Tìm x biết 9x-1=9
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.


(1-3x2)-(x-2)(9x+1)=(3x-4)(3x+4)-9(x+3)2
⇒1-3x2-(9x2+x-18x-2)=9x2-16-9(x2+6x+9)
⇒1-3x2-(9x2-17x-2)= -56x-97
⇒1-3x2-9x2+17x+2=-56x-97
⇒3-12x2+17x=-56x-97
⇒3-12x2+17x+56x+97=0
⇒-12x2+73x+100=0
⇒-(12x2-73x-100)=0

`1,`
`538 - x = 275`
`\Rightarrow x = 538 - 275`
`\Rightarrow x = 263`
Vậy, `x = 263`
`2,`
`45 - 9x = 18`
`\Rightarrow 9x = 45 - 18`
`\Rightarrow 9x = 27`
`\Rightarrow x = 27 \div 9`
`\Rightarrow x = 3`
Vậy, `x = 3`
`3,`
`(5x - 9) \div 3 = 12`
`\Rightarrow 5x - 9 = 12. 3`
`\Rightarrow 5x - 9 = 36`
`\Rightarrow 5x = 36 + 9`
`\Rightarrow 5x = 45`
`\Rightarrow x = 45 \div 5`
`\Rightarrow x = 9`
Vậy, `x = 9.`
1) \(538-x=275\)
\(x=538-275\)
\(x=263\)
2) \(45-9x=18\)
\(9x=27\)
\(x=3\)
3) \(\left(5x-9\right)\div3=12\)
\(\left(5x-9\right)=12\times3\)
\(5x-9=36\)
\(5x=45\)
\(x=9\)

\(\left|x^3+x\right|-\left|9x^2+9\right|=0\)
\(\Leftrightarrow\left|x\left(x^2+1\right)\right|-9\left|x^2+1\right|=0\)
\(\Leftrightarrow\left(\left|x\right|-9\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left|x\right|=9\left(x^2+1\ge1>0\right)\Leftrightarrow x=\pm9\)
Vậy ...
\(\left|x^3+x\right|-\left|9x^2+9\right|=0\)
\(TH1:\left\{{}\begin{matrix}\left|x^3+x\right|=0\\\left|9x^2+9\right|=0\end{matrix}\right.\)
\(\text{Vì }9x^2\ge0\)
\(\Rightarrow9x^2+9\ge9\)
\(TH2:\left|x^3+x\right|=\left|9x^2+9\right|\)
\(\Rightarrow\left[{}\begin{matrix}x^3+x=9x^2-9\\x^3+x=9x^2+9\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^3+x+9x^2+9=0\\x^3+x-9x^2-9=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x.\left(x^2+1\right)+9.\left(x^2+1\right)=0\\x.\left(x^2+1\right)-9.\left(x^2+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=9\end{matrix}\right.\)

=>|x^3+x|=|9x^2+9|
=>x^3+x=9x^2+9 hoặc x^3+x=-9x^2-9
=>x^3-9x^2+x-9=0 hoặc x^3+9x^2+x+9=0
=>x+9=0 hoặc (x-9)(x^2+1)=0
=>x=9 hoặc x=-9

\(a)\)\(\left(x+1\right)\left(x+3\right)-x\left(x-1\right)=8\)
\(\Leftrightarrow x^2+4x+3-x^2+x=8\)
\(\Leftrightarrow5x=5\)
\(\Leftrightarrow x=1\)
Vậy x = 1.
\(b)\)\(9x^2=1-\left(3x+1\right)\left(2x-9\right)\)
\(\Leftrightarrow\left(1-9x^2\right)-\left(3x+1\right)\left(2x-9\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(1-3x\right)-\left(3x+1\right)\left(2x-9\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(1-3x+9-2x\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(10-5x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x+1=0\\10-5x=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x=-1\\5x=10\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{3}\\x=2\end{cases}}\)
Vậy\(x=-\frac{1}{3}\)hoặc\(x=2\)
Dumflinz

a,x+x2+x3+....+x9=459-9
<=>x(1+2+3+...+9)=450
<=>x45=450
<=>x=10
b ,thì tương tự nhé b


Theo bai ta co: 9^x-1=9^1(vi9=9^1)
vi tim so mu nen ta co: x-1=1
x =1+1
x =2
Vay 9^2-1=9 ==> x=2