a) \(36-12x+x^2\) \(=6^2-2.6.x+x^2\)

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

b) \(4x^2+12x+9=\left(2x\right)^2+2.2x.3+3^2\)

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

c) \(-25x^6-y^8+10x^3y^4=-\left[25x^6-10x^3y^4+y^8\right]\)

\(=-\left[\left(5x^3\right)^2-2.5x^3.y^4+\left(y^4\right)^2\right]\)

\(=-\left(5x^3-y^4\right)^2\)

d) \(\dfrac{1}{4}x^2-5xy+25y^2=\left(\dfrac{1}{2}x\right)^2-2.\dfrac{1}{2}x.5y+\left(5y\right)^2\)

\(=\left(\dfrac{1}{2}x-5y\right)^2\)

Học tốt~~~

a. \(36-12x+x^2=6^2-2.6.x+x^2=\left(6-x\right)^2\)

b. \(4x^2+12x+9=\left(2x\right)^2+2.2x.3+3^2=\left(2x+3\right)^2\)

c: \(=-\left(25x^6-10x^3y^4+y^8\right)\)

\(=-\left(5x^3-y^4\right)^2\)

d: \(=\left(\dfrac{1}{2}x\right)^2-2\cdot\dfrac{1}{2}x\cdot5y+\left(5y\right)^2=\left(\dfrac{1}{2}x-5y\right)^2\)

Câu a : \(4x^3-5x^2+6x+9\)

\(=4x^3+3x^2-8x^2-6x+12x+9\)

\(=\left(4x^3+3x^2\right)-\left(8x^2+6x\right)+\left(12x+9\right)\)

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

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

Câu b : \(5x^3-12x^2+14x-4\)

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

\(=\left(5x^3-2x^2\right)-\left(10x^2-4x\right)+\left(10x-4\right)\)

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

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

Câu c : \(x^3-5x^2+2x+8\)

\(=x^3+x^2-6x^2-6x+8x+8\)

\(=\left(x^3+x^2\right)-\left(6x^2+6x\right)+\left(8x+8\right)\)

\(=x^2\left(x+1\right)-6x\left(x+1\right)+8\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2-6x+8\right)\)

\(=\left(x+1\right)\left[x^2-2x-4x+8\right]\)

\(=\left(x+1\right)\left[x\left(x-2\right)-4\left(x-2\right)\right]\)

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

Câu d : \(4x^3+5x^2+10x-12\)

\(=4x^3+8x^2-3x^2+16x-6x-12\)

\(=\left(4x^3-3x^2\right)+\left(8x^2-6x\right)+\left(16x-12\right)\)

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

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

7,      4x mũ 2 - 12x + 9 - y mũ 2 =  -(y-2x+3) (y+2x-3)

8,      16x mũ 2 - 4y mũ 2 + 4y - 1 =   -(2y - 4x - 1) (2y+4x-1)

9,        25 - x mũ 2 - 12x - 36 =  -(x+1) (x+11)

10,        x mũ 2 - 9 - 5 ( x + 3 ) =  (x-8) (x+3)

bạn k cho mình nha 

chúc bạn học tốt :))))

bạn kham khảo link, mình đã làm rồi nhé

Câu hỏi của Phạm Đỗ Bảo Ngọc - Toán lớp 8 - Học trực tuyến OLM 

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

7, \(4x^2-12x+9-y^2=\left(2x-3\right)^2-y^2=\left(2x-3-y\right)\left(2x-3+y\right)\)

8, \(16x^2-4y^2+4y-1=16x^2-\left(2y-1\right)^2=\left(4x-2y+1\right)\left(4x+2y-1\right)\)

9, \(25-x^2-12x-36=25-\left(x+6\right)^2=\left(5-x-6\right)\left(5+x+5\right)=-\left(x+1\right)\left(x+10\right)\)

10, \(x^2-9-5\left(x+3\right)=\left(x-3\right)\left(x+3\right)-5\left(x+3\right)=\left(x+3\right)\left(x-8\right)\)

a. = -1080 - 1111 + 1080 + 1000

= -1111 + 1000

= -111

b. = 8 - 4^2 + 5 - 17

 = 8 - 16 + 5 - 17

= -20

c. Câu này đề có vấn đề em ơi

a. -1080 - (1111 - 1080) + 1000

    = -1080 - 31 + 1000

    =   -1111 + 1000

   =  1000 - 1111

   = -111

b. |-8| - [4² + (-5)] + (-17)

  = 8 - [16 + (-5)] + (-17)

  = 8 - (16 - 5) + (-17)

  = 8 - 11 + (-17)

  = -20

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

2. \(x^6+\dfrac{1}{4}x^3+\dfrac{1}{64}=0\Leftrightarrow\left(x^3\right)^2+2.x^3.\dfrac{1}{8}+\left(\dfrac{1}{8}\right)^2=0\Leftrightarrow\left(x+\dfrac{1}{8}\right)^2=0\Leftrightarrow x=-\dfrac{1}{2}\)4. \(x^3-10x^2+25x=0\Leftrightarrow x^3-5x^2-5x^2+25x=0\)

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

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

5. \(\dfrac{1}{4}x^3-3x^2+9x=0\)

\(\Leftrightarrow x\left(\dfrac{1}{4}x^2-3x+9\right)=0\)

\(\Leftrightarrow x\left[\left(\dfrac{1}{2}x\right)^2-2.\dfrac{1}{2}x.3+3^2\right]=0\)

\(\Leftrightarrow x\left(\dfrac{1}{2}x-3\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

6. \(x^5-16x=0\Leftrightarrow x\left(x^4-16\right)=0\Leftrightarrow x\left(x^2-4\right)\left(x^2+4\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\\x^2=-4\left(l\right)\end{matrix}\right.\)

7. \(4x^2+4x-3=0\Leftrightarrow4x^2-2x^2-6x-3=0\)

\(\Leftrightarrow2x\left(2x-1\right)-3\left(2x-1\right)=0\)

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

8. \(4x^2+28x+48=0\Leftrightarrow4x^2+12x+14x+48=0\)

\(\Leftrightarrow4x\left(x+3\right)+12\left(x+4\right)=0\)

\(\Leftrightarrow\left(4x+12\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-4\end{matrix}\right.\)

9. \(9x^2-12x+3=0\Leftrightarrow9x^2-9x-3x+3=0\Leftrightarrow9x\left(x-1\right)-3\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(9x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)

|2 - x|² + 6x - 3 = 0

<=> (x - 2)² + 6x - 3 = 0

<=> x² - 4x + 4 + 6x - 3 = 0

<=> x² + 2x + 1 = 0

<=> (x + 1)² = 0

<=> x + 1 = 0

<=> x = -1

Bắt phải thể hiện -_-

bn kiểm tra giúp mk đề 2 câu cuối , mk làm ko ra

Tìm x:

1. \(25x^2-20x+4=0\)

⇔ \(\left(5x-2\right)^2=0\)

⇔ \(5x-2=0\)

⇔ \(5x=2\)

⇔ \(x=\dfrac{2}{5}\)

⇒ S = \(\left\{\dfrac{2}{5}\right\}\)

2. \(\left(2x-3\right)^2-\left(2x+1\right).\left(2x-1\right)=0\)

⇔ \(4x^2-12x+9-\left(4x^2-1\right)=0\)

⇔ \(4x^2-12x+9-4x^2+1=0\)

⇔ \(-12x+10=0\)

⇔ \(-12x=-10\)

⇔ \(x=\dfrac{5}{6}\)

⇒ S \(=\left\{\dfrac{5}{6}\right\}\)

3. \(\left(\dfrac{1}{2}x-1\right)\left(\dfrac{1}{2}x+1\right)-\left(\dfrac{1}{2}x-1\right)^2=0\)

⇔ \(\dfrac{1}{4}x^2-1-\left(\dfrac{1}{4}x^2-x+1\right)=0\)

⇔ \(\dfrac{1}{4}x^2-1-\dfrac{1}{4}x^2+x-1=0\)

⇔ \(-2+x=0\)

⇔ \(x=2\)

⇒ S \(=\left\{2\right\}\)

4. \(\left(2x-3\right)^2+\left(2x+5\right)^2=8\left(x+1\right)^2\)

⇔ \(4x^2-12x+9+4x^2+20x+25=8\left(x^2+2x+1\right)\)

⇔ \(8x^2+8x+34=8x^2+16x+8\)

⇔ \(8x+34=16x+8\)

⇔ \(8x-16x=8-34\)

⇔ \(-8x=-26\)

⇔ \(x=\dfrac{13}{4}\)

⇒ S \(=\left\{\dfrac{13}{4}\right\}\)

5.\(4x^2+12x-7=0\)

⇔ \(4x^2+14x-2x-7=0\)

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

⇔ \(\left(2x+7\right)\left(2x-1\right)=0\)

⇔ \(\left[{}\begin{matrix}2x+7=0\\2x-1=0\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x=\dfrac{-7}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

⇒ S \(=\left\{\dfrac{-7}{2};\dfrac{1}{2}\right\}\)

6. \(\dfrac{1}{4}x^2+\dfrac{2}{3}x-\dfrac{5}{9}=0\)

⇔ \(9x^2+24x-20=0\)

⇔ \(9x^2+30x-6x-20=0\)

⇔ \(3x\left(3x+10\right)-2\left(3x+10\right)=0\)

⇔ \(\left(3x+10\right)\left(3x-2\right)=0\)

⇔ \(\left[{}\begin{matrix}3x+10=0\\3x-2=0\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x=\dfrac{-10}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)

⇒ S \(=\left\{\dfrac{-10}{3};\dfrac{2}{3}\right\}\)

7. \(24\dfrac{8}{9}-\dfrac{1}{4}x^2-\dfrac{1}{3}x=0\)

⇔ \(\dfrac{224}{9}-\dfrac{1}{4}x^2-\dfrac{1}{3}x=0\)

⇔ \(896-9x^2-12x=0\)

⇔ \(-896+9x^2+12x=0\)

⇔ \(9x^2+12x-896=0\)

⇔ \(9x^2-84x+96x-896=0\)

⇔ \(3x\left(3x-28\right)+32\left(3x-28\right)=0\)

⇔ \(\left(3x-28\right)\left(3x+32\right)=0\)

⇔ \(\left[{}\begin{matrix}3x-28=0\\3x+32=0\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x=\dfrac{28}{3}\\x=\dfrac{-32}{3}\end{matrix}\right.\)

⇒ S \(=\left\{\dfrac{-32}{3};\dfrac{28}{3}\right\}\)