73/36

73/76

9^x-1 = 9

=> 9^x-1 = 9¹

=> x-1 =1

=> x = 1+1

=> x =2

\(9^{x-1}=9\)

\(9^{x-1}=9^1\)

\(x-1=1\)\(x=2\)

Mình nghĩ bài này nên có thêm \(x\inℤ\) : Lời giải như sau :

Ta có : \(2x+5⋮x+2\)

\(\Leftrightarrow2\left(x+2\right)+1⋮x+2\)

mà \(2\left(x+2\right)⋮x+2\forall x\inℤ\)

Nên \(1⋮x+2\) hay \(x+2\inƯ\left(1\right)\)

\(\Rightarrow x+2\in\left\{-1,1\right\}\)

\(\Rightarrow x\in\left\{-3,-1\right\}\)

Vậy : \(x\in\left\{-3,-1\right\}\) để \(2x+5⋮x+2\)

TA CÓ : 

<br class="Apple-interchange-newline"><div></div>2X+5⋮X+2 

MÀ \(2X+4⋮X+2\)

\(\Rightarrow\left(2X+5\right)-\left(2X+4\right)⋮X+2\)

\(\Rightarrow1⋮X+2\)

\(\Rightarrow X+2\inƯ\left(1\right)=\left\{\pm1\right\}\)\(\Rightarrow X\in\left\{-1;-3\right\}\)

CHÚC BẠN HỌC TỐT NHỚ TI CK CHO MINH NHA:)))

\(\frac{x-12}{3}=\frac{x+1}{4}\)

=>(x-12).4=(x+1)*3

    4x-48=3x+3

    4x-3x=48+3

    x=51

(x-12)/3=(x+1)/4

(x-12)*4=(x+1)*3

x*4-12*4=x*3+1*3

4x-48=3x+3

4x-3x=3+48

x=51

-25/11

\(-3-\frac{-8}{11}\)

\(=\frac{-33}{11}-\frac{-8}{11}\)

\(=\frac{-33-\left(-8\right)}{11}\)

\(=\frac{-33+8}{11}\)

\(=\frac{-25}{11}\)

\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{x\left(x+1\right):2}=1\frac{1991}{1993}\)

\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.....+\frac{2}{x\left(x+1\right)}=1-1\frac{1991}{1993}=\frac{1991}{1993}\)

\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{x\left(x+1\right)}\right)=\frac{1991}{1993}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1991}{1993}:2=\frac{1991}{3986}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{1991}{3986}\)

\(\frac{1}{x+1}=\frac{1}{2}-\frac{1991}{3986}=\frac{1}{1993}\)

=> x + 1 = 1993

=> x = 1993 - 1

=> x = 1992

ccccccccccccccccccccccccccc

có 10 cây cau trồng thành 5 hàng mỗi hàng có 4 cây ????