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Trả lời:
\(x=\frac{9^{11}+2}{9^{11}+3}=\frac{9^{11}+3-1}{9^{11}+3}=\frac{9^{11}+3}{9^{11}+3}-\frac{1}{9^{11}+3}=1-\frac{1}{9^{11}+3}\)
\(y=\frac{9^{12}+2}{9^{12}+3}=\frac{9^{12}+3-1}{9^{12}+3}=\frac{9^{12}+3}{9^{12}+3}-\frac{1}{9^{12}+3}=1-\frac{1}{9^{12}+3}\)
Ta có: \(9^{11}< 9^{12}\)
\(\Leftrightarrow9^{11}+3< 9^{12}+3\)
\(\Leftrightarrow\frac{1}{9^{11}+3}>\frac{1}{9^{12}+3}\)
\(\Leftrightarrow-\frac{1}{9^{11}+3}< -\frac{1}{9^{12}+3}\)
\(\Leftrightarrow1-\frac{1}{9^{11}+3}< 1-\frac{1}{9^{12}+3}\)
\(\Leftrightarrow x< y\)
Vậy x < y
Ta có: x = \(\frac{7^{16}-3}{7^{16}+1}=\frac{7^{16}+1-4}{7^{16}+1}=1-\frac{4}{7^{16}+1}\)
y = \(\frac{7^{17}-3}{7^{17}+1}=\frac{7^{17}+1-4}{7^{17}+1}=1-\frac{4}{7^{17}+1}\)
Do \(7^{16}+1< 7^{17}+1\) => \(\frac{4}{7^{16}+1}>\frac{4}{7^{17}+1}\) => \(-\frac{4}{7^{16}+1}< -\frac{4}{7^{17}+1}\)
=> \(1-\frac{4}{7^{16}+1}< 1-\frac{4}{7^{17}+1}\) => x < y
Trả lời:
\(x=\frac{7^{16}-3}{7^{16}+1}=\frac{7^{16}+1-4}{7^{16}+1}=\frac{7^{16}+1}{7^{16}+1}-\frac{4}{7^{16}+1}=1-\frac{4}{7^{16}+1}\)
\(y=\frac{7^{17}-3}{7^{17}+1}=\frac{7^{17}+1-4}{7^{17}+1}=\frac{7^{17}+1}{7^{17}+1}-\frac{4}{7^{17}+1}=1-\frac{4}{7^{17}+1}\)
Ta có: \(7^{16}< 7^{17}\)
\(\Leftrightarrow7^{16}+1< 7^{17}+1\)
\(\Leftrightarrow\frac{4}{7^{16}+1}>\frac{4}{7^{17}+1}\)
\(\Leftrightarrow-\frac{4}{7^{16}+1}< -\frac{4}{7^{17}+1}\)
\(\Leftrightarrow1-\frac{4}{7^{16}+1}< 1-\frac{4}{7^{17}+1}\)
\(\Leftrightarrow x< y\)
Vậy x < y
a) CÓ: A = (1-1/42).(1-1/52).(1-1/62)......(1-1/2002)
=\(\frac{4^2-1^2}{4^2}\). \(\frac{5^2-1^2}{5^2}\). \(\frac{6^2-1^2}{6^2}\)....... \(\frac{200^2-1^2}{200^2}\)
Ta có công thức sau : a2-b2= a2 -ab+ab-b2
= a(a-b) + b(a-b)
= (a+b)(a-b)
ÁP DỤNG CÔNG THỨC TRÊN VÀO BÀI TOÁN TA ĐƯỢC :
A= \(\frac{3.5}{4^2}\). \(\frac{4.6}{5^2}\). \(\frac{5.7}{6^2}\)......\(\frac{199.201}{200^2}\)
= \(\frac{\left(3.4.5.....199\right)\left(5.6.7....201\right)}{\left(4.5.6......200\right)^2}\)
= \(\frac{\left(3.4.5.......199\right)\left(5.6.7.....200.201\right)}{\left(4.5.6.....199.200\right)\left(4.5.6......200\right)}\)
= \(\frac{3.201}{200.4}\)
= \(\frac{603}{800}\)
b)Từ đề bài ta suy ra : B=\(\frac{1.3}{5.7}\).\(\frac{3.5}{7.9}\). \(\frac{5.7}{9.11}\)...... \(\frac{99.101}{103.105}\)
= \(\frac{1.3^2.5^2.7^2......99^2.101}{5.7^2.9^2.11^2....99^2.101^2.103^2.105}\)
=\(\frac{3^2.5}{101.103^2.105}\)
=\(\frac{3}{7500563}\)
B=(1-2-3+4)+(5-6-7+8)+...+(97-98-99+100)
B=0+0+..+0
B=0
C=2^100-(2^99+2^98+2^97+...+1)
đặt D=2^99+2^98+2^97+...+1
=>D=2^100-1
=>C=2^100-(2^100-1)=1
Ta có: A(1/2) = \(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
2.A(1/2) = \(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(\Rightarrow\) A(1/2) = \(2-\frac{1}{2^{100}}\)
a) (-5x2y + 3xy2 + 7) + (-6x2y + 4xy2 -5)
= -5x2y + 3xy2 + 7 + -6x2y + 4xy2 -5
= -11x2y +7xy2 + 2
b) (2,4a3 - 10a2b) + (7a2b - 2,4a3 + 3ad2)
= 2,4a3 - 10a2b + 7a2b - 2,4a3 + 3ad2
= -3a2b + 3ad2
c) (15x2y - 7xy2 - 6y3) + (2x3 - 12x2y +7xy2)
= 15x2y - 7xy2 - 6y3 + 2x3 - 12x2y +7xy2
= 3x2y- 6y3 + 2x3
5) 32.2+ ( x+52) =102
=> 18 + ( x + 52 )= 102
=> x + 52 = 102 -18
=> x + 52 = 84
=> x = 84 -52
=> x = 32
Ta có:
1) 1 + 3 + 5 + ... + 99 = ( x - 2 ) 2
=> 100 x 50 / 2 = ( x - 2 ) 2
=> 2500 = ( x - 2 ) 2
=> x - 2 = \(\sqrt{2500}=50\)
=> x = 52
2) 3x + 25 = 26.2 + 2.30
3x + 25 = 2.( 26 + 30 )
3x + 25 = 2.56 = 112
3x = 112 - 25 = 87
không tìm được x.Có thể bạn đã sai đề.
3) 49. 7x = 204
=> 7x = 204 : 49 = 204/49
không tìm được x nếu x là số tự nhiên.
4) 22 (x + 32 ) - 5 = 55
22 (x + 32 ) = 60
x + 32 = 15
x = 15 - 9 = 6
5) 32.2 + ( x + 52 ) = 102
x + 52 = 102 - 9.2
x + 52 = 102 - 18 = 84
x = 84 - 52 = 32
32/2x4+52/4x6+...+992/98x100
=9/8+25/24+...+9801/9800
=1+1/8+1+1/24+...+1+1/9800
=1+1+...+1+1/2.4+1/4.6+...+1/98.100
= 49 + A
với A=1/2.4+1/4.6+...+1/98.100
=1/4(1/1.2+1/2.3+...+1/49.50)
=1/4(1-1/2+1/2-1/3+...+1/49-1/50)
=1/4(1-1/50)
=1/4.49/50
=49/200
ta có:32/2x4+52/4x6+...+992/98x100= 49+A= 49+49/200=9849/200
chúc bạn hok tốt