3 Viết các biểu thức sau thành tích
a) 4x^2+4x+1 ; b) x^2-16x+64 ; c) 4x^2-9y^2 ; d) ( x-3).(x^2+3x+9)
K
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DN
1
13 tháng 9 2021
\(a,=\left(x+4\right)^2\\ b,=\left(x-6\right)^2\\ c,=-\left(4x^2-4x+1\right)=-\left(2x-1\right)^2\\ d,=\left(x-1\right)^3\)
YM
0
MH
7 tháng 7 2016
a. \(x^2-4x+4=x^2-2.x.2+2^2=\left(x-2\right)^2\)
b. \(x^2-4y^2=x^2-\left(2y\right)^2=\left(x-2y\right)\left(x+2y\right)\)
c. \(4x^2-4x+1=\left(2x\right)^2-2.2x.1+1^2=\left(2x-1\right)^2\)
d. \(x^3-3x^2+3x-1\)
\(=x^3-1^3-3x^2+3x\)
\(=\left(x-1\right)\left(x^2-x+1\right)-3x\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-x+1-3x\right)\)
\(=\left(x-1\right)\left(x^2-4x+1\right)\)
e. \(4x^2-9=\left(2x\right)^2-3^2=\left(2x-3\right)\left(2x+3\right)\)
g. \(4x^2+12xy+9y^2=\left(2x\right)^2+2.2x.3y+\left(3y\right)^2=\left(2x+3y\right)^2\)
KK
30 tháng 8 2021
a. (a2 - b2)2 - (a2 + b2)2
= (a2 - b2 - a2 - b2)(a2 - b2 + a2 + b2)
= -2b2 . 2a2
b. a6 - b6
<=> (a3)2 - (b3)2
<=> (a3 - b3)(a3 + b3)
30 tháng 8 2021
\(a,\left(a^2-b^2\right)^2-\left(a^2+b^2\right)^2\\ =a^4-2a^2b^2+b^4-a^4-2a^2b^2-b^4\\ =-4a^2b^2\)
\(b,a^6-b^6=a^2\left(a^3-b^3\right)=a^2\left(a-b\right)\left(a^2+ab+b^2\right)\)
\(c,-4x^2+9y^2=\left(3y-2x\right)\left(3y+2x\right)\\
d,\left(x+1\right)^3-\left(2-x\right)^3\\
=\left(x+1-2+x\right)\left[\left(x+1\right)^2+\left(x+1\right)\left(2-x\right)+\left(2-x\right)^2\right]\\
=\left(2x-1\right)\left(x^2+2x+1-x^2+x+2+x^2-4x+4\right)\\
=\left(2x-1\right)\left(x^2-x+7\right)\)
\(e,8+\left(4x-3\right)^3\\ =\left(8+4x-3\right)\left[64-8\left(4x-3\right)+\left(4x-3\right)^2\right]\\ =\left(4x+5\right)\left(64-32x+24+16x^2-24x+9\right)\\ =\left(4x+5\right)\left(16x^2-56x+97\right)\)
\(g,81-\left(9-x^2\right)^2\\ =\left(9-9+x^2\right)\left(9+9-x^2\right)\\ =x^2\left(18-x^2\right)\left[=x^2\left(\sqrt{18}-x\right)\left(\sqrt{18}+x\right)\right]\)
Chỗ trong ngoặc nếu bạn chưa học căn thì ko cần ghi nha
LL
2
SM
4 tháng 9 2018
1) ( 4x + 1 )2 + ( 4x - 1 )2 - 2( 4x + 1 ).( 4x - 1 )
= ( 4x + 1 - 4x - 1 )2
= 22
= 4
2) 4x2 - 9 + ( 2x + 3 )
= ( 2x )2 - 32 + ( 2x + 3 )
= ( 2x + 3 ).( 2x - 3 ) + ( 2x + 3 )
= ( 2x + 3 ). ( 2x - 3 + 1 )
= ( 2x + 3 ) .( 2x - 2 )
= 2.( 2x + 3 ) .( x - 1 )
24 tháng 9 2018
1, (4x+1)^2 + (4x-1)^2 - 2(4x+1)(4x-1)
=[(4x+1)-(4x-1)]^2
=(4x+1-4x+1)^2
=2^2
=4
2, 4x^2 - 9 +(2x+3)
=(4x^2 - 9)+(2x+3)
=(2x+3)(2x-3)+(2x+3)
=(2x+3)(2x-3+1)
=(2x+3)(2x-2)
=2(x-1)(2x+3)
=.= hok tốt!!
NB
2
3 tháng 8 2018
\(x^3+4x^2+4x-16y^2\)
\(=\left(x^3+2x^2\right)+\left(2x^2+4x\right)-16y^2\)
\(=x^2.\left(x+2\right)+2x.\left(x+2\right)-16y^2\)
\(=\left(x+2\right).\left(x^2+2x\right)-16y^2\)
\(=x.\left(x+2\right).\left(x+2\right)-\left(4y\right)^2\)
\(=x.\left(x+2\right)^2-\left(4y\right)^2\)
\(=\left[\sqrt{x}.\left(x+2\right)\right]^2-4y^2\)
\(=\left[\sqrt{x}.\left(x+2\right)-4y\right].\left[\sqrt{x}.\left(x+2\right)+4y\right]\)
Tham khảo nhé~
PN
22 tháng 5 2021
nếu đưa vô căn phải có điều kiện là x > 0
\(x^3+4x^2+4x-16y^2=x\left(x+2\right)^2-\left(4y\right)^2\)
\(=\left(x\sqrt{x}+2\sqrt{x}\right)^2-\left(4y\right)^2=\left(x\sqrt{x}+2\sqrt{x}-4y\right)\left(x\sqrt{x}+2\sqrt{x}+4y\right)\)
21 tháng 8 2021
1: \(4x^2+4x+1=\left(2x+1\right)^2\)
2: \(x^2-20x+100=\left(x-10\right)^2\)
3: \(y^4-14y^2+49=\left(y^2-7\right)^2\)
4: \(125x^3-64y^3=\left(5x-4y\right)\left(25x^2+20xy+16y^2\right)\)
24 tháng 9 2021
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
QD
2
15 tháng 11 2021
a) \(=\left(2x-1\right)^2\)
b) \(=x\left(y^2-x^2+2x-1\right)=x\left[y^2-\left(x-1\right)^2\right]=x\left(y-x+1\right)\left(y+x-1\right)\)
a) 4x2+4x+1
= (2x+1)2
b) x2-16x+64
= (x-8)2
c) 4x2-9y2
= (2x+3y)(2x-3y)
d) ( x-3).(x^2+3x+9)