Bài 4:

a: \(2x^4+18x^2=0\)

=>\(2x^2\left(x^2+9\right)=0\)

=>\(x^2=0\) (Vì \(2\left(x^2+9\right)=2x^2+18\ge18>0\forall x\) )

=>x=0

b: (x-5)(x+5)-15x+75=0

=>(x-5)(x+5)-15(x-5)=0

=>(x-5)(x+5-15)=0

=>(x-5)(x-10)=0

=>\(\left[\begin{array}{l}x-5=0\\ x-10=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5\\ x=10\end{array}\right.\)

c: \(x^4=x^2\)

=>\(x^4-x^2=0\)

=>\(x^2\left(x^2-1\right)=0\)

=>\(\left[\begin{array}{l}x^2=0\\ x^2-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x^2=0\\ x^2=1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=1\\ x=-1\end{array}\right.\)

d: \(12x\left(6x-1\right)-24x^2=0\)

=>12x(6x-1-2x)=0

=>x(4x-1)=0

=>\(\left[\begin{array}{l}x=0\\ 4x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=\frac14\end{array}\right.\)

Bài 2:

a: 4x-16+3y(4-x)

=4(x-4)-3y(x-4)

=(x-4)(4-3y)

b: \(9y^2-6y+1=\left(3y\right)^2-2\cdot3y\cdot1+1^2=\left(3y-1\right)^2\)

c: \(25x^2-4=\left(5x\right)^2-2^2=\left(5x-2\right)\left(5x+2\right)\)

d: \(x^2-12x+36=x^2-2\cdot x\cdot6+6^2=\left(x-6\right)^2\)

e: \(8x^3+36x^2+54x+27\)

\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot3+3\cdot2x\cdot3^2+3^3\)

\(=\left(2x+3\right)^3\)

f: \(\left(2x-5\right)^2-\left(2x+y\right)^2\)

=(2x-5-2x-y)(2x-5+2x+y)

=(-y-5)(4x+y-5)

g: \(\left(2x-y\right)^3+\left(2x+y\right)^3\)

\(=8x^3-12x^2y+6xy^2-y^3+8x^3+12x^2y+6xy^2+y^3\)

\(=16x^3+12xy^2=4x\left(4x^2+3y^2\right)\)

Câu 1:

a: \(6x^2-72x=0\)

=>\(6\left(x^2-12x\right)=0\)

=>\(x^2-12x=0\)

=>x(x-12)=0

=>\(\left[\begin{array}{l}x=0\\ x-12=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=12\end{array}\right.\)

b: \(-2x^4+16x=0\)

=>\(-2x\left(x^3-8\right)=0\)

=>\(x\left(x^3-8\right)=0\)

=>\(\left[\begin{array}{l}x=0\\ x^3-8=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x^3=8\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=2\end{array}\right.\)

c: \(\left(2x-1\right)^3-8x\left(x-3\right)\cdot\left(x+3\right)=-1\)

=>\(8x^3-12x^2+6x-1-8x\cdot\left(x^2-9\right)=-1\)

=>\(8x^3-12x^2+6x-1-8x^3+72x=-1\)

=>\(-12x^2+78x=0\)

=>-6x(2x-13)=0

=>x(2x-13)=0

=>\(\left[\begin{array}{l}x=0\\ 2x-13=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=\frac{13}{2}\end{array}\right.\)

d: \(x\left(x-5\right)-\left(x-3\right)^2=0\)

=>\(x^2-5x-\left(x^2-6x+9\right)=0\)

=>\(x^2-5x-x^2+6x-9=0\)

=>x-9=0

=>x=9

e: \(x\left(x-5\right)+3\left(x-5\right)=0\)

=>(x-5)(x+3)=0

=>\(\left[\begin{array}{l}x-5=0\\ x+3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5\\ x=-3\end{array}\right.\)

f: 2x(x-8)-5(8-x)=0

=>2x(x-8)+5(x-8)=0

=>(x-8)(2x+5)=0

=>\(\left[\begin{array}{l}x-8=0\\ 2x+5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=8\\ x=-\frac52\end{array}\right.\)

g: \(30x-15x^2=0\)

=>15x(2-x)=0

=>x(2-x)=0

=>\(\left[\begin{array}{l}x=0\\ 2-x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=2\end{array}\right.\)

h: \(-4x^3-12x=0\)

=>\(-4x\left(x^2+3\right)=0\)

=>x=0

Tham khảo

\(-5^my+15x^ny\)

\(=5y\left(3x^n-5^{m-1}\right)\)

Tham khảo~

Xét 2 trường hợp : 

Trường hợp 1 :

\(m\ge n\)

\(\Rightarrow-5^my+15x^my\)

\(=-5x^n.x^{m-n}y+15x^ny\)

\(=-5x^ny\left(x^{m-n}-3\right)\)

Trường hợp 2 :

\(m< n\)

\(\Rightarrow-5^my+15x^ny\)

\(=-5x^my+15x^m.x^{n-m}y\)

\(=-5x^my\left(1-3x^{n-m}\right)\)

a) \(15x^ny^{2n}-3x^{n+1}\left(-y\right)^{2n}\)

\(=x^ny^{2n}\left(15-3x\right)\)

\(=3x^ny^{2n}\left(5-x\right)\)

b) \(4x^{2n}y^{n-1}+2\left(-x\right)^{2n+1}y^n\)

\(=4x^{2n}y^{n-1}-2x^{2n+1}y^n\)

\(=2x^{2n}y^{n-1}\left(2-xy\right)\)

1. -5xY(X^m-1-3X^n-1)

Sửa đề: \(AB^2+HC^2=AC^2+HB^2\)

ΔAHB vuông tại H

=>\(HA^2+HB^2=AB^2\)

=>\(HA^2=AB^2-HB^2\left(1\right)\)

ΔAHC vuông tại H

=>\(HA^2+HC^2=AC^2\)

=>\(HA^2=AC^2-HC^2\left(2\right)\)

Từ (1),(2) suy ra \(AB^2-HB^2=AC^2-HC^2\)

=>\(AB^2+HC^2=AC^2+HB^2\)

a: \(\dfrac{x^ny^6}{x^5y^{n-2}}=x^{n-5}y^{8-n}\)

Để đây là phép chia hết thì n-5>=0và 8-n>=0

=>5<=n<=8

b: \(\dfrac{x^6y^{n+2}}{x^ny^4z^{n-3}}=x^{6-n}y^{n-4}z^{3-n}\)

Để đây là phép chia hết thì \(\left\{{}\begin{matrix}6-n>=0\\n-4>=0\\3-n>=0\end{matrix}\right.\Leftrightarrow n\in\varnothing\)

c: \(\dfrac{\left(\dfrac{1}{2}x^5y^{7-n}\right)}{-2x^ny^3}=-\dfrac{1}{4}x^{5-n}y^{4-n}\)

Để đây là phép chia hết thì 5-n>=0 và 4-n>=0

=>n<=4

a.x^4:x^n=x^4-2

b,x^n:x^3=x^n-3

a. x⁴ : xⁿ = x⁴ : x² = x²

b. xⁿ : x³ = x⁴ : x³ = x

c. 5xⁿy³ : 4x²y² = 5x⁴y³ : 4x²y²

d. xⁿyⁿ⁺¹ : x²y⁵ = x⁴y⁶⁺¹ : x²y⁵ = x⁴y⁷ : x²y⁵ = x²y²

\(\frac{36}{6}\) = 6

\(\frac{49}{7}\) = 7

\(\frac{64}{8}\) = 8