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

\(\dfrac{5x^3+10x^2+5x}{x^3+3x^2+3x+1}=\dfrac{5x\left(x+1\right)^2}{\left(x+1\right)^3}=\dfrac{5x}{x+1}\)

bài 1/ 

a) ta có: \(A=\frac{15}{x-1}\)

Để A là phân số \(\Rightarrow x-1\ne0\)

                          \(\Rightarrow x\ne1\)

b) Nếu x = 7

\(\Rightarrow A=\frac{15}{7-1}\)

\(\Rightarrow A=\frac{15}{6}\)

Nếu x = -3

\(\Rightarrow A=\frac{15}{-3-1}\)

\(\Rightarrow A=\frac{15}{-4}\)

Nếu x = 4

\(\Rightarrow A=\frac{15}{4-1}\)

\(\Rightarrow A=\frac{15}{3}=5\)

c) Ta có: \(B=5\)

\(\Leftrightarrow A=\frac{15}{x-1}=5\)

\(\Leftrightarrow x-1=3\)

\(\Leftrightarrow x=4\)

Bài 2/

a) \(\frac{x}{3}=\frac{2}{6}\)

\(\Leftrightarrow6x=6\)

\(\Leftrightarrow x=1\)

b) \(-\frac{x}{14}=\frac{10}{-7}\)

\(\Leftrightarrow7x=140\)

\(\Leftrightarrow x=20\)

hok tốt!!

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

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

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

\(\Leftrightarrow x=-2\)

\(4x^2-12x+5=4x^2-10x-2x+5=2x\left(2x-5\right)-\left(2x-5\right)=\left(2x-1\right)\left(2x-5\right)\)

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

\(\left(x+2\right)\left(-2\right)=0\)\(\Rightarrow x+2=0\) hay \(x=-2\)

\(4x^2-12x+5=4\left(x^2-3x+\frac{5}{4}\right)=4\left(x^2-3x+\frac{9}{4}-\frac{9}{4}+\frac{5}{4}\right)=4\left(x-\frac{3}{2}\right)^2-2=\left(2x-1\right)\left(2x-5\right)\)

= xy(x + y) + yz(y + z) + xyz + xz(x + z) + xyz 

= xy(x + y) + yz(y + z + x) + xz(x + z + y) 

= xy(x + y) + z(x + y + z)(y + x) 

= (x + y)(xy + zx + zy + z²) 

= (x + y)[x(y + z) + z(y + z)] 

= (x + y)(y + z)(z + x)

Bài nào vậy bạn?

\(a,ĐKXĐ\hept{\begin{cases}x-3\ne0\\x+3\ne0\end{cases}\Leftrightarrow x\ne\pm3}\)

Ta có: \(M=\frac{3}{x-3}-\frac{6x}{9-x^2}+\frac{x}{x+3}\)

            \(=\frac{3}{x-3}+\frac{6x}{x^2-9}+\frac{x}{x+3}\)

           \(=\frac{3\left(x+3\right)+6x+x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)

            \(=\frac{3x+9+6x+x^2-3x}{\left(x-3\right)\left(x+3\right)}\)

             \(=\frac{x^2+6x+9}{\left(x-3\right)\left(x+3\right)}\)

              \(=\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}\)

              \(=\frac{x+3}{x-3}\)

\(b,x=\frac{1}{2}\Rightarrow M=\frac{\frac{1}{2}+3}{\frac{1}{2}-3}=-\frac{7}{5}\)

a, \(ĐKXĐ:x^3+8\ne0\Leftrightarrow x\ne-2\)

b, \(C=\frac{2x^2-4x+8}{x^3+8}=\frac{2\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=\frac{2}{x+2}\)

c, \(\left|2x+1\right|=3\Rightarrow\orbr{\begin{cases}2x+1=3\\2x+1=-3\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=-2\left(ktm\right)\end{cases}\Rightarrow x=1}\)

thay vào ta được : \(C=\frac{2}{1+2}=\frac{2}{3}\)

\(\frac{x}{x+2}=2\Leftrightarrow x=2x+4\)

\(\Leftrightarrow x=-4\left(tm\right)\)

câu 1

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

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

\(Q\left(x\right)=2x^4-5x^2-x+1-2x^3\)

\(Q\left(x\right)=2x^4-2x^3-5x^2-x+1\)

b, Ta có A(x)=P(x)+Q(x)

thay số A(x)=\(\left(-2x^4+2x^3+5x^2+3\right)+\left(2x^4-2x^3-5x^2-x+1\right)\)

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

                   \(=-x+4\)

c, A(x)=0 khi 

\(-x+4=0\)

\(x=4\)

vậy no của đa thức là 4

câu 2

tự vẽ hình nhé 

a, xét \(\Delta\) ABC cân tại A có AD là pg 

=> AD vừa là dg cao vừa là đg trung tuyến ( t/c trong tam giác cân )

xét \(\Delta\) ADB vg tại D ( áp dụng định lí Py ta go trong tam giác vg ) có 

\(AB^2=BD^2+AD^2\\ \Rightarrow BD^2=9\Rightarrow BD=3\)

Ta có D là trung đm của BC ( AD là đg trung tuyến ứng vs BC) 

=> BD=CD=\(\dfrac{1}{2}BC\)

=> BC= 6cm

câu b đang nghĩ 

x^2+x-30=0

x^2-5x+6x-30=0

x(x-5)+6(x-5)=0

(x-5)(x+6)=0

=>x-5=0 và x+6=0

x=5           ,x=-6

b)(x+y+z)^3-x^3-y^3-z^3

=(x+y+z)^3-(x^3+y^3+z^3)

=(x+y+z)^3-[(x+y)^3-3xy(x+y)+z^3]

=(x+y+z)^3-[(x+y+z)^3-3(x+y)z(x+y+z)-3xy(x+y)]

=(x+y+z)^3-(x+y+z)^3+3(x+y)z(x+y+z)+3xy(x+y)

=3(x+y)(xz+yz+z^2+xy)