1. a, [x^2.(x-3)-(x-3)] :( x-3) = (x-3 ).(x^2-1) : (x-3) =X^2-1

       2  b, (x-y-z)^5-3 = (x-y-z)^2

       3  c, x^2-1

      4  d, 2x^4 + x^2 - 6x^2 + x^3 - 3 - 3x / x^2 - 3
          = x^2(2x^2 + x + 1) - 3(2x^2 + x + 1) / x^2 - 3
           = (2x^2 + x + 1)(x^2 - 3) / x^2 - 3
           = 2x^2 + x + 1

      5  e, 2.(x-1)

    6   f, (2x³ – 5x² + 6x – 15) : (2x – 5)

     =(2x3−5x2)+(6x−15)=(2x3−5x2)+(6x−15)

     =x2(2x−5)+3(2x−5)=x2(2x−5)+3(2x−5)

     =(x2+3)(2x−5)=(x2+3)(2x−5)

     =(2x3−5x2+6x−15):(2x−5)=x2+3

giúp tui đi

x^3y^4 + 64 = (x^(27y^4)+4)(x^(54y^4)-4x^(27y^4)+16)

4x^4y^4 + 1 = (2x^(128y^4)-2x^(64y^4)+1)(2x^(128y^4)+2x^(64y^4)+1)

32x^4 + 11 = ko biết 

x^4 + 4y^4 = (2y^2-2xy+x^2)(2y^2+2xy+x^2)

x^7 + x^2 + 11 = ko biết

x^8 + x + 1 = (x^2+x+1)(x^6-x^5+x^3-x^2+1)

x^8 + x^7 + 11 = ko biết 

sai ak

\(2x^3+5x=0\Leftrightarrow x\left(2x^2+5\right)=0\Leftrightarrow x=0\)

vì \(2x^2+5\ge5>0\forall x\)

Vậy x = 0 

2x³ + 5x = 0

<=> x ( 2x² + 5 ) = 0

<=> \(\orbr{\begin{cases}x=0\\2x^2+5=0\end{cases}}\). Mà 2x² + 5\(\ge\)5

=> Pt có 1 nghiệm duy nhất là x = 0

câu hỏi đâu bạn?

Ko đăng linh tinh

\(A=a^2+b^2\ge\frac{\left(a+b\right)^2}{2}=\frac{4}{2}=2\)

Dấu ''='' xảy ra khi a = b = 1 

Vậy GTNN của A bằng 2 tại a = b = 1   

\(A=a^2+b^2\)

\(=\left(a+b\right)^2-2ab\)

\(=4-2ab\)

Giả sử \(a;b\ge0\)

Áp dụng bất đẳng thức Cô-si cho hai số a;b dương thì ta có:

\(\frac{a+b}{2}\ge\sqrt{ab}\)

\(\Leftrightarrow\left(\frac{a+b}{2}\right)^2\ge ab\)

\(\Leftrightarrow1\ge ab\)

\(\Rightarrow-2ab\ge-2\)

\(\Leftrightarrow4-2ab\ge2\)

\(\Leftrightarrow A\ge2\)

Vậy \(MinA=2\)

Dấu '' = '' xảy ra khi: \(a=b=1\)

1, \(2x^3-50x=0\Leftrightarrow2x\left(x^2-25\right)=0\Leftrightarrow x=0;x=\pm5\)

2, \(5x^2-4\left(x^2-2x+1\right)-5=0\)

\(\Leftrightarrow5\left(x-1\right)\left(x+1\right)-4\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left[5\left(x+1\right)-4\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+9\right)=0\Leftrightarrow x=-9;x=1\)

3, \(6x\left(x-2\right)=x-2\Leftrightarrow\left(6x-1\right)\left(x-2\right)=0\Leftrightarrow x=\frac{1}{6};x=2\)

4, \(7\left(x-2020\right)^2-x+2020=0\Leftrightarrow7\left(x-2020\right)^2-\left(x-2020\right)=0\)

\(\Leftrightarrow\left(x-2020\right)\left[7\left(x-2020\right)-1\right]=0\Leftrightarrow x=2020;x=\frac{14141}{7}\)

5, \(x^2-10x=-25\Leftrightarrow x^2-10x+25=0\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x=5\)

6, \(x^2-2x-3=0\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\Leftrightarrow x=-1;x=3\)

\(1,\)

\(2x^3-50x=0\)

\(\Leftrightarrow2x\left(x^2-25\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-25=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm5\end{cases}}\)

\(2,\)

\(5x^2-4\left(x^2-2x+1\right)-5=0\)

\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)

\(\Leftrightarrow x^2+8x-9=0\)

\(\Leftrightarrow x^2-x+9x-9=0\)

\(\Leftrightarrow x\left(x-1\right)+9\left(x-1\right)=0\)

\(\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+9=0\\x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-9\\x=1\end{cases}}\)

\(3,\)

\(6x\left(x-2\right)=x-2\)

\(\Leftrightarrow6x\left(x-2\right)-\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(6x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{1}{6}\end{cases}}\)

\(4,\)

\(7\left(x-2020\right)^2-x+2020=0\)

\(\Leftrightarrow7\left(x-2020\right)^2-\left(x-2020\right)=0\)

\(\Leftrightarrow\left(x-2020\right)[7\left(x-2020\right)-1]=0\)

\(\Leftrightarrow\left(x-2020\right)[7x-14141]=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2020\\7x=14141\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2020\\x=\frac{14141}{7}\end{cases}}\)

\(5,\)

\(x^2-10x=-25\)

\(\Leftrightarrow x^2-10x+25=0\)

\(\Leftrightarrow\left(x-5\right)^2=0\)

\(\Leftrightarrow x-5=0\)

\(\Leftrightarrow x=5\)

\(6,\)

\(x^2-2x-3=0\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)

\(1,\)

\(x^2-y^2-2x+2y\)

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

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

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

\(2,\)

\(x^2-25+y^2+2xy\)

\(=\left(x^2+2xy+y^2\right)-25\)

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

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

\(3,\)

\(x^2y-x^3-9y+9x\)

\(=\left(x^2y-x^3\right)-\left(9y-9x\right)\)

\(=x^2\left(y-x\right)-9\left(y-x\right)\)

\(=\left(x^2-9\right)\left(y-x\right)\)

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

\(4,\)

\(x^4+2x^3+x^2\)

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

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

\(5,\)

\(x^4-8x\)

\(=x\left(x^3-8\right)\)

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

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

2, \(x^2-25+y^2+2xy=\left(x+y\right)^2-5^2=\left(x+y-5\right)\left(x+y+5\right)\)

3, \(x^2y-x^3-9y+9x=x^2\left(y-x\right)-9\left(y-x\right)=\left(x-3\right)\left(x+3\right)\left(y-x\right)\)

4, \(x^4+2x^3+x^2=x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)

5, \(x^4+8x=x\left(x^3+8\right)=x\left(x+8\right)\left(x^2-8x+64\right)\)

\(B=\left(x-1\right)^3-\left(x+1\right)^3+6\left(x+1\right)\left(x-1\right)\)

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

\(=x^3-3x^2+3x-1-x^3-3x^2-3x-1+6x^2-6\)

\(=-6x^2-2+6x^2-6\)

\(=-8\)

Vậy biểu thức không phụ thuộc vào biến

\(B=\left(x-1\right)^3-\left(x+1\right)^3+6\left(x+1\right)\left(x-1\right)\)

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

\(=-6x^2-2+6x^2-6=-8\)

Vậy biểu thức ko phụ thuộc vào giá trị biến x 

\(4x^2-12x-7\)

\(=4x^2+2x-14x-7\)

\(=2x\left(2x+1\right)-7\left(2x+1\right)\)

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

Cách 1:

\(4x^2-12x-7\)

\(=\left(4x^2-12x+9\right)-16\)

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

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

\(=\left(2x-7\right)\left(2x+1\right)\)

Cách 2:

\(4x^2-12x-7\)

\(=4x^2+2x-14x-7\)

\(=2x\left(2x+1\right)-7\left(2x+1\right)\)

\(=\left(2x+1\right)\left(2x-7\right)\)