1) \(B\left(x\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(x^2-\frac{9}{16}\right)\left(x^2+3\right)=0\)

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

Mà \(x^2+3>0\left(\forall x\right)\)

=> 2x-5=0 hoặc x-3/4=0 hoặc x+3/4=0

=> x=5/2 hoặc x=3/4 hoặc x=-3/4

Vậy \(x\in\left\{-\frac{3}{4};\frac{3}{4};\frac{5}{2}\right\}\)

2) \(K\left(x\right)=0\)

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

\(\Leftrightarrow\left(x+2\right)\left(2x-5\right)=0\)

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

\(B\left(x\right)=\left(2x-5\right)\left(x^2-\frac{9}{16}\right)\left(x^2+3\right)=0\)

\(\Leftrightarrow x=\frac{5}{2};x=\frac{3}{4};x=-\frac{3}{4}\)

\(K\left(x\right)=2x^2-x-10=0\Leftrightarrow\left(x+2\right)\left(2x-5\right)=0\Leftrightarrow\orbr{\begin{cases}x=-2\\x=\frac{5}{2}\end{cases}}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{19.20}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\)

\(=1-\frac{1}{20}\)

\(=\frac{19}{20}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\) 

\(=\frac{1}{1}-\frac{1}{20}\)             

\(=\frac{19}{20}\)            

a) \(K=\frac{2^{11}.9^2}{3^5.16^2}=\frac{2^{11}.3^4}{2^8.3^5}=\frac{2^3}{3}=\frac{8}{3}\)

b) \(N=\frac{9^3.27^2}{6^2.3^{10}}=\frac{3^6.3^6}{3^2.2^2.3^{10}}=\frac{1}{2^2}=\frac{1}{4}\)

c) \(P=\frac{27^{15}.5^3.8^4}{25^2.81^{11}.2^{11}}=\frac{3^{45}.5^3.2^{12}}{5^4.3^{44}.2^{11}}=\frac{2.3}{5}=\frac{6}{5}\)

a) \(\left(\frac{3}{5}-1\frac{3}{4}\right)\div2\frac{3}{10}=\left(\frac{3}{5}-\frac{7}{4}\right)\div\frac{23}{10}=\frac{-23}{20}.\frac{10}{23}=-\frac{1}{2}\)

b) \(\left(\frac{-2}{3}+\frac{3}{4}\right)^2.\frac{12}{5}-\frac{11}{5}=\left(\frac{1}{12}\right)^2.\frac{12}{5}-\frac{11}{5}=\frac{1}{60}-\frac{11}{5}=-\frac{33}{15}\)

c) \(\frac{-7}{8}\div\frac{21}{16}-\frac{5}{3}\left(\frac{1}{3}-\frac{7}{10}\right)=\frac{-7}{8}\times\frac{16}{21}-\frac{5}{3}.-\frac{11}{30}=-\frac{2}{3}-\frac{11}{18}=-\frac{23}{18}\)

\(\frac{x-1}{13}+\frac{x-2}{12}+\frac{x-3}{11}-3=0\)

\(\Leftrightarrow\left(\frac{x-1}{13}-1\right)+\left(\frac{x-2}{12}-1\right)+\left(\frac{x-3}{11}-1\right)\)

\(\Leftrightarrow\frac{x-14}{13}+\frac{x-14}{12}+\frac{x-14}{11}=0\)

\(\Leftrightarrow\left(x-14\right)\left(\frac{1}{13}+\frac{1}{12}+\frac{1}{11}\right)=0\). Vì \(\left(\frac{1}{13}+\frac{1}{12}+\frac{1}{11}\right)>0\)

\(\Leftrightarrow x-14=0\Rightarrow x=0+14=14\). Vậy \(x=14\)

Bài làm:

Ta có: \(\frac{x-1}{13}+\frac{x-2}{12}+\frac{x-3}{11}-3=0\)

\(\Leftrightarrow\left(\frac{x-1}{13}-1\right)+\left(\frac{x-2}{12}-1\right)+\left(\frac{x-3}{11}-1\right)=0\)

\(\Leftrightarrow\frac{x-14}{13}+\frac{x-14}{12}+\frac{x-14}{11}=0\)

\(\Leftrightarrow\left(x-14\right)\left(\frac{1}{13}+\frac{1}{12}+\frac{1}{11}\right)=0\)

\(\Leftrightarrow x-14=0\)

\(\Rightarrow x=14\)

Ta có: \(\frac{x}{3}=\frac{y}{7}\)\(\Rightarrow\)\(x=\frac{3y}{7}\)

Ta lại có: \(\frac{2}{x}=\frac{y}{-3}\)

         \(\Leftrightarrow xy=-6\)

         \(\Leftrightarrow\frac{3y}{7}.y=-6\)

         \(\Leftrightarrow y^2=-6.\frac{7}{3}\)

         \(\Leftrightarrow y^2=-14\)

Vì \(\hept{\begin{cases}y^2\ge0\forall y\\-14< 0\end{cases}}\)mà \(y^2=-14\)

      \(\Rightarrow\)\(y\in\varnothing\)\(\Rightarrow\)\(x\in\varnothing\)

Vậy x,y không có giá trị 

Bài 2 : 

a, \(P\left(x\right)=2x^5+2-6x^2-3x^3+4x^2-2x+x^3+4x^5=6x^5-2x^3-2x^2+2\)

b, sắp xếp rồi, trên ý 

c, Bậc : 5 

Bài 3 : \(Q\left(x\right)=3x-5=0\Leftrightarrow x=\frac{5}{3}\)

Bài 1 : 

a, \(2xy^2\left(-5x^2y^3\right)=-10x^3y^5\)

b, \(\left(-2x^2yz\right)+\left(-5x^2yz\right)=-7x^2yz\)

Vì\(\frac{121212}{242424}=\frac{12.20202}{24.20202}=\frac{1}{2}\)nên\(\frac{121212}{242424}=\frac{1}{2}\).

Ta có : \(\frac{121212}{242424}\)= \(\frac{1.121212}{2.121212}\)=\(\frac{1}{2}\)

2009 - | x - 2009 | = x 

=>| x - 2009 | = 2009 - x 

=> x = 2009 

\(|x-2009|=2009-x\)         ( 1 ) 

\(ĐK:2009-x\ge0\)                           

\(-x\ge0-2009\)     

\(-x\ge-2009\)    

\(x\le2009\)     

( 1 ) \(\Leftrightarrow\orbr{\begin{cases}x-2009=2009-x\\x-2009=-\left(2009-x\right)\end{cases}}\)   

\(\orbr{\begin{cases}x+x=2009+2009\\x-2009=-2009+x\end{cases}}\)    

\(\orbr{\begin{cases}2x=4018\\x-x=-2009+2009\end{cases}}\)  

\(\orbr{\begin{cases}x=2009\\0x=0\left(llđ\forall x\right)\end{cases}}\)  

Vậy \(x\le2009\) là nghiệm của phương trình