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T
3
18 tháng 7 2019
\(=\frac{5}{3}.\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{301.304}\right)\)
\(=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{304}\right)\)
\(=\frac{5}{3}.\frac{75}{304}\)
\(=\frac{125}{304}\)
18 tháng 7 2019
\(\frac{5}{4×7}+\frac{5}{7×10}+\frac{5}{10×13}+...+\frac{5}{301×304}\)
\(=\frac{5}{4}-\frac{5}{7}+\frac{5}{7}-\frac{5}{10}+\frac{5}{10}-\frac{5}{13}+...+\frac{5}{301}-\frac{5}{304}\)
\(=\frac{5}{4}-\frac{5}{304}\)
\(=\frac{380}{304}-\frac{5}{304}\)
\(=\frac{375}{304}\)
Cbht
HA
Tìm x biết:
a, ( 2x - 5)2 = ( x - 2)2
b, ( x + 1)( 2 - x) - ( 3x + 5)( x + 2) = - 4x2 + 1
Help me!!!
4
22 tháng 10 2017
\(a,\left(2x-5\right)^2=\left(x-2\right)^2\)
\(\Rightarrow\left(2x-5\right)^2-\left(x-2\right)^2=0\)
\(\Rightarrow\left(2x-5-x+2\right)\left(2x-5+x-2\right)=0\)
\(\Rightarrow\left(x-3\right)\left(3x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\3x-7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{7}{3}\end{matrix}\right.\)
\(b,\left(x+1\right)\left(2-x\right)-\left(3x+5\right)\left(x+2\right)=-4x^2+1\)
\(\Rightarrow2x-x^2+2-x-3x^2-6x-5x-10=-4x^2+1\)
\(\Rightarrow-10x-4x^2-12=-4x^2+1\)
\(\Rightarrow-10x-4x^2-12+4x^2-1=0\)
\(\Rightarrow-10x-13=0\)
\(\Rightarrow x=-\dfrac{13}{10}\)
15 tháng 2 2020
\(\frac{3x+2}{x-1}+\frac{2x-4}{x+2}=5\)
\(\Rightarrow\frac{\left(3x+2\right)\left(x+2\right)+\left(2x-4\right)\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}=5\)
\(\Rightarrow\frac{3x^2+2x+6x+4+2x^2-4x-2x+4}{\left(x-1\right)\left(x+2\right)}=5\)
\(\Rightarrow\frac{5x^2+2x+8}{\left(x-1\right)\left(x+2\right)}=5\)
\(\Rightarrow5x^2+2x+8=5\left(x-1\right)\left(x+2\right)\)
\(\Rightarrow5x^2+2x+8=5x^2-5x+10x-10\)
\(\Rightarrow5x^2-5x^2+2x-5x=8-10\)
\(\Rightarrow-3x=-2\)
\(\Rightarrow x=\frac{2}{3}\)
7 tháng 9 2017
\(\left(x+2\right)\left(x+1\right)-\left(x-3\right)\left(x+5\right)\)
\(=x^2+x+2x+2-x^2-5x+3x+15\)
\(=x+15\)
Ta có:
A = \(\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+...+\frac{5}{301.304}\)
A = 5. (\(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{301.304}\))
3A = 3.5. (\(\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{301.304}\))
3A = 5. (\(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{301.304}\))
3A = 5. ( \(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-...-\frac{1}{301}+\frac{1}{301}-\frac{1}{304}\))
3A = 5. ( \(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-...-\frac{1}{301}+\frac{1}{301}-\frac{1}{304}\))
3A = 5. ( \(\frac{1}{4}-\frac{1}{304}\))
3A = \(\frac{5.75}{304}\)
3A = \(\frac{375}{304}\)
A= \(\frac{125}{304}\) . Vậy: A = \(\frac{125}{304}\)