Tìm x hả bạn ?

a ) \(\left(3x+\frac{1}{4}\right)^3=-27\)

\(\left(3x+\frac{1}{4}\right)^3=\left(-3\right)^3\)

\(\Rightarrow3x+\frac{1}{4}=-3\)

\(\Rightarrow3x=-3-\frac{1}{4}=-\frac{13}{4}\)

\(\Rightarrow x=-\frac{13}{4}:3=-\frac{13}{12}\)

Vậy x = \(-\frac{13}{12}\)

cho em hoi câu này xin các anh chị:

10mux x+4y = 2013

7: \(\left(xy+4\right)^2-\left(2x+2y\right)^2\)

=(xy+4-2x-2y)(xy+4+2x+2y)

=[x(y-2)-2(y-2)][x(y+2)+2(y+2)]

=(x-2)(y-2)(x+2)(y+2)

8: \(81x^2-64y^2=\left(9x\right)^2-\left(8y\right)^2\)

=(9x-8y)(9x+8y)

9: \(\left(a^2+b^2+5\right)^2-4\left(ab+2\right)^2\)

\(=\left(a^2+b^2+5\right)^2-\left(2ab+4\right)^2\)

\(=\left(a^2+b^2+5-2ab-4\right)\left(a^2+b^2+5+2ab+4\right)\)

\(=\left\lbrack\left(a-b\right)^2+1\right\rbrack\left\lbrack\left(a+b\right)^2+9\right\rbrack\)

10: \(\left(x-1\right)^2-\left(x+1\right)^2\)

=(x-1-x-1)(x-1+x+1)

=2x*(-2)=-4x

11: \(8x^3-\frac18=\left(2x\right)^3-\left(\frac12\right)^3=\left(2x-\frac12\right)\left(4x^2+2x\cdot\frac12+\frac14\right)\)

\(=\left(2x-\frac12\right)\left(4x^2+x+\frac14\right)\)

12: \(\frac{1}{25}x^2-64y^2=\left(\frac15x\right)^2-\left(8y\right)^2=\left(\frac15x-8y\right)\left(\frac15x+8y\right)\)

13: \(x^3+\frac{1}{27}=x^3+\left(\frac13\right)^3=\left(x+\frac13\right)\left(x^2-\frac13x+\frac19\right)\)

a) \(\left(6x-5y\right)^2=36x^2-60xy+25y^2\)

b) \(\left(4x-1\right)^2=16x^2-8x+1\)

c) \(\left(x+2\right)^2=x^2+4x+4\)

d) \(x^2-64=\left(x-8\right)\left(x+8\right)\)

e) \(4x^2-64=\left(2x-8\right)\left(2x+8\right)\)

f) \(25x^2-4=\left(5x-2\right)\left(5x+2\right)\)

g) \(\left(x+1\right)^3=x^3+3x^2+3x+1\)

h) \(\left(x-3\right)^3=x^3-9x^2+27x-27\)

k) \(x^3+8=\left(x+2\right)\left(x^2-2x+4\right)\)

l) \(x^3-125=\left(x-5\right)\left(x^2+5x+25\right)\)

y) \(27y^3-1=\left(3y-1\right)\left(9y^2+3y+1\right)\)

^{ai Giúp mình với}

a, \(x^3+3x^2-\left(x+3\right)=0\Leftrightarrow x^2\left(x+3\right)-\left(x+3\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+3\right)=0\Leftrightarrow x=1;x=-1;x=-3\)

b, \(15x-5+6x^2-2x=0\Leftrightarrow5\left(3x-1\right)+2x\left(3x-1\right)=0\)

\(\Leftrightarrow\left(2x+5\right)\left(3x-1\right)=0\Leftrightarrow x=-\frac{5}{2};x=\frac{1}{3}\)

c, \(5x-2-25x^2+10x=0\)

\(\Leftrightarrow\left(5x-2\right)-5x\left(5x-2\right)=0\Leftrightarrow\left(1-5x\right)\left(5x-2\right)=0\Leftrightarrow x=\frac{2}{5};x=\frac{1}{5}\)

= 1/5 nha

  Bài 1:

2^\(x\) = 4

2\(^x\) = 2²

 \(x=2\)

Vậy \(x=2\)

Bài 2:

2\(^x\) = 8

2\(^x\) = 2³

\(x=3\)

Vậy \(x=3\)

1, \(3x\left(x-7\right)+2x-14=0\)

\(\Rightarrow3x\left(x-7\right)+2\left(x-7\right)=0\)

\(\Rightarrow\left(x-7\right)\left(3x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=7\\x=\frac{-2}{3}\end{cases}}\)

2, \(x^3+3x^2-\left(x+3\right)=0\)

\(\Rightarrow x^2\left(x+3\right)-\left(x+3\right)=0\)

\(\Rightarrow\left(x+3\right)\left(x^2-1\right)=0\)

\(\Rightarrow\left(x+3\right)\left(x-1\right)\left(x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\pm1\end{cases}}\)

3, \(15x-5+6x^2-2x=0\)

\(\Rightarrow\left(15x-5\right)+\left(6x^2-2x\right)=0\)

\(\Rightarrow5\left(3x-1\right)+2x\left(3x-1\right)=0\)

\(\Rightarrow\left(3x-1\right)\left(5+2x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-1=0\\5+2x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=\frac{-5}{2}\end{cases}}\)

4, \(5x-2-25x^2+10x=0\)

\(\Rightarrow\left(5x-25x^2\right)-\left(2-10x\right)=0\)

\(\Rightarrow5x\left(1-5x\right)-2\left(1-5x\right)=0\)

\(\Rightarrow\left(1-5x\right)\left(5x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}1-5x=0\\5x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{2}{5}\end{cases}}\)

a)     4^x  = 8^12

    (2^2)^x = (2^3)^12

     2^2x   = 2^36

=> 2x = 36

=>   x = 36 : 2 = 18

a) 4^x=8^12

4^x=(2^3)^12

4^x=2^36

4^x=4^18

=> x=18 hoặc -18

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