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27x39+27x63-2x27
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10 tháng 6 2016
a, 27 x 39 + 27 x 63 - 2 x 27
= 27 x ( 39 + 63 - 2 )
= 27 x 100
= 2700
b, 128 x 46 + 128 x 32 + 128x 22
= 128 x ( 46 + 32 + 22 )
= 128 x 100
= 12800
c, 12 x 35 + 35 x 182 - 35 x 94
= 35 x ( 12 + 182 - 94 )
= 35 x 100
= 3500
d, 456 : 2 x 18 + 456 : 3 - 102
= 228 x 18 + 152 - 102
= 4104 + 50
= 4154
HT
10 tháng 6 2016
\(\text{a, 27 x 39 + 27 x 63-2 x 27 =27 x (39+63-2) = 27 x 100 =2700}\)
\(\text{b, 128 x 46 +128 x 32 +128 x 22 =128 x (46+32+22)=128 x 100=12800}\)
\(\text{c , 12 x 35 +35 x 182 -35 x 94 =35 x (12+182-94)=35 x 100=3500.}\)
\(\text{d, 456:2 x 18 +456 :3 -102 =4140+152-102=4190}\)
2,
\(45-\left(x+9\right)=6\Rightarrow45-x-9-6\Rightarrow36-x=6\Rightarrow x=30\)
\(x+7-25=13\Rightarrow x-18=13\Rightarrow x=31\)
\(450:\left(x-19\right)=50\Rightarrow x-19=9\Rightarrow x=28\)
\(4\times\left(x+41\right)=400\Rightarrow x+41=100\Rightarrow x=59\)
\(x+73-26=76\Rightarrow x+47=76\Rightarrow x=31\)
S
18 tháng 5 2017
Ta có:
A=\(\frac{1}{1.101}+\frac{1}{2.102}+...+\frac{1}{25.125}\)
=\(\frac{1}{100}\left(\frac{100}{1.101}+\frac{100}{2.102}+...+\frac{100}{25.125}\right)\)
=\(\frac{1}{100}\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{25}-\frac{1}{125}\right)\)
=\(\frac{1}{100}\left[\left(1+\frac{1}{2}+...+\frac{1}{25}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{125}\right)\right]\)
B=\(\frac{1}{1.26}+\frac{1}{2.27}+...+\frac{1}{100.125}\)
=\(\frac{1}{25}\left(\frac{25}{1.26}+\frac{25}{2.27}+...+\frac{25}{100.125}\right)\)
=\(\frac{1}{25}\left(1-\frac{1}{26}+\frac{1}{2}-\frac{1}{27}+...+\frac{1}{100}-\frac{1}{125}\right)\)
=\(\frac{1}{25}\left[\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{125}\right)\right]\)
=\(\frac{1}{25}\left[\left(1+\frac{1}{2}+...+\frac{1}{25}\right)+\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{100}\right)-\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{100}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{125}\right)\right]\)
= \(\frac{1}{25}\left[\left(1+\frac{1}{2}+...+\frac{1}{25}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{125}\right)\right]\)
=> \(\frac{A}{B}\)=\(\frac{\frac{1}{100}\left[\left(1+\frac{1}{2}+...+\frac{1}{25}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{125}\right)\right]}{\frac{1}{25}\left[\left(1+\frac{1}{2}+...+\frac{1}{25}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{125}\right)\right]}\)=\(\frac{1}{\frac{100}{\frac{1}{25}}}\)=\(\frac{1}{100}\cdot25=\frac{25}{100}=\frac{1}{4}\)
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NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 1
\(\Delta'=\left(m-1\right)^2-\left(2m-3\right)=m^2+4>0;\forall m\)
\(\Rightarrow\) Pt luôn có 2 nghiệm pb với mọi m
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)\\x_1x_2=2m-3\end{matrix}\right.\)
- Với
\(x_1^2-2x_2=7\)
\(\Leftrightarrow x_1\left(x_1+x_2\right)-x_1x_2-2x_2=7\)
\(\Leftrightarrow2\left(m-1\right)x_1-\left(2m-3\right)-2x_2=7\)
\(\Leftrightarrow2mx_1-2\left(x_1+x_2\right)=2m+4\)
\(\Leftrightarrow mx_1-2\left(m-1\right)=m+2\)
\(\Leftrightarrow mx_1=3m\)
- Với \(m=0\) thỏa mãn
- Với \(m\ne0\Rightarrow x_1=3\)
Thế vào \(x_1+x_2=2\left(m-1\right)\Rightarrow x_2=2m-5\)
Thế tiếp vào \(x_1x_2=2m-3\) \(\Rightarrow3\left(2m-5\right)=2m-3\)
\(\Rightarrow m=3\)
Vậy \(\left[{}\begin{matrix}m=0\\m=3\end{matrix}\right.\)
27 × 39 + 27 × 63 - 2 × 27
= 27 × ( 39 + 63 - 2 )
= 27 × 100
= 2700
27 x ( 39 + 63 - 2 )
27 x 100
2700