a. \(9x^2+30x+25=\left(3x+5\right)^2\)

b. \(\dfrac{4}{9}x^4-16x^2=\left(\dfrac{2}{3}x^2-4x\right)\left(\dfrac{2}{3}x^2+4x\right)=x^2\left(\dfrac{2}{3}x-4\right)\left(\dfrac{2}{3}x+4\right)\)

c. \(a^2y^2+b^2x^2-2axby=\left(ay-bx\right)^2\)

d. \(100-\left(3x-y\right)^2=\left(10-3x+y\right)\left(10+3x-y\right)\)

e. \(\dfrac{12}{5}x^2y^2-9x^4-\dfrac{4}{25}y^4=-\left(9x^4-\dfrac{12}{5}x^2y^2+\dfrac{4}{25}y^4\right)=-\left(3x^2-\dfrac{2}{5}y^2\right)^2\)

f. \(64x^2-\left(8a+b\right)^2=\left(8x-8a-b\right)\left(8x+8a+b\right)\)

g. \(27x^3-a^3b^3=\left(3x-ab\right)\left(9x^2+3xab+a^2b^2\right)\)

a) 9x²+30x+25=3²x²+2.3.5x+5²=(3x+5)²

b)12/5x²y²-9x⁴-4/25y⁴=-(9x⁴-12/5x²y²+4/25y⁴)=-(3x-2/5y)²

c)a²y²+b²x²2axby=(ax-by)²

d)64x²-(8a+b)²=(8x-8a-b)(8x+8a+b)

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

b)\(-x^2+10x-25=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)

c)\(\frac{1}{9}x^2+\frac{2}{3}xy+y^2=\left(\frac{1}{3}x+y\right)^2\)

\(a,-25+4x^2=4x^2-25=\left(2x-5\right)\left(2x+5\right)\)

\(b,-x^2+10x-25=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)

\(c,\frac{1}{9}x^2+\frac{2}{3}xy+y^2=\left(\frac{1}{3}x\right)^2+\frac{2.1}{3}xy+y^2=\left(\frac{1}{3}x+y\right)^2\)(sửa đề)

a) \(-25+4x=\left(2x-5\right)\left(2x+5\right)\)

b) \(-x^2+10x-25=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)

c) \(\frac{1}{9}x^2+\frac{2}{3}x+y^2=\left(\frac{1}{3}x+y\right)^2\)

Good luck:3 (bài này dễ mà - _ -)

\(-\left(3x^2-5\right)^2\)

\(- ( 3 x2 - 5)2\)

Phân tích xong ròi nhá

MAGICPENCIL 

HÃY LUÔN :-)

Làm bài 1 thôi !! Mấy bài kia tương tự . Tìm nhân tử chung ra .

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

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

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

c) \(-16+\left(x-3\right)^2=\left(x-3+4\right)\left(x-3-4\right)=x\left(x-7\right)\)

d) \(64+16y+y^2=\left(y+8\right)\left(y+8\right)\)

d, \(\Delta'=225-25.9=0\)pt có nghiệm kép 

\(x_1=x_2=\dfrac{-15}{9}=-\dfrac{5}{3}\)

e, \(\Delta'=4.5-4=16>0\)pt có 2 nghiệm pb 

\(x_1=2\sqrt{5}-4;x_2=2\sqrt{5}+4\)

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

=>3x+5=0

hay x=-5/3

e: \(\text{Δ}=\left(4\sqrt{5}\right)^2-4\cdot1\cdot4=80-16=64>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{4\sqrt{5}-8}{2}=2\sqrt{5}-4\\x_2=2\sqrt{5}+4\end{matrix}\right.\)

d, \(\Delta=30^2-9.4.25=0\)

Vậy pt có nghiệm kép:\(x_{1,2}=\dfrac{-b}{2a}=\dfrac{-30}{2.9}=\dfrac{-30}{18}=\dfrac{-5}{3}\)

e, \(\Delta=\left(-4\sqrt{5}\right)^2-4.1.4=80-16=64\)

\(x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{4\sqrt{5}+\sqrt{64}}{2.1}=\dfrac{4\sqrt{5}+8}{2}=4+2\sqrt{5}\)

\(x_1=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{4\sqrt{5}-\sqrt{64}}{2.1}=\dfrac{4\sqrt{5}-8}{2}=-4+2\sqrt{5}\)

\(\sqrt{9x^2-6x+1}+\sqrt{25-30+9x^2}\)

=\(\sqrt{\left(3x-1\right)^2}+\sqrt{\left(5-3x\right)^2}\)

=|3x-1|+|5-3x| ≥ |3x-1+5-3x|

<=> |3x-1|+|5-3x| ≥ |4|

=> Min A =4 khi (3x-1)(5-3x) ≥ 0

ta có bảng

=> x ≤ 1/3 hoặc x ≥ 5/3

vậy .....

`9x^2+4y^2-12xy+6x-4y+1`

`=(3x)^2-2.3x.2y+(2y)^2+2(3x-2y)+1`

`=(3x-2y)^2+2(3x-2y)+1`

`=(3x-2y+1)^2`

9x²+4y²-12xy+6x-4y+1=(3x-2y+1)²