1/3x^2-5/6=1
3x√ x-1 +2√ (x-1)^3=0 ......... square root symbol is over the x-1 and x-1^3
x^3-26x^3/2=27
2x^2(2x-1)^1/3=4x(2x-1)^4/3
15/2^x-1=7
3^x-1 + 3^x+1 =30
Any help on any of thse problems would be appreciated!
I can't read the first, third, fourth, fifth and sixth questions. Some brackets would be nice... LaTeX would be even nicer...
$\displaystyle \displaystyle \begin{align*} 3x\sqrt{x-1} + 2\sqrt{(x - 1)^3} &= 0 \\ 3x\sqrt{x - 1} + 2\sqrt{x-1}\sqrt{(x - 1)^2} &= 0 \\ 3x\sqrt{x - 1} + 2\sqrt{x - 1}|x - 1| &= 0 \\ \sqrt{x - 1}\left(3x + 2|x - 1|\right) &= 0 \\ \sqrt{x - 1} = 0 \textrm{ or } 3x + 2|x - 1| &= 0 \end{align*}$
Go from here...
Hello, etran3358!
I'll guess what the last one means . . .
3^x-1+3^x+1=30
3^x/3+3.3^x=30
10/3.3^x=30
3^x=9
3^x=3^2
x=2
If my solution is hard to read and understand,
it's because I typed it the way you did ...
no parentheses, no spaces.
This is what I meant . . .
We are given: .3^{x-1} + 3^{x+1} = 30
which equals: .(3^x)/3 + 3(3^x) = 30
add: .(10/3)(3^x) = 30
Multiply by 3/10: .3^x = 9
Hence: .3^x = 3^2
Therefore: .x = 2
Hello, etran3358!
You still need parentheses,
or learn to use LaTeX more accurately.
$\displaystyle 1/3 x^2-5/6x=1 $ is #1.
Can you see that this looks like: .$\displaystyle \frac{1}{3x^2} -= \frac{5}{6x} \:=\:1\,?$
But it might be: .$\displaystyle \frac{1}{3}x^2 - \frac{5}{6}x \:=\:1$
Hello, etran3358!
$\displaystyle 2x^2 (2x-1) ^{\frac{1}{3}} \:=\: 4x(2x-1) ^{\frac{4}{3}}$
We have: .$\displaystyle 2x^2(2x-1)^{\frac{1}{3}} - 4x(2x-1)^{\frac{4}{3}} \:=\:0$
Factor: . .$\displaystyle 2x\cdot(2x-1)^{\frac{1}{3}}\big[x - 2(2x-1)\big] \:=\:0$
. . . . . . . .$\displaystyle 2x\cdot(2x-1)^{\frac{1}{3}}\cdot(2-3x) \:=\:0$
Therefore:
. . $\displaystyle 2x \:=\:0 \quad\Rightarrow\quad x \:=\:0$
. . $\displaystyle (2x-1)^{\frac{1}{3}} \:=\:0 \quad\Rightarrow\quad 2x-1 \:=\:0 \quad\Rightarrow\quad x \:=\:\tfrac{1}{2}$
. . $\displaystyle 2 - 3x \:=\:0 \quad\Rightarrow\quad x \:=\:\tfrac{2}{3}$