3⁹⁸ >2¹⁴⁷   

ta có:

3⁹⁸=(3x2)⁴⁹=6⁴⁹

2¹⁴⁷=(2x3)⁴⁹=6⁴⁹

vậy 3⁹⁸= 2¹⁴⁷

so sánh \(3^{98}\)và \(2^{147}\)

\(3^{98}=3^{2\cdot49}=\left(3^2\right)^{49}=9^{49}\)

\(2^{147}=2^{3\cdot49}=\left(2^3\right)^{49}=8^{49}\)

Ta có \(9^{49}>8^{49}\)

\(\Rightarrow3^{98}>2^{147}\)

`A=3/4+8/9+.............+9999/10000`

`=1-1/4+1-1/9+,,,,,,,,,,+1-1/10000`

`=99-(1/4+1/9+.........+1/10000)<99-0=99`

`=>A<99`

 địt mẹ con ngu t khinh

Giải:

\(A=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+...+\dfrac{9999}{10000}\) 

\(A=\left(1-\dfrac{1}{4}\right)+\left(1-\dfrac{8}{9}\right)+\left(1-\dfrac{1}{16}\right)+...+\left(1-\dfrac{1}{10000}\right)\) 

\(A=99-\left(\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{19}+...+\dfrac{1}{10000}\right)< 99\) 

\(\Rightarrow A< 99\left(đpcm\right)\) 

Chúc bạn học tốt!

Ta có :

( 98⁴ )³ = 98^4*3 = 98¹²

Vì 98 = 98

mà 12 < 15

Nên 98¹² < 98¹⁵

Hay ( 98⁴ )³ < 98¹⁵

Vậy (98⁴)³ < 98¹⁵

3:

a: \(\sqrt{\dfrac{2}{3}}=\sqrt{\dfrac{6}{9}}=\dfrac{\sqrt{6}}{3}\)

b: \(\dfrac{x}{y}\cdot\sqrt{\dfrac{y}{x}}=\sqrt{\dfrac{x^2}{y^2}\cdot\dfrac{y}{x}}=\sqrt{\dfrac{x}{y}}=\dfrac{\sqrt{xy}}{y}\)

2:

a: 2căn 7=căn 28

3căn 2=căn 18

mà 28>18

nên 2*căn 7>3*căn 2

b: 5=2+3

mà 3>căn 2

nên 2+3>2+căn 2

=>5>2+căn 2

1) a) \(\sqrt{98}-\sqrt{72}+0,5\sqrt{8}\)

\(=\sqrt{49.2}-\sqrt{36.2}+0,5\sqrt{4.2}\)

\(=7\sqrt{2}-6\sqrt{2}+0,5.2\sqrt{2}\)

\(=7\sqrt{2}-6\sqrt{2}+\sqrt{2}=2\sqrt{2}\)

b) \(\sqrt{9a}-\sqrt{16a}+\sqrt{49}\)

\(=3\sqrt{a}-4\sqrt{a}+7=7-\sqrt{a}\)

2. a) \(2\sqrt{7}=\sqrt{4.7}=\sqrt{28}\)

\(3\sqrt{2}=\sqrt{9.2}=\sqrt{18}\)

Mà \(\sqrt{28}>\sqrt{18}\Rightarrow2\sqrt{7}>3\sqrt{2}\)

b) \(5=2+3=2+\sqrt{9}\)

Vì \(\sqrt{9}>\sqrt{2}\Rightarrow2+\sqrt{9}>2+\sqrt{2}\Rightarrow5>2+\sqrt{2}\)

3. a) \(\sqrt{\dfrac{2}{3}}=\sqrt{\dfrac{6}{9}}=\dfrac{\sqrt{6}}{3}\)

b) \(\dfrac{x}{y}.\sqrt{\dfrac{y}{x}}=\sqrt{\dfrac{x^2}{y^2}.\dfrac{y}{x}}=\sqrt{\dfrac{x}{y}}=\dfrac{\sqrt{xy}}{y}\)

A = \(\frac{3}{2}+\frac{7}{6}+\frac{13}{12}+...+\frac{91}{90}+\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{6}\right)+\left(1+\frac{1}{12}\right)+...+\left(1+\frac{1}{90}\right)\)

\(=\left(1+\frac{1}{1.2}\right)+\left(1+\frac{1}{2.3}\right)+\left(1+\frac{1}{3.4}\right)+...+\left(1+\frac{1}{9.10}\right)\)

\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)+\left(1+1+1+...+1\right)\)(9 số hạng 1)

\(=\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)+1.9\)

\(=\left(1-\frac{1}{10}\right)+9=10-\frac{1}{10}=\frac{99}{10}>\frac{98}{11}\)

a=1+1/1.2+1+1+1/2.3+....+1+1/9.10

a=1+1+...+1(9 chữ số 1)+1/1-1/2+1/2-1/3+...+1/9-1/10

a=9+1-1/10

a=9+9/10=9+0.9=9.9

b=98/11<98/10=9.8<9.9

=>vậy a>b

b)39/52 và 98/112
98/112=98:14/112:14=7/8
39/52=39:13/52:13=3/4=3x2/4x2=6/8
vì 7<8=>39/52<98/112
đúng thì k cho mình nha

\(2^{98}=4^{49}\text{ thấy ngay sự chênh lệch giữa:}\)

\(5^{47}\text{ và: }4^{49}\text{ nên: }5^{47}>4^{49}\text{ hay: }5^{47}>2^{98}\)

fkmvjj 

Trần Đức ?