a) cho g(x) = x - 1/7 = 0

=> x = 1/7

vậy x = 1/7 là nghiệm của đa thức g(x)

b) cho h(x)  = 2x + 5 = 0

=> 2x = -5

=> x = -5/2

vậy x = -5/2 là nghiệm của đa thức h(x)

wekhos vậy

Mẫu chung là 60 r làm bthg thôi bn

1) ĐKXĐ: x \(\ne\)1; x \(\ne\)0

Ta có: A = \(\frac{4x^2-3x+17}{x^3-1}+\frac{2x-1}{x^2+x+1}+\frac{6x}{x-x^2}\)

A = \(\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{\left(2x-1\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{6x}{x\left(x-1\right)}\)

A = \(\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x^2-2x-x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

A = \(\frac{4x^2-3x+17+2x^2-3x+1-6x^2-6x-6}{\left(x-1\right)\left(x^2+x+1\right)}\)

A = \(\frac{-12x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)

A = \(\frac{-12\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=-\frac{12}{x^2+x+1}\)

b) Ta có: B = \(\frac{x+9y}{x^2-9y^2}-\frac{3y}{x^2+3xy}\)

B = \(\frac{x+9y}{\left(x-3y\right)\left(x+3y\right)}-\frac{3y}{x\left(x+3y\right)}\)

B = \(\frac{x\left(x+9y\right)}{x\left(x-3y\right)\left(x+3y\right)}-\frac{3y\left(x-3y\right)}{x\left(x+3y\right)\left(x-3y\right)}\)

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

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

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

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

\(A=\frac{4x^2-3x+17}{x^3-1}+\frac{2x-1}{x^2+x+1}+\frac{6x}{x-x^2}\)

\(A=\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x-1}{x^2+x+1}+\frac{6x}{x\left(1-x\right)}\)

\(A=\frac{4x^2-3x+17}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x-1}{x^2+x+1}-\frac{6x}{x\left(x-1\right)}\)

\(A=\frac{x\left(4x^2-3x+17\right)+x\left(x-1\right)\left(2x-1\right)-6x\left(x^2+x+1\right)}{x\left(x-1\right)\left(x^2+x+1\right)}\)

\(A=\frac{4x^3-3x^2+17x+x\left(2x^2-x-2x+1\right)-6x^3-6x^2-6x}{x\left(x-1\right)\left(x^2+x+1\right)}\)

\(A=\frac{\left(4x^3+2x^3-6x^3\right)-3x^2-3x^3-6x^2+17x+x-6x}{x\left(x-1\right)\left(x^2+x+1\right)}\)

\(A=\frac{-12x^2+12x}{x\left(x-1\right)\left(x^2+x+1\right)}\)

\(A=\frac{-12x\left(x-1\right)}{x\left(x-1\right)\left(x^2+x+1\right)}=\frac{-12}{x^2+x+1}\)

\(a\frac{x^2-49}{x+5}:\left(x-7\right)\)

\(=\frac{\left(x-7\right)\left(x+7\right)}{x+5}.\frac{1}{\left(x-7\right)}\)

\(=\frac{x+7}{x+5}\)

\(b,\frac{2x+7}{x+2}-\frac{x+8}{2x+4}\)

\(=\frac{2\left(2x+7\right)}{2\left(x+2\right)}-\frac{x+8}{2\left(x+2\right)}=\frac{4x+14-x+8}{2\left(x+2\right)}\)

\(=\frac{3x+22}{2\left(x+2\right)}\)

a) \(\frac{x^2-49}{x+5}\div\left(x-7\right)=\frac{\left(x-7\right)\left(x+7\right)}{x+5}.\frac{1}{x-7}=\frac{x+7}{x+5}\)

b) \(\frac{2x+7}{x+2}-\frac{x+8}{2x+4}=\frac{2\left(2x+7\right)}{2\left(x+2\right)}-\frac{x+8}{2\left(x+2\right)}=\frac{\left(4x+14\right)-\left(x+8\right)}{2\left(x+2\right)}\)

\(=\frac{4x+14-x-8}{2\left(x+2\right)}=\frac{3x+6}{2\left(x+2\right)}=\frac{3\left(x+2\right)}{2\left(x+2\right)}=\frac{3}{2}\)

Bài 1:

a)    \(x^3-5x^2+8x-4\)

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

b) Ta có:  \(\frac{A}{M}=\frac{10x^2-7x-5}{2x-3}=5x+4+\frac{7}{2x-3}\)

   Với \(x\in Z\)thì  \(A⋮M\)khi \(\frac{7}{2x-3}\in Z\)\(\Rightarrow7⋮\left(2x-3\right)\)\(\Rightarrow2x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

\(\Rightarrow=\left\{1;5;\pm2\right\}\)thì khi đó \(A⋮M\)

Các bài làm này có đúng ko ạ, ai đó duyệt giúp em, em cảm ơn.

Bài 1:

a)x³-5x²+8x-4=x³-4x²+4x-x²+4x-4

=x(x²-4x-4)-(x²-4x+4)

=(x-1) (x-2)²

b)Xét:

\(\frac{a}{b}-\frac{10x^2-7x-5}{2x-3}\)

=\(5x+4+\frac{7}{2x-3}\)

Với x thuộc Z thì A /\ B khi \(\frac{7}{2x-3}\) thuộc  Z => 7 /\ (2x-3)

Mà Ư(7)={-1;1;-7;7} => x=5;-2;2;1 thì A /\ B

c)Biến đổi \(\frac{x}{y^3-1}-\frac{x}{x^3-1}=\frac{x^4-x-y^4+y}{\left(y^3-1\right)\left(x^3-1\right)}\)

=\(\frac{\left(x^4-y^4\right)\left(x-y\right)}{xy\left(y^2+y+1\right)\left(x^2+x+1\right)}\)(do x+y=1=>y-1=-x và x-1=-y)

=\(\frac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)-\left(x-y\right)}{xy\left[x^2y^2+y^2x+y^2+xy^2+xy+y+x^2+x+1\right]}\)

=\(\frac{\left(x-y\right)\left(x^2+y^2-1\right)}{xy\left[x^2y^2+xy\left(x+y\right)+x^2+y^2+xy+2\right]}\)

=\(\frac{\left(x-y\right)\left(x^2-x+y^2-y\right)}{xy\left[x^2y^2+\left(x+y\right)^2+2\right]}=\frac{\left(x-y\right)\left[x\left(x-1\right)+y\left(y-1\right)\right]}{xy\left(x^2y^2+3\right)}\)

=\(\frac{\left(x-y\right)\left[x\left(-y\right)+y\left(-x\right)\right]}{xy\left(x^2y^2+3\right)}=\frac{\left(x-y\right)\left(-2xy\right)}{xy\left(x^2y^2+3\right)}\)

=\(\frac{-2\left(x-y\right)}{x^2y^2+3}\)Suy ra điều phải chứng minh

Bài 2 )

a)(x²+x)²+4(x²+x)=12 đặt y=x²+x

   y²+4y-12=0 <=>y²+6y-2y-12=0

<=>(y+6)(y-2)=0 <=> y=-6;y=2

>x²+x=-6 vô nghiệm vì x²+x+6 > 0 với mọi x

>x²+x=2 <=> x²+x-2=0 <=> x²+2x-x-2=0

<=>x(x+2)-(x+2)=0 <=>(x+2)(x-1) <=>  x=-2;x-1

Vậy nghiệm của phương trình x=-2;x=1

b)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}+\frac{x+4}{2005}+\frac{x+5}{2004}\)\(+\frac{x+6}{2003}\)

=\(\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)+\left(\frac{x+4}{2005}+1\right)\)\(+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)

<=>\(\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}\)\(+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)

<=>\(\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}\)\(-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x+2009}{2003}=0\)

Nhờ OLM xét giùm em vs ạ !

a) \(\frac{6}{x^2+4x}+\frac{3}{2x+8}=\frac{6.2}{2x\left(x+4\right)}+\frac{3x}{2x\left(x+4\right)}=\frac{12+3x}{2x\left(x+4\right)}=\frac{3\left(x+4\right)}{2x\left(x+4\right)}=\frac{3}{2x}\)

c) \(\frac{-5}{4+2y}+\frac{y-2}{2y+y^2}=\frac{-5.y}{2y\left(y+2\right)}+\frac{2\left(y-2\right)}{2y\left(y+2\right)}=\frac{-5y+2y-4}{2y\left(y+2\right)}=\frac{-3y-4}{2y\left(y+2\right)}\)

d) \(\frac{x-1}{x^2-2xy}+\frac{3}{2xy-x^2}=\frac{x-1}{x\left(x-2y\right)}-\frac{3}{x\left(x-2y\right)}=\frac{x-1-3}{x\left(x-2y\right)}=\frac{x-4}{x\left(x-2y\right)}\)

ĐKXĐ : \(x\ne1\)biến đổi phương trình zề dạng

\(x^2+5x+4=0=>\left(x+1\right)\left(x+4\right)=0=>x=-1hx=-4\)

loại x=-1 . => x=-4

cụ thể hơn ik bạn

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