D=\(\dfrac{abc+a+b+c-1-ab-bc-ca}{a^2b+1-a^2-b}\)

\(=\dfrac{\left(abc-bc\right)-\left(ca-c\right)-\left(ab-b\right)+\left(a-1\right)}{\left(a^2b-a^2\right)+\left(1-b\right)}\)

\(=\dfrac{bc\left(a-1\right)-c\left(a-1\right)-b\left(a-1\right)+\left(a-1\right)}{a^2\left(b-1\right)+\left(1-b\right)}\)

\(=\dfrac{\left(a-1\right)\left(bc-c-b+1\right)}{a^2\left(b-1\right)-\left(b-1\right)}=\dfrac{\left(a-1\right)\left[\left(bc-c\right)-\left(b-1\right)\right]}{\left(b-1\right)\left(a^2-1\right)}\)

\(=\dfrac{\left(a-1\right)\left[c\left(b-1\right)-\left(b-1\right)\right]}{\left(b-1\right)\left(a-1\right)\left(a+1\right)}=\dfrac{\left(a-1\right)\left(b-1\right)\left(c-1\right)}{\left(b-1\right)\left(a-1\right)\left(a+1\right)}\)

\(=\dfrac{c-1}{a+1}\)

d, \(\Delta'=225-25.9=0\)pt có nghiệm kép 

\(x_1=x_2=\dfrac{-15}{9}=-\dfrac{5}{3}\)

e, \(\Delta'=4.5-4=16>0\)pt có 2 nghiệm pb 

\(x_1=2\sqrt{5}-4;x_2=2\sqrt{5}+4\)

d: \(\Leftrightarrow\left(3x+5\right)^2=0\)

=>3x+5=0

hay x=-5/3

e: \(\text{Δ}=\left(4\sqrt{5}\right)^2-4\cdot1\cdot4=80-16=64>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{4\sqrt{5}-8}{2}=2\sqrt{5}-4\\x_2=2\sqrt{5}+4\end{matrix}\right.\)

d, \(\Delta=30^2-9.4.25=0\)

Vậy pt có nghiệm kép:\(x_{1,2}=\dfrac{-b}{2a}=\dfrac{-30}{2.9}=\dfrac{-30}{18}=\dfrac{-5}{3}\)

e, \(\Delta=\left(-4\sqrt{5}\right)^2-4.1.4=80-16=64\)

\(x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{4\sqrt{5}+\sqrt{64}}{2.1}=\dfrac{4\sqrt{5}+8}{2}=4+2\sqrt{5}\)

\(x_1=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{4\sqrt{5}-\sqrt{64}}{2.1}=\dfrac{4\sqrt{5}-8}{2}=-4+2\sqrt{5}\)

\(a,C=\dfrac{2x^2-x-x-1+2-x^2}{x-1}\left(x\ne1\right)\\ C=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\\ b,D=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\left(a>0;a\ne1\right)\\ D=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)

Có 

\(\left(\dfrac{\dfrac{x}{x+1}}{\dfrac{x^2}{x^2+x+1}}-\dfrac{2x+1}{x^2+x}\right)\dfrac{x^2-1}{x-1}\)ĐK : \(x\ne\pm1\)

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

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

à xin lỗi mình nhầm dòng cuối 

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

Để biểu thức trên nhận giá trị dương khi 

\(\dfrac{\left(x+1\right)\left(x-2\right)}{x}>0\)bạn tự xét TH cả tử và mẫu nhé, mình đánh trên này bị lỗi 

Bài 1: 

a: \(Q=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\left(x+\sqrt{x}\right)\)

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

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

Ta có :

\(D=\dfrac{1}{5}-\dfrac{1}{5^2}+\dfrac{1}{5^3}-\dfrac{1}{5^4}+\dfrac{1}{5^5}-..........-\dfrac{1}{5^{100}}+\dfrac{1}{5^{101}}\)

\(5D=1-\dfrac{1}{5}+\dfrac{1}{5^2}-\dfrac{1}{5^3}+\dfrac{1}{5^4}-\dfrac{1}{5^5}+..........+\dfrac{1}{5^{100}}\)

\(5D+D=\left(1-\dfrac{1}{5}+\dfrac{1}{5^2}-\dfrac{1}{5^3}+.........+\dfrac{1}{5^{100}}\right)+\left(\dfrac{1}{5}-\dfrac{1}{5^2}+..............-\dfrac{1}{5^{100}}+\dfrac{1}{5^{101}}\right)\)\(6D=1-\dfrac{1}{5^{101}}\)

\(D=\dfrac{1-\dfrac{1}{5^{101}}}{6}\)

Bài 1: 

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

b: Để A=3 thì 3x-9=x+1

=>2x=10

hay x=5

Bài 2: 

a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+2-x}{x+2}\)

\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{2}=\dfrac{-3}{x-2}\)

b: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)

hay \(x\in\left\{3;1;5;-1\right\}\)

`(x+1)^2-(x-1)^2+3(x+1)(x-1)`

`=(x+1+x-1)(x+1-x+1)+3(x^2-1)`

`=2x.2+3x^2-3`

`=3x^2+4x-3`

Lời giải:

$(x+1)^2-(x-1)^2-3(x+1)x(x-1)$

$=(x+1-x+1)(x+1+x-1)-3x(x^2-1)$

$=4x-3x(x^2-1)=x[4-3(x^2-1)]=x(7-3x^2)$

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

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

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

a.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)

\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)