Giải gúp mk tí nhé
THU GỌN:(x^2+1)(x^4-x^2+1)
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DP
1
9 tháng 3 2017
1/2=2/4 , 1/14=2/28 , 1/35=2/70
ta có: 2/4=2/1.4 , 2/28=2/4.7 , 2/70=2/7.10 ................
thay vào ta có: 2/1.4+2/4.7+2/7.10+........+2/x(x+3)=1340/2011
2.(1/1.4+1/4.7+1/7.10+....+1/x(x+3)=1340/2011
2.3.(1/1.4+1/4.7+1/7.10+....+1/x(x+3)=(1340/2011).3
2.(3/1.4+3/4.7+3/7.10+......+3/x(x+3)=4020/2011
2.(1- 1/4 + 1/4- 1/7 + 1/7- 1/10 +.......+1/x - 1/x+3)=4020/2011
1-1/x+3 =4020/2011 :2= 2010/2011
1- 2010/2011=1/x+3
1/2011=1/x+3
x+3= 2011
x=2008
24 tháng 11 2019
ta có:
1-1/2+1/2-1/3+1/3-1/4+....+1/x -1/x+1 =499/500
1-1/x+1 =499/500
1/x+1 =1/500
x+1=500
x=499
24 tháng 11 2019
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{X\times\left(X+1\right)}=\frac{499}{500}\)
\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{X}-\frac{1}{X+1}=\frac{499}{500}\)
\(\Leftrightarrow1-\frac{1}{X+1}=\frac{499}{500}\)
\(\Leftrightarrow\frac{1}{X+1}=\frac{1}{500}\)
\(\Leftrightarrow X+1=500\)
\(\Leftrightarrow X=499\)
MT
1
9 tháng 11 2021
\(1,\Delta=\left(-11\right)^2-4\cdot30=1\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11-1}{2}=5\\x=\dfrac{11+1}{2}=6\end{matrix}\right.\\ 2,\Delta=\left(-1\right)^2-4\left(-20\right)=81\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{81}}{2}=-4\\x=\dfrac{1+\sqrt{81}}{2}=5\end{matrix}\right.\\ 3,\Delta=14^2-4\cdot24=100\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-14-\sqrt{100}}{2}=-12\\x=\dfrac{-14+\sqrt{100}}{2}=-2\end{matrix}\right.\\ 4,\Delta=8^2-4\left(-2\right)3=88\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-8-\sqrt{88}}{6}=\dfrac{-4+\sqrt{22}}{3}\\x=\dfrac{-8+\sqrt{88}}{6}=\dfrac{-4-\sqrt{22}}{3}\end{matrix}\right.\)
18 tháng 2 2020
Ta có :
\(B=\left(\frac{1}{x-4}-\frac{1}{x+4\sqrt{x}+4}\right).\frac{x+2\sqrt{x}}{\sqrt{x}}\)
\(=\left(\frac{1}{\left(\sqrt{x}+2\right)\left(\sqrt{x-2}\right)}-\frac{1}{\left(\sqrt{x}+2\right)^2}\right).\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}}\)
\(=\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}-\frac{\sqrt{x}-2}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}\right).\left(\sqrt{x}+2\right)\)
\(=\frac{\sqrt{x}+2-\sqrt{x}+2}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}.\left(\sqrt{x}+2\right)\)
\(=\frac{4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(\left(x^2+1\right)\left(x^4-x^2+1\right)\)
\(=x^4x^2+1.x^4-\left(x^2\right)^2+1.x^2-1.x^2+1.1\)
\(=x^6+x^4-x^4+1\)
\(=x^6+1\)
bạn áp dụng HĐT: \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(\left(x^2+1\right)\left(x^4-x^2+1\right)\)
\(=\left(x^2\right)^3+1^3\)
\(=x^6+1\)