1, \(x^3+4x^2+4x=0\Leftrightarrow x\left(x^2+4x+4\right)=0\)

\(\Leftrightarrow x\left(x+2\right)^2=0\Leftrightarrow x=-2;x=0\)

2, \(\left(x+3\right)^2-4=0\Leftrightarrow\left(x+3-2\right)\left(x+3+2\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\Leftrightarrow x=-5;x=1\)

3, \(x^4-9x^2=0\Leftrightarrow x^2\left(x^2-9\right)=0\)

\(\Leftrightarrow x^2\left(x-3\right)\left(x+3\right)=0\Leftrightarrow x=0;\pm3\)

4, \(x^2-6x+9=81\Leftrightarrow\left(x-3\right)^2=9^2\)

\(\Leftrightarrow\left(x-3-9\right)\left(x-3+9\right)=0\Leftrightarrow\left(x-12\right)\left(x+6\right)=0\Leftrightarrow x=-6;x=12\)

5, em xem lại đề nhé

à lag tý @@

5, \(x^3+6x^2+9x-4x=0\Leftrightarrow x^3+6x^2+5x=0\)

\(\Leftrightarrow x\left(x^2+6x+5\right)=0\Leftrightarrow x\left(x^2+x+5x+5\right)=0\)

\(\Leftrightarrow x\left(x+1\right)\left(x+5\right)=0\Leftrightarrow x=-5;x=-1;x=0\)

a)\(x^3+4x^2+4x=0\)

\(\Leftrightarrow x\left(x^2+4x+4\right)=0\)

\(\Leftrightarrow x\left(x+2\right)^2=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\\left(x+2\right)^2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}}\)

b)\(\left(x+3\right)^2-4=0\)

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

c)\(x^4-9x^2=0\)

\(\Leftrightarrow x^2\left(x^2-9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-9=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm3\end{cases}}}\)

d)\(x^2-6x+9=81\)

\(\Leftrightarrow\left(x-3\right)^2=81\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=9\\x-3=-9\end{cases}\Leftrightarrow\orbr{\begin{cases}x=12\\x=-6\end{cases}}}\)

e)\(x^3+6x^2+9x-4x=0\)

\(\Leftrightarrow x^3+6x^2+5x=0\)

\(\Leftrightarrow\left(x^2+5x\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+5x=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0;x=-5\\x=-1\end{cases}}}\)

#H

b) (5/2-3x)=25/9

            3x = 5/2-25/9

            3x =-5/18

              x =-5/18:3

              x=-5/54

\(e.\left(x-1\right)^5=-32\)

  \(\left(x-1\right)^5=\left(-2\right)^5\)

   \(x-1=-2\)

   \(x\)      \(=-2+1\)

   \(x\)        \(=-1\)

Vậy \(x=-1\)

Từ đề bài, suy ra:

(x+1)²²=0 hoặc (y-3)¹²=0

<=> x+1=0 hoặc y-3=0

<=> x=-1 hoặc y=3

Vậy x\(\in\){-1;3}

woa! thanks Gà Mờ nha

TL ;

helu kni do di mn

ht

@ giang ??????????

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x² + 2x = 0

=> x(x + 2) = 0

=> \(\orbr{\begin{cases}x=0\\x+2=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)

(x - 2) + 3.x² - 6x = 0

=> (x - 2) + 3x² - 3x . 2 = 0

=> (x - 2) + 3x.(x - 2) = 0

=> (1 + 3x)(x - 2) = 0

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

=> \(\orbr{\begin{cases}x=-\frac{1}{3}\\x=2\end{cases}}\)

a) \(81^{40}=\left(3^4\right)^{40}=3^{160}\)

   \(27^{14}=\left(3^3\right)^{14}=3^{42}\)

Vì \(3^{160}>3^{42}\) => \(81^{40}>27^{14}\)

b) \(5^{64}=5^{4.16}=625^{16}\)

   \(3^{96}=3^{6.16}=729^{16}\)

Vì \(625^{16}< 729^{16}\)=> \(5^{64}< 3^{96}\)

c) \(125^{12}=\left(5^3\right)^{12}=5^{36}\)

   \(25^{10}=\left(5^2\right)^{10}=5^{20}\)

Vì \(5^{36}>5^{20}\)=> \(125^{12}>25^{10}\)

T_i_c_k nha,mơn bạn nhìu ^^

a, 81⁴⁰>27¹⁴

b, 5⁶⁴<3⁹⁶

c, 125¹²<25¹⁰

a. 81⁴⁰  lớn hơn 27¹⁴

b. 5⁶⁴ bé hơn 3⁹⁶

^c. 125¹² lớn hơn 25¹⁰

k cho mk nhá