14/24 : 2 = 7/12

10/36 : 2 = 5/18

\(\dfrac{5}{27}=\dfrac{5}{27}\)

\(\dfrac{14}{24}=\dfrac{14:2}{24:2}=\dfrac{7}{12}\)

\(\dfrac{10}{36}=\dfrac{10:2}{36:2}=\dfrac{5}{18}\)

5/27 = 0.(185)

14/24 = 7/12

10/36 = 5/18

147 -( x x 36) = 29

X x 36 = 147 - 29

X x 36 = 118

X = 118 : 36

X = ...

Chúc bạn hok tốt

oki 

147- ( X x 36 ) =29

X x 36=147 - 29

X x 36 =118

X        =118 : 36

X       = .....

bn ơi, 118:36 bằng mấy vj

ta có

(1/3+1/6+1/36) +(1/10+1/15+1/45)+(1/21+1/28)

=(\(\frac{12+6+1}{36}\)+\(\frac{9+6+2}{90}\)+\(\frac{4+3}{84}\)

19/36+17/90+1/12

=(19/36+1/12)+17/90

=7/12+17/90

=105/180+34/180

=139/180

1/3 +1/6+1/10+1/15+1/21+1/28+1/36+1/45

=1/1x3+1/3x2+1/2x5+1/3x5+1/3x7+1/7x4+1/4x9+1/9x5

=1/1-1/3+1/3-1/2....+1/9-1/5

=1/1

bạn ơ sao mình tính trong máy tính đc 4/5 mà

Lời giải:

a. $145-135(x-2)^2=10$

$135(x-2)^2=145-10=135$

$(x-2)^2=135:135=1=1^2=(-1)^2$
$\Rightarrow x-2=1$ hoặc $x-2=-1$

$\Rightarrow x=3$ hoặc $x=1$

--------------------------

b.

$x\in Ư(36); x\geq 12$

$\Rightarrow x\in\left\{12; 18; 36\right\}$

c.

$x-1=B(9)\Rightarrow x-1\in\left\{0; 9; 18; 27; 36; 45;54;.....\right\}$

$\Rightarrow x\in\left\{1; 10; 19; 28; 37;46; 55;....\right\}$

Mà $25< x< 50$ nên $x\in\left\{28; 37; 46\right\}$

roi minh se giai [500 nam nua nhe]

9864 nha bạn

Chọn C

Ta có

 \(a^2+1=a^2+ab+bc+ca=a\left(a+b\right)+c\left(a+b\right)=\left(a+b\right).\left(a+c\right)\\ Cmtt:b^2+1=\left(b+a\right).\left(b+c\right)\\ c^2+1=\left(c+a\right).\left(c+b\right)\)

Nên

 \(\dfrac{b-c}{a^2+1}+\dfrac{c-a}{b^2+1}+\dfrac{a-b}{c^2+1}\\ =\dfrac{\left(b-c\right)}{\left(a+b\right)\left(a+c\right)}+\dfrac{\left(c-a\right)}{\left(b+c\right)\left(b+a\right)}+\dfrac{\left(a-b\right)}{\left(c+a\right)\left(c+b\right)}\\ =\dfrac{\left(b-c\right)\left(b+c\right)+\left(c-a\right)\left(c+a\right)+\left(a-b\right)\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\\ =\dfrac{b^2-c^2+c^2-a^2+a^2-b^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\\ =0\)

\(\dfrac{b-c}{a^2+1}+\dfrac{c-a}{b^2+1}+\dfrac{a-b}{c^2+1}\)

\(=\dfrac{b-c}{a^2+ab+bc+ac}+\dfrac{c-a}{b^2+ab+bc+ca}+\dfrac{a-b}{c^2+ab+bc+ca}\)

\(=\dfrac{b-c}{a\left(a+b\right)+c\left(a+b\right)}+\dfrac{c-a}{b\left(a+b\right)+c\left(a+b\right)}+\dfrac{a-b}{c\left(c+a\right)+b\left(a+c\right)}\)

\(=\dfrac{b-c}{\left(a+c\right)\left(a+b\right)}+\dfrac{c-a}{\left(b+c\right)\left(a+b\right)}+\dfrac{a-b}{\left(b+c\right)\left(a+c\right)}\)

\(=\dfrac{\left(b-c\right)\left(b+c\right)+\left(c-a\right)\left(a+c\right)+\left(a-b\right)\left(a+b\right)}{\left(a+c\right)\left(a+b\right)\left(b+c\right)}\)

\(=\dfrac{b^2-c^2+c^2-a^2+a^2-b^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\) 

a:3 dư 3 nên a-3: hết cho 4 nên a-23 chia hết cho 4

a:9 dư 5 nên a-5 chia hết cho 9 nên a-23 chia hết cho 9

nên a-23 chia hết cho 9 và 4 mà UCLN(9;4)=1 nên a-23 chia hết cho 36

a chia 36 dư 23

k nha bạn