\(\dfrac{8^{11}\cdot3^{17}}{27^{10}\cdot9^{15}}=\dfrac{2^{33}\cdot3^{17}}{3^{30}\cdot3^{30}}=\dfrac{2^{33}\cdot3^{17}}{3^{60}}=\dfrac{2^{33}}{3^{43}}\)

\(\dfrac{8^{11}.3^{17}}{27^{10}.9^{15}}\\ =\dfrac{\left(2^3\right)^{11}.3^{17}}{\left(3^3\right)^{10}.\left(3^2\right)^{15}}\\ =\dfrac{2^{33}.3^{17}}{3^{30}.3^{30}}\\ =\dfrac{2^{33}.3^{17}}{3^{60}}\\ =\dfrac{2^{33}}{3^{43}}\)

a: \(-\dfrac{2}{3}\cdot x=\dfrac{4}{15}\)

=>\(x=\dfrac{4}{15}:\dfrac{-2}{3}=\dfrac{4}{15}\cdot\dfrac{-3}{2}=\dfrac{-12}{30}=-\dfrac{2}{5}\)

b: \(-\dfrac{7}{19}\cdot x=\dfrac{-13}{24}\)

=>\(x=\dfrac{13}{24}:\dfrac{7}{19}=\dfrac{13}{24}\cdot\dfrac{19}{7}=\dfrac{247}{168}\)

\(-\dfrac{2}{3}x=\dfrac{4}{15}\)

<=> \(x=\dfrac{4}{15}:\left(-\dfrac{2}{3}\right)\)

<=> \(x=\dfrac{4}{15}.\left(-\dfrac{3}{2}\right)\)

<=> \(x=-\dfrac{2}{5}\)

\(-\dfrac{7}{19}.x=-\dfrac{13}{24}\)

=> \(x=\left(-\dfrac{13}{24}\right):\left(-\dfrac{7}{19}\right)\)

=> \(x=\dfrac{13}{24}.\dfrac{19}{7}\)

=> \(x=\dfrac{247}{168}\)

`A(x) + B(x) = 6x^4 - 3x^2 - 5`

`A(x) - B(x) = 4x^4 - 6x^3 + 7x^2 + 8x - 9`

Áp dụng bài toán tổng hiệu ta có: 

`A(x) = [(6x^4 - 3x^2 - 5) + (4x^4 - 6x^3 + 7x^2 + 8x - 9)] : 2`

`= (6x^4 - 3x^2 - 5 + 4x^4 - 6x^3 + 7x^2 + 8x - 9) : 2`

`= (10x^4 - 6x^3 + 4x^2 + 8x - 14) : 2`

`= 5x^4 - 3x^3 + 2x^2 + 4x - 7`

`B(x) = (6x^4 - 3x^2 - 5) - (5x^4 - 3x^3 + 2x^2 + 4x - 7)`

`= 6x^4 - 3x^2 - 5 - 5x^4 + 3x^3 - 2x^2 - 4x + 7`

`= x^4 + 3x^3 - 5x^2 - 4x + 2`

Vậy ....

\(2A\left(x\right)=\left(6x^4-3x^2-5\right)+\left(4x^4-6x^3+7x^2+8x-9\right)\\ =\left(6x^4+4x^4\right)-6x^3+\left(-3x^2+7x^2\right)+8x+\left(-5-9\right)\\ =10x^4-6x^3+4x^2+8x-14\\ =>A\left(x\right)=5x^4-3x^3+2x^2+4x-7\)

\(=>B\left(x\right)=\left(6x^4-3x^2-5\right)-A\left(x\right)\\ =\left(6x^4-3x^2-5\right)-\left(5x^4-3x^3+2x^2+4x-7\right)\\ =\left(6x^4-5x^4\right)+3x^3+\left(-3x^2-2x^2\right)-4x+\left(-5+7\right)\\ =x^4+3x^3-5x^2-4x+2\)

\(\left(3x-2\right)^{2004}=\left(3x-2\right)^{2006}\)

=>\(\left(3x-2\right)^{2006}-\left(3x-2\right)^{2004}=0\)

=>\(\left(3x-2\right)^{2004}\left[\left(3x-2\right)^2-1\right]=0\)

=>\(\left(3x-2\right)^{2004}\cdot\left(3x-3\right)\left(3x-1\right)=0\)

=>\(\left[{}\begin{matrix}3x-2=0\\3x-3=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(\left(3x-2\right)^{2004}=\left(3x-2\right)^{2006}\\ =>\left(3x-2\right)^{2006}-\left(3x-2\right)^{2004}=0\\ =>\left(3x-2\right)^{2004}\left[\left(3x-2\right)^2-1\right]=0\)

+) \(\left(3x-2\right)^{2004}=0=>3x-2=0=>x=\dfrac{2}{3}\)

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

\(TH1:3x-2=1=>3x=1+2=3=>x=\dfrac{3}{3}=1\\ TH2:3x-2=-1=>3x=-1+2=1=>x=\dfrac{1}{3}\)

Xét ΔABC có \(\widehat{ABC}+\widehat{ACB}+\widehat{BAC}=180^0\)

=>\(2\cdot\left(\widehat{IBC}+\widehat{ICB}\right)=180^0-\widehat{BAC}\)

=>\(\widehat{IBC}+\widehat{ICB}=90^0-\dfrac{1}{2}\cdot\widehat{BAC}\)

Xét ΔBIC có \(\widehat{BIC}+\widehat{IBC}+\widehat{ICB}=180^0\)

=>\(\widehat{BIC}+90^0-\dfrac{1}{2}\widehat{BAC}=180^0\)

=>\(\widehat{BIC}=180^0-90^0+\dfrac{1}{2}\cdot\widehat{BAC}=90^0+\dfrac{1}{2}\cdot\widehat{BAC}\)

A và B cùng dấu nên AB>0

=>\(2x^3\cdot\left(-3\right)x^4>0\)

=>\(x^7< 0\)

=>x<0

có nhân 2 với -3 k ạ

\(-\dfrac{2}{5}+\dfrac{5}{6}x=-\dfrac{4}{15}\)

=>\(\dfrac{5}{6}x=-\dfrac{4}{15}+\dfrac{2}{5}=\dfrac{2}{15}\)

=>\(x=\dfrac{2}{15}:\dfrac{5}{6}=\dfrac{2}{15}\cdot\dfrac{6}{5}=\dfrac{12}{75}=\dfrac{4}{25}\)

\(\left(x+\dfrac{5}{3}\right)\left(x-\dfrac{5}{4}\right)=0\)

<=> \(\left[{}\begin{matrix}x+\dfrac{5}{3}=0\\x-\dfrac{5}{4}=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=\dfrac{5}{4}\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=\dfrac{5}{4}\end{matrix}\right.\)

\(\left(x+\dfrac{5}{3}\right)\left(x-\dfrac{5}{4}\right)=0\\ TH1:x+\dfrac{5}{3}=0\\ =>x=\dfrac{-5}{3}\\ TH2:x-\dfrac{5}{4}=0\\ =>x=\dfrac{5}{4}\)

Vậy: ...

\(\left(\dfrac{3}{4}x-\dfrac{9}{16}\right)\left(1.5+\dfrac{-3}{5}:x\right)=0\left(x\ne0\right)\\ TH1:\dfrac{3}{4}x-\dfrac{9}{16}=0\\ =>\dfrac{3}{4}x=\dfrac{9}{16}\\ =>x=\dfrac{9}{16}:\dfrac{3}{4}=\dfrac{3}{4}\left(tm\right)\\ TH2:1,5+\dfrac{-3}{5}:x=0\\ =>\dfrac{3}{5}:x=\dfrac{3}{2}\\ =>x=\dfrac{3}{5}:\dfrac{3}{2}=\dfrac{2}{5}\left(tm\right)\)

\(x=\dfrac{3}{4}\) và \(x=\dfrac{2}{5}\)