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4 tháng 10 2021
1) \(\dfrac{1}{27}+a^3=\left(\dfrac{1}{3}+a\right)\left(\dfrac{1}{9}-\dfrac{a}{3}+a^2\right)\)
2) \(=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
3) \(=\left(\dfrac{1}{2}x+2y\right)\left(\dfrac{1}{4}x-xy+4y^2\right)\)
4) \(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)
5) \(=\left(x^3+1\right)\left(x^6-x^3+1\right)\)
6) \(=\left(x-4\right)\left(x^2+4x+16\right)\)
7) \(=\left(x-5\right)\left(x^2+5x+25\right)\)
8) \(=\left(2x^2-3y\right)\left(4x^4+6x^2y+9y^2\right)\)
9) \(=\left(\dfrac{1}{4}x^2-5y\right)\left(\dfrac{1}{16}x^4+\dfrac{5}{4}x^2y+25y^2\right)\)
10) \(=\left(\dfrac{1}{2}x-2\right)\left(\dfrac{1}{4}x^2+x+4\right)\)
11) \(=\left(x+2\right)^3\)
12) \(=\left(x+3\right)^3\)
Y
14 tháng 8 2018
a) \(\left(5x-1\right)^6=729\)
\(\Rightarrow\left[{}\begin{matrix}\left(5x-1\right)^6=3^6\\\left(5x-1\right)^6=\left(-3\right)^6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}5x-1=3\\5x-1=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}5x=4\\5x=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\)
b) \(\dfrac{8}{25}=\dfrac{2^x}{5^{x-1}}\)
\(\Rightarrow\left[{}\begin{matrix}2^x=2^3\\5^{x-1}=5^2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x-1=2\end{matrix}\right.\)
\(\Rightarrow x=3\)
Vậy x = 3
c) \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{2}\right)^{10}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{3x}=\left(\dfrac{1}{2}\right)^{10}\)
\(\Rightarrow3x=10\)
\(\Rightarrow x=\dfrac{10}{3}\)
d) \(9^x:3^x=3\)
\(\Rightarrow\left(9:3\right)^x=3\)
\(\Rightarrow3^x=3^1\)
\(\Rightarrow x=1\)
11 tháng 5 2021
`(x-4)/(x-3)-5/(3+x)=-27/(9-x^2)(x ne +-3)`
`<=>(x-4)/(x-3)-5/(x+3)-27/(x^2-9)=0`
`<=>(x-4)(x+3)-5(x-3)-27=0`
`<=>x^2-x-12-5x+15-27=0`
`<=>x^2-6x-16=0`
`<=>x^2+2x-8x-16=0`
`<=>(x+2)(x-8)=0`
`<=>x=-2\or\x=8`
`S={-2,8}`
28 tháng 7 2021
a) Ta có: \(\dfrac{x}{x-3}-\dfrac{6}{x}-\dfrac{9}{x^2-3x}\)
\(=\dfrac{x^2}{x\left(x-3\right)}-\dfrac{6\left(x-3\right)}{x\left(x-3\right)}-\dfrac{9}{x\left(x-3\right)}\)
\(=\dfrac{x^2-6x+18-9}{x\left(x-3\right)}\)
\(=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)
b) Ta có: \(\dfrac{7}{x}-\dfrac{x}{x+6}+\dfrac{36}{x^2+6x}\)
\(=\dfrac{7\left(x+6\right)-x^2+36}{x\left(x+6\right)}\)
\(=\dfrac{7x+42-x^2+36}{x\left(x+6\right)}\)
\(=\dfrac{-\left(x^2-7x-78\right)}{x\left(x+6\right)}\)
\(=\dfrac{-\left(x^2-13x+6x-78\right)}{x\left(x+6\right)}\)
\(=\dfrac{-\left[x\left(x-13\right)+6\left(x-13\right)\right]}{x\left(x+6\right)}\)
\(=\dfrac{13-x}{x}\)
c) Ta có: \(\dfrac{6}{x-3}-\dfrac{2x-6}{x^2-9}-\dfrac{4}{x+3}\)
\(=\dfrac{6\left(x+3\right)-2x+6-4\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{6x+18-2x+6-4x+12}{\left(x-3\right)\left(x+3\right)}=\dfrac{36}{\left(x-3\right)\left(x+3\right)}\)
NV
Nguyễn Việt Lâm
Giáo viên
4 tháng 12 2018
Xét \(x^{27}+x^{24}+x^{21}+x^{18}+x^{15}+x^{12}+x^9+x^6+x^3+1\)
\(=\left(x^{27}+x^{21}+x^{15}+x^9+x^3\right)+\left(x^{24}+x^{18}+x^{12}+x^6+1\right)\)
\(=x^3\left(x^{24}+x^{18}+x^{12}+x^6+1\right)+\left(x^{24}+x^{18}+x^{12}+x^6+1\right)\)
\(=\left(x^3+1\right)\left(x^{24}+x^{18}+x^{12}+x^6+1\right)\)
Vậy ta có
\(VT=\dfrac{x^{24}+x^{18}+x^{12}+x^6+1}{\left(x^3+1\right)\left(x^{24}+x^{18}+x^{12}+x^6+1\right)}=\dfrac{1}{x^3+1}\) (đpcm)
TH
2
3 tháng 2 2019
b) x-45/55 + x-47/53 = x-55/45 + x-53/47
<=>x-45/55 -1 + x-47/53 -1= x-55/45 -1 + x-53/47 - 1
<=>x-100/55 + x-100/53 = x-100/45 + x-100/47
<=>(x-100)(1/55+1/53-1/45-1/47)=0
<=>x-100=0
<=>x=100
Vậy x = 100
24 tháng 11 2021
\(\dfrac{2}{36a^2b^2-1}=\dfrac{2}{\left(6ab-1\right)\left(6ab+1\right)}\\ \dfrac{1}{6ab+1}=\dfrac{6ab-1}{\left(6ab-1\right)\left(6ab+1\right)};\dfrac{1}{6ab-1}=\dfrac{6ab+1}{\left(6ab-1\right)\left(6ab+1\right)}\)
\(\dfrac{x}{x^3-27}=\dfrac{x\left(x-3\right)}{\left(x-3\right)^2\left(x^2+3x+9\right)}\\ \dfrac{2x}{x^2-6x+9}=\dfrac{2x\left(x^2+3x+9\right)}{\left(x-3\right)^2\left(x^2+3x+9\right)}\\ \dfrac{1}{x^2+3x+9}=\dfrac{\left(x-3\right)^2}{\left(x-3\right)^2\left(x^2+3x+9\right)}\)
\(\dfrac{x^2-x}{x^2-1}=\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x}{x+1}=\dfrac{x\left(x+1\right)}{\left(x+1\right)^2}\\ \dfrac{3x}{x^3+2x^2+x}=\dfrac{3x}{x\left(x^2+2x+1\right)}=\dfrac{3}{\left(x+1\right)^2}\\ 2x=\dfrac{2x\left(x+1\right)^2}{\left(x+1\right)^2}\)
\(\dfrac{x^3}{27}+\dfrac{x^6}{729}-x^9=\dfrac{x}{3}+\dfrac{x}{3}-\dfrac{3x^9}{3}=\dfrac{2x-3x^9}{3}=\dfrac{x\left(2-3x^8\right)}{3}\)