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\(3^{x+1}=9^x\\ \Leftrightarrow3^{x+1}=3^{2x}\\ \Leftrightarrow x+1=2x\\ \Leftrightarrow x=1\)
\(VP=9^x=\left(3^2\right)^x=3^{2x}\\ Vì:3^{x+1}=9^x=3^{2x}\\ Nên:x+1=2x\\ \Rightarrow2x-x=1\\ Vậy:x=1\)
`1)` Yêu cầu là gì ạ?
`2)`
`P(x)-Q(x)=`\((6x^3-3x^2+5x-1)-(-6x^3+3x^2-2x+7)\)
`= 6x^3-3x^2+5x-1+6x^3-3x^2+2x-7`
`= (6x^3+6x^3)+(-3x^2-3x^2)+(5x+2x)+(-1-7)`
`= 12x^3-6x^2+7x-8`
`3)`
`(-3x^3+15x^2+81x):(-3x)`
`= (-3x^3) \div (-3x) + 15x^2 \div (-3x) + 81x \div (-3x)`
`= x^2-5x-27`
\(\hept{\begin{cases}9x=4y\\3x-2y=-54\end{cases}\Leftrightarrow}\hept{\hept{\begin{cases}9x-4y=0\\3x-2y=-54\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}9x-4y=0\\6x-4y=-108\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}3x=108\\9x-4y=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=36\\9.36-4y=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=36\\324-4y=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=36\\4y=324\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=36\\y=81\end{cases}}\)
Vậy hệ phương trình có nghiệm (x,y)= \(\left(36,81\right)\)
ko ghi lại đề
<=> 36x 2-12 x -36x2+27x=30
<=>15x=30
<=> x=30:15
=>x=2
\(\left(3x^2-4\right)-9x\left(4x-3\right)=30\)
\(\Rightarrow3x^2-4-36x^2+27x=30\)
\(\Rightarrow-33x^2+27x-34=0\)
tớ chịu
{[(9x+3x-128) x 3] x 10} + 1000 =760
{[(9x+3x-128) x 3] x 10} =760 - 1000
{[(9x+3x-128) x 3] x 10} =-240
{[(9x+3x-128) x 3] x 10} =-240
{[x(9+3)-128] x 3} =-240 : 10
{[x(9+3)-128] x 3} =-24
{[12x-128] x 3} =-24
[12x-128] =-24 : 3
[12x-128] =-8
12x =-8+128
12x =120
x =120:12
x =10
{[(9x+3x-128) x 3] x 10} + 1000 =760
[(x(9+3)-128)x3]x10=760-1000
[(x.12-128)x3]x10=-240
(x.12-128)x3=-240:10
(x.12-128)x3=-24
(x.12-128)=-24:3
x.12-128=-8
x.12=-8+128
x.12=120
x=120:12
x=10
vậy x=10
a)\(\left|x^3+x\right|-\left|9x^2+9\right|=0\)
Mà \(\hept{\begin{cases}x^3+x\ge0\\9x^2+9\ge0\end{cases}}\) và \(\left|x^3+x\right|-\left|9x^2+9\right|=0\)
\(\Rightarrow\hept{\begin{cases}x^3+x=0\\9x^2+9=0\end{cases}}\)
Mà \(9x^2\ge0\Leftrightarrow9x^2+9>0\)
Vậy \(x\in\left\{\varnothing\right\}\)
b) \(\left(3x+2\right)-\left(x-1\right)=4\left(x+1\right)\)
\(\Leftrightarrow3x+2-x+1=4x+4\)
\(\Leftrightarrow\left(3x-x\right)+\left(2+1\right)=4x+4\)
\(\Leftrightarrow2x+3=4x+4\)
\(\Leftrightarrow2x-4x=4-3\)
\(\Leftrightarrow-2x=1\)
\(\Leftrightarrow x=\frac{-1}{2}\)
Vậy\(x=\frac{-1}{2}\)
c) \(2\left(x-1\right)-5\left(x+2\right)=-10\)
\(\Leftrightarrow2-2-5x-10=-10\)
\(\Leftrightarrow2-2-5x=0\)
\(\Leftrightarrow0-5x=0\)
\(\Leftrightarrow5x=0\)
\(\Leftrightarrow x=0\)
Vậy x = 0
\(3^{x+1}.9^{x+2}.27^x=81^x\)
\(3^x.3.3^{2\left(x+2\right)}.3^{3x}=3^{4x}\)
\(3^{x+1+2x+4+3x}=3^{4x}\)
\(\Rightarrow x+1+2x+4+3x=4x\)
\(6x+5-4x=0\)
\(2x+5=0\)
\(2x=-5\Rightarrow x=-\dfrac{5}{2}\)