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2) Bạn làm phép chia đa thức cho đa thức, kẻ hẳn dấu chia ra như tiểu học ấy. Được kết quả là \(\left(4y^2+1\right)\) dư (-2y+6) nhé.
3) a) \(x^2-9=0\Leftrightarrow x^2=9\Leftrightarrow x=\pm3\)
b) \(\left(x^2+1\right)\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow x^2+1=0\) hoặc x-3=0 hoặc x+2=0
Trường hợp 1 loại vì \(x^2\) không âm, hai trường hợp còn lại tìm được x=3 và x = -2.
4) a)\(x^2-y^2+2y-1=x^2-\left(y^2-2y+1\right)=x^2-\left(y-1\right)^2=\left(x-y+1\right)\left(x+y-1\right)\)
b) \(5x^2-10xy-20z^2+5y^2\)
= \(5\left(x^2-2xy-4z^2+y^2\right)\)
= \(5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
= 5 ( x-y-2z ) ( x-y+2z )
5) \(x^3=x\Leftrightarrow x=\pm1\)
Bài 4 :
a) \(x^3+x^2y-xy^2-y^3=x^2\left(x+y\right)-y^2\left(x+y\right)=\left(x^2-y^2\right)\left(x+y\right)=\left(x-y\right)\left(x+y\right)^2\)
b)\(x^2y^2+1-x^2-y^2=\left(x^2y^2-x^2\right)-\left(y^2-1\right)=x^2\left(y^2-1\right)-\left(y^2-1\right)=\left(x^2-1\right)\left(y^2-1\right)=\left(x-1\right)\left(x+1\right)\left(y-1\right)\left(y+1\right)\)
c) \(x^2-y^2-4x+4y=\left(x^2-y^2\right)-\left(4x-4y\right)=\left(x-y\right)\left(x+y\right)-4\left(x-y\right)=\left(x-y\right)\left(x+y-4\right)\)
d)
\(x^2-y^2-2x-2y=\)\(\left(x^2-y^2\right)-\left(2x+2y\right)=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)=\left(x+y\right)\left(x-y-2\right)\)
e) Trùng câu d
f) \(x^3-y^3-3x+3y=\left(x-y\right)\left(x^2-xy+y^2\right)-3\left(x-y\right)=\left(x-y\right)\left(x^2-xy+y^2-3\right)\)
Bài 5:
a) \(x^3-x^2-x+1=0\)
\(\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy ...
b) Sửa đề : \(\left(2x-3\right)^2-\left(4x^2-9\right)=0\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x-3-2x-3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(-6\right)=0\)\
\(\Leftrightarrow2x-3=6\)
\(\Leftrightarrow x=\frac{9}{2}\)
vậy........
c) \(x^4+2x^3-6x-9=0\)
\(\Leftrightarrow\left(x^4-9\right)+\left(2x^3-6x\right)=0\)
\(\Leftrightarrow\left(x^2-3\right)\left(x^2+3\right)+2x\left(x^2-3\right)=0\)
\(\Leftrightarrow\left(x^2-3\right)\left(x^2+2x+3\right)=0\)
\(\Leftrightarrow x^2-3=0\Leftrightarrow x^2=3\Leftrightarrow x=\pm\sqrt{3}\)
Vậy
d) \(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}2-x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
Vậy ........
Bài 2:
a: \(=x\left(x^2-6x+9\right)=x\left(x-3\right)^2\)
b: \(=\left(x-1\right)\left(x-7\right)\)
c: \(=x\left(x-2y\right)+3\left(x-2y\right)=\left(x-2y\right)\left(x+3\right)\)
a: \(x^2+6xy+9y^2=\left(x+3y\right)^2\)
b: \(4a^4-4a^2b^2+b^4=\left(2a^2-b^2\right)^2\)
\(x^6-2x^3y+y^2=\left(x^3-y\right)^2\)
b: \(\left(x+y\right)^3-\left(x-y\right)^3\)
\(=\left(x+y-x+y\right)\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)\)
\(=2y\left(3x^2+y^2\right)\)
\(25x^4-10x^2y^2+y^4=\left(5x^2-y^2\right)^2\)
\(-a^2-2a-1=-\left(a+1\right)^2\)
\(a.\left(8x^4-4x^3+x^2\right):2x^2=4x^2-2x+\frac{1}{2}\)
\(b.\left(2x^4-x^3+3x^2\right):\left(-\frac{1}{3x^2}\right)=-6x^6+3x^5-9x^4\)
\(c.\left(-18x^3y^5+12x^2y^2-6xy^3\right):6xy=-3x^2y^4+2xy-y^2\)
\(d.\left(\frac{3}{4x^3y^6}+\frac{6}{5x^4y^5}-\frac{9}{10x^5y}\right):-\frac{3}{5x^3y}=-\frac{5}{4y^5}-\frac{2}{xy^4}-\frac{3}{2x^2}\)
a: \(=7x\left(xy-3\right)\)
d: \(=\left(x+1\right)\left(10x-8y\right)\)
\(=2\left(5x-4y\right)\left(x+1\right)\)
e: \(=\left(x-100\right)\cdot7x\)
f: \(=x\left(x^2-4\right)=x\left(x-2\right)\left(x+2\right)\)
Bài 12:
1) A = x2 - 6x + 11
= (x2 - 6x + 9) + 2
= (x - 3)2 + 2
Ta có: (x - 3)2 ≥ 0 ∀ x
Dấu ''='' xảy ra khi x - 3 = 0 ⇔ x = 3
Do đó: (x - 3)2 + 2 ≥ 2
Hay A ≥ 2
Dấu ''='' xảy ra khi x = 3
Vậy Min A = 2 tại x = 3
2) B = x2 - 20x + 101
= (x2 - 20x + 100) + 1
= (x - 10)2 + 1
Ta có: (x - 10)2 ≥ 0 ∀ x
Dấu ''='' xảy ra khi x - 10 = 0 ⇔ x = 10
Do đó: (x - 10)2 + 1 ≥ 1
Hay B ≥ 1
Dấu ''='' xảy ra khi x = 10
Vậy Min B = 1 tại x = 10
Bài 1:
a) \((x-5)(3x+3)-3x(x-3)+3x+7\)
\(=3(x-5)(x+1)-3x(x-3)+3x+7\)
\(=3(x^2+x-5x-5)-(3x^2-9x)+3x+7\)
\(=3(x^2-4x-5)-(3x^2-9x)+3x+7\)
\(=-8\)
b) \((x-3)(x^2+3x+9)-(54+x^3)\)
\(=(x-3)(x^2-3.x+3^2)-(54+x^3)\)
\(=x^3-3^3-(54+x^3)=-81\)
c) Sửa đề:
\((3x+y)(9x^2-3xy+y^2)-(3x-y)(9x^2+3xy+y^2)\)
\(=(3x+y)[(3x)^2-3x.y+y^2]-(3x-y)[(3x)^2+3x.y+y^2]\)
\(=(3x)^3+y^3-[(3x)^3-y^3]\)
\(=2y^3\)
Câu 2:
\(a)14x^2y^2-21xy^2+28x^2y\)
\(=7xy(2xy-3y+4x)\)
b) \((x+y)^2-4x^2\)\(=(x+y)^2-(2x)^2=(x+y-2x)(x+y+2x)\)
\(=(y-x)(3x+y)\)
c) \(2x^2-2xy-5x+5y\)
\(=(2x^2-2xy)-(5x-5y)\)
\(=2x(x-y)-5(x-y)=(x-y)(2x-5)\)
d) \(2xy-x^2-y^2+16\)
\(=16-(x^2+y^2-2xy)=4^2-(x-y)^2\)
\(=[4-(x-y)][4+(x-y)]=(4-x+y)(4+x-y)\)
Bài 1:
a: \(M=3\left[\left(x+y\right)^2-2xy\right]-\left[\left(x+y\right)^3-3xy\left(x+y\right)\right]+1\)
\(=3\left(4-2xy\right)-\left[8-6xy\right]+1\)
\(=12-6xy-8+6xy+1=5\)
b: \(N=\left(2x-y\right)^3+3\left(2x-y\right)^2+3\left(2x-y\right)+11\)
\(=9^3+3\cdot9^2+3\cdot9+11\)
=729+243+27+11
=729+270+11=1010