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23 tháng 7 2021

a)\(\left(5x-1\right)^2-196=0\)

\(\Leftrightarrow\left(5x-1\right)^2=196\)

\(\Leftrightarrow5x-1=14\)

\(\Leftrightarrow x=3\)

b)\(4x^2+\frac{1}{4}=2x\)

\(\Leftrightarrow4x^2+\frac{1}{4}-2x=0\)

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

\(\Leftrightarrow2x+\frac{1}{2}=0\)

\(\Leftrightarrow x=-\frac{1}{4}\)

c)\(x^2-12x=-36\)

\(\Leftrightarrow x^2-12x+36=0\)

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

\(\Leftrightarrow x-6=0\)

\(\Leftrightarrow x=6\)

#H

23 tháng 7 2021

a) (5x - 1)2 - 196 = 0

<=> (5x - 1 - 14)(5x - 1 + 14) = 0

<=> (5x - 15)(5x + 13) = 0

<=> \(\orbr{\begin{cases}5x-15=0\\5x+13=0\end{cases}}\) <=> \(\orbr{\begin{cases}x=3\\x=-\frac{13}{5}\end{cases}}\)

Vậy S = {3; -13/5}

b) Ta có: 4x2 + 1/4 = 2x

<=> 16x2 - 8x + 1  = 0

<=> (4x - 1)2 = 0

<=> 4x-  1 = 0

<=> x = 1/4

Vậy S = {1/4}

c) x2 - 12x = -36

<=> x2 - 12x + 36 = 0 

<=> (x - 6)2 0 

<=> x - 6 = 0

<=> x = 6

Vậy S = {6}

21 tháng 8 2018

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~~~

26 tháng 7 2018

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

23 tháng 7 2021

a) \(x^2-\frac{1}{49}=0\)

<=> \(\left(x-\frac{1}{7}\right)\left(x+\frac{1}{7}\right)=0\)

<=> \(\orbr{\begin{cases}x-\frac{1}{7}=0\\x+\frac{1}{7}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{7}\\x=-\frac{1}{7}\end{cases}}\)

Vậy x = \(\pm\frac{1}{7}\)

b) \(64-\frac{1}{4}x^2=0\)

<=> \(\left(8-\frac{1}{2}x\right)\left(8+\frac{1}{2}x\right)=0\)

<=> \(\orbr{\begin{cases}8-\frac{1}{2}x=0\\8+\frac{1}{2}x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=16\\x=-16\end{cases}}\)

Vậy \(x=\pm16\)

c) 9x2 + 12x + 4 = 0

<=> (3x + 2)2 = 0

<=> 3x + 2 = 0 

<=> x = -2/3

Vậy x = -2/3

e) \(x^2+\frac{1}{4}=x\) 

<=> \(x^2-x+\frac{1}{4}=0\)

<=> \(\left(x-\frac{1}{2}\right)^2=0\)

<=> \(x=\frac{1}{2}\)

Vậy \(x=\frac{1}{2}\)

23 tháng 7 2021

d, sửa đề : \(x^2+4=4x\Leftrightarrow x^2-4x+4=0\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x=2\)

i, \(4-\frac{12}{x}+\frac{9}{x^2}=0\)ĐK : \(x\ne0\)

Vì \(x\ne0\)Nhân 2 vế với \(x^2\)phương trình có dạng 

\(4x^2-12x+9=0\Leftrightarrow\left(2x-3\right)^2=0\Leftrightarrow x=\frac{3}{2}\)

8 tháng 8 2021

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

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 : \(4x^3+5x^2+10x-12\)

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

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

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

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

30 tháng 7 2021

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

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

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

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

4, sửa đề : 

 \(6x^2-12x-7x+14=6x\left(x-2\right)-7\left(x-2\right)=\left(6x-7\right)\left(x-2\right)\)

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

30 tháng 7 2021

a) x2(x-3)-4x+12

=x2(x-3)-4(x-3)

=(x-3)(x2-4)

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

b) 2a(x+y)-x-y

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

=(x+y)(2a-1)

c) 2x-4+5x2-10x

=2(x-2)+5x(x-2)

=(x-2)(2+5x)

d) 5x2-12x-7x+14

=5x2-19x+14

e) xy-y2-3x+3y

=y(x-y)-3(x-y)

=(x-y)(y-3)

#H

24 tháng 12 2020

Bài 1: 

a) Ta có: \(\left(15x^2\cdot y^2\cdot z\right):3xyz\)

\(=\dfrac{15x^2y^2z}{3xyz}\)

\(=5xy\)

b) Ta có: \(3x^2\cdot\left(5x^2-4x+3\right)\)

\(=3x^2\cdot5x^2-3x^2\cdot4x+3x^2\cdot3\)

\(=15x^4-12x^3+9x^2\)

c) Ta có: \(\left(2x^2-3x\right):\left(x-4\right)\)

\(=\dfrac{2x^2-8x+5x-20+20}{x-4}\)

\(=\dfrac{2x\left(x-4\right)+5\left(x-4\right)+20}{x-4}\)

\(=2x+5+\dfrac{20}{x-4}\)

d) Ta có: \(-5xy\cdot\left(3x^2y-5xy+y^2\right)\)

\(=-5xy\cdot3x^2y+5xy\cdot5xy-5xy\cdot y^2\)

\(=-15x^3y^2+25x^2y^2-5xy^3\)

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 :))))

8 tháng 8 2021

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