2}\)

\(\Rightarrow x=\frac{7}{10}\cdot3\)

\(\Rightarrow x=\frac{21}{10}=2,1\)

\(a,\frac{1}{9}\)=\(\frac{3}{27}\)

\(b,\frac{4}{3}\)=\(\frac{8}{6}\)

\(c,\frac{x}{3}\)-\(\frac{1}{2}=\frac{1}{5}\)

\(\frac{x}{3}=\frac{1}{5}+\frac{1}{2}\)

\(\frac{x}{3}=\frac{7}{10}\)

\(\)

làm luông di

đây nhá , chị nêu phương pháp cho 

\(\frac{A}{B}\)= \(\frac{C}{D}\) 

<=> A*D=B*C 

a: ĐKXĐ: x<>-3

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

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

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

c) \(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)+\frac{5}{9}=\frac{23}{27}\)

\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{23}{27}-\frac{5}{9}\)

\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{23}{27}-\frac{15}{27}\)

\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{8}{27}\)

\(2+\frac{3}{4}x=\frac{7}{9}\div\frac{8}{27}\)

\(2+\frac{3}{4}x=\frac{7}{9}.\frac{27}{8}\)

\(2+\frac{3}{4}x=\frac{21}{8}\)

\(\frac{3}{4}x=\frac{21}{8}-2\)

\(\frac{3}{4}x=\frac{21}{8}-\frac{16}{8}\)

\(\frac{3}{4}x=\frac{5}{8}\)

\(x=\frac{5}{8}\div\frac{3}{4}\)

\(x=\frac{5}{8}.\frac{4}{3}\)

\(x=\frac{5}{6}\)

Vậy \(x=\frac{5}{6}\).

d) \(\left|x-\frac{1}{3}\right|-\frac{3}{4}=\frac{5}{3}\)

\(\left|x-\frac{1}{3}\right|=\frac{5}{3}+\frac{3}{4}\)

\(\left|x-\frac{1}{3}\right|=\frac{20}{12}+\frac{9}{12}\)

\(\left|x-\frac{1}{3}\right|=\frac{29}{12}\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{29}{12}\\x-\frac{1}{3}=-\frac{29}{12}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{11}{4}\\x=-\frac{25}{12}\end{cases}}\)

Vậy \(x\in\left\{\frac{11}{4};-\frac{25}{12}\right\}\).

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

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

     \(\frac{x}{-6}=\frac{y}{4}=\frac{z}{-\frac{1}{2}}\)\(=\frac{4x+y-2z}{-6.4+4-\left(-\frac{1}{2}.2\right)}\)\(=\frac{27}{-24+4+1}=\frac{27}{-19}\)

        \(\frac{x}{-6}=\frac{-27}{19}\)\(=>x=\frac{\left(-27\right).\left(-6\right)}{19}=\frac{162}{19}\)

Ta có:

\(\frac{x}{-6}\)=\(\frac{4x}{-24}\)

\(\frac{z}{\frac{-1}{2}}\)=\(\frac{2z}{-1}\)

Theo tính chất của dãy tỉ số = nhau:

\(\frac{4x}{-24}\)= \(\frac{y}{4}\)=\(\frac{2z}{-1}\)= \(\frac{4x+y-2z}{-24+4-\left(-1\right)}\)= \(\frac{-27}{19}\)

\(\frac{x}{-6}\)= \(\frac{-27}{19}\)nên x= \(\frac{162}{19}\)= \(8\frac{10}{19}\)

\(\frac{y}{4}\)= \(\frac{-27}{19}\)nên y= \(\frac{-108}{19}\)=\(-5\frac{13}{19}\)

\(\frac{z}{-\frac{1}{2}}\)= \(\frac{-27}{19}\)nên z= \(\frac{27}{38}\)

\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)

\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)

\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)

\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)

\(\Leftrightarrow x=-2020\)

khó lắm

bây h thì bạn giải đc chưa

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