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16 tháng 7 2019

a. 16a2 - 49.( b - c )2

= ( 4a )2 - 72.( b - c )2

= ( 4a )2 - [ 7.( b - c ) ]2

= ( 4a )2 - ( 7b - 7c )2

= ( 4a - 7b + 7c ).( 4a + 7b - 7c )

b. ( ax + by )2 - ( ax - by )2

=( ax + by + ax - by ).( ax + by - ax + by )

= 2ax . 2by

= 2.( ax + by )

c.a6 - 1

= ( a3 )2 - 1

= ( a3 - 1 ).( a3 + 1 )

= ( a - 1 ).( a2 + a + 1 ).( a + 1 ).( a2 - a + 1 )

d. a8 - b8

= ( a4 )2 - ( b4 )2

= ( a4 - b4 ).( a4 + b4 )

= [ ( a2 )2 - ( b2 )2 ].( a4 + b4 )

= ( a2 - b2 ).( a2 + b2 ).( a4 + b4 )

= ( a - b ).( a + b ).( a2 + b2 ).( a4 + b4 )

B2

( x - 4 )2 - 36 = 0

\(\Leftrightarrow\) ( x - 4 )2 = 36

\(\Leftrightarrow\) ( x - 4 )2 = 62

\(\Leftrightarrow\) x + 4 = \(\pm\) 6

\(\Leftrightarrow\)\(\orbr{\begin{cases}x+4=6\\x+4=-6\end{cases}}\) \(\Leftrightarrow\)\(\orbr{\begin{cases}x=10\\x=-2\end{cases}}\)

Vậy x = 10 , x = -2

b. ( x - 8 )2 = 121

\(\Leftrightarrow\) ( x - 8 )2 = 112

\(\Leftrightarrow\) x - 8 = \(\pm\)11

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-8=11\\x-8=-11\end{cases}}\) \(\Leftrightarrow\)\(\orbr{\begin{cases}x=19\\x=-3\end{cases}}\)

Vậy x = 19 , x = -3

c. x2 + 8x + 16 = 0

\(\Leftrightarrow\)x2 + 2.4x + 42 = 0

\(\Leftrightarrow\) ( x + 4 )2 = 0

\(\Leftrightarrow\) x + 4 = 0

\(\Leftrightarrow\) x = -4

Vậy x = -4

d. 4x2 - 12x = - 9

\(\Leftrightarrow\)( 2x )2 - 2.2.x.3 + 32 = 0

\(\Leftrightarrow\) ( 2x - 3 )2 = 0

\(\Leftrightarrow\) 2x - 3 = 0

\(\Leftrightarrow\) 2x = 3

\(\Leftrightarrow\) \(x=\frac{3}{2}\)

Vậy x = \(\frac{3}{2}\)

25 tháng 7 2017

Bài 1 : 

a ) \(x^2-6x-y^2+9=\left(x^2-6x+9\right)-y^2=\left(x-3\right)^2-y^2=\left(x-3+y\right)\left(x-3-y\right)\)

b)  \(25-4x^2-4xy-y^2=5^2-\left(4x^2+4xy+y^2\right)=5^2-\left(2x+y\right)^2=\left(5+2x+y\right)\left(5-2x-y\right)\)

c)  \(x^2+2xy+y^2-xz-yz=\left(x+y\right)^2-z.\left(x+y\right)=\left(x+y\right)\left(x+y-z\right)\)

d)   \(x^2-4xy+4y^2-z^2+4tz-4t^2=\left(x^2-4xy+4y^2\right)-\left(z^2-4tz+4t^2\right)\)

\(=\left(x-2y\right)^2-\left(z-2t\right)^2=\left(x-2y+z-2t\right).\left(x-2y-z+2t\right)\)

BÀi 2 : 

a)   \(ax^2+cx^2-ay+ay^2-cy+cy^2=\left(ax^2+cx^2\right)-\left(ay+cy\right)+\left(ay^2+cy^2\right)\)

\(=x^2.\left(a+c\right)-y\left(a+c\right)+y^2.\left(a+c\right)=\left(a+c\right).\left(x^2-y+y^2\right)\)

b)   \(ax^2+ay^2-bx^2-by^2+b-a=\left(ax^2-bx^2\right)+\left(ay^2-by^2\right)-\left(a-b\right)\)

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

c)  \(ac^2-ad-bc^2+cd+bd-c^3=\left(ac^2-ad\right)+\left(cd+bd\right)-\left(bc^2+c^3\right)\)

\(=-a.\left(d-c^2\right)+d.\left(b+c\right)-c^2.\left(b+c\right)=\left(b+c\right).\left(d-c^2\right)-a\left(d-c^2\right)\)

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

BÀi 3 : 

a)  \(x.\left(x-5\right)-4x+20=0\) \(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\) \(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x-5=0\\x-4=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=5\\x=4\end{cases}}}\)

b)  \(x.\left(x+6\right)-7x-42=0\)\(\Leftrightarrow x.\left(x+6\right)-7.\left(x+6\right)=0\) \(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x+6=0\\x-7=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-6\\x=7\end{cases}}}\)

c)   \(x^3-5x^2+x-5=0\) \(\Leftrightarrow x^2.\left(x-5\right)+\left(x-5\right)=0\) \(\Leftrightarrow\left(x-5\right)\left(x^2+1\right)\)

\(\Leftrightarrow\hept{\begin{cases}x^2+1=0\\x-5=0\end{cases}\Leftrightarrow\hept{\begin{cases}x^2=-1\left(KTM\right)\\x=5\end{cases}}}\)

d)   \(x^4-2x^3+10x^2-20x=0\) \(\Leftrightarrow x.\left(x^3-2x^2+10x-20\right)=0\)\(\Leftrightarrow x.\left[x^2.\left(x-2\right)+10.\left(x-2\right)\right]=0\)  \(\Leftrightarrow x.\left(x-2\right)\left(x^2+10=0\right)\)

\(\Leftrightarrow\hept{\begin{cases}x=0\\x-2=0\\x^2+10=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=2\\x^2=-10\left(KTM\right)\end{cases}}}\)

23 tháng 9 2017

Bài1:

\(â,\left(x-4\right)^2-36=0\\ \Leftrightarrow\left(x-4\right)^2=36\\ \Leftrightarrow x-4\in\left\{-6;6\right\}\\ \Leftrightarrow x\in\left\{-2;10\right\}\)

Vậy...

b<Tương tự

c,\(x^2+8x+16=0\\ \Leftrightarrow\left(x+4\right)^2=0\\ \Leftrightarrow x+4=0\\ \Leftrightarrow x=-4\)

Vậy...

d,Tương tự

Bài2:

\(a,75^2-25^2\\ =\left(75-25\right)\left(75+25\right)\\ =100.50=5000\)

\(b,53^2-47^2\\ =\left(53-47\right)\left(53+47\right)\\ =6.100=600\)

23 tháng 9 2017

Bài 1:

a) (x-4)^2 - 36 = 0

=> (x-4)^2 = 36

=> (x-4)^2 = 6^2

=> x-4 = 6

=>x = 2

b) (x-8)^2 = 121

=> (x-8)^2 = 11^2

=> x-8 = 11

=> x = 19

c) x^2 + 8x +16 = 0

=> x( x +8) = -16

=> x = -4

d) 4x^2 - 12x = -9( Mk chưa nghĩ ra !)

1 tháng 7 2019

Bài 1 :

\(a,\)\(\left(x-4\right)^2-36=0\)\(\Rightarrow\left(x-4-6\right)\left(x-4+6\right)=0\)

\(\Rightarrow\left(x-10\right)\left(x-2\right)=0\)\(\Rightarrow x\in\left\{10;2\right\}\)

\(b,\)\(\left(x+8\right)^2=121\)\(\Rightarrow\left(x+8\right)^2-11^2=0\)

\(\Rightarrow\left(x+8+11\right)\left(x+8-11\right)=0\)\(\Rightarrow\left(x+19\right)\left(x-3\right)=0\)\(\Rightarrow x\in\left\{-19;3\right\}\)

\(c,x^2+8x+16=0\)\(\Rightarrow\left(x+4\right)^2=0\)

\(\Rightarrow x+4=0\)\(\Leftrightarrow x=-4\)

\(d,4x^2-12x=-9\)\(\Rightarrow4x^2-12x+9=0\)

\(\Rightarrow\left(2x-3\right)^2=0\)\(\Rightarrow2x-3=0\)\(\Rightarrow x=\frac{3}{2}\)

1 tháng 7 2019

Bài 1 a) (x-4)^2 -36=0

=>  (x-4)^2 = 36

=> x-4 = 6

=> x= 10

b) (x+8)^2 = 121

=> x+8 = 11

=> x=3

c) x^2 + 8x +16=0

=> (x+4)^2 =0

=> x+ 4 =0 => x= -4

d) 4x^2 - 12x= -9

=> 4x^2 -12x+9=0

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

=> 2x-3 =0

=> x= 3/2

3 tháng 11 2017

A) \(\left(x-3\right)^2-\left(x+2\right)^2\)

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

\(=-5.\left(2x-1\right)\)

B) \(\left(4x^2+2xy+y^2\right)\left(2x-y\right)-\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)

\(=\left(2x\right)^3-y^3-\left[\left(2x\right)^3+y^3\right]\)

\(=8x^3-y^3-8x^3-y^3\)

\(=-2y^3\)

C) \(x^2+6x+8\)

\(=x^2+6x+9-1\)

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

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

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

bài 3 A) \(x^2-16=0\)

\(\left(x-4\right)\left(x+4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-4=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)

vậy \(\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)

B) \(x^4-2x^3+10x^2-20x=0\)

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

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

\(\Rightarrow\orbr{\begin{cases}x^3+10x=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x\left(x^2+10\right)=0\\x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

vậy \(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

3 tháng 8 2021

x=0

x=2

16 tháng 10 2016

hoa mắt chóng mặt

16 tháng 10 2016

Nhờ bạn làm cho mik ít câu cũng dc