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1/
a, x2+36=12x
<=>x2-12x+36=0
<=>(x-6)2=0
<=>x-6=0
<=>x=6
b, 5x(x-3)+3-x=0
<=>5x(x-3)-(x-3)=0
<=>(5x-1)(x-3)=0
<=>\(\orbr{\begin{cases}5x-1=0\\x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=3\end{cases}}}\)
2/ Sửa đề x2z2 = y2z2
Đặt \(A=4x\left(x+y\right)\left(x+y+z\right)\left(x+z\right)+y^2z^2=4x\left(x+y+z\right)\left(x+y\right)\left(x+z\right)+y^2z^2\)
\(=4\left(x^2+xy+xz\right)\left(x^2+xz+xy+yz\right)+y^2z^2\)
Đặt x2+xy+xz=t, ta có
\(A=4t\left(t+yz\right)+y^2z^2=4t^2+4tyz+y^2z^2=\left(2t+yz\right)^2=\left(2x^2+2xy+2xz+y^2z^2\right)^2\ge0\)
mk viết đáp án, ko biết biến đổi ib mk
a) \(x^3+3x^2y-9xy^2+5y^3=\left(x+5y\right)\left(x-y\right)^2\)
b) \(x^4+x^3+6x^2+5x+5=\left(x^2+5\right)\left(x^2+x+1\right)\)
c) \(x^4-2x^3-12x^2+12x+36=\left(x^2-6\right)\left(x^2-2x-6\right)\)
d) \(x^8y^8+x^4y^4+1=\left(x^2y^2-xy+1\right)\left(x^2y^2+xy+1\right)\left(x^4y^4-x^2y^2+1\right)\)
a, \(x^4+6x^3+7x^2-6x+1\)
\(=x^4-2x^2+1+6x^3+9x^2+6x\)
\(=\left(x^2-1\right)^2+6x\left(x^2-1\right)+9x^2\)
\(=\left(x^2-1+3x\right)^2\)
b, \(x^4-7x^3+14x^2-7x+1\)
\(=x^4+2x^2+1+7x^3+12x^2-7x\)
\(=\left(x^2+1\right)^2-7x\left(x^2+1\right)+12^2\)
\(=\left(x^2-1+3x\right)^2\)
c, \(12x^2-11x-36\)
\(=12x^2-27x+16x-36\)
\(=3x\left(4x-9\right)+4\left(4x-9\right)\)
\(=\left(4x-9\right)\left(3x+4\right)\)
a)2(x-3)+12-4x
=x2(x-3)-4(x-3)
=(x2-4)(x-3)
=(x2-22)(x-3)
=(x+2)(x-2)(x-3)
b)x3-4x2-12x+27
=x3-7x2+9x+3x2-21x+27
=x(x2-7x+9)+3(x2-7x+9)
=(x+3)(x2-7x+9)
a)\(x^2\left(x-3\right)+12-4x\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x^2-2^2\right)\left(x-3\right)\)
\(=\left(x+2\right)\left(x-2\right)\left(x-3\right)\)
Với x = -3 ta có -27-4*9+ 36+27=0 do đó đa thức chứa nhân tử x+3
Ta có: x^3 -4x^2-12x+27 = x^3 +3x^2 -7x^2-21x+9x+27 =(x^3 +3x^2)-(7x^2+21x) + (9x+27) =x^2(x+3) -7x(x+3)+ 9(x+3)=(x+3)(X^2 - 7x+9)
* Xét x^2 -7x + 9 = x^2 - 2x.7/2 +49/4-49/4+9 = (x-7/2)^2 -13/4 =(x-7/2- √13/2)(x-7/2+√13/2)
Vậy: x^3 -4x^2-12x+27 = (x+3)(x-7/2)^2 -13/4 =(x-7/2- √13/2)(x-7/2+√13/2)
k cho mình nha
\(x^3+2x^2+2x+1=\left(x^3+1\right)+\left(2x^2+2x\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1+2x\right)=\left(x+1\right)\left(x^2+x+1\right)\)
\(x^3-4x^2+12x-27=x^3-3x^2-x^2+3x+9x-27\)
\(=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-x+9\right)\)
\(x^4+2x^3+2x^2+2x+1=x^4+x^2+2x^3+x^2+2x+1\)
\(=x^2\left(x^2+1\right)+2x\left(x^2+1\right)+\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^2+2x+1\right)\)
\(=\left(x^2+1\right)\left(x+1\right)^2\)
\(x^4-2x^3+2x-1=\left(x^4-1\right)-2x\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2+1-2x\right)=\left(x^2-1\right)\left(x-1\right)^2\)
\(x^3+2x^2+2x+1=\left(x^3+x^2\right)+\left(x^2+x\right)+\left(x+1\right)\)
\(=x^2.\left(x+1\right)+x.\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right).\left(x^2+x+1\right)\)
\(x^3-4x^2+12x-27\)
\(=\left(x^3-x^2\right)-\left(3x^2-3x\right)+\left(9x-27\right)\)
\(=x^2.\left(x-1\right)-3x.\left(x-1\right)+9.\left(x-3\right)\)
\(=\left(x-1\right).\left(x^2-3x\right)+9.\left(x-3\right)\)
\(=x.\left(x-1\right).\left(x-3\right)+9.\left(x-3\right)\)
\(=\left(x-3\right)\left[x.\left(x-1\right)+9\right]\)
câu a đặt chung x ra là xong
câu b
x^3 + 3x^2 - 7x^2 - 21x + 9x+ 27 còn lại tự làm nhé
a) x3 - 2x2 + x - xy2
= x (x2 - 2x + 1 - y2)
= x [(x2 - 2x + 1) - y2]
= x [(x - 1)2 - y2]
= x [(x - 1) + y] [(x - 1) - y]
= x (x - 1 + y) (x - 1 - y)
b) x3 - 4x2 - 12x + 27
= (x3 + 27) - (4x2 + 12x)
= (x3 + 33) - 4x (x + 3)
= (x + 3) (x2 - 3x + 32) - 4x (x + 3)
= (x + 3) [(x2 - 3x + 9) - 4x]
= (x + 3) (x2 - 3x + 9 - 4x)
= (x + 3) (x2 - 7x + 9)
#Học tôt!!!
~NTTH~
a: \(x^2+12x+36=0\)
=>\(x^2+2\cdot x\cdot6+6^2=0\)
=>\(\left(x+6\right)^2=0\)
=>x+6=0
=>x=-6
b: \(4x^2-4x+1=0\)
=>\(\left(2x\right)^2-2\cdot2x\cdot1+1^2=0\)
=>\(\left(2x-1\right)^2=0\)
=>2x-1=0
=>2x=1
=>x=1/2
c: \(x^3+6x^2+12x+8=0\)
=>\(x^3+3\cdot x^2\cdot2+3\cdot x\cdot2^2+2^3=0\)
=>\(\left(x+2\right)^3=0\)
=>x+2=0
=>x=-2