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a) \(x^2+4x+4=x^2+2.2x+2^2=\left(x+2\right)^2\)
\(\left(x^2+4x+4\right)\div\left(x+2\right)=x+2\)
b) \(x^3-1=\left(x-1\right)\left(x^2+x+1\right)\)
\(\left(x^3-1\right)\div\left(x-1\right)=x^2+x+1\)
c) \(x^3+6x^2+12x+8=x^3+3.x^2.2+3.x.2^2+2^3=\left(x+2\right)^3\)
\(\left(x^3+6x^2+12x+8\right)\div\left(x+2\right)=\left(x+2\right)^2\)
Bài 3:
a) \(\left(4x^2+4xy+y^2\right):\left(2x+y\right)=\left(2x+y\right)^2:\left(2x+y\right)=2x+y\)
b) \(\left(27x^3+1\right):\left(3x+1\right)=\left(3x+1\right)\left(9x^2-9x+1\right):\left(3x+1\right)=9x^2-9x+1\)
c) \(\left(x^2-6xy+9y^2\right):\left(3y-x\right)=\left(x-3y\right)^2:\left(3y-x\right)=\left(3y-x\right)^2:\left(3y-x\right)=3y-x\)
d) \(\left(8x^3-1\right):\left(4x^2+2x+1\right)=\left(2x-1\right)\left(4x^2+2x+1\right):\left(4x^2+2x+1\right)=2x-1\)
Bài 4: Tương tự bài 3 '-'
Bài 4 :
a ) ( 4x4 - 9 ) : ( 2x2 - 3 )
= ( 2x2 + 3 )( 2x2 - 3 ) : ( 2x2 - 3 )
= 2x2 + 3
b ) ( 8x3 - 27 ) : ( 4x2 + 6x + 9 )
= ( 2x - 3 )( 4x2 + 6x + 9 ) : ( 4x2 + 6x + 9 )
= 2x - 3
1, a, ( x2 - 1 )*( x2 + 2x)
= x4 + 2x3 - x2 - 2x
b, ( 2x - 1 )*( 3x + 2 )*( 3 - x )
= ( 6x2 + 4x - 3x - 2 )*( 3 - x )
= ( 6x2 + x - 2 )*( 3 - x )
= 18x2 - 6x3 + 3x - x2 - 6 + 2x
= -6x3 - 17x2 + 5x - 6
2, b, B = \(8x^3+48x^2+96x+64\)
= \(\left(2x+4\right)^3\)
Thay x = 8 vào B ta có:
B = \(\left(2\cdot8+4\right)^3\)
= ( 16 + 4 )3
= 203
= 8000
mình chỉ làm đc câu b thôi câu a bạn xem lại đề đi hình như câu a sai đề rồi
a) \(\left(3x-4\right)^2+2\left(3x-4\right)\left(2x+4\right)+\left(2x+4\right)^2\)\(=\left(3x-4+2x+4\right)^2=\left(5x\right)^2=25x^2\)
b)\(\left(3x+4\right)^2+\left(7+3x\right)^2-\left(6x+8\right)\left(3x+7\right)\)
\(=\left(3x+4\right)^2-2\left(3x+4\right)\left(7+3x\right)+\left(7+3x\right)^2\)
\(=\left[3x+4-\left(7+3x\right)\right]^2=\left(3x+4-7-3x\right)^2=\left(-3\right)^2=9\)
c)\(\left(2x+1\right)^2+2\left(4x^2-1\right)+\left(2x-1\right)^2\)
\(=\left(2x+1\right)^2+2\left(\left(2x\right)^2-1^2\right)+\left(2x-1\right)^2\)
\(=\left(2x+1\right)^2+2\left(2x+1\right)\left(2x-1\right)+\left(2x-1\right)^2\)
\(=\left(2x+1+2x-1\right)^2=\left(4x\right)^2=14x^2\)
xong rồi đấy,bạn k cho mình nhé
a,(5x3-4x2+7x):x
=\(5x^2-4x+7\)
b, (x5+12x3-9x2):4x2
=\(\dfrac{1}{4}x^3+3x-\dfrac{9}{4}\)
c,d tương tự
các bài khác bn tự lm nhé mk bận rồi xl nhìu nha
Ta có : \(\left(x-3\right)^3+3.\left(x+1\right)^2=\left(x^2-2x+4\right)\left(x+2\right)\)
\(\Leftrightarrow x^3-9x^2+27x-27+3.\left(x^2+2x+1\right)=x^3+8\)
\(\Leftrightarrow x^3-6x^2+33x-24=x^3+8\)
\(\Leftrightarrow-6x^2+33x-32=0\)
\(\Leftrightarrow6x^2-33x+32=0\)
\(\Leftrightarrow x=\frac{33\pm\sqrt{321}}{12}\)
\(\left(a^2-1\right)\left(a^2-a+1\right)\left(a^2+a+1\right)=\left(a-1\right)\left(a^2+a+1\right)\left(a+1\right)\left(a^2-a+1\right)\)
\(=\left(a^3-1\right)\left(a^3+1\right)=a^6-1\)
(2x4-8x2+8) : (4-2x2)
= 2(x4-4x2+4) : 2(2-x2)
= (x4-4x2+4) : (2-x2)
= (x2 - 2) : (2-x2)
= - 1
\(2x^4+8x^2+8=2\left(x^4+4x^2+4\right)=2\left(x^2+2\right)^2\)
\(\left(4-2x^2\right)=2\left(2-x^2\right)\Rightarrow\frac{2x^4+8x^2+8}{4-2x^2}=\frac{2\left(x^2+2\right)^2}{2\left(2-x^2\right)}=\frac{\left(x^2+2\right)^2}{2-x^2}\)
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