`a)(x-2)(x^2+2x+4)-x(x-3)(x+3)=26`

`<=>x^3-2^3-x(x^2-9)=26`

`<=>x^3-8-x^3+9x=26`

`<=>9x-8=26`

`<=>9x=34`

`<=>x=34/9`

`b)(x-3)(x^2+3x+9)-x(x+4)(x-4)=21`

`<=>x^3-3^3-x(x^2-16)=21`

`<=>x^3-27-x^3+16x=21`

`<=>16x-27=21`

`<=>16x=48`

`<=>x=3`

\(a,\Rightarrow 2x^2-10x-3x-2x^2=26\\ \Rightarrow -13x=26\\ \Rightarrow x=-2\\ b, \Rightarrow -2x^2+3x+3-3x-3+2x^2-x=18\\ \Rightarrow -x=18\Rightarrow x=-18\)

\(a,\Rightarrow2x^2-18x-2x^2=0\\ \Rightarrow-18x=0\Rightarrow x=0\\ b,\Rightarrow2x^2-5x-12+x^2-7x+10=3x^2-17x+20\\ \Rightarrow5x=22\Rightarrow x=\dfrac{22}{5}\)

a)5^x=125=5³

=>x=3

b)3^2x=81=3⁴

=>2x=4

x=4/2

x=2

c)5^2x-3-2*5²=5²*3

5^2x-3=5²*3+2*5²=5²(3+2)=5²*5

5^2x-3=5³

=>2x-3=3

2x=3+3

2x=6

x=6/2

x=3

lần sau nếu ghi dẫu mũ như vậy thì bạn nhớ ngoặc vào 

chịu mới lớp 4 mà !!!

hỏi làm chi

ai hoi may 

a/ 2x(x-5)-x(3+2x)=26

 =>2x²-10x-2x²-3x=26

=>-13x=26

=>x=-2

b/ 49x²-81=0

=>49x²=81

=>x²=\(\frac{49}{81}\)

\(\Rightarrow x^2=\left(-\frac{9}{7}\right)^2\)hoặc \(\left(\frac{9}{7}\right)^2\)

\(\Rightarrow x=\pm\frac{9}{7}\)

2x(x - 5) - x(3 + 2x) = 26

2x² - 10x - 3x - 2x² = 26

- 13x = 26

x = 26 : (- 13)

x = - 2

49x² - 81 = 0

(7x - 9)(7x + 9) = 0

Th1:

7x - 9 = 0

7x = 9

x = 9/7

Th2:

7x + 9 = 0

7x = - 9 

x = - 9/7

Vậy x = 9/7 hoặc x = - 9/7

a) 2³.(20 - 2x) = 4.2⁵

8.(20 - 2x) = 4.32

8.(20 - 2x) = 128

20 - 2x = 128 : 8

20 - 2x = 16

2x = 20 - 16

2x = 4

x = 4 : 2

x = 2

b) 3²⁺ˣ = 81

3²⁺ˣ = 3⁴

2 + x = 4

x = 4 - 2

x = 2

1)

`7x^2 -49x=0`

`<=>x(7x-49)=0`

\(< =>\left[{}\begin{matrix}x=0\\7x-49=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=7\end{matrix}\right.\)

2)

`8x^2 -16x=0`

`<=>x(8x-16)=0`

\(< =>\left[{}\begin{matrix}x=0\\8x-16=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

3)

`2x^3 +40x=0`

`<=>x(2x^2 +40)=0`

`<=>x=0` hoặc`2x^2 +40=0`

`<=>x=0` hoặc `2x^2 =-40` (vô lí vì `2x^2` luôn lớn hơn hoặc bằng 0)

`<=>x=0`

4)

`-x^3 +16x=0`

`<=>x^3 -16x=0`

`<=>x(x^2 -16)=0`

\(< =>\left[{}\begin{matrix}x=0\\x^2-16=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x^2=16\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

3^x-5 . 3^2x = 81

3^(x-5)+2x = 3⁴

\(\Rightarrow\)( x - 5 ) + 2x = 4

\(\Rightarrow\)3x - 5 = 4

\(\Rightarrow\)3x = 4 + 5

\(\Rightarrow\)3x = 9

\(\Rightarrow\)x = 9 : 3

\(\Rightarrow\)x = 3