\(5^{x-2}+5^{x+2}=3130\)

\(\Rightarrow5^x\cdot\dfrac{1}{25}+5^x\cdot25\)

\(\Rightarrow5^x\cdot\left(\dfrac{1}{25}+25\right)=3130\)

\(\Rightarrow5^x\cdot\dfrac{626}{25}=3130\)

\(\Rightarrow5^x=3130:\dfrac{626}{25}\)

\(\Rightarrow5^x=125\)

\(\Rightarrow5^x=5^3\)

\(\Rightarrow x=3\)

Vậy: x=3

\(5^{x-2}+5^{x+2}=3130\\ \Leftrightarrow5^{-2}.5^x+5^2.5^x=3130\\ \Leftrightarrow5^x\left(\dfrac{1}{25}+25\right)=3130\\ \Leftrightarrow5^x.\dfrac{626}{25}=3130\\ \Leftrightarrow5^x=125\\ \Leftrightarrow5^x=5^3\\ \Leftrightarrow x=3\)

Vậy x = 3

\(5^{x-2}+5^{x+2}=3130\)

\(5^{x-2}+5^{x-2+4}=3130\)

\(5^{x-2}+5^{x-2}.5^4=3130\)

\(5^{x-2}.\left(1+5^4\right)=3130\)

\(5^{x-2}.626=3130\)

\(5^{x-2}=3130:626\)

\(5^{x-2}=5\)

\(x-2=1\)

\(x=1+2\)

\(x=3\)

a, \(-4x+5+2x-1=3\Leftrightarrow-2x=-1\Leftrightarrow x=\dfrac{1}{2}\)

b, \(-2x+2=2\Leftrightarrow x=0\)

c, \(-2x-6=-8\Leftrightarrow x=1\)

Ta có x² + 5x - 3√(x² + 5x + 2) - 2 = 0

<=> x² + 5x - 2 = 3√(x² + 5x + 2)

<=> (x² + 5x - 2)² = [3√(x² + 5x + 2)]²

<=> x⁴ + 25x² + 4 + 10x³ - 20x - 4x³ = 9(x² + 5x + 2)

<=> x⁴ + 6x³ + 25x² - 20x + 4 = 9x² + 45x + 18

<=> x⁴ + 6x³ + 25x² - 20x + 4 - (9x² + 45x + 18) = 0

<=> x⁴ + 6x³ + 25x² - 9x² - 20x - 45x + 4 - 18 = 0

<=> x⁴ + 6x³ + 16x² - 65x - 14 = 0

Đến đây, ta phân tích đa thức thành nhân tử bằng phương pháp hệ số bất định

Ta có x⁴ + 6x³ + 16x² - 65x - 14 sau khia phân tích có dạng (x² + ax + b)(x² + cx + d) = x⁴ + (a+c)x³ + (ac+b+d)x² + (ad+bc)x + db

=> x⁴ + 6x³ + 16x² - 65x - 14 = x⁴ + (a+c)x³ + (ac+b+d)x² + (ad+bc)x + db

<=> a+c = 6 ; ac+b+d = 16 ; ad+dc = -65 ; db = -14

Sau đó bạn tìm ra a,b,c,d và giải ra phương trình.

Mình chỉ mới lớp 7 nên chưa tìm ra đươc a,b,c,d.Mong bạn thông cảm cho mình

`-10x(2-x)-5x(x+2)=5x(x+3)`

`<=> -20x + 10x^2 - 5x^2 - 10x = 5x^2 +15x`

`<=> 5x^2 - 30x = 5x^2 + 15x`

`<=> -30x = 15x`

`<=> -45x = 0`

`<=> x = 0`

Vậy `S = {0}`

\(-10x\left(2-x\right)-5x\left(x+2\right)=5x\left(x+3\right)\)

\(\text{⇔}10x\left(x-2\right)+5x\left(x-2\right)=-5x\left(x-3\right)\)

\(\text{⇔}\left(x-2\right)\left(10x+5x\right)=-5x\left(x-3\right)\)

\(\text{⇔}15x\left(x-2\right)=-5x^2+15\)

\(\text{⇔}15x^2-30=-5x^2+15\)

\(\text{⇔}15x^2+5x^2=30+15\)

\(\text{⇔}20x^2=45\)

\(\text{⇔}x=\sqrt{\dfrac{45}{20}}=\dfrac{3}{2}\)

Vậy: \(x=\dfrac{3}{2}\)

Ta có: \(-10x\left(2-x\right)-5x\left(x+2\right)=5x\left(x+3\right)\)

\(\Leftrightarrow-20x+10x^2-5x^2-10x-5x^2-15x=0\)

\(\Leftrightarrow-45x=0\)

hay x=0

1 , <=>  25x^2 + 10x + 1 - ( 25x^2 - 9) = 30 

      <=> 25x^2 + 10x + 1 - 25x^2 + 9 = 30 

      <=> 10x + 10                 = 30 

     <=> 10 ( x + 1)                 = 30

       <=>  x + 1                      = 3 

        <=>  x                           = 2 

2, ( x + 3)(x^2 - 3x + 9 ) - x(x+2)(x-2) = 15 

<=> x^3 - 27 - x(x^2 - 4)                      = 15

<=> x^3 - 27 - x^3 + 4x                        = 15 

<=> 4x -27                                          = 15 

<=> 4x                                                = 15 + 27

<=> 4x                                                 =42

<=> x                                                     = 42/4 = 21/2 

******************

\(A\left(x\right)=5x^2-5x+3=5\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>0,\forall x\)

⇒ pt vô nghiệm

\(B\left(x\right)=4x^2-3x+7=4\left(x-\dfrac{3}{8}\right)^2+\dfrac{103}{16}>0,\forall x\)

⇒ pt vô nghiệm

\(C\left(x\right)=5x^2-11x+6=\left(5x^2-5x\right)-\left(6x-6\right)\)

\(=5x\left(x-1\right)-6\left(x-1\right)=\left(5x-6\right)\left(x-1\right)\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\\x=1\end{matrix}\right.\)

Vậy ...

a, Ta có : 

\(A\left(x\right)=5x^2-5x+1+2=0\Leftrightarrow5x^2-6x+3=0\)

\(\Leftrightarrow5\left(x^2-\dfrac{2.3}{5}+\dfrac{9}{25}-\dfrac{9}{25}\right)+3=0\Leftrightarrow5\left(x-\dfrac{3}{5}\right)^2+\dfrac{6}{5}=0\)( vô lí )

vậy đa thức ko có nghiệm 

b, \(B\left(x\right)=4x^2-3x+7=0\Leftrightarrow4\left(x^2-\dfrac{2.3}{8}+\dfrac{9}{64}-\dfrac{9}{64}\right)+7=0\)

\(\Leftrightarrow4\left(x-\dfrac{3}{8}\right)^2+\dfrac{103}{64}=0\)( vô lí ) 

Vậy đa thức ko có nghiệm 

c, \(C\left(x\right)=5x^2-11x+6=0\Leftrightarrow5x^2-6x-5x+6=0\)

\(\Leftrightarrow5x\left(x-1\right)-6\left(x-1\right)=0\Leftrightarrow\left(5x-6\right)\left(x-1\right)=0\Leftrightarrow x=\dfrac{6}{5};x=1\)

`(5x-1)^2-(5x-4)(5x+4)=7`

`\Leftrightarrow 25x^2-10x+1-25x^2+16-7=0`

`\Leftrightarrow  -10x+10=0`

`\Leftrightarrow  x=1`