CMR:
3^(3n+2) + 5.[2^(3n+1)] chia hết cho 19
Mọi người giúp mình nhe. thanh kiu hihi
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+)\(\left(x+y+z\right)^2=0^2=0=x^2+y^2+z^2+2xy+2yz+2zx\)
\(\Leftrightarrow x^2+y^2+z^2=-2\left(xy+yz+xz\right)\)
+)\(\frac{9\left(x^2+y^2+z^2\right)}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=\frac{-18\left(xy+yz+xz\right)}{2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)}\)
\(=\frac{-18\left(xy+yz+zx\right)}{-6\left(xy+yz+zx\right)}=3\)
Bài 1:
Vì a chia cho 3 dư 1 \(\Rightarrow a\equiv1\left(mod3\right)\)
b chia cho 3 dư 2 \(\Rightarrow b\equiv2\left(mod3\right)\)
\(\Rightarrow ab\equiv2\left(mod3\right)\)
Vậy ab chia cho 3 dư 2
Cách 2: ( hướng dẫn)
a chia 3 dư 1 nên a=3k+1(k thuộc N ) b chia 3 dư 2 nên b=3k+2 ( k thuộc N )
Từ đó nhân ra ab=(3k+1)(3k+2) rồi chứng minh
Bài 2:
Ta có: \(n\left(2n-3\right)-2n\left(n+1\right)\)
\(=2n^2-3n-2n^2-2n\)
\(=-5n\)
Vì \(n\)nguyên \(\Rightarrow-5n⋮5\)
\(\Rightarrow n\left(2n-3\right)-2n\left(n+1\right)⋮5\forall n\in Z\left(đpcm\right)\)
\(A=3\left(x^2-\frac{2}{3}x-\frac{1}{3}\right)\)
\(A=3\left(x^2-2\cdot\frac{1}{3}x+\frac{1}{9}-\frac{4}{9}\right)\)
\(A=\left(x-\frac{1}{3}\right)^2-\frac{4}{3}\)\(\supseteq-\frac{4}{3}\)
Dấu = xr khi x=1/3
Vậy Min A=-4/3 tại x=1/3
\(A=3x^2-2x-1\)
\(=3\left(x^2-\frac{2}{3}x-\frac{1}{3}\right)\)
\(=3\left(x^2-2.x.\frac{1}{3}+\frac{1}{9}-\frac{1}{9}-\frac{1}{3}\right)\)
\(=3\left(x-\frac{1}{3}\right)^2-\frac{4}{3}\)
Vì \(3\left(x-\frac{1}{3}\right)^2\ge0;\forall x\)
\(\Rightarrow3\left(x-\frac{1}{3}\right)^2-\frac{4}{3}\ge0-\frac{4}{3};\forall x\)
Hay \(A\ge\frac{-4}{3};\forall x\)
Dấu"=" xảy ra \(\Leftrightarrow\left(x-\frac{1}{3}\right)^2=0\)
\(\Leftrightarrow x=\frac{1}{3}\)
Vậy MIN \(A=\frac{-4}{3}\)\(\Leftrightarrow x=\frac{1}{3}\)
\(K=\sqrt{\frac{1}{a^2+b^2}+\frac{1}{\left(a+b\right)^2}+\sqrt{\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{\left(a^2+b^2\right)^2}}}\)
\(=\sqrt{\frac{1}{a^2+b^2}+\frac{1}{\left(a+b\right)^2}+\sqrt{\left(\frac{1}{a^2}+\frac{1}{b^2}\right)^2-\frac{2}{a^2+b^2}\left(\frac{1}{a^2}+\frac{1}{b^2}\right)+\frac{1}{\left(a^2+b^2\right)^2}}}\)
\(=\sqrt{\frac{1}{a^2+b^2}+\frac{1}{\left(a+b\right)^2}+\sqrt{\left(\frac{1}{a^2}+\frac{1}{b^2}-\frac{1}{a^2+b^2}\right)^2}}\)
\(=\sqrt{\frac{1}{\left(a+b\right)^2}+\frac{1}{a^2}+\frac{1}{b^2}}\)
\(=\sqrt{\frac{1}{\left(a+b\right)^2}+\left(\frac{1}{a}+\frac{1}{b}\right)^2-\frac{2}{\left(a+b\right)}\left(\frac{1}{a}+\frac{1}{b}\right)}\)
\(=\sqrt{\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\right)^2}=\left|\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\right|\)
Chúc bạn học tốt !!!
\(A=5x^2-25x+35+7y^8\)
\(=5\left(x^2-5x+7\right)+7y^8\)
\(=5\left(x^2-5x+\frac{25}{4}+\frac{3}{4}\right)+7y^8\)
\(=5\left[\left(x-\frac{5}{2}\right)^2+\frac{3}{4}\right]+7y^8\)
\(=5\left(x-\frac{5}{2}\right)^2+\frac{15}{4}+7y^8\ge\frac{15}{4}\)
\(\Leftrightarrow x=\frac{5}{2};y=0\)