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Yêu cầu của bài là gì vậy. Tính A? hay Chứng minh A < 2 hoặc chứng minh A không phải là số nguyên

Chứng minh A < 2

\(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\)

\(< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\)

\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=2-\frac{1}{50}< 2\)

Vậy A < 2

Đặt \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1990^2}=A\)

ta có :\(\frac{1}{2^2}=\frac{1}{2\cdot2}=\frac{1}{4}\)

\(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)

\(...\)

\(\frac{1}{1990^2}=\frac{1}{1990\cdot1990}< \frac{1}{1989\cdot1990}\)

\(\Rightarrow A< \frac{1}{4}+\frac{1}{2\cdot3}+...+\frac{1}{1989\cdot1990}\)

\(\Rightarrow A< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1989}-\frac{1}{1990}\)

\(\Rightarrow A< \frac{1}{4}+\frac{1}{2}-\frac{1}{1990}=\frac{3}{4}-\frac{1}{1990}< \frac{3}{4}\)

\(\Rightarrow A< \frac{3}{4}\left(ĐPCM\right)\)

Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1990^2}< \frac{3}{4}\)

hk tốt #

Ta có \(\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{1990^2}< \frac{1}{1989.1990}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1990^2}< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1989.1990}\)

                                                                     \(< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1989}-\frac{1}{1990}\)

                                                                    \(< \frac{1}{4}+\frac{1}{2}-\frac{1}{1990}=\frac{3}{4}-\frac{1}{1990}< \frac{3}{4}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1990^2}< \frac{3}{4}\)

\(\Rightarrow\)Bài toán được chứng minh

( 4 ^x - 13 ) ⁴ + 4 ³ = 145

( 4 ^x - 13 ) ⁴ + 64 = 145

( 4 ^x - 13 ) ⁴         = 81

( 4 ^x - 13 ) ⁴         = 3 ⁴

=> 4 x - 13 = 3

    4 x         = 16

       x         = 4

3 ^{x + 2} - 3 ^x    = 72

3 ^x ( 3 ² - 1 ) =  72

3 ^x . 8          = 72

3 ^x               = 9

3 ^x               = 3 ²

=> x = 2

2 ^{x + 2} - 2 ^{x - 1} = 224

2 ^x ( 2 ² - 2 ^-1 ) = 224

2 ^x . 3 , 5         = 224

2 ^x                   = 64

2 ^x                   = 2 ⁶

=> x = 6

( 4 ^x - 13 ) ⁴ + 4 ³ = 145

( 4 ^x - 13 ) ⁴ + 64 = 145

( 4 ^x - 13 ) ⁴         = 81

( 4 ^x - 13 ) ⁴         = 3 ⁴

=> 4 x - 13 = 3

    4 x         = 16

       x         = 4

3 ^{x + 2} - 3 ^x    = 72

3 ^x ( 3 ² - 1 ) =  72

3 ^x . 8          = 72

3 ^x               = 9

3 ^x               = 3 ²

=> x = 2

2 ^{x + 2} - 2 ^{x - 1} = 224

2 ^x ( 2 ² - 2 ^-1 ) = 224

2 ^x . 3 , 5         = 224

2 ^x                   = 64

2 ^x                   = 2 ⁶

=> x = 6

\(M=\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+...+\dfrac{1}{1+2+3+...+59}\\ =\dfrac{1}{\dfrac{3\cdot4}{2}}+\dfrac{1}{\dfrac{4\cdot5}{2}}+...+\dfrac{1}{\dfrac{59\cdot60}{2}}\\ =\dfrac{2}{3\cdot4}+\dfrac{2}{4\cdot5}+...+\dfrac{2}{59\cdot60}\\ =2\left(\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{59\cdot60}\right)\\ =2\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{59}-\dfrac{1}{60}\right)\\ =2\cdot\dfrac{19}{60}\\ =\dfrac{38}{60}< \dfrac{40}{60}=\dfrac{2}{3}\)

\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{2007^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{2006.2007}\)

\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{2006}-\dfrac{1}{2007}\)

\(=\dfrac{1}{4}-\dfrac{1}{2007}< \dfrac{1}{4}\)

\(\Rightarrow\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{2007^2}< \dfrac{1}{4}\left(đpcm\right)\)

Vậy...

1) 120 + 20 : 4 - 2².3

=> 120 + 5 - 12

=> 125-12

=>113

2) ( 120 + 20 ) : 4 -

3) 120 + ( 20 : 4 - ).3

=> 120 + 1 . 3

=> 120 + 3

=> 123

1)120+20:4-2\(^2\) .3=120+5-4.3=120+5-12=113

2)(120+20):4-2\(^2\) .3=140:4-4.3=35-12=23

3)120+(20:4-2\(^2\) ).3=120+(5-4).3=120+1.3=120+3=123

sai đề rùi

\(4^{3x+1}-8.4^2=8.4^2\)

\(\Rightarrow4^{3x}.4=8.4^2+8.4^2=16.4^2\)

\(\Rightarrow4^{3^x}=4.4.4=4^3\)

\(\Rightarrow3x=3\)

\(\Rightarrow x=1\)

\(7^{5x-1}=7^4\)

\(\Rightarrow5x-1=4\)

\(\Rightarrow5x=5\)

\(\Rightarrow x=1\)