ta có :

\(\hept{\begin{cases}0,\left(27\right)=\frac{27}{99}\\0,\left(72\right)=\frac{72}{99}\end{cases}\Rightarrow0,\left(27\right)+0,\left(72\right)=\frac{27}{99}+\frac{72}{99}=\frac{99}{99}=1}\)

. \(\hept{\begin{cases}0,\left(317\right)=\frac{317}{999}\\0,\left(682\right)=\frac{682}{999}\end{cases}\Rightarrow0,\left(317\right)+0,\left(682\right)=\frac{317}{999}+\frac{682}{999}=\frac{999}{999}=1}\)

Đặt \(a=2k;b=-3k;c=-4,5k\)

Thay vào P ta được

 \(P=\frac{3.2k-2.\left(-3k\right)}{8.2k+3k+3\left(-4,5k\right)}=\frac{6k+6k}{16k+3k-13,5k}=\frac{12k}{5,5k}=\frac{12}{5.5}=12.\frac{2}{11}=\frac{24}{11}\)

đây nha bạn

-14 /37; 4 /3 ; (-14) /33 ; 17 /20; 18 /19 ; 0

chúc bạn học tốt

sao mọi người tl mà bạn ko k thế 

Tìm x, y, z biết 3x= 4y, 2y =5z và x+y+z=3

x = 60/41, y = 45/41, z = 18/41;

\(\left(x^2-8x\right)-3x+24=0\)

\(\Leftrightarrow x\left(x-8\right)-3\left(x-8\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-8\right)=0\Leftrightarrow x=3;x=8\)

bài 59; tìm x biết

2, ( x mũ 2 - 8x ) - 3x + 24 = 0

x= 3 ; x =  8 

TL: 

/2x-1/ + /1-2x/=0

=> 2x-1 =0 và 1-2x=0

=> x=1/2

Trả lời:

a, \(\left(5x+1\right)^2=36\)

\(\Leftrightarrow\orbr{\begin{cases}\left(5x+1\right)^2=6^2\\\left(5x+1\right)^2=\left(-6\right)^2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}5x+1=6\\5x+1=-6\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{7}{5}\end{cases}}}\)

Vậy x = 1; x = - 7/5

b, \(\left(x-2\right)^3=2^6\)

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

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

\(\Leftrightarrow x-2=4\)

\(\Leftrightarrow x=6\)

Vậy x = 6

c, \(\left(8x-1\right)^{2x+1}=5^{2x+1}\)

\(\Leftrightarrow8x-1=5\)

\(\Leftrightarrow8x=6\)

\(\Leftrightarrow x=\frac{3}{4}\)

d, \(\left(x-3,5\right)^2+\left(y-1\right)^4\le0\)

Mà \(\left(x-3,5\right)^2\ge0\forall x;\left(y-1\right)^4\ge0\forall y\)

\(\Rightarrow\hept{\begin{cases}x-3,5=0\\y-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=3,5\\y=1\end{cases}}}\)

Vậy x = 3,5; y = 1

a. ta có 2y+3 là số lẻ nên

\(\left|2y+3\right|\in\left\{1,3\right\}\)

\(TH1:\left|2y+3\right|=1\Rightarrow\hept{\begin{cases}\left|2y+3\right|=1\\\left|x+5\right|=14\end{cases}}\) vậy (x,y) = (-19,-2) , (-19,-1)  (9,-2) , (9,-1)

TH2: \(\left|2y+3\right|=3\Rightarrow\hept{\begin{cases}\left|2y+3\right|=3\\\left|x+5\right|=6\end{cases}}\)Vậy (x,y) =( -11,-1) , (-11,0) , (1,-1), (1,0)

b. ta có \(\left(2x\right)^2+\left|y+3\right|=9\)

\(TH1:\left|2x\right|=2\Rightarrow\hept{\begin{cases}\left|2x\right|=2\\\left|y+3\right|=5\end{cases}}\) vậy (x,y) = (-1,-8) ,(-1,2) ,(1,-8), (1,2)

\(TH2:\left|2x\right|=0\Leftrightarrow\hept{\begin{cases}\left|2x\right|=0\\\left|y+3\right|=9\end{cases}}\)vậy (x,y=(0,-12) , (0.6)