a: \(=\dfrac{3b+4a}{6ab}\)

b: \(=\dfrac{x^2-2x+1-x^2-2x-1}{x^2-1}=\dfrac{-4x}{x^2-1}\)

c: \(=\dfrac{xz+yz-xy-xz}{xyz}=\dfrac{yz-xy}{xyz}=\dfrac{z-x}{xz}\)

d: \(=\dfrac{2x+6-12}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x-6}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x+3}\)

e: \(=\dfrac{x-2+2}{\left(x-2\right)^2}=\dfrac{x}{\left(x-2\right)^2}\)

`a, a/(a-3) - 3/(a+3) = (a(a+3) - 3(a-3))/(a^2-9)`

`= (a^2+9)/(a^2-9)`

`b, 1/(2x) + 2/x^2 = x/(2x^2) + 4/(2x^2) = (x+4)/(2x^2)`

`c, 4/(x^2-1) - 2/(x^2+x) = (4x)/(x(x-1)(x+1)) - (2(x-1))/(x(x+1)(x-1))`

`= (2x+2)/(x(x-1)(x+1)`

`= 2/(x(x-1))`

\(a,=\dfrac{4y.5x^3}{3x^2.2y^3}=\dfrac{20x^3y}{6x^2y^3}=\dfrac{10x}{3y^2}\\ b,=\dfrac{\left(x-1\right)^2.x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x-1\right)}=\dfrac{\left(x-1\right)^2.x.\left(x+1\right)}{\left(x-1\right)^2.\left(x+1\right)}=x\)

\(c,=\dfrac{x\left(2+x\right).3\left(x^3+1\right)}{\left(x^2-x+1\right).3.\left(x+2\right)}=\dfrac{3x.\left(x+2\right).\left(x+1\right)\left(x^2-x+1\right)}{\left(x^2-x+1\right).3\left(x+2\right)}=x\left(x+1\right)\)

a) Tại x=-1

\(\Rightarrow x^5-5=\left(-1\right)^5-5=-6\)

\(a,\)Thay \(x=-1\) vào \(x^5-5\)

\(\Rightarrow\left(-1\right)^5-5=-6\)

\(b,\) 

+ TH1:

Thay \(x=1\) vào \(x^2-3x-5\)

\(\Rightarrow1^2-3.1-5=-7\)

+TH2:

Thay \(x=-1\) vào \(x^2-3x-5\)

\(\Rightarrow\left(-1\right)^2-3.\left(-1\right)-5=-1\)

\(\left\{{}\begin{matrix}\dfrac{3x+1}{6xy^4}\\\dfrac{x^2-5}{4x^2y^3}\end{matrix}\right.\)

⇒ \(\left\{{}\begin{matrix}\dfrac{2x\left(3x+1\right)}{12x^2y^4}\\\dfrac{3y\left(x^2-5\right)}{12x^2y^4}\end{matrix}\right.\)

a) \(P\left(x\right)-x\left(x+5\right)-\left(2x-3\right)+x^2\left(3x-2\right)\)

\(P\left(x\right)=-x^2-5x-2x+3+3x^3-2x^2\)

\(P\left(x\right)=3x^3+\left(-x^2-2x^2\right)-\left(5x+2x\right)+3\)

\(P\left(x\right)=3x^3-3x^2-7x+3\)

b) \(Q\left(x\right)=2x\left(x+1\right)+3x\left(5-x\right)-7\left(x-5\right)\)

\(Q\left(x\right)=2x^2+2x+15x-3x^2-7x+35\)

\(Q\left(x\right)=-x^2+10x+35\)

a: P(x)=-x^2-5x-2x+3+3x^3-2x^2

=3x^3-3x^2-7x+3

b: Q(x)=2x^2+2x+15x-3x^2-7x+35

=-x^2+10x+35

a: Ta có: \(\left(x+5\right)^2-4x\left(2x+3\right)^2-\left(2x-1\right)\left(x+3\right)\left(x-3\right)\)

\(=x^2+10x+25-4x\left(4x^2+12x+9\right)-\left(2x-1\right)\left(x^2-9\right)\)

\(=x^2+10x+25-16x^3-48x^2-36x-2x^3+18x+x^2-9\)

\(=-18x^3-46x^2-8x+16\)

a: \(\left(x+1\right)^5\)

\(=x^5+C_5^1\cdot x^4\cdot1^1+C_5^2\cdot x^3\cdot1^2+C_5^3\cdot x^2\cdot1^3+C_5^4\cdot x\cdot1^4+1\)

\(=x^5+5x^4+10x^3+10x^2+5x+1\)

b: \(\left(x-2y\right)^6\)

\(=x^6-C_6^1\cdot x^5\cdot\left(2y\right)^1+C_6^2\cdot x^4\cdot\left(2y\right)^2-C_6^3\cdot x^3\cdot\left(2y\right)^3+C_6^4\cdot x^2\cdot\left(2y\right)^4-C_6^5\cdot x\cdot\left(2y\right)^5+\left(2y\right)^6\)

\(=x^6-12x^5y+60x^4y^2-160x^3y^3+960x^2y^4-192xy^5+64y^6\)

c: \(\left(x^2+\frac{1}{x}\right)^5\)

\(=\left(x^2\right)^5+C_5^1\cdot\left(x^2\right)^4\cdot\left(\frac{1}{x}\right)^1+C_5^2\cdot\left(x^2\right)^3\cdot\left(\frac{1}{x}\right)^2+C_5^3\cdot\left(x^2\right)^2\cdot\left(\frac{1}{x}\right)^3+C_5^4\cdot x^2\cdot\left(\frac{1}{x}\right)^4+\left(\frac{1}{x}\right)^5\)

\(=x^{10}+5x^7+10x^4+10x+\frac{5}{x^2}+\frac{1}{x^5}\)

d: \(\left(x^3-\frac{2}{x}\right)^6\)

\(=\left(x^3\right)^6-C_6^1\cdot\left(x^3\right)^5\cdot\left(\frac{2}{x}\right)^1+C_6^2\cdot\left(x^3\right)^4\cdot\left(\frac{2}{x}\right)^2-C_6^3\left(x^3\right)^3\cdot\left(\frac{2}{x}\right)^3+C_6^4\cdot\left(x^3\right)^2\cdot\left(\frac{2}{x}\right)^4-C_6^5\cdot\left(x^3\right)\cdot\left(\frac{2}{x}\right)^5+\left(\frac{2}{x}\right)^6\)

\(=x^{18}-12\cdot x^{14}+60x^{10}-160x^6+960x^2-\frac{192}{x^2}+\frac{64}{x^6}\)

\(a,2\left(x-1\right)\left(x+1\right)+\left(x-1\right)^2+\left(x+1\right)^2\)

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

\(=2x^2-2+2x^2+2=4x^2\)

\(b,\left(x-y+1\right)^2+\left(1-y\right)^2+2\left(x-y+1\right)\left(y-1\right)\)

\(=\left(x-y+1\right)^2+2\left(x-y+1\right)\left(y-1\right)+\left(y-1\right)^2\)

\(=\left[\left(x-y+1\right)+\left(y-1\right)\right]^2\)

\(=\left[x-y+1+y-1\right]^2=x^2\)

đề cuối phải sửa cái cuối thành \(\left(3x+5\right)^2\) 

\(c,\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)

\(=\left[\left(3x+1\right)-\left(3x+5\right)\right]^2=\left[3x+1-3x-5\right]^2=16\)