bài 1:

a. \((x+1)(x+3) - x(x+2)=7 \)

    \(x^2+ 3x +x +3 - x^2 -2x =7\)

    \(x^2+4x+3-x^2-2x=7\)

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

    \(2x=4\)

    \(x = 2\)

Bài 2:

a)

\((3x-5)(2x+11) -(2x+3)(3x+7) \)

\(= 6x^2 +33x-10x-55-6x^2-14x-9x-10\)

\(= (6x^2-6x^2)+(33x-10x-14x-9x)-(55+10)\)

\(=-65\)

\(\)

1.

a. \((x+1)(x^2-x+1)-(x-1)(x^2+x+1)\)

\(=x^3 + 1-(x^3-1) = 2 \)

b.

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

2.

a. \(x^2-4y^2+12y-9=x^2-\left[\left(2y\right)^2-2\cdot2y\cdot3+3^2\right]=x^2-\left(2y-3\right)^2=\left(x-2y+3\right)\left(x+2y-3\right)\)

b.

\(5x^2+3\left(x+y\right)^2-5y^2\)

\(=3\left(x+y\right)^2+5\left(x^2-y^2\right)\)

\(=3\left(x+y\right)^2+5\left(x+y\right)\left(x-y\right)\)

\(=\left(x+y\right)\left[3\left(x+y\right)+5\left(x-y\right)\right]\)

\(=\left(x+y\right)\left(3x+3y+5x-5y\right)\)

\(=\left(x+y\right)\left(8x-2y\right)=2\left(x+y\right)\left(4x-y\right)\)

3.

a.

\(x^3-x=0\)

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

\(\Leftrightarrow x\left(x+1\right)\left(x-1\right)\)

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

b.

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

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

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=x+1\\3x+1=-x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)

B1:

a, 2x(x-3)+2(2-x²)

=2x²-6x+4-2x²

= -6x+4

b, (4x-1)(3-x)+(2x+1)²

=(12x-4x²-3+x)+(4x²+4x+1)

=16x-2

B2

a, 5x²-2x(x-3)-3x²=12

=>2x²-2x²+6x=12.

=>6x=12

=>x=6

b,(3x+1)²=25

=>(3x+1)²=5²

=>3x+1=5

=>3x=4

=>x=\(\frac{4}{3}\)

a: \(A=\dfrac{5x-15+2x+6-3x^2+2x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{-3x^2+9x}{\left(x-3\right)\left(x+3\right)}=\dfrac{-3x}{x+3}\)

\(ĐK:x\ne\pm3\\ a,A=\dfrac{5x-15+2x+6-3x^2+2x+9}{\left(x+3\right)\left(x-3\right)}\\ A=\dfrac{-3x^2+9x-1}{\left(x-3\right)\left(x+3\right)}\\ b,\left|x-2\right|=1\Leftrightarrow x=1\left(x\ne3\right)\\ \Leftrightarrow A=\dfrac{-3+9-1}{\left(-2\right)\cdot4}=\dfrac{5}{-8}\)

Bài 1:

\(\left(\dfrac{x}{x^2-49}-\dfrac{x-7}{x^2+7x}\right):\dfrac{2x-7}{x^2+7x}+\dfrac{x}{7-x}\)

\(=\left(\dfrac{x}{\left(x-7\right)\left(x+7\right)}-\dfrac{x-7}{x\cdot\left(x+7\right)}\right)\cdot\dfrac{x^2+7x}{2x-7}+\dfrac{x}{-\left(x-7\right)}\)

\(=\dfrac{x^2-\left(x-7\right)^2}{x\cdot\left(x-7\right)\left(x+7\right)}\cdot\dfrac{x\cdot\left(x+7\right)}{2x-7}-\dfrac{x}{x-7}\)

\(=\dfrac{\left(x-\left(x-7\right)\right)\cdot\left(x+x-7\right)}{x-7}\cdot\dfrac{1}{2x-7}-\dfrac{x}{x-7}\)

\(=\dfrac{\left(x-x+7\right)\cdot\left(2x-7\right)}{x-7}\cdot\dfrac{1}{2x-7}-\dfrac{x}{x-7}\)

\(=\dfrac{7}{x-7}-\dfrac{x}{x-7}\)

\(=\dfrac{7-x}{x-7}\)

\(=\dfrac{-\left(x-7\right)}{x-7}\)

\(=-1\)

A = \(\left(\dfrac{x}{x^2-49}-\dfrac{x-7}{x^2+7x}\right):\dfrac{2x-7}{x^2+7x}+\dfrac{x}{7-x}\)

A = \(\left(\dfrac{x}{\left(x+7\right)\left(x-7\right)}-\dfrac{x-7}{x\left(x+7\right)}\right):\dfrac{2x-7}{x\left(x+7\right)}+\dfrac{x}{7-x}\)

A = \(\left(\dfrac{x^2-\left(x-7\right)^2}{\left(x+7\right)\left(x-7\right)x}\right):\dfrac{2x-7}{x\left(x+7\right)}-\dfrac{x}{x-7}\)

A = \(\left(\dfrac{x^2-\left(x^2-14x+49\right)}{\left(x+7\right)\left(x-7\right)x}\right):\dfrac{\left(2x-7\right)\left(x-7\right)-\left(x^3+7x^2\right)}{\left(x+7\right)\left(x-7\right)x}\)

A = \(\dfrac{14x-49}{\left(x+7\right)\left(x-7\right)x}:\dfrac{-x^3-5x^2-21x+49}{\left(x+7\right)\left(x-7\right)x}\)

A = \(\dfrac{14x-49}{\left(x+7\right)\left(x-7\right)x}.\dfrac{\left(x+7\right)\left(x-7\right)x}{-x^3-5x^2-21x+49}\)

A = \(\dfrac{14x-49}{-x^3-5x^2-21x+49}\)

Bài 2:

\(B=\left[\dfrac{3}{x+1}+\left(\dfrac{3}{x}-\dfrac{x}{x^2+2x+1}\right):\dfrac{2x^2+3x}{x+1}\right]:\dfrac{1+3x}{x^2+x}\)

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

\(=\left(\dfrac{3}{x+1}+\dfrac{3x^2+6x+3-x^2}{x\left(x+1\right)^2}\cdot\dfrac{x+1}{2x^2+3x}\right)\cdot\dfrac{x\left(x+1\right)}{1+3x}\)

\(=\left(\dfrac{3}{x+1}+\dfrac{2x^2+6x+3}{x\left(x+1\right)}\cdot\dfrac{1}{2x^2+3x}\right)\cdot\dfrac{x\left(x+1\right)}{1+3x}\)

\(=\left(\dfrac{3}{x+1}+\dfrac{2x^2+6x+3}{x\left(x+1\right)\left(2x^2+3x\right)}\right)\cdot\dfrac{x\left(x+1\right)}{1+3x}\)

\(=\dfrac{3x\cdot\left(2x^2+3x\right)+2x^2+6x+3}{x\left(x+1\right)\left(2x^2+3x\right)}\cdot\dfrac{x\left(x+1\right)}{1+3x}\)

\(=\dfrac{6x^3+9x^2+2x^2+6x+3}{2x^2+3x}\cdot\dfrac{1}{1+3x}\)

\(=\dfrac{6x^3+11x^2+6x+3}{2x^2+3x}\cdot\dfrac{1}{1+3x}\)

\(=\dfrac{6x^3+11x^2+6x+3}{\left(2x^2+3x\right)\left(1+3x\right)}\)

\(=\dfrac{6x^3+11x^2+6x+3}{2x^2+6x^3+3x+9x^2}\)

\(=\dfrac{6x^3+11x^2+6x+3}{11x^2+6x^3+3x}\)

a, \(P+\left(5x^2+9xy\right)=6x^2+9xy-x\)

\(\Rightarrow P=x^2-x\)

Gỉa sử : x = 1 là nghiệm của đa thức 

Thay x = 1 vào P ta được : \(1-1=0\)*đúng*

Vậy x = 1 là nghiệm của đa thức trên 

b, Với \(x\ge\frac{1}{7}\)đa thức có dạng : \(A=2x^2+7x-1-5+x-2x^2=8x-6\)(1) 

Với \(x< \frac{1}{7}\)đa thức có dạng : \(A=2x^2-7x+1-5+x-2x^2=-6x-4\)(2) 

TH1 : Với đa thức (1) ta có : \(8x-6=2\Leftrightarrow x=1\)

TH2 : Với đa thức (2) ta có : \(-6x-4=2\Leftrightarrow x=-1\)

a) Thu gọn và sắp xếp:

f(x)= \(5x^4-4x^3-2x^2-9x+7\)

g(x)=\(-5x^4+4x^3+3x^2+9x-11\)

b) f(x) + g(x)= \(5x^4-4x^3-2x^2-9x+7\) + ( \(-5x^4+4x^3+3x^2+9x-11\))

= \(5x^4-4x^3-2x^2-9x+7\) \(-5x^4+4x^3+3x^2+9x-11\)

= \(5x^4-5x^4-4x^3+4x^3-2x^2+3x^2+7-11\)

= \(x^2-4\)

Vậy H(x) = \(x^2-4\)

f(x) - g(x)= \(5x^4-4x^3-2x^2-9x+7\) - ( \(-5x^4+4x^3+3x^2+9x-11\))

= \(5x^4-4x^3-2x^2-9x+7\) \(+5x^4-4x^3-3x^2-9x+11\)

= \(5x^4+5x^4-4x^3-4x^3-2x^2-3x^2-9x-9x+7+11\)

= \(10x^4-8x^3-5x^2-18x+18\)

Vậy P(x) = \(10x^4-8x^3-5x^2-18x+18\)

c) Đa thức H(x) có nghiệm khi:

\(x^2-4=0\)

x.x-4=0

x.x=4

\(x^2\) =4

=> x= \(\pm2\)

Vậy x=2 hoặc x=-2 là nghiệm của đa thức H(x)

trong sản xuất, con người đã làm gì để tận dụng sự đa đạng của điều kiện môi trường sống.

mọi người giúp em với, mai em thi rồi

a) * Thu gän

Bài 1: Thu gọn

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

\(=\left[\left(x+3\right)-\left(x+2\right)\right]\cdot\left[\left(x+3\right)+\left(x+2\right)\right]\)

\(=\left(x+3-x-2\right)\left(x+3+x+2\right)\)

\(=2x+5\)

b) Ta có: \(B=\left(2x-1\right)\left(2x+1\right)-\left(2x+3\right)^2\)

\(=4x^2-1-\left(4x^2+12x+9\right)\)

\(=4x^2-1-4x^2-12x-9\)

\(=-12x-10\)

Bài 2: Tìm x

a) Ta có: \(x^2+2x+1=9\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

Vậy: \(x\in\left\{-4;2\right\}\)

Bài 1:

a: \(A=\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{5-x}{1-x^2}\right):\dfrac{1-2x}{x^2-1}\)

\(=\dfrac{-x-1+2x-2-x+5}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)

\(=\dfrac{2}{1-2x}\)

b: Để A>0 thì 1-2x>0

=>2x<1

=>x<1/2