\(5^{x+1}+5^{x-1}=130\)

\(5^x\cdot5^1+5^x\div5^1=130\)

\(5^x\cdot5^1+5^x\cdot\dfrac{1}{5}=130\)

\(5^x\cdot\left(5+\dfrac{1}{5}\right)=130\)

\(5^x\cdot\dfrac{26}{5}=130\)

\(5^x=130\div\dfrac{26}{5}\)

\(5^x=130\cdot\dfrac{5}{26}\)

\(5^x=25\)

\(\Rightarrow5^x=5^2\)

\(\Rightarrow x=2\)

Mọi người còn câu trả lời nào khác không cứ trả lời đi mik tick cho

5^x+1.5^x-1=130

5^x-1.5²+5^x-1.1=130

5^x-1.(5²+1)=130

5^x-1.(25+1)=130

5^x-1.26=130

5^x-1=130:26

5^x-1=5¹

^{=> x-1=1}

^x=1+1

^x=2

^{Tick cho mình nha}

Lời giải:

$5^x+5^{x+1}+5^{x+2}+5^{x+3}=1+2+3+...+87+88-4^2$

$5^x(1+5+5^2+5^3)=88.89:2-16$

$5^x.156=3900$

$5^x=3900:156=25=5^2$

$\Rightarrow x=2$

- \(x^2\) + 5\(x\) - 4 = 0

-\(x^2\) + \(x\) + 4\(x\) - 4 = 0

(- \(x^2\) + \(x\)) + (4\(x\) - 4) = 0

-\(x\)(\(x-1\)) + 4\(\times\)( \(x\) -1) = 0

(\(x-1\))( -\(x\) +4) = 0

\(\left[{}\begin{matrix}x-1=0\\-x+4=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)

\(x\) \(\in\) { 1; 4}

`-x^2+5x-4 =0`

`\Rightarrow x^2-5x+4=0`

`\Rightarrow x^2-4x-x+4=0`

`\Rightarrow (x^2-4x)-(x-4)=0`

`\Rightarrow x(x-4)-(x-4)=0`

`\Rightarrow (x-4)(x-1)=0`

\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\end{matrix}\right.\)

`\Rightarrow `\(\left[{}\begin{matrix}x=0+4\\x=0+1\end{matrix}\right.\)

``\Rightarrow `\(\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\)

Vậy, nghiệm của đa thức là `x={4; 1}.`

Bậc 4

Nếu sai ai đó sửa lại nha

_Hết_

P = 5x ( 4x2 - 2x + 1 ) - 2x ( 10x2 - 5x - 2 )

P = 20x3 - 10x2 + 5x - 20x3 + 10x2 + 4x

P = 5 x + 4x = 9x

Thay x vào biểu thúc ta có :

9 . 15 = 135

Vậy giá trị của biểu thức là 135 khi x = 15

p=135

cái đồ copy

\(\left(x^2-5x+8\right)^2-\left(5x-17\right)^2=0\)

\(\Leftrightarrow\left(x^2-5x+8-5x+17\right)\left(x^2-5x+8+5x-17\right)=0\)

\(\Leftrightarrow\left(x^2-10x+25\right)\left(x^2-9\right)=0\)

\(\Leftrightarrow\left(x^2-5x-5x+25\right)\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[x\left(x-5\right)-5\left(x-5\right)\right]\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left(x-5\right)^2.\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-5\right)^2=0\\x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=3\\x=-3\end{matrix}\right.\)

\(5x.3.\left(x+1\right)=15x\left(x+1\right)=15x^2+15\)

`sqrt{5x+3}+sqrt{10x-1}+5x^2-6x-2=0`

`đk:x>=1/10`

`pt<=>sqrt{5x+3}-2+sqrt{10x-1}-1+5x^2-6x+1=0`

`<=>(5x-1)/(sqrt{5x+3}+2)+(10x-2)/(sqrt{10x-1}+1)+(5x-1)(x-1)=0`

`<=>(5x-1)(1/(sqrt{5x+3}+2)+2/(sqrt{10x-1}+1)+x-1)=0`

`<=>5x-1=0`

`<=>x=1/5`

Nham nghiem thi ta thay pt co 1 ngiem x=2/3

=> tach nhu sau :

\(3x^3-2x^2-3x^2+2x+3x-2\) 

\(3x^2\left(x-\frac{2}{3}\right)-3x\left(x-\frac{2}{3}\right)+3\left(x-\frac{2}{3}\right)\)

\(\left(x-\frac{2}{3}\right)\left(3x^2-3x+3\right)\)

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