a) x² - 4y² 

= x² - ( 2y )²

= ( x - 2y )( x + 2y )

b) x² + x - 12

= x² - 3x + 4x - 12

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

= ( x - 3 )( x + 4 )

c) x² + 2xy + y² - 11

= ( x² + 2xy + y² ) - 11

= ( x + y )² - ( √11 )²

= ( x + y - √11 )( x + y + √11 )

d) x⁴ + 1 

= ( x⁴ + 2x² + 1 ) - 2x²

= ( x² + 1 )² - ( √2x )²

= ( x² - √2x + 1 )( x² + √2x + 1 )

a) \(x^2-4y^2\)

\(=x^2-\left(2y\right)^2\)

\(=\left(x-2y\right).\left(x+2y\right)\)

b) \(x^2+x-12\)

\(=x^2+4x-3x-12\)

\(=\left(x^2+4x\right)-\left(3x+12\right)\)

\(=x.\left(x+4\right)-3.\left(x+4\right)\)

\(=\left(x+4\right).\left(x-3\right)\)

c) \(x^2+2xy+y^2-11\)

\(=\left(x^2+2xy+y^2\right)-11\)

\(=\left(x+y\right)^2-11\)

\(=\left(x+y\right)^2-\left(\sqrt{11}\right)^2\)

\(=\left(x+y-\sqrt{11}\right).\left(x+y+\sqrt{11}\right)\)

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

a) Điều kiện xác định:

\(\hept{\begin{cases}x+1\ne0\\x-1\ne0\\2x\ne0\end{cases}}\Rightarrow\hept{\begin{cases}x\ne0-1\\x\ne0+1\\x\ne0\end{cases}}\Rightarrow\hept{\begin{cases}x\ne-1\\x\ne1\\x\ne0\end{cases}}\)

Vậy để P có nghĩa thì \(x\ne-1;x\ne1\) và \(x\ne0.\)

b) Rút gọn:

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

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

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

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

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

\(P=\frac{2x}{\left(x-1\right).\left(x+1\right)}.\frac{x-1}{2x}\)

\(P=\frac{2x.\left(x-1\right)}{2x.\left(x-1\right).\left(x+1\right)}\)

\(P=\frac{1}{x+1}.\)

a. \(\frac{2x+3}{15}=\frac{7}{5}\)

\(\Leftrightarrow5\left(2x+3\right)=15.7\)

\(\Leftrightarrow10x+15=105\)

\(\Leftrightarrow10x=90\)

\(\Leftrightarrow x=9\)

b. \(\frac{x-2}{9}=\frac{8}{3}\)

\(\Leftrightarrow3\left(x-2\right)=9.8\)

\(\Leftrightarrow3x-6=72\)

\(\Leftrightarrow3x=78\)

\(\Leftrightarrow x=26\)

c. \(\frac{-8}{x}=\frac{-x}{18}\)

\(\Leftrightarrow-x^2=-144\)

\(\Leftrightarrow x^2=12^2\)

\(\Leftrightarrow\orbr{\begin{cases}x=12\\x=-12\end{cases}}\)

Mấy câu kia tương tự

d, \(\frac{2x+3}{6}=\frac{x-2}{5}\Leftrightarrow10x+15=6x-12\Leftrightarrow4x=-27\Leftrightarrow x=-\frac{27}{4}\)

e, \(\frac{x+1}{22}=\frac{6}{x}\Leftrightarrow x^2+x=132\Leftrightarrow x^2+x-132=0\Leftrightarrow\left(x-11\right)\left(x+12\right)=0\Leftrightarrow\orbr{\begin{cases}x=11\\x=-12\end{cases}}\)

f, \(\frac{2x-1}{2}=\frac{5}{x}\Leftrightarrow2x^2-x=10\Leftrightarrow2x^2-x-10=0\Leftrightarrow\left(x+2\right)\left(2x-5\right)=0\Leftrightarrow\orbr{\begin{cases}x=-2\\x=\frac{5}{2}\end{cases}}\)

g, \(\left(2x-1\right)\left(2x+1\right)=63\Leftrightarrow4x^2+2x-2x-1=63\Leftrightarrow4x^2-64=0\)

\(\Leftrightarrow x^2=16\Leftrightarrow x=\pm4\)

h, \(\frac{10x+5}{6}=\frac{5}{x+1}\Leftrightarrow\left(10x+5\right)\left(x+1\right)=30\Leftrightarrow10x^2+10x+5x+5=30\)

\(\Leftrightarrow10x^2+15x-25=0\Leftrightarrow5\left(2x+5\right)\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{2}\\x=1\end{cases}}\)

1.students of our school are friendly with one another

từ cần điền là.friendly        

1. encourage
2. necessity
3. fashionable
4. Modernization

1. where

2 over

3knew

4.although

5.when

6.a

7.notes

8.count

9. said

10.afford

tích cho mik nhé

1.I study at International Secondary School.

2.I am in grade six and I am in class 6A.

3.There are three buildings and a swimming pool in my school.

4.My class is on the first floor of the building A.

5.My favorite subjects are Vietnamese and Maths.

     nhớ tít

study well

1. I study at International Secondary school. 

2. I am in grade six and I am in class 6A .

3,There are three buildings and a swimming pool in my school.

4. my class is on the first floor of the building A.

5.My favourite subject are Vietnames and Maths.