viết các bieetr thức sau dưới dạng tích
a. 9x^2+30x+25;\(\dfrac{4}{9}\)x^4-16x^2
b. \(\dfrac{12}{5}\)x^2y^2-9x^4-\(\dfrac{4}{25}\)y^4
c. a^2y^2+b^2x^2-2axby
d. 64x^2-(8a+b)^2
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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)
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:
\(\left\{{}\begin{matrix}x_1=\dfrac{4\sqrt{5}-8}{2}=2\sqrt{5}-4\\x_2=2\sqrt{5}+4\end{matrix}\right.\)
\(\sqrt{9x^2-6x+1}+\sqrt{25-30+9x^2}\)
=\(\sqrt{\left(3x-1\right)^2}+\sqrt{\left(5-3x\right)^2}\)
=|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`
a: \(9x^2+30x+25=\left(3x+5\right)^2\)
b: \(\dfrac{4}{9}x^4-16x^2=x^2\left(\dfrac{4}{9}x^2-16\right)=x^2\left(\dfrac{2}{3}x-4\right)\left(\dfrac{2}{3}x+4\right)\)
c: \(\dfrac{12}{5}x^2y^2-9x^4-\dfrac{4}{25}y^4\)
\(=-\left(9x^4-\dfrac{12}{5}x^2y^2+\dfrac{4}{25}y^4\right)\)
\(=-\left(3x^2-\dfrac{2}{5}y^2\right)^2\)