tính giá trị biểu thức 1+3+5+...+2009/1+2+3+..+2010.giúp minh nha ^-^
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136.48+16.2.136+68.2.20
=136.48+32.136+136.20
=136.( 48+32+20)
=136.100
=13600
\(136\cdot48+16\cdot272+68\cdot20\cdot2\)
\(=136\cdot48+16\cdot2\cdot136+136\cdot20\)
\(=136\cdot48+136\cdot32+136\cdot20\)
\(=136\cdot\left(48+32+20\right)\)
\(=136\cdot100\)
\(=13600\)
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\(a,\sqrt{x}>1\Leftrightarrow\sqrt{x}>\sqrt{1}\Leftrightarrow x>1.\)
\(b,\sqrt{x}< 3\Rightarrow\sqrt{x}< \sqrt{9}\Rightarrow x< 9\)
\(c,\sqrt{x}=14\Rightarrow\sqrt{x}=\sqrt{196}\Leftrightarrow x=196\)
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a) x - 26 : 13 = 2017
x - 2 = 2017
x = 2017 + 2
x = 2019
b) \(\frac{2}{3}+\frac{1}{3}.\left(x+28\right)=18\)
\(\frac{1}{3}.\left(x+28\right)=18-\frac{2}{3}\)
\(\frac{1}{3}.\left(x+28\right)=\frac{54}{3}-\frac{2}{3}\)
\(\frac{1}{3}.\left(x+28\right)=\frac{52}{3}\)
\(x+28=\frac{52}{3}:\frac{1}{3}\)
\(x+28=52\)
\(x=52-28\)
\(x=24\)
c) \(\frac{7}{8}.\left(x-27\right)=\frac{9}{8}-0,125\)
\(\frac{7}{8}.\left(x-27\right)=1\)
\(x-27=1:\frac{7}{8}\)
\(x-27=\frac{8}{7}\)
\(x=\frac{8}{7}+27\)
\(x=\frac{8}{7}+\frac{189}{7}\)
\(x=\frac{197}{7}\)
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a) \(x^3-30x^2-31x+1\)
\(=x^3-31x^2+x^2-31x+1\)
\(=x^2\left(x-31\right)+x\left(x-31\right)+1\)
\(=\left(x-31\right)\left(x^2+x\right)+1\)(1)
Thay x=31 hay x-31=0 vào (1) ta được :
\(0.\left(x^2+x\right)+1\)
\(=1\)
b) Vì \(x=14\)
\(\Rightarrow15=x+1\)
\(16=x+2\)
\(29=2x+1\)
\(13=x-1\)( nhớ ngoặc kí hiệu "và " 4 dòng này lại )
Thay vào biểu thức ta được :
\(x^5-\left(x+1\right)x^4+\left(x+2\right)x^3-\left(2x+1\right)x^2+\left(x-1\right)x\)
\(=x^5-x^5-x^4+x^3+2x^3-2x^3-x^2+x^2-x\)
\(=-x\)thay x=14 vào bt ta được :
\(=-14\)
\(\frac{1+3+5+...+2009}{1+2+3+...+2010}\)
\(=\frac{\frac{\left[\left(2009-1\right):2+1\right]\left(2009+1\right)}{2}}{\frac{\left(2010+1\right)2010}{2}}\)
\(=\frac{\frac{1005.2010}{2}}{\frac{2011.2010}{2}}\)
\(=\frac{1005.2010}{2}.\frac{2}{2011.2010}\)
\(=\frac{1005}{2011}\)