R(x) =           2x² + 3x - 1

^-  M(x) =   -x³ + x² 

                x³ + x² + 3x - 1

Vậy R(x) - M(x) = x³ + x² + 3x - 1

Bài 1:

a: ĐKXĐ: x∉{2;3;-1;-2}

b: \(B=\left(\frac{x-2}{x^2-5x+6}-\frac{x+3}{2-x}-\frac{x+2}{x-3}\right):\left(2-\frac{x}{x+1}\right)\)

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

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

\(=\frac{x-2+x^2-9-x^2+4}{\left(x-2\right)\left(x-3\right)}\cdot\frac{x+1}{x+2}=\frac{\left(x-7\right)\left(x+1\right)}{\left(x-2\right)\left(x-3\right)\left(x+2\right)}\)

c: B=0

=>(x-7)(x+1)=0

=>x=7(nhận) hoặc x=-1(loại)

Bài 2:

a: ĐKXĐ: x∉{1;-1}

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

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

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

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

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

c: \(C=\frac25\)

=>\(\frac{\left(x-1\right)\left(x+1\right)}{2x^2+x+1}=\frac25\)

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

=>\(5x^2-5=4x^2+2x+2\)

=>\(x^2-2x-7=0\)

=>\(x^2-2x+1-8=0\)

=>\(\left(x-1\right)^2=8\)

=>\(\left[\begin{array}{l}x-1=2\sqrt2\\ x-1=-2\sqrt2\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\sqrt2+1\left(nhận\right)\\ x=-2\sqrt2+1\left(nhận\right)\end{array}\right.\)

giúp mình vớiiii

B(3)=2*3^2-4*3+3=18-12+3=9

B(-1/2)=2*1/4-4*(-1/2)+3=1/2+3+2=1/2+5=11/2

Bài 2:

a: a>b

=>2a>2b

=>2a+1>2b+1

mà 2b+1>2b-3

nên 2a+1>2b-3

b: 3x-1<=x+2

=>3x-x<=1+2

=>2x<=3

=>\(x\le\frac32\)

c: 2x+3>0

=>2x>-3

=>\(x>-\frac32\)

3x+1<x-4

=>3x-x<-4-1

=>2x<-5

=>\(x<-\frac52\)

2(x+1)+3>=3(5-x)

=>2x+2+3>=15-3x

=>2x+5>=15-3x

=>5x>=10

=>x>=2

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

=>\(\frac{5x-3\left(x+1\right)}{15}>1\)

=>5x-3(x+1)>15

=>5x-3x-3>15

=>2x>18

=>x>9

Bài 3:

Gọi độ dài quãng đường AB là x(km)

(Điều kiện: x>0)

Thời gian ô tô đi từ A đến B là \(\frac{x}{50}\) (giờ)

Thời gian ô tô đi từ B về A là \(\frac{x}{60}\) (giờ)

Tổng thời gian cả đi và về là 6h30p-1h=5h30p=5,5 giờ nên ta có:

\(\frac{x}{50}+\frac{x}{60}=5,5\)

=>\(\frac{6x}{300}+\frac{5x}{300}=5,5\)

=>\(x\cdot\frac{11}{300}=\frac{11}{2}\)

=>x=150(nhận)

Vậy: Độ dài quãng đường AB là 150km

Bài 1:

a: 2x-1=0

=>2x=1

=>\(x=\frac12\)

b: 3x-2=5+x

=>3x-x=2+5

=>2x=7

=>\(x=\frac72\)

c: 2(x-3)-4=3(1+x)-5x

=>2x-6-4=3+3x-5x

=>2x-10=3-2x

=>4x=13

=>\(x=\frac{13}{4}\)

d: \(\frac{x+1}{2}-\frac{2x}{3}=1\)

=>\(\frac{3\left(x+1\right)-4x}{6}=1\)

=>3(x+1)-4x=6

=>3x+3-4x=6

=>3-x=6

=>x=3-6=-3

e: x(x-2)+3(x-2)=0

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

=>\(\left[\begin{array}{l}x-2=0\\ x+3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=-3\end{array}\right.\)

f: ĐKXĐ: x<>0; x<>1

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

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

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

=>4x=5

=>x=5/4(nhận)

bai dai dong qua

a) (x-2)^3-x(x+1)(x-1)+6x(x-3)=0

\(x^3-6x^2+12x-8-x\left(x^2-1\right)+6x\left(x-3\right)=0\)

\(x^3-6x^2+12x-8-x^3+x+6x^2-18x=0\)

\(-5x-8=0\)

\(x=-\frac{8}{5}\)

Mai mik làm mấy bài kia sau