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\(-3x\left(x+2\right)^2+\left(x+3\right)\left(x-1\right)\left(x+1\right)-\left(2x-3\right)^2\)
\(=-3x\left(x^2+4x+4\right)+\left(x+3\right)\left(x^2-1\right)-\left(4x^2-12x+9\right)\)
\(=-3x^3-12x^2-12x+x^3-x+3x^2-3-4x^2+12x-9\)
\(=-2x^3-13x^2-x-12\)
a) (1+3x)^4 = 256
=> (1+3x)^4= 4^4
=> 1+3x=4
=> 3x=3
=> x=1
b) \(\frac{3^{15}.29+3^{15}.88}{3^{13}.81}=\frac{3^{15}.\left(29+88\right)}{3^{13}.81}=\frac{3^{15}.117}{3^{13}.81}=\frac{3^2.13}{9}=13\)
\(\frac{4}{3x7}\)+\(\frac{4}{7x11}\)+....+\(\frac{4}{\left(3x-1\right)\left(3x+3\right)}\)= \(\frac{3}{10}\)
=\(\frac{1}{3}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{11}\)+...+\(\frac{1}{3x-1}\)- \(\frac{1}{3x+3}\)= \(\frac{3}{10}\)
=> \(\frac{1}{3}\)- \(\frac{1}{3x+3}\)=\(\frac{3}{10}\)
=> \(\frac{1}{3}\)-\(\frac{3}{10}\)=\(\frac{1}{3x+3}\)
=> \(\frac{1}{30}\)=\(\frac{1}{3x+3}\)
nhân chéo ta có: 3x+3=30
=> 3x=30-3=27
=> x= 27:3
=>x=9
nhé ^_^ mơn nhiều
\(x=\dfrac{5}{a-1}< 0\)
\(\Leftrightarrow a-1< 0\Leftrightarrow a< 1\left(1\right)\)
Và \(x=\dfrac{5}{a-1}\in Z\)
\(\Rightarrow a-1\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\left(2\right)\)
\(\Rightarrow a\left\{2;0;6;-4\right\}\)
\(\left(1\right),\left(2\right)\Rightarrow a\in\left\{-4;0\right\}\)
Vì \(\left|x+2\right|+\left|2x+3\right|+\left|3x+4\right|\ge0\)
=> \(7x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow\left|x+2\right|+\left|2x+3\right|+\left|3x+4\right|=x+2+2x+3+3x+4\)
\(\Rightarrow6x+7=7x\)
=> x=7
x10 : x7 =\(\dfrac{1}{27}\)
= x3 = \(\dfrac{1}{27}\)
= \(\dfrac{1}{3}\)
Trả lời :
\(\left(3x-3\right)^3=-27\)
\(\left(3x-3\right)^3=\left(-3\right)^3\)
\(3x-3=-3\)
\(3x=-3+3\)
\(3x=0\)
\(x=0\)