LN
1
K
Khách
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22 tháng 12 2021
Bài 2:
\(\Leftrightarrow\left(x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
21 tháng 10 2021
Bài 4:
\(x^3-2x^2+x=x\left(x-1\right)^2\)
\(5\left(x-y\right)-y\left(x-y\right)=\left(x-y\right)\left(5-y\right)\)
\(x^2-12x+36=\left(x-6\right)^2\)
TV
2
KN
8 tháng 3 2020
\(\frac{x^2+6x+9}{\left(x-1\right)^2}.\frac{2x^2-4x-2}{4x^2+24x+36}\)
\(=\frac{x^2+6x+9}{\left(x-1\right)^2}.\frac{2x^2-4x-2}{4\left(x^2+6x+9\right)}\)
\(=\frac{1}{\left(x-1\right)^2}.\frac{2x^2-4x-2}{4}\)
\(=\frac{2x^2-4x-2}{4x^2-8x+4}\)
8 tháng 3 2020
\(\frac{x^2+6x+9}{\left(x-1\right)^2}.\frac{2x^2-4x-2}{4x^2+24x+36}\)
\(=\frac{x^2+2\left(x\right)\left(3\right)+3^2}{\left(x-1\right)^2}.\frac{2x^2-4x-2}{4x^2+24x+36}\)
\(=\frac{\left(x+3\right)^2}{\left(x-1\right)^2}.\frac{2x^2+4x-2}{4x^2+24x+36}\)
\(=\frac{\left(x+3\right)^2}{\left(x-1\right)^2}.\frac{2\left(x^2-2x-1\right)}{4x^2+24x+36}\)
\(=\frac{\left(x+3\right)^2}{\left(x-1\right)^2}.\frac{2\left(x^2-2x-1\right)}{4\left(x^2+2\left(x\right)\left(3\right)+3^2\right)}\)
\(=\frac{1}{\left(x-1\right)^2}.\frac{2\left(x^2-2x-1\right)}{4}\)
\(=\frac{1.2\left(x^2-2x-1\right)}{\left(x-1\right)^2.4}\)
\(=\frac{2\left(x^2-2x-1\right)}{4\left(x-1\right)^2}\)
\(=\frac{x^2-2x-1}{2\left(x-1\right)^2}\)
PK
0
13 tháng 11 2021
\(6x\left(4x-5\right)-24x^2=24x^2-30x-24x^2=-30x\)
ý B
PK
14 tháng 8 2016
Ta có: \(\frac{\left(x^2\right)^2-10x^2+9}{x^4+6x^3+9x^2+2x^3+12x^2+18x+x^2+6x+9}\)
= \(\frac{\left(x^2-1\right)\left(x^2-3\right)}{x^2\left(x^2+6x+9\right)+2x\left(x^2+6x+9\right)+\left(x^2+6x+9\right)}\)
= \(\frac{\left(x-1\right)\left(x+1\right)\left(x-3\right)\left(x+3\right)}{\left(x^2+6x+9\right)\left(x^2+2x+1\right)}\)
= \(\frac{\left(x-1\right)\left(x+1\right)\left(x-3\right)\left(x+3\right)}{\left(x+3\right)^2.\left(x+1\right)^2}\)
= \(\frac{\left(x-1\right)\left(x+1\right)\left(x-3\right)\left(x+3\right)}{\left(x+3\right)\left(x+3\right)\left(x+1\right)\left(x+1\right)}\)
= \(\frac{\left(x-1\right)\left(x-3\right)}{\left(x+1\right)\left(x+3\right)}\)
TH
1
YN
2 tháng 3 2022
`Answer:`
a, `4x^2-24x+36=(x-3)^3`
`<=>4(x^2-6x+9)-(x-3)^3=0`
`<=>4(x-3)^2-(x-3)^3=0`
`<=>(x-3)^2.(4-x+3)=0`
`<=>(x-3)^2.(7-x)=0`
`<=>x-3=0` hoặc `7-x=0`
`<=>x=3` hoặc `x=7`
b, `(8x^3-7x^2):x^2=3x+\sqrt{\frac{9}{25}}`
`<=>8x^3:x^2-7x^2:x^2=3x+\sqrt{\frac{9}{25}}`
`<=>8x-7=3x+\sqrt{\frac{9}{25}}`
`<=>8x-7=3x+3/5`
`<=>8x=3x+\frac{38}{5}`
`<=>8x-3x=3x+\frac{38}{5}-3x`
`<=>5x=\frac{38}{5}`
`<=>x=\frac{38}{25}`
Đkxđ : \(x\ne3;-3\)
Ta có :
\(\frac{4x^2-24x+36}{x^2-9}\)
\(=\frac{4\left(x^2-6x+9\right)}{x^2-3^2}\)
\(=\frac{4\left(x^2-2.3.x+3^2\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{4\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{4\left(x-3\right)}{x+3}\)