Thu gọn biểu thức:
A= 1/21 + 1/23 + 1/25 + .... + 1/299
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\(\dfrac{7}{19}x\dfrac{8}{23}+\dfrac{7}{19}x\dfrac{15}{23}+1\dfrac{7}{19}\)
= \(\dfrac{7}{19}x\left(\dfrac{8}{23}+\dfrac{15}{23}\right)+1+\dfrac{7}{19}\)
=\(\dfrac{7}{19}x1+1+\dfrac{7}{19}\)
= \(\dfrac{7}{19}+1+\dfrac{7}{19}=1\dfrac{14}{19}\) = \(\dfrac{33}{19}\)
\(\dfrac{75}{100}+\dfrac{18}{21}+\dfrac{49}{32}+\dfrac{1}{4}+\dfrac{3}{21}-\dfrac{17}{32}\)
= \(\dfrac{3}{4}+\dfrac{6}{7}+\dfrac{49}{32}+\dfrac{1}{4}+\dfrac{1}{7}-\dfrac{17}{32}\)
= \(\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+\left(\dfrac{6}{7}+\dfrac{1}{7}\right)+\left(\dfrac{49}{32}-\dfrac{17}{32}\right)\)
= 1 + 1 + 1 = 3
\(\dfrac{8}{9}x\dfrac{15}{16}x\dfrac{24}{25}x\dfrac{35}{36}x\dfrac{48}{49}x\dfrac{63}{64}\)
= \(\dfrac{3}{4}\) *Câu này bạn tự sử dụng gạch nhé!
`1,`
`a,`
`7/19 \times 8/23 + 7/19 \times 15/23 + 1 7/19`
`= 7/19 \times 8/23 + 7/19 \times 15/23 + 1 + 7/19`
`= 7/19 \times (8/23 + 15/23 + 1) + 1`
`= 7/19 \times 2 + 1`
`=14/19 + 1`
`= 33/19`
`b,`
`75/100 + 18/21 + 49/32 + 1/4 + 3/21 - 17/32`
`= 75/100 + (18/21 + 3/21) + (49/32 - 17/32) + 1/4`
`= 0,75 + 1 + 1 + 0,25`
`= (0,75 + 0,25) + 1 + 1`
`= 1+1+1=3`
`c,`
`8/9 \times 15/16 \times 24/25 \times 35/36 \times 48/49 \times 63/64`
`=` \(\dfrac{2\times3}{3\times3}\times\dfrac{3\times5}{4\times4}\times\dfrac{3\times4\times2}{5\times5}\times\dfrac{5\times7}{6\times6}\times\dfrac{6\times8}{7\times7}\times\dfrac{7\times9}{8\times8}\)
`= 3/4` (bạn sử dụng gạch, rút gọn các số là được nhé).
B1
Số nhóm biểu thức nhò là
(2001+3)/2+1=1003(số)
mà giá trị mỗi biểu thức là 1
=> 1+1*1003=1004
-23*63+23*21-58*23\(=23.\left(-63+21-58\right)=23.\left(-100\right)=-2300\)
HTDT
-23 x 63 + 23 x 21 - 58 x 23
= (-23) x ( 63 + 58 ) - 21
=(-23) x 121 - 21
= (-23) x 100
= 77
Hk tốt
k nhé
\(a,A=\left(x^2-x\right)\left(x^2-x-12\right)\\ A=\left(x^2-x\right)^2-12\left(x^2-x\right)\\ A=\left(x^2-x\right)^2-12\left(x^2-x\right)+36-36\\ A=\left(x^2-x+6\right)^2-36\ge-36\\ A_{min}=-36\Leftrightarrow x^2-x+6=0\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\\ b,B=4x^4+4x^3+5x^2+4x+3\\ B=\left(4x^4+4x^3+x^2\right)+\left(x^2+4x+4\right)-1\\ B=x^2\left(2x+1\right)^2+\left(x+2\right)^2-1\ge-1\\ B_{min}=-1\Leftrightarrow\left\{{}\begin{matrix}x\left(2x+1\right)=0\\x+2=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
Vậy dấu \("="\) không xảy ra
ta có: \(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{x}.\)
\(A=1+\frac{1}{2}+\frac{1}{2.2}+\frac{1}{2.2.2}+...+\frac{1}{x}\)
\(\Rightarrow2A=2+1+\frac{1}{2}+\frac{1}{2.2}+...+\frac{1}{x:2}\)
\(\Rightarrow2A-A=2-\frac{1}{x}\)
\(A=2-\frac{1}{x}=\frac{4095}{2048}\)
=> 1/x = 1/2048
=> x = 2048 ( 2048 = 211 )
#)Giải :
\(\left(\frac{1}{2}\right)^{15}\div\left(\frac{1}{4}\right)^{20}=\left(\frac{1}{2}\right)^{15}\div\left(\frac{1}{2}\right)^{40}=\left(\frac{1}{2}\right)^{-25}\)
Ta có : \(A=\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+.....+\frac{1}{2^{99}}\)
\(\Rightarrow2^2A=2+\frac{1}{2}+\frac{1}{2^3}+.....+\frac{1}{2^{97}}\)
\(\Rightarrow4A-A=2-\frac{1}{2^{99}}\)
\(\Rightarrow3A=2-\frac{1}{2^{99}}\)
\(\Rightarrow A=\frac{2-\frac{1}{2^{99}}}{3}\)