Bài 7: Phân tích đa thức thành nhân tử bằng phương pháp dùng hằng đẳng thức

a: \(x^2-8x+16x=x^2+8x=x\left(x+8\right)\)

b: \(4x^2-8xyz+4y^2=4\left(x^2-2xyz+y^2\right)\)

c: \(ab^2+\dfrac{1}{4}a^2b^4+1=\left(\dfrac{1}{2}ab^2+1\right)^2\)

\(9x^2y^2+15x^2y-21xy^2\\ =3xy\left(3xy+5x-7y\right)\)

Ta có: \(9x^2y^2+15x^2y-21xy^2\)

\(=3xy\left(3xy+5x-7y\right)\)

d) \(4x^4-x^2=x^2\left(4x^2-1\right)=x^2\left(2x-1\right)\left(2x+1\right)\)

e) Ta có: \(6x^2-7x-5\)

\(=6x^2-10x+3x-5\)

\(=2x\left(3x-5\right)+\left(3x-5\right)\)

\(=\left(3x-5\right)\left(2x+1\right)\)

f: Ta có: \(-4x^2+23x-15\)

\(=-4x^2+20x+3x-15\)

\(=-4x\left(x-5\right)+3\left(x-5\right)\)

\(=\left(x-5\right)\left(-4x+3\right)\)

a) \(\left(n+3\right)^2-\left(n-1\right)^2=\left(n+3-n+1\right)\left(n+3+n-1\right)\)

\(=4\left(2n+2\right)=8\left(n+1\right)\)

=> \(\left(n+3\right)^2-\left(n-1\right)^2\) chia hết cho 8

b) \(\left(n+6\right)^2-\left(n-6\right)^2=\left(n+6-n+6\right)\left(n+6+n-6\right)\)

\(=12\left(2n\right)=24n\) 

\(=>\left(n+6\right)^2-\left(n-6\right)^2\)chia hết cho 24

a) \(\left(n+3\right)^2-\left(n-1\right)^2=\left(n+3-n+1\right)\left(n+3+n-1\right)=4\left(2n+2\right)=4.2\left(n+1\right)=8\left(n+1\right)⋮8\)

b) \(\left(n+6\right)^2-\left(n-6\right)^2=\left(n+6-n+6\right)\left(n+6+n-6\right)=12.2n=24n⋮24\)

Mik ra đáp án là

a) Th1:x=3

Th2:x=2

b) Th1:x=0

Th2:x=5/2

c) x=2

a) \(\left(2x-1\right)^2-25=0\)

\(\Leftrightarrow\left(2x-1\right)^5=25\)

\(\Leftrightarrow2x-1=5\)

\(\Leftrightarrow x=3\)

b) \(8x^3-50x=0\)

\(\Leftrightarrow2x\left(4x^2-25\right)=0\)

\(\Leftrightarrow2x\left(x^2-\dfrac{25}{4}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-\dfrac{25}{4}=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm\dfrac{5}{2}\end{matrix}\right.\)

c) \(\left(x-2\right)\left(x^2+2x+7\right)+2\left(x^2-4\right)-5\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+7+2x+4-5\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+4x+6\right)=0\) (*)

mà \(x^2+4x+6=\left(x+2\right)^2+2\ge2\forall x\)

(*) \(\Rightarrow x-2=0\Rightarrow x=2\)

a) \(x^2-6xy+9y^2=x^2-2.3.x.y+\left(3y\right)^2=\left(x-3y\right)^2\)

b) \(x^3+6x^2y+12xy^2+8y^3=x^3+3.2.x^2y+3.x.\left(2y\right)^2+\left(2y\right)^3=\left(x+2y\right)^3\)

c) \(x^3-64=x^3-4^3=\left(x-4\right)\left(x^2+4x+16\right)\)

d) \(125x^3+y^6=\left(5x\right)^3+\left(y^2\right)^3=\left(5x+y^2\right)\left(25x^2-5xy^2+y^4\right)\)

e)

 \(0,125\left(a+1\right)^3-1\\ =\left[0,5.\left(a+1\right)\right]^3-1^3\\ =\left(0,5a+0,5\right)^3-1^3\\ =\left(0,5a+0,5-1\right)\left[\left(0,5a+0,5\right)^2+\left(0,5a+0,5\right).1+1^2\right]\\ =\left(0,5a-0,5\right)\left(0,25a^2+0,5a+0,25+0,5a+0,5+1\right)\\ =\left(0,5a-0,5\right)\left(0,25a^2+a+1,75\right)\)

Bài 5: 

a) Ta có: \(\left(2x+1\right)^2-\left(x-1\right)^2\)

\(=\left(2x+1-x+1\right)\left(2x+1+x-1\right)\)

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

b: Ta có: \(9\left(x+5\right)^2-\left(x-7\right)^2\)

\(=\left(3x+15\right)^2-\left(x-7\right)^2\)

\(=\left(3x+15-x+7\right)\left(3x+15+x-7\right)\)

\(=\left(2x+22\right)\left(4x+8\right)\)

\(=8\left(x+11\right)\left(x+2\right)\)

c: Ta có: \(25\left(x-y\right)^2-16\left(x+y\right)^2\)

\(=\left(5x-5y\right)^2-\left(4x+4y\right)^2\)

\(=\left(5x-5y-4x-4y\right)\left(5x-5y+4x+4y\right)\)

\(=\left(x-9y\right)\left(9x-y\right)\)

d: Ta có: \(49\left(y-4\right)^2-9\left(y+2\right)^2\)

\(=\left(7y-28\right)^2-\left(3y+6\right)^2\)

\(=\left(7y-28-3y-6\right)\left(7y-28+3y+6\right)\)

\(=\left(4y-34\right)\left(10y-22\right)\)

\(=4\left(2y-17\right)\left(5y-11\right)\)

Bài 4:

a) \(2x\left(x+1\right)+2\left(x+1\right)=2\left(x+1\right)^2\)

b: \(y^2\left(x^2+y\right)-x^2z-yz\)

\(=y^2\left(x^2+y\right)-z\left(x^2+y\right)\)

\(=\left(x^2+y\right)\left(y^2-z\right)\)

c: \(4x\left(x-2y\right)+8y\left(2y-x\right)\)

\(=4x\left(x-2y\right)-8y\left(x-2y\right)\)

\(=4\left(x-2y\right)^2\)

b: \(\left(a-b\right)^2-c^2=\left(a-b-c\right)\left(a-b+c\right)\)

c: \(4x^2+12x+9=\left(2x+3\right)^2\)

d: \(25x^2-20xy+4y^2=\left(5x-2y\right)^2\)

e: \(8x^6-27y^3=\left(2x^2-3y\right)\left(4x^2+6x^2y+9y^2\right)\)

bài 3

a) \(x^2-6xy+9y^2=\left(x-3y\right)^2\)

b) \(x^3+6x^2y+12xy^2+8y^3=\left(x+2y\right)^3\)

c)\(x^3-64=x^3-4^3=\left(x-4\right)\cdot\left(x+4x+16\right)\)

d) 125x^3 +y^6= (5x)^3+(y^2)^3= (5x+y^2) *( 25x^2- 10xy+y^4) (cái này mà còn rút gọn được nữa thì bạn rút nhé)

e) \(0,5^3\left(a+1\right)^3-1=\left(0,5a+0,5\right)^3-1^3\)

(máy mik lỗi k gõ đc tiếp)

4/

a) \(2x\left(x+1\right)+2\left(x+1\right)=\left(x+1\right)\left(2x+2\right)=\left(x+1\right)\left(x+1\right)\cdot2\)

b) \(y^2\left(x^2+y\right)-zx^2-zy=y^2\left(x^2+y\right)-z\left(x^2+y\right)=\left(x^2+y\right)\left(y^2-z\right)\)

C)\(4x\left(x-2y\right)+8y\left(2y-x\right)=4x\left(x-2y\right)-8y\left(x-2y\right)=\left(4x-8y\right)\left(x-2y\right)=2\cdot\left(2x-4y\right)\left(x-2y\right)\)

d) (bạn đặt nhân tử chung x+1 ra ngoài rồi rút gọn tiếp  tự làm là tốt)

chúc bạn ngày càng tiến bộ

x(x^2-16)=0

=>x^2-16=0

=>x^2=16

=>x=+-4