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2)
a) Ta có: \(4n-5⋮2n-1\)
\(\Rightarrow\left(4n-2\right)-3⋮2n-1\)
\(\Rightarrow2\left(2n-1\right)-3⋮2n-1\)
\(\Rightarrow-3⋮2n-1\)
\(\Rightarrow2n-1\in\left\{1;3\right\}\) ( Vì \(n\in N\) )
\(\Rightarrow\left\{{}\begin{matrix}2n-1=1\Rightarrow n=1\\2n-1=3\Rightarrow n=2\end{matrix}\right.\)
Vậy n=1 hoặc n=2
b) Ta có: \(3n+2⋮n-1\)
\(\Rightarrow\left(3n-3\right)+5⋮n-1\)
\(\Rightarrow3\left(n-1\right)+5⋮n-1\)
\(\Rightarrow5⋮n-1\)
\(\Rightarrow n-1\in\left\{1;5\right\}\) ( Vì \(n\in N\) )
\(\Rightarrow\left\{{}\begin{matrix}n-1=1\Rightarrow n=2\\n-1=5\Rightarrow n=6\end{matrix}\right.\)
Vậy n=2 hoặc n=6
1. vì (2x-1)(y-1)=29 mà \(x,y\in N\)\(\Rightarrow\left\{{}\begin{matrix}2x-1>0\\y-1>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>\dfrac{1}{2}\\y>1\end{matrix}\right.\)
ta có:\(\left(2x-1\right)\left(y-1\right)=29\Rightarrow2x-1=\dfrac{29}{y-1}\)
vì: \(x\in N\Rightarrow\dfrac{29}{y-1}\in N\)
\(\Rightarrow29⋮y-1\Rightarrow y\in\left\{2;30\right\}\)
với y=2 => x=15
với y=30 => x=1
5.
(x^2 -1)(x^2 +9) <0
(x+3)(x+1)(x-1)(x-3)<0
x \(\in\)(-3;-1)U(1;3)
\(B=1!+2.2!+3.3!+...+k.k!\)
\(=1!+\left(3-1\right)2!+\left(4-1\right)3!+...+\left(k+1-1\right)k!\)
\(=1!+3!-2!+4!-3!+...+\left(k+1\right)!-k!\)
\(=\left(k+1\right)!-1\)
\(C=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{n-1}{n!}\)
\(=\frac{2}{2!}-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+\frac{4}{4!}-\frac{1}{4!}+...+\frac{n}{n!}-\frac{1}{n!}\)
\(=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{\left(n-1\right)!}-\frac{1}{n!}\)
\(=1-\frac{1}{n!}\)
2.
Với \(n=0\Rightarrow1\ge\frac{1}{2}\) đúng
Với \(n=1\Rightarrow1\ge1\) đúng
Giả sử BĐT đúng với \(n=k\ge2\) hay \(k!\ge2^{k-1}\)
Ta cần chứng minh nó cũng đúng với \(n=k+1\) hay \(\left(k+1\right)!\ge2^k\)
Thật vậy, ta có:
\(\left(k+1\right)!=k!\left(k+1\right)\ge2^{k-1}.\left(k+1\right)>2^{k-1}.2=2^k\) (đpcm)
Câu 1:
Áp dụng BĐT Cauchy:
\(1+x^3+y^3\geq 3\sqrt[3]{x^3y^3}=3xy\)
\(\Rightarrow \frac{\sqrt{1+x^3+y^3}}{xy}\geq \frac{\sqrt{3xy}}{xy}=\sqrt{\frac{3}{xy}}\)
Hoàn toàn tương tự:
\(\frac{\sqrt{1+y^3+z^3}}{yz}\geq \sqrt{\frac{3}{yz}}; \frac{\sqrt{1+z^3+x^3}}{xz}\geq \sqrt{\frac{3}{xz}}\)
Cộng theo vế các BĐT thu được:
\(\text{VT}\geq \sqrt{\frac{3}{xy}}+\sqrt{\frac{3}{yz}}+\sqrt{\frac{3}{xz}}\geq 3\sqrt[6]{\frac{27}{x^2y^2z^2}}=3\sqrt[6]{27}=3\sqrt{3}\) (Cauchy)
Ta có đpcm
Dấu bằng xảy ra khi $x=y=z=1$
Câu 4:
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{2}{x}+\frac{3}{y}\right)(x+y)\geq (\sqrt{2}+\sqrt{3})^2\)
\(\Leftrightarrow 1.(x+y)\geq (\sqrt{2}+\sqrt{3})^2\Rightarrow x+y\geq 5+2\sqrt{6}\)
Vậy \(A_{\min}=5+2\sqrt{6}\)
Dấu bằng xảy ra khi \(x=2+\sqrt{6}; y=3+\sqrt{6}\)
------------------------------
Áp dụng BĐT Cauchy:
\(\frac{ab}{a^2+b^2}+\frac{a^2+b^2}{4ab}\geq 2\sqrt{\frac{ab}{a^2+b^2}.\frac{a^2+b^2}{4ab}}=1\)
\(a^2+b^2\geq 2ab\Rightarrow \frac{3(a^2+b^2)}{4ab}\geq \frac{6ab}{4ab}=\frac{3}{2}\)
Cộng theo vế hai BĐT trên:
\(\Rightarrow B\geq 1+\frac{3}{2}=\frac{5}{2}\) hay \(B_{\min}=\frac{5}{2}\). Dấu bằng xảy ra khi $a=b$
1)
a) \(5n-8⋮4-n\)
\(\Rightarrow-20+5n+12⋮4-n\)
\(\Rightarrow-5\left(4-n\right)+12⋮4-n\)
\(\Rightarrow12⋮4-n\)
\(\Rightarrow4-n\in\left\{-1;1;-2;2;-3;3;-4;4;-6;6;-12;12\right\}\)
+) \(4-n=-1\Rightarrow n=5\)
+) \(4-n=1\Rightarrow n=3\)
+) \(4-n=-2\Rightarrow n=6\)
+) \(4-n=2\Rightarrow n=2\)
+) \(4-n=-3\Rightarrow n=7\)
+) \(4-n=3\Rightarrow n=1\)
+) \(4-n=-4\Rightarrow n=8\)
+) \(4-n=4\Rightarrow n=0\)
+) \(4-n=-6\Rightarrow n=10\)
+) \(4-n=6\Rightarrow n=-2\)
+) \(4-n=-12\Rightarrow n=16\)
+) \(4-n=12\Rightarrow n=-8\)
Vậy \(n\in\left\{5;3;6;2;7;1;8;0;10;-2;16;-8\right\}\)
b) Ta có:\(n^2+3n+6⋮n+3\)
\(\Rightarrow n\left(n+3\right)+6⋮n+3\)
\(\Rightarrow6⋮n+3\)
\(\Rightarrow n+3\in\left\{-1;1;-2;2;-3;3;-6;6\right\}\)
+) \(n+3=-1\Rightarrow n=-4\)
+) \(n+3=1\Rightarrow n=-2\)
+) \(n+3=-2\Rightarrow n=-5\)
+) \(n+3=2\Rightarrow n=-1\)
+) \(n+3=-3\Rightarrow n=-6\)
+) \(n+3=3\Rightarrow n=0\)
+) \(n+3=-6\Rightarrow n=-9\)
+) \(n+3=6\Rightarrow n=3\)
Vậy \(n\in\left\{-4;-2;-5;-1;-6;0;-9;3\right\}\)