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MT
1
TN
11 tháng 12 2016
16ab + 4b2 - 9 + 16a2
= (16a2 + 16ab + 4b2) - 9
= (4a+2b)2 - 32
= (4a+2b-3)(4a+2b+3)
NT
2
N
25 tháng 11 2021
\(16a^2b-16ab+4b=4b\left(4a^2-4a+1\right)=4b\left(2a-1\right)^2\\ 5a^3-10a=5a\left(a^2-2\right)=5a\left(a-\sqrt{2}\right)\left(a+\sqrt{2}\right)\\ 3x-3z+x^2-2xz+z^2\\ =3\left(x-z\right)+\left(x-z\right)^2=\left(x-z\right)\left(3+x-z\right)\)
R
25 tháng 11 2021
\(16a^2b-16ab+4b=4b\left[\left(4a^2\right)-4a+1\right]=4b\left[\left(2a\right)^2-4a+1\right]=4b\left(2a-1\right)^2\)
\(5a^3-10a=5a\left(a^2-2\right)\)
\(3x-3z+x^2-2xz+z^2=\left(3x-3z\right)+\left(x^2-2xz+z^2\right)=3\left(x-z\right)+\left(x-z\right)^2=\left(x-z\right)\left[3+\left(x-z\right)\right]=\left(x-z\right)\left(3+x-z\right)\)
10 tháng 1 2018
a) \(x^6+x^4+x^2y^2+y^4-y^6\)
\(=x^6-y^6+x^4+x^2y^2+y^4\)
\(=\left[\left(x^3\right)^2-\left(y^3\right)^2\right]+x^4+2x^2y^2+y^4-x^2y^2\)
\(=\left(x^3-y^3\right)\left(x^3+y^3\right)+\left(x^2+y^2\right)^2-x^2y^2\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)+\left(x^2+y^2\right)^2-x^2y^2\)
\(=\left(x-y\right)\left(x+y\right)\left[\left(x^2+y^2\right)^2-\left(xy\right)^2\right]+\left(x^2+y^2\right)^2-x^2y^2\)
\(=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)^2-\left(x-y\right)\left(x+y\right)x^2y^2+\left(x^2+y^2\right)^2-x^2y^2\)
\(=\left(x^2+y^2\right)^2\left[\left(x-y\right)\left(x+y\right)+1\right]-x^2y^2\left[\left(x-y\right)\left(x+y\right)+1\right]\)
\(=\left[\left(x-y\right)\left(x+y\right)+1\right]\left[\left(x^2+y^2\right)^2-\left(xy\right)^2\right]\)
\(=\left(x^2-y^2+1\right)\left(x^2-xy+y^2\right)\left(x^2+xy+y^2\right)\)
10 tháng 1 2018
c) \(\left(2a+b\right)^3+6a+3b-4\)
\(=\left(2a+b\right)^3+3\left(2a+b\right)-4\)
Đặt 2a + b = t.
Ta có: \(t^3+3t-4\)
\(=t^3-t^2+t^2-t+4t-4\)
\(=t^2\left(t-1\right)+t\left(t-1\right)+4\left(t-1\right)\)
\(=\left(t-1\right)\left(t^2+t+4\right)\)
Thay t = 2a + b vào biểu thức:
\(\left(t-1\right)\left(t^2+t+4\right)=\left(2a+b-1\right)\left(4a^2+4ab+b^2+2a+b+4\right)\)
NT
0
NN
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 = \(\frac{3}{2}\)
HQ
Hà Quang Minh
Giáo viên
12 tháng 1 2024
a) \({x^2} + 4x + 4 = {x^2} + 2.x.2 + {2^2} = {\left( {x + 2} \right)^2}\)
b) \(16{a^2} - 16ab + 4{b^2} = {\left( {4a} \right)^2} - 2.4a.2b + {\left( {2b} \right)^2} = {\left( {4a - 2b} \right)^2}\)
AH
Akai Haruma
Giáo viên
25 tháng 7 2021
Lời giải:
a.
\(-16a^4b^6-24a^5b^5-9a^6b^4=-[(4a^2b^3)^2+2.(4a^2b^3).(3a^3b^2)+(3a^3b^2)^2]\)
\(=-(4a^2b^3+3a^3b^2)^2=-[a^2b^2(4b+3a)]^2\)
\(=-a^4b^4(3a+4b)^2\)
b.
$x^3-6x^2y+12xy^2-8x^3$
$=x^3-3.x^2.2y+3.x(2y)^2-(2y)^3=(x-2y)^3$
c.
$x^3+\frac{3}{2}x^2+\frac{3}{4}x+\frac{1}{8}$
$=x^3+3.x^2.\frac{1}{2}+3.x.\frac{1}{2^2}+(\frac{1}{2})^3$
$=(x+\frac{1}{2})^3$
25 tháng 7 2021
a) Ta có: \(-16a^4b^6-24a^5b^5-9a^6b^4\)
\(=-a^4b^4\left(16b^2+24ab+9a^2\right)\)
\(=-a^4b^4\cdot\left(4b+3a\right)^2\)
b) Ta có: \(x^3-6x^2y+12xy^2-8y^3\)
\(=x^3-3\cdot x^2\cdot2y+3\cdot x\cdot\left(2y\right)^2-\left(2y\right)^3\)
\(=\left(x-2y\right)^3\)
c) Ta có: \(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}\)
\(=x^3+3\cdot x^2\cdot\dfrac{1}{2}+3\cdot x\cdot\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3\)
\(=\left(x+\dfrac{1}{2}\right)^3\)
`a,x^2 + 4x + 4`
`=x^2 + 2 . x . 2 + 2^2`
`=(x+2)^2`
`b,16a^{2} - 16ab + 4b^{2}`
`=(4a)^{2} - 2 . 4a . 2b + (2b)^{2}`
`=(4a-2b)^{2}`
a)\(x^2+4x+4\)
\(=x^2+2\cdot x\cdot2+2^2\)
\(=\left(x+2\right)^2\)
b) \(16a^2-16ab+4b^2\)
\(=\left(4a\right)^2-2\cdot4a\cdot2b+\left(2b\right)^2\)
\(=\left(4a-2b\right)^2\)