3x-1=9 giúp mik với
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f(x)=9x3-1/3x+3x2-3x+1/3x2-1/9x3-3x2-9x+27+3x
= 9x3-1/9x3+3x2+1/3x2-3x2-1/3-3x-9x+3x+27
= 80/9x3+1/3x2-28/3x+27
x+y=9 nên x=9-y
\(M=\dfrac{4\left(9-y\right)-9}{3\left(9-y\right)+y}-\dfrac{4y+9}{3y+9-y}\)
\(=\dfrac{36-4y-9}{27-3y+y}-\dfrac{4y+9}{2y+9}\)
\(=\dfrac{4y-27}{2y-27}-\dfrac{4y+9}{2y+9}\)
\(=\dfrac{8y^2+36y-54y-243-\left(8y^2-108y+18y-243\right)}{\left(2y-27\right)\left(2y+9\right)}\)
\(=\dfrac{8y^2-18y-243-8y^2+90y+243}{\left(2y-27\right)\left(2y+9\right)}=\dfrac{72y}{\left(2y-27\right)\left(2y+9\right)}\)
a: =>4/3x=7/9-4/9=1/3
=>x=1/4
b: =>5/2-x=9/14:(-4/7)=-9/8
=>x=5/2+9/8=29/8
c: =>3x+3/4=8/3
=>3x=23/12
hay x=23/36
d: =>-5/6-x=7/12-4/12=3/12=1/4
=>x=-5/6-1/4=-10/12-3/12=-13/12
\(|2x-7|=12\Leftrightarrow\orbr{\begin{cases}2x-7=12\\2x-7=-12\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=19\\2x=-5\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{19}{2}\\x=-\frac{5}{2}\end{cases}}}\)
\(|4x+3|=|3x-1|\Leftrightarrow\orbr{\begin{cases}4x+3=3x-1\\4x+3=1-3x\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-4\\7x=-2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-4\\x=-\frac{2}{7}\end{cases}}}\)
\(|3x+5|=2x+9\left(ĐKXĐ:2x+9\ge0\Leftrightarrow x\ge-\frac{9}{2}\right)\)
\(\Leftrightarrow\orbr{\begin{cases}3x+5=2x+9\\3x+5=-2x-9\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\5x=-14\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=4\left(tm\right)\\x=-\frac{14}{5}\end{cases}}}\left(tm\right)\)
Tự KL cho mỗi phần
\(\left|4x+8\right|+\left|3x+6\right|+\left|2x+4\right|=9\)
\(\Leftrightarrow4\left|x+2\right|+3\left|x+2\right|+2\left|x+2\right|=9\)
\(\Leftrightarrow9\left|x+2\right|=9\)
\(\Leftrightarrow\left|x+2\right|=1\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=1\\x+2=-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-3\end{cases}}\)
\(3^{x-1}+3^x+3^{x+1}=39\)
\(3^{x-1}+3^{x-1}.3+9.3^{x-1}=39\)
\(13.3^{x-1}=39\)
\(3^{x-1}=39:13=3\)
\(x-1=1\)
\(x=2\)
Sửa đề: 3ˣ⁻¹ + 3ˣ + 3ˣ⁺¹ = 39
3ˣ⁻¹ + 3ˣ + 3ˣ⁺¹ = 39
3ˣ⁻¹.(1 + 3 + 3²) = 39
3ˣ⁻¹ . 13 = 39
3ˣ⁻¹ = 39 : 13
3ˣ⁻¹ = 3
x - 1 = 1
x = 1 + 1
x = 2
a) \(3^{x+1}=9^x\)
\(\Rightarrow3^{x+1}=\left(3.3\right)^x\)
\(\Rightarrow3^{x+1}=3^{3x}\)
\(\Rightarrow x+1=3x\)
\(\Rightarrow3x-x=2x=1\)
\(\Rightarrow x=1\)
b) \(2^{3x+2}=4^{x+5}\)
\(2^{3x+2}=\left(2.2\right)^{x+5}\)
\(\Rightarrow2^{3x+2}=2^{2\left(x+5\right)}=2^{2x+10}\)
\(\Rightarrow3x+2=2x+10\Rightarrow3x+2=2x+2+8\)
\(\Rightarrow3x=2x+8\Rightarrow3x-2x=8\)
\(\Rightarrow1x=8\Rightarrow x=8\)
\(x\left(1-3x\right)\left(4-3x\right)-\left(x-4\right)\left(3x+5\right)=4x-15x^2+9x^3-3x^2+7x+20=9x^3-18x^2+11x+20\)
x(1 - 3x)(4 - 3x) - (x - 4)(3x + 5)
= (x - 3x2)(4 - 3x) - 3x2 - 5x + 12x + 20
= 4x - 3x2 - 12x2 + 9x3 - 3x2 - 5x + 12x + 20
= 9x3 - 18x2 + 11x + 20
3x-1 = 9
=> 3x-1 = 32
=> x - 1 = 2
=> x = 3
Vậy : x = 3
\(3^{x-1}=9\)
\(\Rightarrow3^{x-1}=3^2\)
\(\Rightarrow x-1=2\)
\(\Rightarrow x=3\)