A=\(\dfrac{5^{2}}{1.6}+\dfrac{5^{2}}{6.11}+...+\dfrac{5^{2}}{26.31}\)
Chứng tỏ A<5
Tính giá trị của biểu thức: \(A=\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+...+\dfrac{5^2}{26.31}\)
`A = ( 5^2 )/( 1*6)+(5^2)/(6*11)+.....+(5^2)/(26*31)`
`= 5*( 5/( 1*6)+ 5/(6*11)+.....+5/(26*31))`
`= 5*( 1 - 1/6 + 1/6 - 1/11 +....+1/26 - 1/31 )`
`= 5*( 1 - 1/31 )`
`= 5 * 30/31 = 150/31`
\(A=\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+...+\dfrac{5^2}{26.31}\)
\(=5.\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+...+\dfrac{5}{26.31}\right)\)
\(=5.\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{26}-\dfrac{1}{31}\right)\)
\(=5.\left(1-\dfrac{1}{31}\right)=5.\dfrac{30}{31}=\dfrac{150}{31}\)
cho A=5*2/1.6+5*2/6.11+...+5*2/26.31
chứng tỏ A>1
A = 52 /1.6+52 /6.11+.....+52 /26.31
chứng tỏ A>1
A= 52 /1.6 + 52 /6.11 +...+ 52 /26.31
..
=> A= 5.( 5/ 1.6 + 5/ 6.11 +...+ 5 /26.31)
=> A= 5.( 1- 1/6 + 1/6 - 1/11 +...+ 1/26 - 1/31)
=> A= 5.( 1 - 1/31 )
=> A= 5. 30/31 = 150/31 > 1
=>A=52(1/1.6+1/6.11+...+1/26.31)
=>A=25/2(2/1.6+2/6.11+...+2/26.31)
=>A=25/2(1/1-1/6+1/6-1/11+....+1/26-1/31)
=>A=25/2(1+0+0+.....+1/31)
=>A=25/2X34/31
=>A=850/62
=>A=425/31
=>A>1(425>31=>A<1)
cho biểu thức A= 5 mũ 2 phần 1.6 + 5 mũ 2 phần 6.11 + ...+ 5 mũ 2 phần 26.31 . Chứng tỏ A > 1
Có: \(A=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{26.31}\)
\(=5.\left(\frac{6-1}{1.6}+\frac{11-6}{6.11}+...+\frac{31-26}{26.31}\right)\)
\(=5.\left(\frac{6}{1.6}-\frac{1}{1.6}+\frac{11}{6.11}-\frac{6}{6.11}+...+\frac{31}{26.31}-\frac{26}{26.31}\right)\)
\(=5.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(=5.\left(1-\frac{1}{31}\right)\)
\(=5.\frac{30}{31}=\frac{150}{31}>\frac{31}{31}=1\)
\(\Rightarrow A>1\)
Ta có: A=\(\frac{5^2}{1.6}\)+\(\frac{5^2}{6.11}\)+...+\(\frac{5^2}{26.31}\)
=5.(\(\frac{5}{1.6}\)+\(\frac{5}{6.11}\)+...+\(\frac{5}{26.31}\))
=5.(1-\(\frac{1}{6}\)+\(\frac{1}{6}\)-\(\frac{1}{11}\)+\(\frac{1}{11}\)+...+\(\frac{1}{26}\)-\(\frac{1}{30}\))
=5.(1-\(\frac{1}{30}\))
=5.\(\frac{29}{30}\)
=\(\frac{29}{6}\)>1
Hay A>1
=> đpcm
Bài 1 : Tính
a) A = \(\left(\dfrac{2}{3}+\dfrac{3}{4}-\dfrac{7}{12}\right):\left(\dfrac{55}{123}+\dfrac{555}{1234}+\dfrac{5555}{12345}\right)\)
b) B = \(\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+\dfrac{5^2}{11.16}+...+\dfrac{5^2}{101.106}\)
c) C = \(\dfrac{2x^2+3x-1}{3x-2}\) với \(\left|x-1\right|=2\)
a, bạn tự làm
b, \(B=\dfrac{5^2}{5}\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{101}-\dfrac{1}{106}\right)\)
\(=5\left(1-\dfrac{1}{106}\right)=\dfrac{5.105}{106}=\dfrac{525}{106}\)
c, đk : \(x\ne\dfrac{2}{3}\)
Ta có : \(\left|x-1\right|=2\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)(tm)
Với x = 3 suy ra \(C=\dfrac{2.9+9-1}{3.3-2}=\dfrac{26}{7}\)
Với x = -1 suy ra \(C=\dfrac{2-3-1}{-3-2}=\dfrac{-2}{-5}=\dfrac{2}{5}\)
chứng tỏ rằng 14n+3/21n+5 là phân số tối giản với moi thuộc z
A=5^2/1.6+5^2/6.11+.....+5^2/26.31>1
A=\(\frac{5^2}{1.6}+\frac{5^2}{6.11}+....+\frac{5^2}{26.31}\)
=>A=5.(\(\frac{5}{1.6}+\frac{5}{6.11}+....+\frac{5}{26.31}\))
=>A=5.(\(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\))
=>A=5.(\(\frac{1}{1}-\frac{1}{31}\))
=>A=5.\(\frac{30}{31}\)
=>A=\(\frac{150}{31}\)
=>A>1( vì tử của A lớn hơn mẫu )
a, gọi ƯCLN(14n+3;21n+5)=d
=> \(\left\{{}\begin{matrix}14n+3\\21n+5\end{matrix}\right.\)⋮d =>\(\left\{{}\begin{matrix}3\left(14n+3\right)\\2\left(21n+5\right)\end{matrix}\right.\)⋮d=>\(\left\{{}\begin{matrix}42n+9\\42n+10\end{matrix}\right.\)⋮d
=>(42n+10)-(42n+9)⋮d
=>1⋮d
=>d=1
Do ƯCLN của 14n+3 ; 21n+5 là 1
=> 2 số trên là hai số nguyên tố cùng nhau
=>hai số đó nếu chia cho nhau thì sẽ ko chia hết
=> hai số đó khi biểu diễn ở dạng phân số thì sẽ thành phân số tối giản
gọi ƯCLN(14n+3;21n+5)=d
=>\(\left\{{}\begin{matrix}14n+3\\21n+5\end{matrix}\right.\)⋮d => \(\left\{{}\begin{matrix}21\left(14n+3\right)\\14\left(21n+5\right)\end{matrix}\right.\)⋮d
A=\(\frac{5^2}{1.6}\)+\(\frac{5^2}{6.11}\)+...........+\(\frac{5^2}{26.31}\)
chứng tỏ A>1
#)Giải :
Ta có :
\(A=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{26.31}=5\left(\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{26.31}\right)\)
\(=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)=5\left(1-\frac{1}{31}\right)=5\times\frac{30}{31}=\frac{150}{31}>1\)
\(\Rightarrow A>1\)
A= \(\frac{5^2}{1.6^{ }}\) + \(\frac{5^2}{6.11}\) + ... + \(\frac{5^2}{26.31}\)
Chứng tỏ A > 1
Ta có: \(A=\frac{5^2}{1\cdot6}+\frac{5^2}{6\cdot11}+...+\frac{5^2}{26\cdot31}\)
\(=5\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+...+\frac{5}{26\cdot31}\right)\)
\(=5\cdot\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(=5\cdot\left(1-\frac{1}{31}\right)=5\cdot\frac{30}{31}=\frac{150}{31}>1\)
hay A>1(đpcm)
tìm X
\(\dfrac{5x}{1.6}+\dfrac{5x}{6.11}+\dfrac{5x}{11.16}+\dfrac{5x}{16.21}+\dfrac{5x}{21.26}+\dfrac{5x}{26.31}\)
\(\dfrac{5x}{1.6}+\dfrac{5x}{6.11}+\dfrac{5x}{11.16}+\dfrac{5x}{16.21}+\dfrac{5x}{21.26}+\dfrac{5x}{26.31}=1\)
\(=x\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+\dfrac{5}{16.21}+\dfrac{5}{21.26}+\dfrac{5}{26.31}\right)=1\)
\(=x\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{31}\right)=1\)
\(=x\left(1-\dfrac{1}{31}\right)=1\)
\(\Rightarrow x=1:\left(1-\dfrac{1}{31}\right)=\dfrac{31}{30}\)