tìm tích
3/22x8/32 x 15/42 .... x 99/102
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\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
a, \(x\) + 99: 3 = 55
\(x\) + 33 = 55
\(x\) = 55 - 33
\(x\) = 22
b, (\(x\) - 25):15 = 20
\(x\) - 25 = 20 x 15
\(x\) - 25 = 300
\(x\) = 300 + 25
\(x\) = 325
c, (3\(x\) - 15).7 = 42
3\(x\) - 15 = 42:7
3\(x\) - 15 = 6
3\(x\) = 6 + 15
3\(x\) = 21
\(x\) = 21: 3
\(x\) = 7
a) 43+(9-21)=317-(x+317)
->43+(-12)=317-(x+317)
-> 31 =317-(x+317)
-> 317-(x+317)=31
-> x+317 =317-31=286
-> x = 286 - 317
-> x = -31
Ung ho minh nha
a)có người làm rồi
b)(15-x)+(x-12)=7-(-5+x)
=>15-x+x-12=7-(-5)-x
=>(-x+x)+15-12=12-x
=>3=12-x
=>-9=-x
=>x=9
c)đề sai x0 là sao
d)|x|+|y|=1
\(\Rightarrow\hept{\begin{cases}x=\pm1\\y=0\end{cases}}\).Vì x đạt GTTĐ
\(\Rightarrow\hept{\begin{cases}x=0\\y=\pm1\end{cases}}\).Vì y đạt GTTĐ
e)(x+1) +(x+3)+(x+5)+...+(x+99)=0
=>(x+x+...+x)+(1+3+...+99)=0
=>50x+2500=0
=>50x=-2500
=>x=-50
Bài 3:
a: Ta có: \(23\left(42-x\right)=23\)
\(\Leftrightarrow42-x=1\)
hay x=41
b: Ta có: 15(x-3)=30
nên x-3=2
hay x=5
Bài 1:
a: 32+89+68=100+89=189
b: 64+112+236=300+112=412
c: \(1350+360+650+40=2000+400=2400\)
a) \(B=\left(x^2+2x+1\right)+\left(y^2-2.2.y+2^2\right)=\left(x+1\right)^2+\left(y-2\right)^2\)
thay x=99 và y=102 vào B ta có:
\(B=\left(99+1\right)^2+\left(102-2\right)^2=100^2-100^2=0\)
b)
b) \(2x^2+16x+32-2y^2=2\left(x^2+8x+16-y^2\right)=2\left(\left(x+4\right)^2-y^2\right)=2\left(x+4-y\right)\left(x+4+y\right)\)
a. 64 x 25 + 35 x 25 + 25
= 64 x 25 + 35 x 25 + 25 x 1
= 25 x ( 64 + 35 + 1 )
= 25 x 100
= 2500
b. 58 x 42 + 32 x 8 + 5 x 16
c. ( 42 x 43 + 57 x 43 + 43 ) - 360 : 4
= 42 x 43 + 57 x 43 + 43 x 1 - 90
= 43 x ( 42 + 57 +1 ) - 90
= 43 x 100 - 90
= 4300 - 90
= 4210
d. 456 : 2 x 181456 : 3 - 102
Trả lời:
a,\(64\times25+35\times25+25\)
\(=25\times\left(64+35+1\right)\)
\(=25\times100\)
\(=2500\)
b,\(58\times42+32\times8+5\times16\)
\(=2436+256+80\)
\(=2772\)
c,\(\left(42\times43+57\times43+43\right)-360\div4\)
\(=43\times\left(42+57+1\right)-90\)
\(=43\times100-90\)
\(=4300-90\)
\(=4210\)
d,\(456\div2\times181456\div3-102\)
\(=228\div3\times181456-102\)
\(=76\times181456-102\)
\(=13790656-102\)
\(=13790554\)
Đặt A= \(\frac{3}{2^2}\) . \(\frac{8}{3^2}\) . \(\frac{15}{4^2}\). ... . \(\frac{99}{10^{10}}\)
= \(\frac{1.3}{2.2}\) . \(\frac{2.4}{3.3}\) . \(\frac{3.5}{4.4}\) . ... . \(\frac{9.11}{10.10}\)
= \(\frac{1.2.3.4.....9}{2.3.4.5.6.....9.10}\) . \(\frac{3.4.....9.10.11}{2.3.4.5.6.....9.10}\)
= \(\frac{1}{10}\) . \(\frac{11}{2}\) = \(\frac{11}{20}\)