Bài 3: 

a) Đặt f(x)=0

\(\Leftrightarrow x^2-4x+3=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

b) Đặt f(x)=0

\(\Leftrightarrow x^2-7x+12=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

Bài 3:

c) Đặt f(x)=0

\(\Leftrightarrow x^2+2x+1=0\)

\(\Leftrightarrow\left(x+1\right)^2=0\)

\(\Leftrightarrow x+1=0\)

hay x=-1

d) Đặt f(x)=0

\(\Leftrightarrow x^4+2=0\)

\(\Leftrightarrow x^4=-2\)(Vô lý)

Bài 1:

a: A(x)=M(x)-N(x)

\(=-x^3+2,5x^2-0,5x-1\frac12+x^3-2,5x^2-2x+6=-2,5x+4,5\)

Đặt A(x)=0

=>-2,5x+4,5=0

=>-2,5x=-4,5

=>x=1,8

b: B(x)=M(x)+N(x)

\(=-x^3+2,5x^2-0,5x+1,5-x^3+2,5x^2+2x-6=-2x^3+5x^2+1,5x-4,5\)

Bậc là 3

Bài 3:

a: Đặt f(x)=0

=>\(x^2-4x+3=0\)

=>(x-1)(x-3)=0

=>x=1 hoặc x=3

b: Đặt f(x)=0

=>\(x^2-7x+12=0\)

=>(x-3)(x-4)=0

=>x=3 hoặc x=4

c: Đặt f(x)=0

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

=>\(\left(x+1\right)^2=0\)

=>x+1=0

=>x=-1

d: \(x^4\ge0\forall x\)

=>\(x^4+2\ge2>0\forall x\)

=>f(x) vô nghiệm

a: 

Sửa đề: \(P=\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{4x^2}{x^2-9}\right):\left(\dfrac{5}{3-x}-\dfrac{4x+2}{3x-x^2}\right)\)\(P=\left(\dfrac{-\left(x+3\right)}{x-3}+\dfrac{x-3}{x+3}-\dfrac{4x^2}{\left(x-3\right)\left(x+3\right)}\right):\dfrac{5x-4x-2}{x\left(3-x\right)}\)

\(=\dfrac{-x^2-6x-9+x^2-6x+9-4x^2}{\left(x-3\right)\left(x+3\right)}:\dfrac{x-2}{x\left(3-x\right)}\)

\(=\dfrac{-4x^2-12x}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x\left(3-x\right)}{x-2}\)

\(=\dfrac{-4x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{-x\left(x-3\right)}{x-2}=\dfrac{4x^2}{x-2}\)

b: x^2-4x+3=0

=>x=1(nhận) hoặc x=3(loại)

Khi x=1 thì \(P=\dfrac{4\cdot1^2}{1-2}=-4\)

c: P>0

=>x-2>0

=>x>2

d: P nguyên

=>4x^2 chia hết cho x-2

=>4x^2-16+16 chia hết cho x-2

=>x-2 thuộc {1;-1;2;-2;4;-4;8;-8;16;-16}

=>x thuộc {1;4;6;-2;10;-6;18;-14}

1. (x - 1)^3 + 3.(x - 3)^2 - (x + 2).(x^2 - 2x + 4) = (x + 2)^3 - (x - 3).(x^2 + 9) - 6x^2 + 5 
<=> x^3 - 3x^2 + 3x - 1 + 3(x^2 - 6x + 9) - (x^3 + 2^3) 
= x^3 + 6x^2 + 12x + 8 - (x^3 - 3x^2 + 9x -27) - 6x^2 + 5 
<=> x^3 - 3x^2 + 3x - 1 + 3x^2 - 18x + 27 - x^3 - 8 
= x^3 + 6x^2 + 12x + 8 - x^3 + 3x^2 - 9x + 27 - 6x^2 + 5 
<=> 3x - 18x -12x - 3x^2 + 9x = 27 + 5 + 8 + 8 + 1 - 27 
<=> - 3x^2 - 18x - 22 = 0 
<=> 3x^2 + 18x + 22 = 0 

Nửa chu vi mảnh đất là: 

                                               120 : 2 = 60 (m)

Chiều dài hơn chiều rộng là:

                                               5 + 5 = 10 (m)

Chiều rộng là:

                                          ( 60 - 10 ) : 2 = 25 (m)

Chiều dài là:

                                                25 + 10 = 35 (m)

Diện tích là:

                                               25  35 = 875 ( )

a) \(\left(2x+1\right)^2-4\left(x+2\right)^2=9\\ \Rightarrow\left[2x+1-2\left(x+2\right)\right]\left[2x+1+2\left(x+2\right)\right]=9\\ \Rightarrow\left(2x+1-2x-4\right)\left(2x+1+2x+4\right)=9\\ \Rightarrow-3\left(4x+5\right)=9\\ \Rightarrow4x+5=-3\\ \Rightarrow4x=-8\\ \Rightarrow x=-2\)

b c d nữa bạn

Thiên Hương đẹp quá đi mất?

 Cho hoi dap de hoi chi khong duoc noi lung tung day la pham loi trong hoi dap

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

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

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

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

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

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

Bài 1: 

c) ĐKXĐ: \(x\notin\left\{\dfrac{1}{4};-\dfrac{1}{4}\right\}\)

Ta có: \(\dfrac{3}{1-4x}=\dfrac{2}{4x+1}-\dfrac{8+6x}{16x^2-1}\)

\(\Leftrightarrow\dfrac{-3\left(4x+1\right)}{\left(4x-1\right)\left(4x+1\right)}=\dfrac{2\left(4x-1\right)}{\left(4x+1\right)\left(4x-1\right)}-\dfrac{6x+8}{\left(4x-1\right)\left(4x+1\right)}\)

Suy ra: \(-12x-3=8x-2-6x-8\)

\(\Leftrightarrow-12x-3-2x+10=0\)

\(\Leftrightarrow-14x+7=0\)

\(\Leftrightarrow-14x=-7\)

\(\Leftrightarrow x=\dfrac{1}{2}\)(nhận)

Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)

Đề câu a có nhầm không nhỉ chứ lớp 8 chưa học phương trình vô tỉ ;-;

Không nhầm tag mình làm tiếp

a, Ta có : \(\left(x+2\right)^3-2\left(x-3\right)^2+18=0\)

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

\(\Leftrightarrow x^3+4x^2+24x+8=0\)

b, Ta có : \(\left(x-5\right)^3-x\left(x-2\right)\left(x+2\right)+125=0\)

\(\Leftrightarrow x^3+75x-15x^2-125-x^3+4x+125=0\)

\(\Leftrightarrow-15x^2+79x=0\)

\(\Leftrightarrow x\left(-15x+79\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{71}{15}\end{matrix}\right.\)

Vậy ...

Câu a xem lại giúp ạ nghiệm rất xấu ;V

`b)(x-5)^3-x(x-2)(x+2)=-125`

`<=>x^3-15x^2+75x-125+125-x(x^2-4)=0`

`<=>x^3-15x^2+75x-x^3+4x^2=0`

`<=>75x-11x^2=0`

`<=>x(75-11x)=0`

`<=>` \(\left[ \begin{array}{l}x=0\\x=\dfrac{75}{11}\end{array} \right.\) 

a) Mình ko rõ 

b) \(\left(x-1\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+3\left(x^2-4\right)=2\)

\(\Leftrightarrow x^3-3x^2+3x-1-\left(x^3+27\right)+3x^2-12=2\)

\(\Leftrightarrow x^3+3x-13-x^3-27=2\)

\(\Leftrightarrow3x-40=2\)

\(\Leftrightarrow3x=42\)

\(\Leftrightarrow x=14\)

\(x\)(\(x\) - 2)^? vậy em ha