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\(a,=\left(4x^2\right)^2\left(x-y\right)-\left(x-y\right)\)
\(=\left[\left(4x^2\right)^2-1^2\right]\left(x-y\right)\)
\(=\left(4x^2+1\right)\left(4x^2-1\right)\left(x-y\right)\)
\(=\left(4x^2+1\right)\left(2x+1\right)\left(2x-1\right)\left(x-y\right)\)
\(2x^3y-2xy^3-4xy^2-2xy\)
\(=2xy.\left(x^2-y^2-2y-1\right)\)
\(=2xy.[x^2-\left(y^2+2y+1\right)]\)
\(=2xy.[x^2-\left(y+1\right)^2]\)
\(=2xy.\left(x+y+1\right).\left(x-y-1\right)\)
Vậy chọn đáp án A
a) \(x^2-2x-4y^2-4y=\left(x^2-2x+1\right)-\left(4y^2+4y+1\right)\)
\(=\left(x-1\right)^2-\left(2y+1\right)^2=\left(x-1-2y-1\right)\left(x-1+2y+1\right)\)
\(=\left(x-2y-3\right)\left(x+2y\right)\)
b) \(x^2-4x^2y^2+y^2+2xy=\left(x^2+2xy+y^2\right)-4x^2y^2\)
\(=\left(x+y\right)^2-4x^2y^2=\left(x+y-2xy\right)\left(x+y+2xy\right)\)
c) \(x^6-x^4+2x^3+2x^2=\left(x^6+2x^3+1\right)-\left(x^4-2x^2+1\right)\)
\(=\left(x^3+1\right)^2-\left(x^2-1\right)^2=\left(x^3+1-x^2+1\right)\left(x^3+1+x^2-1\right)=x^2\left(x^3-x^2+2\right)\left(x+1\right)\)
d) \(x^3+3x^2+3x+1-8y^3=\left(x+1\right)^3-8y^3=\left(x+1-2y\right)\left(x^2+2x+1+2xy+2y+4y^2\right)\)
Bài 2:
a: \(3x^2-3xy=3x\left(x-y\right)\)
b: \(x^2-4y^2=\left(x-2y\right)\left(x+2y\right)\)
c: \(3x-3y+xy-y^2=\left(x-y\right)\left(3+y\right)\)
d: \(x^2-y^2+2y-1=\left(x-y+1\right)\left(x+y-1\right)\)
a) \(x^3-16x=0\)
\(\Leftrightarrow x\left(x^2-16\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-16=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}\)
Vậy tập nghiệm \(S=\left\{-4;0;4\right\}\)
b) \(x^4-2x^3+10x^2-20x=0\)
\(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^3+10x\right)\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x^2+10\right)\left(x-2\right)=0\)
Mà \(x^2+10>0\)nên \(\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
Vậy tập nghiệm S = { 0;2}
1
a, 2x2+4x+2-2y2 = 2(x2+2x+1-y2)= 2[(x+1)2-y2 ] = 2(x-y+1)(x+y+1)
b, 2x - 2y - x2 + 2xy - y2= 2(x -y) - (x2 - 2xy + y2) = 2(x-y)-(x-y)2=(x-y)(2-x+y)
c, x2-y2-2y-1=x2-(y2+2y+1)=x2-(y+1)2=(x-y-1)(x+y+1)
d, x2-4x-2xy-4y+y2= x2-2xy+y2-4x-4y=(x-y)
2.
a, x2-3x+2=x2-x-2x+2=x(x-1)-2(x-1)=(x-2)(x-1)
b, x2+5x+6=x2+2x+3x+6=x(x+2)+3(x+2)=(x+3)(x+2)
c, x2+6x-6=
a)x^2-(a+b)x+ab
= x^2 - ax - bx + ab
= (x^2 - ax) - (bx - ab)
= x(x-a) - b(x-a)
= (x-b)(x-a)
b)7x^3-3xyz-21x^2+9z
=
c)4x+4y-x^2(x+y)
= 4(x + y) - x^2(x+y)
= (4-x^2) (x+y)
= (2-x)(2+x)(x+y)
d) y^2+y-x^2+x
= (y^2 - x^2) + (x+y)
= (y-x)(y+x)+ (x+y)
= (y-x+1) (x+y)
e)4x^2-2x-y^2-y
= [(2x)^2 - y^2] - (2x +y)
= (2x-y)(2x+y) - (2x+y)
= (2x -y -1)(2x+y)
f)9x^2-25y^2-6x+10y
=
Bài 1, dạng 1:
a) Biểu thức không phân tích được thành nhân tử.
b)
\(x^4y^4+64=(x^2y^2)^2+8^2=(x^2y^2)^2+8^2+2.x^2y^2.8-16x^2y^2\)
\(=(x^2y^2+8)^2-(4xy)^2=(x^2y^2+8-4xy)(x^2y^2+8+4xy)\)
c)
\(x^4y^4+4=(x^2y^2)^2+2^2=(x^2y^2)^2+2^2+2.x^2y^2.2-4x^2y^2\)
\(=(x^2y^2+2)^2-(2xy)^2=(x^2y^2+2-2xy)(x^2y^2+2+2xy)\)
f)
\(x^8+x+1=x^8-x^2+x^2+x+1\)
\(=x^2(x^6-1)+(x^2+x+1)=x^2(x^3-1)(x^3+1)+(x^2+x+1)\)
\(=x^2(x-1)(x^2+x+1)(x^3+1)+(x^2+x+1)\)
\(=(x^2+x+1)[x^2(x-1)(x^3+1)+1]=(x^2+x+1)(x^6-x^5+x^3-x^2+1)\)
g)
\(x^8+x^7+1=x^8-x^2+x^7-x+x^2+x+1\)
\(=x^2(x^6-1)+x(x^6-1)+x^2+x+1\)
\(=(x^6-1)(x^2+x)+x^2+x+1\)
\(=(x^3-1)(x^3+1)(x^2+x)+x^2+x+1\)
\(=(x-1)(x^2+x+1)(x^3+1)(x^2+x)+(x^2+x+1)\)
\(=(x^2+x+1)[(x-1)(x^3+1)(x^2+x)+1]=(x^2+x+1)(x^6-x^4+x^3-x+1)\)
h)
Biểu thức không phân tích được thành nhân tử.
k)
\(x^4+4y^4=(x^2)^2+(2y^2)^2+2x^2.2y^2-4x^2y^2\)
\(=(x^2+2y^2)^2-(2xy)^2=(x^2+2y^2-2xy)(x^2+2y^2+2xy)\)
l)
\(4x^4+1=(2x^2)^2+1^2+2.2x^2.1-4x^2\)
\(=(2x^2+1)^2-(2x)^2=(2x^2+1-2x)(2x^2+1+2x)\)
Bài 2 dạng 4
a)
\(a^2-b^2-2x(a-b)=(a^2-b^2)-2x(a-b)=(a-b)(a+b)-2x(a-b)\)
\(=(a-b)(a+b-2x)\)
b)
\(a^2-b^2-2x(a+b)=(a^2-b^2)-2x(a+b)\)
\(=(a-b)(a+b)-2x(a+b)=(a+b)(a-b-2x)\)