\(^{\left(3^4\right)^{x+1}=81^{x+3}\Rightarrow81^{x+1}=81^{x+3}\Rightarrow x+1=x+3}\)vô nghiệm vì 1 khác 3

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

a) (3x-7)⁵=32

=> (3x-7)⁵=2⁵

=> 3x-7=2

=> 3x=2+7=9

=>x=9:3=3

b) (4x-1)³=27.125

=> (4x-1)³=3³.5³

=> (4x-1)³=(3.5)³

=> (4x-1)³=15³

=> 4x-1=15

(Các bước còn lại tương tự câu a)

a)(3x-7)^5=32

(3x-7)^5=2^5

3x-7=2

3x=2+7

3x=9

x=9:3

x=3

2^x+2^2x+1=96

2^x+2^2x.2¹=96

2^x+2^2x.2=96

2^x+2^2x=96:2

2^x+2^2x=48

a/ 3^{4x + 4} = 81^{x + 3}

=> 81^{x + 4} = 81^{x + 3}

=> Đề sai!

b/ (2x - 1)⁵ = 81² : 9

=> (2x - 1)⁵ = 9³

Đề sai tiếp

Xem lại đi ==

Cô giáo tớ giao như thế!

 a)3^4x+4 = 81^x+3

\(\Leftrightarrow3^{4x+4}=\left(3^3\right)^{x+3}\)

\(\Leftrightarrow3^{4x+4}=3^{3x+9}\)

\(\Leftrightarrow4x+4=3x+9\)

\(\Leftrightarrow x=5\)

b)( 2x-1 )⁵ = 81² : 9

\(\Leftrightarrow\left(2x-1\right)^5=\left(9^2\right)^2:9\)

\(\Leftrightarrow\left(2x-1\right)^5=9^4:9\)

\(\Leftrightarrow\left(2x-1\right)^5=9^3\)

Hơi lạ xem lại đề

a)\(4x\left(x-5\right)-\left(x-1\right)\left(4x-3\right)=5\)

    \(4x^2-20x-\left(4x^2-7x+3\right)=5\)

    \(4x^2-20x-4x^2+7x-3=5\)

     \(-13x=8\)

      \(x=-\frac{8}{13}\)

b)\(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)

     \(48x^2-32x+5+3x-48x^2-7+112x=81\)

     \(83x-2=81\)

     \(x=1\)

Bài 1: 

\(\left(2x-5\right)^2-4\left(2x-5\right)+4=0\)

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

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

\(2x-5-2=0\)

\(2x-7=0\)

\(2x=0+7\)

\(2x=7\)

\(x=\frac{7}{2}\)

Bài 3: 

\(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=46\)

\(\left(4x\right)^2-3^2-16x^2+40x-25=46\)

\(4^2x^2-3^2-16x^2+40x-25=46\)

\(16x^2-9-16x^2+40x-25=46\)

\(-34+40x=46\)

\(40x-34=46\)

\(40x=46+34\)

\(40x=80\)

\(x=2\)

bài 2:

a) \(81^2=\left(80+1\right)^2=80^2+2.80+1=6400+160+1=6561\)

b) \(99^2=\left(100-1\right)^2=100^2-2.100+1=10000-200+1=8801\)

Bài 1 :

\(C=\frac{1}{\left|x-2\right|+3}\)

\(C\le\frac{1}{3}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Vậy....

Bài 2 :

a) \(\left(\frac{1}{2}\right)^{3x-1}=\frac{1}{32}\)

\(\left(\frac{1}{2}\right)^{3x-1}=\left(\frac{1}{2}\right)^5\)

\(\Rightarrow3x-1=5\)

\(\Rightarrow3x=6\)

\(\Rightarrow x=2\)

b) \(2\cdot3^{x-405}=3^{x-1}\)

\(2=3^{x-1}:3^{x-405}\)

\(2=3^{x-1-x+405}\)

\(2=3^{404}\)( vô lí )

=> x thuộc rỗng

c) \(\frac{1}{81}\cdot27^{2x}=\left(-9\right)^4\)

\(\frac{27^{2x}}{81}=9^4\)

\(\frac{\left(3^3\right)^{2x}}{3^4}=\left(3^2\right)^4\)

\(\frac{3^{6x}}{3^4}=3^8\)

\(3^{6x-4}=3^8\)

\(\Rightarrow6x-4=8\)

\(\Rightarrow6x=12\)

\(\Rightarrow x=2\)

d) \(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)

\(\left(4x-1\right)^{30}-\left(4x-1\right)^{20}=0\)

\(\left(4x-1\right)^{20}\cdot\left[\left(4x-1\right)^{10}-1\right]=0\)

\(\Rightarrow\orbr{\begin{cases}4x-1=0\\4x-1=\left\{\pm1\right\}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=\left\{\frac{1}{2};0\right\}\end{cases}}\)