CMR a mũ 2 -a chia hết cho 2 (a thuộc N)
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Ta có:
\(1-\frac{1}{1+2+...+k}=1-\frac{1}{\frac{k\left(k+1\right)}{2}}=\frac{k\left(k+1\right)-2}{k\left(k+1\right)}=\frac{\left(k-1\right)\left(k+2\right)}{k\left(k+1\right)}\)
Áp dụng biểu thức trên ta được:
\(P=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+3+...+2018}\right)\)
\(P=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.....\frac{2017.2020}{2018.2019}\)
\(P=\frac{1}{2018}.\frac{2020}{3}=\frac{1010}{3027}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x-y-z+3=0\Leftrightarrow x=y+z-3\)
\(x^2-y^2-z^2=\left(y+z-3\right)^2-y^2-z^2=y^2+z^2+9+2yz-6y-6z-y^2-z^2\)
\(=2yz-6y-6z+9=1\)
\(\Leftrightarrow yz-3y-3z+4=0\)
\(\Leftrightarrow\left(y-3\right)\left(z-3\right)=5=1.5=\left(-1\right).\left(-5\right)\)
Xét bảng:
y-3 | 1 | 5 | -1 | -5 |
z-3 | 5 | 1 | -5 | -1 |
y | 4 | 8 | 2 | -2 |
z | 8 | 4 | -2 | 2 |
x | 9 | 9 | -3 | -3 |
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có
\(MN\perp BC;AB\perp BC\) => MN//AB \(\Rightarrow\frac{MN}{AB}=\frac{CM}{CA}\) (Talet trong tam giác)
\(MP\perp AD;CD\perp AD\) => MP//CD \(\Rightarrow\frac{MP}{CD}=\frac{AM}{CA}\) (Talet trong tam giác)
\(\Rightarrow\frac{MN}{AB}+\frac{MP}{CD}=\frac{CM}{CA}+\frac{AM}{CA}=\frac{CA}{CA}=1\left(dpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)(vì \(a+b+c\ne0\))
\(\Leftrightarrow a^2+b^2+c^2=ab+bc+ca\)
\(N=\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{a^2+b^2+c^2}{a^2+b^2+c^2+2ab+2bc+2ca}=\frac{a^2+b^2+c^2}{a^2+b^2+c^2+2\left(a^2+b^2+c^2\right)}=\frac{1}{3}\)
Ta có : \(a^2-a=a\left(a-1\right)\)
Xét :
\(\Rightarrow a\left(a-1\right)=2k\left(2k-1\right)⋮2\)
\(\Rightarrow a\left(a-1\right)=\left(2k+1\right)\left(2k+1-1\right)=\left(2k+1\right)2k⋮2\)
Suy ra : \(a\left(a-1\right)⋮2\forall a\inℕ\)
hay \(a^2-a⋮2\forall a\inℕ\)