Bài 5: 

a. 1 - 2y + y²

= (1 - y)²

b. (x + 1)² - 25

= (x + 1)² - 5²

= (x + 1 - 5)(x + 1 + 5)

= (x - 4)(x + 6)

c. 1 - 4x²

= 1² - (2x)²

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

d. 8 - 27x³

= 2³ - (3x)³

= (2 - 3x)(4 + 6x + 9x²)

e. (đề hơi khó hiểu ''x3'' !?)

g. x³ + 8y³

= (x + 2y)(x² - 2xy + y²)

$=x^3-2x^5$

b) $(x^2+1)(5-x)$

$=5x^2-x^3+5-x$

$=-x^3+5x^2-x+5$

$=x^3+3x^2-4x-2x^2-6x+8$

$=x^3+x^2-10x+8$

$=x^2-x^3+4x-2x+2x^2-8$

$=-x^3+3x^2+2x-8$

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

Trước hết:

$(3x+2)(3-x)=9x+6-3x^2-2x$

$=-3x^2+7x+6$

Do đó:

$(2x-1)(-3x^2+7x+6)$

$=-6x^3+14x^2+12x+3x^2-7x-6$

$=-6x^3+17x^2+5x-6$

g) $(x+3)(x^2+3x-5)$

$=x^3+3x^2-5x+3x^2+9x-15$

$=x^3+6x^2+4x-15$

$=x^4y-2x^2y-6xy-2x^3+4x+12$

$=20x^5-5x^4+10x^3-4x^4+x^3-2x^2+8x^3-2x^2+4x-12x^2+3x-6$

$=20x^5-9x^4+19x^3-16x^2+7x-6$

a: ĐKXĐ: \(\frac{3x+1}{x-1}\ge0\)

=>x>1 hoặc x<=-1/3

Đặt \(t=\sqrt{\frac{3x+1}{x-1}}\)

=>\(\frac{1}{t}=\sqrt{\frac{x-1}{3x+1}}\)

Phương trình ban đầu trở thành: \(t+6\cdot\frac{1}{t}=5\)

=>\(\frac{t^2+6}{t}=5\)

=>\(t^2+6=5t\)

=>\(t^2-5t+6=0\)

=>(t-2)(t-3)=0

=>t=2 hoặc t=3

t=2

=>\(\sqrt{\frac{3x+1}{x-1}}=2\)

=>\(\frac{3x+1}{x-1}=4\)

=>4(x-1)=3x+1

=>4x-4=3x+1

=>x=5(nhận)

t=3

=>\(\sqrt{\frac{3x+1}{x-1}}=3\)

=>\(\frac{3x+1}{x-1}=9\)

=>9(x-1)=3x+1

=>9x-9=3x+1

=>6x=10

=>x=5/3(nhận)

b: ĐKXĐ: x>=2

Đặt \(t = \sqrt{x-2} + \sqrt{x+1}\) (Điều kiện: t>0)

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

=>\(2x-7+2\sqrt{x^2 - x - 2}=t^2-6\)

Phương trình ban đầu trở thành: \(t^2-6=t\)

=>\(t^2-t-6=0\)

=>(t-3)(t+2)=0

=>t-3=0

=>t=3

=>\(\sqrt{x-2} + \sqrt{x+1} = 3\)

=>\(2x - 1 + 2\sqrt{x^2 - x - 2} = 9\)

=>\(2\sqrt{x^2 - x - 2}=10-2x\)

=>\(\sqrt{x^2 - x - 2}=5-x\)

=>\(\begin{cases}5-x\ge0\\ x^2-x-2=\left(5-x\right)^2=x^2-10x+25\end{cases}\Rightarrow\begin{cases}x\le5\\ 9x=25+2=27\end{cases}\)

=>x=3

1. A

2. C 

Câu 1: A

Câu 2: C

Câu 2: 

Theo Vi-ét \(\hept{\begin{cases}x_1+x_2=-\frac{5}{3}\\x_1x_2=-2\end{cases}}\)

Ta có \(S=y_1+y_2=x_1+x_2+\frac{1}{x_1}+\frac{1}{x_2}=\left(x_1+x_2\right)+\frac{x_1+x_2}{x_1x_2}\)

                                                                           \(=-\frac{5}{3}+\frac{\frac{-5}{3}}{-2}=-\frac{5}{6}\)

       \(P=x_1x_2=\left(x_1+\frac{1}{x_2}\right)\left(x_2+\frac{1}{x_1}\right)=x_1x_2+1+1+\frac{1}{x_1x_2}=-2+2+\frac{1}{-2}=-\frac{1}{2}\)

Khi đó y₁ ; y₂ là nghiệm của pt

\(Y^2-SY+P=0\) 

\(\Leftrightarrow Y^2+\frac{5}{6}Y-\frac{1}{2}=0\)

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{5}{3}\\x_1x_2=-2\end{matrix}\right.\)

Ta có: \(\left\{{}\begin{matrix}y_1+y_2=2x_1-x_2+2x_2-x_1\\y_1y_2=\left(2x_1-x_2\right)\left(2x_2-x_1\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}y_1+y_2=x_1+x_2\\y_1y_2=-2x_1^2-2x_2^2+5x_1x_2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}y_1+y_2=-\dfrac{5}{3}\\y_1y_2=-2\left(x_1+x_2\right)^2+9x_1x_2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}y_1+y_2=-\dfrac{5}{3}\\y_1y_2=-2.\left(-\dfrac{5}{3}\right)^2+9.\left(-2\right)=-\dfrac{212}{9}\end{matrix}\right.\)

\(\Rightarrow y_1;y_2\) là nghiệm của:

\(y^2+\dfrac{5}{3}y-\dfrac{212}{9}=0\Leftrightarrow9y^2+10y-212=0\)

\(y_1+y_2=\left(x_1+x_2\right)+\dfrac{x_1+x_2}{x_1x_2}\)

\(=\dfrac{-5}{3}+\dfrac{-5}{3}:\left(-2\right)=\dfrac{-5}{3}+\dfrac{5}{6}=\dfrac{-5}{6}\)

\(y_1y_2=\left(x_1+\dfrac{1}{x_2}\right)\left(x_2+\dfrac{1}{x_1}\right)\)

\(=x_1x_2+2+\dfrac{1}{x_1x_2}=\left(-2\right)+2+\dfrac{1}{\left(-2\right)}=\dfrac{-1}{2}\)

Pt cần tìm có dạng là \(y^2+\dfrac{5}{6}y-\dfrac{1}{2}=0\)