a) \(5^{x-1}+5^{x-3}=650\)

\(\Rightarrow5^x\left(\frac{1}{5}+\frac{1}{125}\right)=650\)

\(\Rightarrow5^x=650:\frac{26}{125}\)

\(\Rightarrow5^x=3125\)

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

\(\Rightarrow x=5\)

\(x\left(x-5\right)\left(x+5\right)-\left(x-2\right)\left(x^2+2x+4\right)=-17\)

\(\Leftrightarrow x^3-25x-x^3+8=-17\)

\(\Leftrightarrow-25x=-25\)

hay x=1

\(5^{x-1}+5^{x-2}+5^{x-3}=775\\ 5^{x-3}.\left(25+5+1\right)=775\\ 5^{x-3}.31=775\\ 5^{x-3}=\dfrac{775}{31}=25=5^2\\ x-3=2\\ x=5\)

Bài 5:

a)x+37=50

⇔x=13

b)2x-3=11

⇔2x=14

⇔x=7

c)(2+x):5=6

⇔2+x=30

⇔x=28

d)2+x:5=6

⇔x:5=4

⇔x=20

a)x+37=50
   x      =50-37
  x       =13
b)2.x-3=11
   2.x   =11+3
   2.x   =14
      x   =14:2
      x   =7
c)(2+x):5=6
   2+x =6.5
   2+x =30
  x=30-2
  x=28
d)2+x:5=6
   x:5=6-2
   x:5=4
   x=4.5
   x=20
Câu trả lời nhớ ghi đầy đủ như này nha bạn

   

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

=>\(x\left(x^2-25\right)-\left(x^3+8\right)=3\)

=>\(x^3-25x-x^3-8=3\)

=>-25x-8=3

=>-25x=11

=>\(x=\dfrac{11}{-25}=\dfrac{-11}{25}\)

x(x+5)(x−5)−(x+2)(x2−2x+4)=3

=>x3−25x−x3+2x2−4x−2x2+4x−8=3

=>−25x= 11

=> x= −1125

a: \(\left(x+5\right)^2-\left(x-5\right)^2-2x+1=0\)

=>\(x^2+10x+25-\left(x^2-10x+25\right)-2x+1=0\)

=>\(x^2+8x+26-x^2+10x-25=0\)

=>18x+1=0

=>\(x=-\dfrac{1}{18}\)

b: \(\left(2x-7\right)^2-\left(x+3\right)^2=3x^2+6\)

=>\(4x^2-28x+49-\left(x^2+6x+9\right)-3x^2-6=0\)

=>\(x^2-28x+43-x^2-6x-9=0\)

=>34-34x=0

=>34x=34

=>x=1

c: \(\left(3x+2\right)^2-9\left(x-5\right)\left(x+5\right)=225-5x\)

=>\(9x^2+12x+4-9\left(x^2-25\right)-225+5x=0\)

=>\(9x^2+17x+4-225-9x^2+225=0\)

=>17x+4=0

=>x=-4/17

Lời giải:

$(x-5)^2=(x-5)$

$(x-5)^2-(x-5)=0$

$(x-5)(x-5-1)=0$

$(x-5)(x-6)=0$

Suy ra $x-5=0$ hoặc $x-6=0$

Suy ra $x=5$ hoặc $x=6$