|x+3/4|=x+3/4 hoặc -x-3/4

+) x+3/4-3/4=x

=>x=x

+) -x-3/4-3/4=x

=>-x-3/2=x

=>-2x=3/2

=>x=-3/4

\(\left|x+\frac{3}{4}\right|=x+\frac{3}{4}\)

\(\Rightarrow\text{ }x+\frac{3}{4}\ge0\)                   \(\Rightarrow\text{ }x=-\frac{3}{4}\) ( Vì \(-\frac{3}{4}+\frac{3}{4}=0\) )

\(\Rightarrow\text{ }x\ge0\)                             \(\Rightarrow\text{ }x=0\)

Vậy ta có : \(x+\frac{3}{4}=x+\frac{3}{4}\)

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

                   \(2x=\frac{3}{2}\)  

                   \(x=\frac{3}{2}\text{ : }2\)

                   \(x=\frac{3}{4}\)

Vậy \(x\in\left\{0\text{ ; }-\frac{3}{4}\text{ ; }\frac{3}{4}\right\}\)

650 = 2 . 5² . 13

1236 = 2² . 3 . 103

492 = 2² . 3 . 41

1353 = 3.11.41

650= 2.5.5.13

1236=2.2.3.103

492=2.2.3.41

1353=3.11.41

4^3 * 32^5 - 8^8 k chia hết cho 5

đáp án là 3 

=3

75/100+18/21+19/32+1/4+3/21+13/32

= 18/21 + 3/21 + 19/32 + 13/32 + 75/100 + 1/4

= (18/21 + 3/21) + (19/32 + 13/32) + (3/4 + 1/4)

= 21/21 + 32/32 + 4/4

= 1 + 1 + 1

= 3

A=[(100+2).50:2]-[(99+1).50:2]

  = 2550-2500=50

A = ( 2 + 4 + ... + 100 ) - ( 1 + 3 + ... + 99 )

A = 2550 - 2500

A = 50

=))

= 5^2017( 1+5-5^2)

=5^2017. (-19) chia hết cho 19

\(5^{2017}+5^{2018}-5^{2019}=5^{2017}\left(1+5-5^2\right)=5^{2017}\left(-19\right)⋮19\)

a) 5^270 < 5^300

b) 13^100 < 13^102

c) 3^4995 > 3^4000

So sánh :

a) 125^90 và 25^150

125^90 = (5^3)^90 = 5^270

25^150 = (5^2)^150 = 5^300

125^90 < 25^150

B) 13^100 và 169^51

169^51 = (13^2)^51 = 13^102

13^100 < 169^51

c,243^999 và 9^2000

243^999 = (3^5)^999 = 3^4995

9^2000 = (3^2)^2000 = 3^4000

243^999 > 9^2000

231-(x-6)=1339/3

x=231-1339/3+6

x=-628/ 3

1339:13 à nhon

#)Giải : (Đg rảnh nên làm lun :v)

Ta có : \(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{50}-\frac{1}{51}=1-\frac{1}{51}=\frac{50}{51}< 2\)

\(\Rightarrow A< \frac{50}{51}< 2\)

\(\Rightarrow A< 2\left(đpcm\right)\)