\(P=\dfrac{x^3+8y^3}{4^3+4^3}=\dfrac{\left(x+2y\right)^3-3\cdot x\cdot2y\cdot\left(x+2y\right)}{128}\)

\(=\dfrac{\left(-8\right)^3-6\cdot\left(-6\right)\cdot\left(-8\right)}{128}=\dfrac{128-6\cdot48}{128}=-\dfrac{5}{4}\)

a: \(M=\dfrac{5x+5-8-4x+4}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x-1}\)

b: \(N=\dfrac{5x+5-8-4x+4}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x-1}\)

1. \(M=\dfrac{5}{x-1}-\dfrac{8}{x^2-1}-\dfrac{4}{x+1}\left(x\ne\pm1\right).\)

\(M=\dfrac{5\left(x+1\right)-8-4\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)\(M=\dfrac{5x+5-8-4x+4}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}\)

\(M=\dfrac{1}{x-1}.\)

2. \(N=\dfrac{5}{x-1}+\dfrac{8}{1-x^2}-\dfrac{4}{x+1}\left(x\ne\pm1\right).\)

\(N=\dfrac{5\left(x+1\right)-8-4\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{5x+5-8-4x+4}{\left(x-1\right)\left(x+1\right)}\)

\(N=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x-1}.\)

3. \(Q=\dfrac{1}{2x-1}-\dfrac{4}{4x^2-1}-\dfrac{2}{2x+1}\left(x\ne\pm\dfrac{1}{2}\right).\)

\(Q=\dfrac{2x+1-4-2\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x-3-4x+2}{\left(2x-1\right)\left(2x+1\right)}\)

\(Q=\dfrac{-2x-1}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{-1}{2x-1}.\)

4. \(F=\dfrac{x+3}{x-2}+\dfrac{x+2}{3-x}+\dfrac{x+2}{x^2-5x+6}\left(x\ne2,x\ne3\right).\)

\(F=\dfrac{x+3}{x-2}-\dfrac{x+2}{x-3}+\dfrac{x+2}{\left(x-3\right)\left(x-2\right)}\)

\(F=\dfrac{\left(x+3\right)\left(x-3\right)-\left(x+2\right)\left(x-2\right)+x+2}{\left(x-2\right)\left(x-3\right)}\)

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

\(F=\dfrac{1}{x-2}.\)

a: Ta có: \(8x\left(x-2017\right)-2x+4034=0\)

\(\Leftrightarrow\left(x-2017\right)\left(4x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)

c: Ta có: \(4-x=2\left(x-4\right)^2\)

\(\Leftrightarrow2\left(x-4\right)^2+x-4=0\)

\(\Leftrightarrow\left(x-4\right)\left(2x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{7}{2}\end{matrix}\right.\)

\(a,=x^2+x+4x+4=\left(x+1\right)\left(x+4\right)\\ b,=x^2+2x-3x-6=\left(x-3\right)\left(x+2\right)\\ c,=x^2-2x-3x+6=\left(x-2\right)\left(x-3\right)\\ d,=3\left(x^2-2x+5x-10\right)=3\left(x-2\right)\left(x+5\right)\\ e,=-3x^2+6x-x+2=\left(x-2\right)\left(1-3x\right)\\ f,=x^2-x-6x+6=\left(x-1\right)\left(x-6\right)\\ h,=4\left(x^2-3x-6x+18\right)=4\left(x-3\right)\left(x-6\right)\\ i,=3\left(3x^2-3x-8x+5\right)=3\left(x-1\right)\left(3x-8\right)\\ k,=-\left(2x^2+x+4x+2\right)=-\left(2x+1\right)\left(x+2\right)\\ l,=x^2-2xy-5xy+10y^2=\left(x-2y\right)\left(x-5y\right)\\ m,=x^2-xy-2xy+2y^2=\left(x-y\right)\left(x-2y\right)\\ n,=x^2+xy-3xy-3y^2=\left(x+y\right)\left(x-3y\right)\)

Bào quan riboxom trong chất tế bào có chức năng gì? 

ĐKXĐ: \(x\notin\left\{0;-9\right\}\)

Ta có: \(\dfrac{1}{x+9}-\dfrac{1}{x}=\dfrac{1}{5}+\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{20x}{20x\left(x+9\right)}-\dfrac{20\left(x+9\right)}{20x\left(x+9\right)}=\dfrac{4x\left(x+9\right)+5x\left(x+9\right)}{20x\left(x+9\right)}\)

Suy ra: \(4x^2+36x+5x^2+45x=20x-20x-180\)

\(\Leftrightarrow9x^2+81x+180=0\)

\(\Leftrightarrow x^2+9x+20=0\)

\(\Leftrightarrow x^2+4x+5x+20=0\)

\(\Leftrightarrow x\left(x+4\right)+5\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\left(nhận\right)\\x=-5\left(nhận\right)\end{matrix}\right.\)

Vậy: S={-4;-5}

a. (3x - 1)² - 16

= (3x - 1)² - 4²

= (3x - 1 - 4)(3x - 1 + 4)

= (3x - 5)(3x + 3)

= 3(3x - 5)(x + 1)

b. (5x - 4)² - 49x²

= (5x - 4)² - (7x)²

= (5x - 4 - 7x)(5x - 4 + 7x)

= \(\left(-2x-4\right)\left(2x-4\right)\)

= 2(-x - 2) . 2(x - 2)

= 4(-x - 2)(x - 2)

c. (2x + 5)² - (x - 9)²

= (2x + 5 - x + 9)(2x + 5 + x - 9)

= (x + 14)(3x - 4)

2. 

a)(3x-1)²-16 =  (3x-1)²-4²

                        =(3x-1-4).(3x-1+4)= (3x-5).(3x+3)

b)(5x-4)²-49x²  =(5x-4)²-(7x)²

                        =  (5x-4-7x).(5x-4+7x) = (-2x-4).(12x-4)

c)(2x+5)²-(x-9)²  =(2x+5-x+9).(2x+5+x-9)  =(x+14).(3x-4)

d)(3x+1)²-4.(x-2)² =(3x+1)²-[2.(x-2)]²  = [3x+1-2.(x-2)].[3x+1+2.(x-2)]

                                                            = (3x+1-2x+4).(3x+1+2x-4)

                                                           = (x+5).(5x-3)

e)9.(2x+3)²-4.(x+1)²  =[3.(2x+3)]²-[2.(x+1)]² 

 =[3.(2x+3)-2.(x+1)].[3.(2x+3)+2.(x+1)]

=(6x+9-2x-2).(6x+9+2x+2)

=(4x+7).(8x+11)