chiu bai nay ko biet lam

Ta có:

\(\frac{a+b}{b+c}=\frac{c+d}{d+a}=\frac{a+b+c+d}{a+b+c+d}=1.\)

=> \(\hept{\begin{cases}a+b=b+c\\c+d=d+a\end{cases}}\)=> a=c (đpcm) 

Ta có: 2P-3Q=1 <=> 2/x-3/-3/1-2x/=1

Xét các trường hợp:

1/ \(\hept{\begin{cases}x-3\ge0\\1-2x\ge0\end{cases}}\)<=> \(\hept{\begin{cases}x\ge3\\\frac{1}{2}\ge x\end{cases}}\) => Không xảy ra

2/ \(\hept{\begin{cases}x-3\ge0\\1-2x< 0\end{cases}}\)<=> \(\hept{\begin{cases}x\ge3\\x>\frac{1}{2}\end{cases}}\) => \(x\ge3\)Khi đó ta có: 

2(x-3)-3(2x-1)=1 <=> 2x-6-6x+3=1  <=> 4x=-4 => x=-1 (Loại vì \(x\ge3\))

3/ \(\hept{\begin{cases}x-3< 0\\1-2x\ge0\end{cases}}\)<=> \(\hept{\begin{cases}x< 3\\x\le\frac{1}{2}\end{cases}}\) => \(x\le\frac{1}{2}\)Khi đó ta có: 

2(3-x)-3(1-2x)=1 <=> 6-2x-3+6x=1  <=> 4x=-2 => x=-1/2 (Nhận)

4/ \(\hept{\begin{cases}x-3< 0\\1-2x< 0\end{cases}}\)<=> \(\hept{\begin{cases}x< 3\\x>\frac{1}{2}\end{cases}}\) => \(\frac{1}{2}< x< 3\)Khi đó ta có: 

2(3-x)-3(2x-1)=1 <=> 6-2x-6x+3=1  <=> 8x=8 => x=1 (Nhận)

Đáp số: x=1 và \(x=-\frac{1}{2}\)

5^x + 5^x+2 =650

=> 5^x  +5².5^x = 650

=> 5^x . ( 1 + 5²) = 650

=> 5^x .26 =650

=> 5^x = 650 :26 = 25

=> 5^x = 5² 

=> x = 2

Vậy x =2

Ta có :

5^x + 5^x+2 = 650

5^x + 5^x x 5² = 650

5^x + 5^x x 25 = 650

5^x x (1 +25 ) = 650

5^x x 26 = 650

     5^x   =  650 : 26

     5^x    =  25

       5x   =   5²

      x    =   2

    Vậy x = 2

222222

bố thằng nào biết đc

\(\frac{x-1}{13}+\frac{x-2}{14}+\frac{x-3}{15}+3=0\)

\(\frac{x-1}{13}+1+\frac{x-2}{14}+1+\frac{x-3}{15}+1=0\)

\(\frac{x-1}{13}+\frac{13}{13}+\frac{x-2}{14}+\frac{14}{14}+\frac{x-3}{15}+\frac{15}{15}=0\)

\(\frac{x+12}{13}+\frac{x+12}{14}+\frac{x+12}{15}=0\)

\(\left(x+12\right)\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)=0\)

\(x+12=0\)hay \(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}=0\)(vô lý)

=>x+12=0

x=0-12

x=-12

Vậy x=-12

\(\frac{x-1}{13}\)+    \(\frac{x-2}{14}\) +  \(\frac{x-3}{15}\)+  3  =  0

=> (\(\frac{x-1}{13}\)+  1 )   +  (\(\frac{x-2}{14}\)+  1)  +  (\(\frac{x-3}{15}\) +  1)  =  0

=>  \(\frac{x+12}{13}\)+  \(\frac{x+12}{14}\)+  \(\frac{x+12}{15}\)=  0

=> (x + 12) . (\(\frac{1}{13}\)+\(\frac{1}{14}\)+  \(\frac{1}{15}\))  = 0

=>(x + 12) . \(\frac{587}{2730}\)= 0

=> x + 12 = 0

=> x = -12

Vậy x = -12