x^2+2x+1−y^2

(a+b)^2=a^2+2ab+b^2

x^2+2x+1=(x+1)^2

(x+1)^2−y^2

a^2−b^2=(a−b)(a+b)

(x+1)^2−y^2=(x+1−y)(x+1+y)

=(x−y+1)(x+y+1)

\(x^2\) + 2\(x\) + 1 - y\(^2\)

= (\(x^2\) + 2.\(x\).1 + 1\(^2\)) - y\(^2\)

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

= (\(x+1-y\))(\(x+1+y\))

45opkik

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

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

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

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

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

Ta có: \(x^5+x^4+x^3+x^2+x+1\)

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

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

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

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

1) Biến đổi A, ta được:

\(A=\frac{x-2+7}{x-2}=1+\frac{7}{x-2}\)

Do đó:

\(A< 1\Rightarrow1+\frac{7}{x-2}< 1\Rightarrow\frac{7}{x-2}< 0\left(1\right)\)

Mà 7>0 nên:

\(\left(1\right)\Rightarrow x-2< 0\Rightarrow x< 2\)

2)

+) Biến đổi B, ta được:

\(B=\frac{3\left(x-2\right)+2x^2-x-19-x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\\ =\frac{3x-6+2x^2-x-19-x^2-2x}{\left(x-2\right)\left(x+2\right)}=\frac{x^2-25}{x^2-4}\left(đpcm\right)\)

+) Từ 1) và 2), ta suy ra:

\(P=\frac{B}{A}=\frac{\frac{x+5}{x-2}}{\frac{\left(x-5\right)\left(x+5\right)}{\left(x-2\right)\left(x+2\right)}}=\frac{1}{\frac{x-5}{x+2}}=\frac{x+2}{x-5}\)

3) Biến đổi P, ta được:

\(P=\frac{x-5+3}{x-5}=1+\frac{3}{x-5}\)

P nguyên khi và chỉ khi \(\frac{3}{x-5}\) nguyên, hay \(x-5\inƯ\left(3\right)\)

Ta có bảng:

Vậy ta có 4 giá trị của x trên thoả mãn đề bài.

Chúc bạn học tốt nha

a) \(\frac{x^2}{x-1}-\frac{2x}{x-1}+\frac{1}{x-1}\)

\(=\frac{x^2-2x+1}{x-1}\)

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

b) \(\left(\frac{1}{1-2x}+\frac{1}{1+2x}\right):\frac{1}{1-2x}\)

\(=\left(\frac{1+2x}{\left(1-2x\right)\left(1+2x\right)}+\frac{1-2x}{\left(1+2x\right)\left(1-2x\right)}\right):\frac{1}{1-2x}\)

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

\(=\frac{2}{1+2x}\)

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

\(\text{b) }\dfrac{x}{xy-y^2}+\dfrac{2x-y}{xy-x^2}\\ =\dfrac{x}{y\left(x-y\right)}+\dfrac{2x-y}{x\left(y-x\right)}\\ =\dfrac{x}{y\left(x-y\right)}-\dfrac{2x-y}{x\left(x-y\right)}\\ =\dfrac{x^2}{y\left(x-y\right)x}-\dfrac{\left(2x-y\right)y}{x\left(x-y\right)y}\\ =\dfrac{x^2-\left(2x-y\right)y}{xy\left(x-y\right)}\\ =\dfrac{x^2-2xy+y^2}{xy\left(x-y\right)}\\ =\dfrac{\left(x-y\right)^2}{xy\left(x-y\right)}\\ =\dfrac{x-y}{xy}\)

Ta có:  MTC=36

Quy đồng 

\(x=\frac{x.36}{36}\)

\(\frac{1-2x}{9}=\frac{\left(1-2x\right).4}{36}\)

\(\frac{3x-2}{12}=\frac{\left(3x-2\right).3}{36}\)

Ta có

:\(\frac{36x+4-8x+9x-6}{36}=\frac{37x-2}{36}\)

Tick mình nha

a) x^2+2x+1-4y^2

= (x+1)^2-(2y)^2

=(x-2y+1)(x+2y+1)

b) (x^3-2x^2+5x-1):(x-5)

Đặt tính chia ta dc thương là x^2+3x (dư 20x-1), vì phép chia có dư cho nên nhân tử là (x^3-2x^2+5x-1).(1/x-5)