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\(\left(3x-2\right)^5=-243\)
\(\left(3x-2\right)^5=-3^5\)
\(=>3x-2=-3\)
\(=>3x=-3+2\)
\(=>3x=-1\)
\(=>x=-\frac{1}{3}\)
Vậy \(x=-\frac{1}{3}\)
Đề trước đó:
(x-7)(x+1)-(x-3)^2=(3x-5)(3x+5)-(3x+1)^2+(x-2)^2-x
<=>x^2+x-7x-7-x^2+6x-9=9x^2-25-9x^2-6x-1+x^2-4x+4-x
<=>x^2-11x-6=0
<=>x^2-2x. 11/2 + 121/4-145/4=0
<=>(x-11/2)^2=145/4
<=>|x-11/2|=căn(145)/2
<=>x=[11+-căn(145)]/2
a) \(E=|3x-7|+|3x+2|+8\)
\(E=|7-3x|+|3x+2|+8\)
Do : \(|a|\ge a\)
\(\Rightarrow E_{min}\text{=}7-3x+3x+2+8\)
\(\Rightarrow E_{min}\text{=}17\)
Dấu '' = '' xảy ra : \(\Leftrightarrow\dfrac{-2}{3}\le x\le\dfrac{7}{3}\)
Ta có :
\(7.3^{x-1}-3^{x+2}=-540\)
\(\Leftrightarrow\)\(7.3^x:3-3^x.3^2=-540\)
\(\Leftrightarrow\)\(\frac{7}{3}.3^x-9.3^x=-540\)
\(\Leftrightarrow\)\(3^x\left(\frac{7}{3}-9\right)=-540\)
\(\Leftrightarrow\)\(3^x.\frac{-20}{3}=-540\)
\(\Leftrightarrow\)\(3^x=\left(-540\right):\frac{-20}{3}\)
\(\Leftrightarrow\)\(3^x=81\)
\(\Leftrightarrow\)\(3^x=3^4\)
\(\Leftrightarrow\)\(x=4\)
Vậy \(x=4\)
a) (1+3x)^4 = 256
=> (1+3x)^4= 4^4
=> 1+3x=4
=> 3x=3
=> x=1
b) \(\frac{3^{15}.29+3^{15}.88}{3^{13}.81}=\frac{3^{15}.\left(29+88\right)}{3^{13}.81}=\frac{3^{15}.117}{3^{13}.81}=\frac{3^2.13}{9}=13\)