a) Phân thức M xác định khi và chỉ khi :

+) \(2x-2\ne0\Leftrightarrow x\ne1\)

+) \(2x+2\ne0\Leftrightarrow x\ne-1\)

+) \(1-\frac{x-3}{x+1}\ne0\)

\(\Leftrightarrow x-3\ne x+1\)

\(\Leftrightarrow0x\ne4\left(\text{luôn đúng}\right)\)

Vậy \(x\ne\left\{1;-1\right\}\)

b) \(M=\left(\frac{x-2}{2x-2}-\frac{x+3}{2x+2}+\frac{3}{2x-2}\right):\left(1-\frac{x-3}{x+1}\right)\)

\(M=\left(\frac{\left(x-2\right)\left(2x+2\right)}{\left(2x-2\right)\left(2x+2\right)}-\frac{\left(x+3\right)\left(2x-2\right)}{\left(2x-2\right)\left(2x+2\right)}+\frac{3\left(2x+2\right)}{\left(2x-2\right)\left(2x+2\right)}\right):\left(\frac{x+1-x+3}{x+1}\right)\)

\(M=\left(\frac{2x^2-2x-4-2x^2-4x+6+6x+6}{\left(2x-2\right)\left(2x+2\right)}\right):\left(\frac{4}{x+1}\right)\)

\(M=\frac{8}{2\left(x-1\right)2\left(x+1\right)}\cdot\frac{x+1}{4}\)

\(M=\frac{8\left(x+1\right)}{4\left(x-1\right)\left(x+1\right)\cdot4}\)

\(M=\frac{8\left(x+1\right)}{8\left(x+1\right)\left(x-1\right)}\)

\(M=\frac{1}{x-1}\)

\(M=\left(\frac{x-2}{2x-2}-\frac{x+3}{2x+2}+\frac{3}{2x-2}\right):\left(1-\frac{x-3}{x+1}\right)\)

\(=\left(\frac{x+1}{2x-2}-\frac{x+3}{2x+2}\right):\left(\frac{4}{x+1}\right)=\left[\frac{\left(x+1\right)\left(2x+2\right)-\left(x+3\right)\left(2x-2\right)}{\left(2x-2\right)\left(2x+2\right)}\right]:\left(\frac{4}{x+1}\right)\)

\(=\left[\frac{2x^2+4x+2-2x^2+2x+6-6x+6}{4x^2-4}\right]:\left(\frac{4}{x+1}\right)\)

\(=\left[\frac{6x+8-6x+6}{4x^2-4}\right]:\left(\frac{4}{x+1}\right)\)

\(=\frac{14}{4x^2-4}:\left(\frac{4}{x+1}\right)=\frac{14x+14}{16x^2-16}=\frac{7x+7}{8x^2-8}\)

Thực hiện phép chia đa thức ta được :

3x⁵ - x⁴ - 2x³ + x² + 4x + 5 : ( x² - 2x + 2 ) = ( 3x³ + 5x² + 2x - 5 ) dư ( -10x + 15 )

Vậy để dư bằng 0 thì -10x + 15 = 0 <=> 3/2

Vậy x = 3/2

a, ĐKXĐ: \(\hept{\begin{cases}x^3+1\ne0\\x^9+x^7-3x^2-3\ne0\\x^2+1\ne0\end{cases}}\)

b, \(Q=\left[\left(x^4-x+\frac{x-3}{x^3+1}\right).\frac{\left(x^3-2x^2+2x-1\right)\left(x+1\right)}{x^9+x^7-3x^2-3}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)

\(Q=\left[\frac{\left(x^3+1\right)\left(x^4-x\right)+x-3}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)

\(Q=\left[\left(x^7-3\right).\frac{\left(x-1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)

\(Q=\frac{x-1+x^2+1-2x-12}{x^2+1}\)

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

5.\(C\text{ó}x^2-12=0\Rightarrow x^2=12\Rightarrow x=\sqrt{12}ho\text{ặc}x=-\sqrt{12}\)

Mà x>0\(\Rightarrow x=\sqrt{12}\)

6.Vì x-y=4\(\Rightarrow\left(x-y\right)^2=x^2-2xy+y^2=x^2-10+y^2=4^2=16\Rightarrow x^2+y^2=26\)

Có \(\left(x+y\right)^2=x^2+2xy+y^2=26+10=36=6^2=\left(-6\right)^2\)

Vì xy>0 và x>0 =>y>0=>x+y>0=>x+y=6

7. \(3x^2+7=\left(x+2\right)\left(3x+1\right)\)

\(3x^2+7=3x^2+7x+2\)

\(3x^2+7-3x^2-7x-2=0\)

-7x+5=0

-7x=-5

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

8.\(\left(2x+1\right)^2-4\left(x+2\right)^2=9\)

\(\left(2x+1\right)^2-\left(2x+4\right)^2=9\)

(2x+1-2x-4)(2x+1+2x+4)=9

-3(4x+5)=9

4x+5=-3

4x=-8

x=-2

Còn câu 9 và 10 để mình nghiên cứu đã

biet x+y =2 tinh min 3x^2 + y^2

1: \(y=\frac{x+m}{x-1}\)

=>y'=\(\frac{\left(x+m\right)^{\prime}\left(x-1\right)-\left(x+m\right)\left(x-1\right)^{\prime}}{\left(x-1\right)^2}=\frac{\left(x-1\right)-\left(x+m\right)}{\left(x-1\right)^2}=\frac{-m-1}{\left(x-1\right)^2}\)

Khi x=2 thì \(y=\frac{2+m}{2-1}=m+2\)

Khi x=4 thì \(y=\frac{4+m}{4-1}=\frac{m+4}{3}\)

TH1: Hàm số đồng biến trên khoảng [2;4]

=>y'>0

=>-m-1>0

=>-m>1

=>m<-1

Khi Hàm số đồng biến trên khoảng [2;4] thì \(y_{\min}=y\left(2\right)=m+2\)

=>m+2=3

=>m=1(loại)

TH2: Hàm số nghịch biến trên khoảng [2;4]

=>y'<0

=>-m-1<0

=>-m<1

=>m>-1

Khi Hàm số nghịch biến trên khoảng [2;4] thì \(y_{\min}=y\left(4\right)=\frac{m+4}{3}\)

Theo đề, ta có: \(\frac{m+4}{3}=3\)

=>m+4=9

=>m=5(nhận)

TH3: Hàm số là hàm hằng

=>-m-1=0

=>m+1=0

=>m=-1

=>\(y=\frac{x-1}{x-1}=1<>3\)

=>Loại

2: \(y=2x^3-3x^2-m\)

=>y'=\(2\cdot3x^2-3\cdot2x=6x^2-6x=6x\left(x-1\right)\)

Đặt y'=0

=>6x(x-1)=0

=>x=0(nhận) hoặc x=1(nhận)

Khi x=-1 thì \(y=2\cdot\left(-1\right)^3-3\cdot\left(-1\right)^2-m=-2-3-m=-m-5\)

Khi x=0 thì \(y=2\cdot0^3-3\cdot0^2-m=-m\)

Khi x=1 thì \(y=2\cdot1^3-3\cdot1^2-m=2-3-m=-1-m\)

Vì - m-5<-m-1<-m

nên y min=-m-5

=>-5-m=1

=>m=-5-1=-6