1. Cho A = 1/2 × (b - 2/5) ÷ 7/5
a) Tính giá trị của a khi B = 2/3
b) Tính giá trị của B khi a = 3/7
2.Tính nhanh: 25×4-0,5×40×5×0,2×20×0,25
1+2+4+8+...+128+256
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Đặt \(A=25\times4-0,5\times40\times5\times0,2\times20\times0,25\)
\(\Leftrightarrow A=100-\left(0,5\times0,2\right)\times\left(40\times0,25\right)\times\left(5\times20\right)\)
\(\Leftrightarrow A=100-0,1\times10\times100\)
\(\Leftrightarrow A=100-100=0\)
Vậy theo đề bài, ta có :
\(\frac{25\times4-0,5\times40\times5\times0,2\times20\times0,25}{1+2+8+...+128+256}=\frac{0}{1+2+8+...+128+256}=0\)
Vậy ..................
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\(\frac{25\times4-0,5\times40\times5\times0,2\times20\times0,25}{1+2+4+8+...+128+256}\)
\(=\frac{100-\left(20\times5\right)\times\left(0,5\times0,2\times0,25\times40\right)}{1+2+4+8+...+128+256}\)
\(=\frac{100-100\times\left(\frac{1}{40}\times40\right)}{1+2+4+8+...+128+256}\)
\(=\frac{100\times\left(1-1\right)}{1+2+4+8+...+128+256}\)
\(=0\)
\(\frac{25\times4-0,5\times40\times5\times0,2\times20\times0,25}{1+2+4+8+...128+256}\)
\(=\frac{100-\left(0,5\times20\right)\times\left(40\times0,25\right)\times\left(5\times0,2\right)}{1+2+4+8+...+128+256}\)
\(=\frac{100-10\times10\times1}{1+2+4+8+...+128+256}\)
\(=\frac{100-100}{1+2+4+8+...+128+256}\)
\(=\frac{0}{1+2+4+8+...+128+256}=0\)
`A)đk:x>=0,x ne 25`
`A=9=>A=(3+2)/(3-5)=-5/2`
`B)B=(3sqrtx-15+20-2sqrtx)/(x-25)`
`=(sqrtx+5)/(x-25)`
`=1/(sqrtx-5)`
`A=B.|x-4|`
`<=>A/B=|x-4|`
`<=>\sqrtx+2=|x-4|`
`<=>\sqrtx+2=(sqrtx+2)|sqrtx-2|`
`<=>|sqrtx-2|=1`
`+)sqrtx-2=1<=>x=9(tm)`
`+)sqrtx-2=-1<=>x=1(tm)`
Vậy `S={1,9}`
a, Thay x=9 vào biểu thức A ta có
\(A=\dfrac{\sqrt{9}+2}{\sqrt{9}-5}\)
\(A=\dfrac{3+2}{3-5}=\dfrac{5}{-2}=-2,5\)
Vậy A =-2,5 khi x=9
thay b=2/3, ta có: A=1/3x(2/3-2/5):7/5=4/63
thay A= 3/7, ta có: 3/7=1/3x(b-2/5):7/5
1/3x(b-2/5):7/5=3/7
1/3x(b-2/5) =3/7x7/5
1/3x(b-2/5) = 3/5
b-2/5 = 3/5:1/3
b-2/5 = 9/5
b = 9/5+2/5
b = 11/5
2:
=1-1+1-1=0
3:
a: =>34*(100+1)/2:a=17
=>a=101
b: =>5/3(x-1/2)=5/4
=>x-1/2=5/4:5/3=3/4
=>x=5/4
1a, \(\dfrac{2005}{2001}\) = 1+\(\dfrac{4}{2001}\); \(\dfrac{2009}{2005}\)=1+\(\dfrac{4}{2005}\)vì\(\dfrac{4}{2001}\)>\(\dfrac{4}{2005}\)nên\(\dfrac{2005}{2001}\)>\(\dfrac{2009}{2005}\)
1b,\(\dfrac{1313}{1515}\)=\(\dfrac{1313:101}{1515:101}\)= \(\dfrac{13}{15}\); \(\dfrac{131313}{151515}\)=\(\dfrac{131313:10101}{151515:10101}\)=\(\dfrac{13}{15}\)
Vậy \(\dfrac{13}{15}\)=\(\dfrac{1313}{1515}\)=\(\dfrac{131313}{151515}\)