\(A=x^3-15x^4+16x^3-29x^2\)

\(A=\left(x^3+16x^3\right)-15x^4-29x^2\)

\(A=17x^3-15x^4-29x^2\)

\(A=-15x^4+17x^3-29x^2\)

\(A=-x^2\left(15x^2-17x+29\right)\)

a: =>9(2x+1)=6(3-x)

=>3(2x+1)=2(3-x)

=>6x+3=6-2x

=>8x=3

=>x=3/8

b: =>-3x^2-2+3x^2-18x=-26

=>-18x=-24

=>x=4/3

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

2. \(x^2+5x+\dfrac{25}{4}=x^2+2.x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2=\left(x+\dfrac{5}{2}\right)^2\)

3. \(16x^2-8x+1=\left(4x-1\right)^2\)

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

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

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

7. \(4x^2-20x+25-y^2=\left(2x-5\right)^2-y^2=\left(2x-5-y\right)\left(2x-5+y\right)\)

Kết quả nè

1.(x^2-1)^2×(x^2+1)^2

2.(2x-5)^2÷4

3.(4x-1)^2

4.(x+y)^2×(x-y)

5.(1/4x-4/9y)×(1/4x+1/9y)

1.(x^2-1)^2=[(x-1).(x+1]^2

2). (x+5/2)^2

3). {4x-1)^2

A=(12x³y⁴-15x²y³-9x³y²):3x²y

  =(12x³y⁴:3x²y)-(15x²y³:3x²y)- (9x³y²:3x²y)

  =4x²y³-5y²-3x²

B, sai đề bạn ơi

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

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 

Trả lời:

7, 4x² - 12x + 9 - y²

= ( 4x² - 12x + 9 ) - y² 

= ( 2x - 3 )² - y² 

= ( 2x - 3 - y )( 2x - 3 + y )

8, 16x² - 4y² + 4y - 1

= 16x² - ( 4y² - 4y + 1 )

= ( 4x )² - ( 2y - 1 )²

= ( 4x - 2y + 1 )( 4x + 2y - 1 )

9, 25 - x² - 12x - 36

= 25 - ( x² + 12x + 36 )

= 5² - ( x + 6 )²

= ( 5 - x - 6 )( 5 + x + 6 )

= ( - 1 - x )( 11 + x )

10, x² - 9 - 5 ( x + 3 )

= ( x - 3 )( x + 3 ) - 5 ( x + 3 )

= ( x + 3 )( x - 3 - 5 )

= ( x + 3 )( x - 8 )

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 -_-

`Answer:`