|6x-x^2-9|=6x-x^2-9
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KV
0
KV
1
NH
0
NN
1
8 tháng 7 2016
\(\frac{1}{x^2+6x+9}+\frac{1}{6x-x^2-9}+\frac{x}{x^2-9}\)
\(=\frac{1}{\left(x+3\right)^2}+-\frac{1}{\left(x-3\right)^2}+\frac{x}{\left(x+3\right)\left(x-3\right)}\)
\(=\frac{\left(x-3\right)^2-\left(x+3\right)^2+x\left(x+3\right)\left(x-3\right)}{\left(x+3\right)^2\left(x-3\right)^2}\)
\(=\frac{x^2-6x+9-x^2-6x-9+x^3-9x}{\left(x+3\right)^2\left(x-3\right)^2}\)
\(=\frac{x^3-21x}{\left(x+3\right)^2\left(x-3\right)^2}\)
QA
2
7 tháng 8 2016
\(\frac{1}{x^2+6x+9}+\frac{1}{6x-x^2-9}+\frac{x}{x^2-9}\)
\(=\frac{1}{\left(x+3\right)^2}+\frac{-1}{\left(x-3\right)^2}+\frac{x}{\left(x+3\right)\left(x-3\right)}\)
\(=\frac{\left(x-3\right)^2-\left(x+3\right)^2+x\left(x+3\right)\left(x-3\right)}{\left(x+3\right)^2\left(x-3\right)^2}\)
\(=\frac{x^2-6x+9-x^2-6x-9+x^3-9x}{\left(x+3\right)^2\left(x-3\right)^2}\)
\(=\frac{x^3-21x}{\left(x+3\right)^2\left(x-3\right)^2}\)
7 tháng 8 2016
\(\frac{1}{x^2+6x+9}+\frac{1}{6x-x^2-9}+\frac{x}{x^2-9}\)
\(=\frac{1}{\left(x+3\right)^2}-\frac{1}{\left(x-3\right)^2}+\frac{x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{\left(x-3\right)^2}{\left(x+3\right)^2\left(x-3\right)^2}-\frac{\left(x+3\right)^2}{\left(x+3\right)^2\left(x-3\right)^2}+\frac{x\left(x+3\right)\left(x-3\right)}{\left(x+3\right)^2\left(x-3\right)^2}\)
\(=\frac{x^2-6x+9-x^2-6x-9+x^3-9x}{\left(x+3\right)^2\left(x-3\right)^2}\)
\(=\frac{x^3-21x}{\left(x+3\right)^2\left(x-3\right)^2}\)
TC
21 tháng 5 2023
a) `sqrt(x^2-6x _9) = 4-x`
`<=> sqrt[(x-3)^2] =4-x`
`<=> |x-3| =4-x ( đk :x<=4)`
`<=> |x-3| = |4-x|`
`<=> [(x-3 =4-x),(x-3 = x-4):}`
`<=>[(x = 7/2(t//m)),(0=-1(vl)):}`
Vậy `S = {7/2}`
b) `sqrt(x^2 -9) + sqrt(x^2 -6x +9) =0(đk : x>=3(hoặc) x<=-3)`
`<=>sqrt(x^2 -9) =- sqrt(x^2 -6x +9) `
`<=>(sqrt(x^2 -9))^2 =(- sqrt(x^2 -6x +9))^2`
`<=> x^2 -9 = x^2 -6x +9`
`<=> 6x = 9+9 =18`
`<=> x=3(t//m)`
Vậy `S={3}`
TC
21 tháng 5 2023
c) `sqrt(x^2 -2x+1) + sqrt(x^2-4x+4) =3`
`<=> sqrt[(x-1)^2] +sqrt[(x-2)^2] =3`
`<=> |x-1| +|x-2| =3`
xét `x<1 =>{(|x-1| =1-x ),(|x-2|=2-x):}`
`=> 1-x +2-x =3`
`=> x = 0(t//m)`
xét `1<=x<2 => {(|x-1|=x-1),(|x-2|= 2-x):}`
`=> x-1 +2-x =3`
`=>1=3 (vl)`
xét `x>=2 => {(|x-1| =x-1),(|x-2|=x-2):}`
`=> x-1+x-2 =3`
`=> x=3(t//m)`
Vậy `S = {0;3}`
H
1 tháng 12 2018
\(C=\frac{1}{x^2+6x+9}+\frac{1}{6x-x^2-9}+\frac{x}{x^2-9}.\)
\(C=\frac{1}{\left(x+3\right)^2}+\frac{-1}{-\left(6x-x^2-9\right)}+\frac{x}{\left(x+3\right)\left(x-3\right)}\)
\(C=\frac{1}{\left(x+3\right)^2}+\frac{-1}{-6x+x^2+9}+\frac{x}{\left(x+3\right)\left(x-3\right)}\)
\(C=\frac{x-3}{\left(x+3\right)\left(x-3\right)}+\frac{-\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{x}{\left(x+3\right)\left(x-3\right)}\)
\(C=\frac{x-3.-x-3.x}{\left(x+3\right).\left(x-3\right)}=\frac{-6x}{\left(x+3\right)\left(x-3\right)}=\frac{-6x}{\left(x^2-9\right)}\)
TN
0
\(\left|6x-x^2-9\right|=6x-x^2-9\Leftrightarrow6x-x^2-9\ge0\)
\(\Leftrightarrow x^2-6x+9\le0\Leftrightarrow\left(x-3\right)^2\le0\)
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Tập nghiệm của phương trình là \(S=\left\{3\right\}\)