a,  +Xét tam giác ABM và ACM có:
  AB=AC(Giả thiết)  --
  AM là cạnh chung)  I  =>tam giác ABM=ACM (C-C-C)
  MB=MC(Giả thiết) --
b, +Ta có: tam giác ABM=ACM
 => góc AMB=góc AMC (2 góc tương ứng)
    +Ta có:
góc AMB+AMC=180 ( 2 góc kề bù)
      AMB+AMB=180
      AMB = 90(độ)
=>AM vuông góc với BC
c, +Ta có: tam giác ABM=ACM
     => góc BAM=góc CAM(2 góc tương ứng)
     =>AM là tia phân giác của góc BAC
         hay AM là tia phân giác của góc A
Vậy a,tam giác ABM=ACM
       b,AM vuông góc với BC
       c,AM là tia phân giác của góc A

bạn Hà Anh làm đúng rồi

2x−49=5.3²

⇔2x−49=5.9

⇔2x−49=45

⇔2x=94

⇔x=47

200−(2x+6)=4³

⇔2x+6=136

⇔2x=130

⇔x=65

mik chỉ biết 2 ý đầu thui

xin lỗi

c) \(5.\left(x-9\right)=300\)

\(\Leftrightarrow\left(x-9\right)=300:5\)

\(\Leftrightarrow x-9=60\)

\(\Leftrightarrow x=60+9\)

\(\Leftrightarrow x=69\)

Vậy \(x=69.\)

d) x19 là sao bạn?

e) \(198-\left(x+4\right)=120\)

\(\Leftrightarrow\left(x+4\right)=198-120\)

\(\Leftrightarrow x+4=78\)

\(\Leftrightarrow x=78-4\)

\(\Leftrightarrow x=74\)

Vậy \(x=74.\)

f) \(140:\left(x-8\right)=7\)

\(\Leftrightarrow\left(x-8\right)=140:7\)

\(\Leftrightarrow x-8=20\)

\(\Leftrightarrow x=20+8\)

\(\Leftrightarrow x=28\)

Vậy \(x=28.\)

g) \(11.\left(x-7\right)=77\)

\(\Leftrightarrow\left(x-7\right)=77:11\)

\(\Leftrightarrow x-7=7\)

\(\Leftrightarrow x=7+7\)

\(\Leftrightarrow x=14\)

Vậy \(x=14.\)

h) \(\left(6x-39\right):3=281\)

\(\Leftrightarrow\left(6x-39\right)=281.3\)

\(\Leftrightarrow6x-39=843\)

\(\Leftrightarrow6x=843+39\)

\(\Leftrightarrow6x=882\)

\(\Leftrightarrow x=882:6\)

\(\Leftrightarrow x=147\)

Vậy \(x=147.\)

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