3^x + 3^x+1 + 3^x+2 + 3^x+3 = 360

<=> 3^x + 3^x.3 + 3^x.3² + 3^x.3³ = 360

<=> 3^x( 1 + 3 + 3² + 3³ ) = 360

<=> 3^x.40 = 360

<=> 3^x = 9

<=> 3^x = 3²

<=> x = 2

a) \(x^2-\left(x-3\right)\left(3x+1\right)=9\)

\(\Leftrightarrow x^2-9-\left(x-3\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+3-3x-1\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(2-2x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\2x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

Vậy nghiệm của pt x = 3 hoặc x = 1

Giải phương trình

a, x² - (x-3)(3x+1) = 9

\(\Leftrightarrow\) x² - 3x² + 8x +3 = 9

\(\Leftrightarrow\) -2x² + 8x - 6 = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

b, (x+14)³ - (x+12)³ =1352

\(\Leftrightarrow\) (x+14-x-12)[(x+14)² + (x+14)(x+12) + (x+12)² ] = 1352

\(\Leftrightarrow\) 6(x² + 28x + 196 + x² + 26x + 168 + x² +24x +144) =1352

\(\Leftrightarrow\) 18x² +468x + 3048 = 1352

Pt nghiệm vô tỉ

b) Ta có: \(x^3+4x+5=0\)

\(\Leftrightarrow x^3-x+5x+5=0\)

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

\(\Leftrightarrow x\left(x+1\right)\left(x-1\right)+5\left(x+1\right)=0\)

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

mà \(x^2-x+5>0\forall x\)

nên x+1=0

hay x=-1

Vậy: S={-1}

a)x²-(x+3)(3x+1)=9

⇔(x-3)(x+3)-(x+3)(3x+1)=0

⇔x+3=0 hoặc 3x+1=0 

1.x+3=0 ⇔x=-3

2.3x+1=0⇔x=-1/3

phương trình có 2 nghiệm x=-3 và x=-1/3

\(a,5.\left(x+35\right)=515\)

\(x+35=103\)

\(x=103-35\)

\(x=68\)

\(b,12.x-33=3^2.3^3\)

\(12x-33=243\)

\(12x=276\)

\(x=23\)

\(d,12.\left(x-1\right):3=4^3-2^3\)

\(12.\left(x-1\right):3=56\)

\(12.\left(x-1\right)=168\)

\(x-1=14\)

\(x=15\)

 b. a0a0a : a x 3 - 20202 = 10101

Em xin tham khảo:

a, aaa : a : 3 - 360 : 10

= 111 : 3 - 36

= 37 - 36

= 1

b, b. a0a0a : a x 3 - 20202 = 10101

1/2 x 1/3 x 1/4 x 1/5 x 1/6 x ( x - 1,010 ) = 1/360 - 1/720

1/2 x 1/3 x 1/4 x 1/5 x 1/6 x ( x - 1,010) = 1/720

( x - 1,010 ) x 1/2 x 1/3 x 1/4 x 1/5 x 1/6 = 1/720

( x - 1,010 ) x 1/720 = 1/720

x - 1,010 = 1/720 : 1/720

x - 1,010 = 1

x            = 1 + 1,010

x            = 2,01

(x+1)(x+2)(x+3)(x+4)-360

=(x²+5x+4)(x²+5x+6)-360

Đặt x²+5x+4=t

Ta có : t(t+2)-360=t²+2t-360=(t²+2t+1)-361=(x+1)²-19²=(t+20)(t-18)

= (x²+5x+24)(x²+5x-14)

= (x²+5x+24)(x-2)(x+7)

\(12\left(x+5\right)+2x=130\\\Leftrightarrow 12x+60+2x=130\\ \Leftrightarrow14x=70\\ \Leftrightarrow x=5\\ ----\\ 23\left(x-5\right)-12x=138\\ \Leftrightarrow23x-115-12x=138\\ \Leftrightarrow23x-12x=138+115\\ \Leftrightarrow11x=253\\ \Leftrightarrow x=\dfrac{253}{11}=23\\ ----\\ 360-12x+23\left(x-5\right)=278\\ \Leftrightarrow360-12x+23x-115=278\\ \Leftrightarrow-12x+23x=278+115-360\\ \Leftrightarrow11x=33\\ \Leftrightarrow x=\dfrac{33}{11}=3\)

\(6\left(x+3\right)+3\left(x-5\right)=278\\ \Leftrightarrow6x+18-3x-15=278\\ \Leftrightarrow6x-3x=278+15-18\\ \Leftrightarrow3x=275\\ \Leftrightarrow x=\dfrac{275}{3}\\ ---\\ \left(7-x\right)\left(3x-90\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}7-x=0\\3x-90=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=30\end{matrix}\right.\)

\(\left(x-2\right)\left(4x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\4x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\4x=2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\end{matrix}\right.\\ ---\\ \left(2x-18\right)\left(3x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-18=0\\3x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=18\\3x=9\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=9\\x=3\end{matrix}\right.\)