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A.\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}\) \(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)\left(n+1-n\right)}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}\)
=\(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
b. ap dungtinh B =\(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}=1-\frac{1}{10}=\frac{9}{10}\)
1.
\(\Leftrightarrow x-3=\dfrac{1}{8}\)
\(\Leftrightarrow x=\dfrac{25}{8}\)
\(a.\left(\sqrt{3}-1\right)^2=4-2\sqrt{3}\) ( sửa đề )
\(VP=4-2\sqrt{3}=3-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2=VT\)
⇒ ĐPCM.
\(b.\left(\sqrt{3}+1\right)^2=4+2\sqrt{3}\) ( sửa đề )
\(VP=4+2\sqrt{3}=3+2\sqrt{3}+1=\left(\sqrt{3}+1\right)^2=VT\)
⇒ ĐPCM.
1/ \(a+1=\sqrt[4]{\frac{\left(\sqrt{3}+1\right)^2}{\left(\sqrt{3}-1\right)^2}}-\sqrt[4]{\frac{\left(\sqrt{3}-1\right)^2}{\left(\sqrt{3}+1\right)^2}}=\sqrt{\frac{\sqrt{3}+1}{\sqrt{3}-1}}-\sqrt{\frac{\sqrt{3}-1}{\sqrt{3}+1}}\)
\(=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}}=\frac{\sqrt{3}+1-\sqrt{3}+1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)
2/ \(a+b=5\Leftrightarrow\left(a+b\right)^3=125\)
\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=125\)
\(\Rightarrow a^3+b^3=125-3ab\left(a+b\right)=125-3.1.5=110\)
3/ \(mn\left(mn+1\right)^2-\left(m+n\right)^2.mn\)
\(=mn\left(\left(mn+1\right)^2-\left(m+n\right)^2\right)\)
\(=mn\left(mn+1-m-n\right)\left(mn+1+m+n\right)\)
\(=mn\left(m-1\right)\left(n-1\right)\left(m+1\right)\left(n+1\right)\)
\(=\left(m-1\right)m\left(m+1\right)\left(n-1\right)n\left(n+1\right)\)
Do \(\left(m-1\right)m\left(m+1\right)\) và \(\left(n-1\right)n\left(n+1\right)\) đều là tích của 3 số nguyên liên tiếp nên chúng đều chia hết cho 3 \(\Rightarrow\) tích của chúng chia hết cho 36
4/
Do \(0\le x\le1\Rightarrow\left\{{}\begin{matrix}x\ge0\\x-1\le0\end{matrix}\right.\) \(\Rightarrow x\left(x-1\right)\le0\)
\(\Leftrightarrow x^2-x\le0\Leftrightarrow x^2\le x\)
Dấu "=" xảy ra khi \(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
5/ Đặt \(\left\{{}\begin{matrix}\sqrt{5a+4}=x\\\sqrt{5b+4}=y\\\sqrt{5c+4}=z\end{matrix}\right.\)
Do \(a+b+c=1\Rightarrow0\le a;b;c\le1\)
\(\Rightarrow2\le x;y;z\le3\) và \(x^2+y^2+z^2=5\left(a+b+c\right)+12=17\)
Khi đó ta có:
Do \(2\le x\le3\Rightarrow\left(x-2\right)\left(x-3\right)\le0\)
\(\Leftrightarrow x^2-5x+6\le0\Leftrightarrow x\ge\frac{x^2+6}{5}\)
Tương tự: \(y\ge\frac{y^2+6}{5}\) ; \(z\ge\frac{z^2+6}{5}\)
Cộng vế với vế:
\(A=x+y+z\ge\frac{x^2+y^2+z^2+18}{5}=\frac{17+18}{5}=7\)
\(\Rightarrow A_{min}=7\) khi \(\left(x;y;z\right)=\left(2;2;3\right)\) và các hoán vị hay \(\left(a;b;c\right)=\left(0;0;1\right)\) và các hoán vị
\(A=\sqrt{4+\sqrt{4+\sqrt{4}+...}}\\ \)>0
a)
\(A=\sqrt{4+A}\Leftrightarrow A^2=4+A\Leftrightarrow A^2-A-4=0\)
\(\Delta=1+16=17\)
\(A_1=\dfrac{1+\sqrt{17}}{2}< \dfrac{1+5}{2}=3\)
\(A_2=\dfrac{1-\sqrt{17}}{2}\)<0 loại
Vậy A < 3
b) Chứng minh quy nạp
(13+23+.....+n3)=(1+2+3+...+n)2=> KL
b).đặt \(A=\sqrt{1^3+2^3+3^3+...+n^3}\)
ta có hằng đẳng thức: \(x^3-x=\left(x-1\right)x\left(x+1\right)\)
\(1^3+2^3+3^3+...+n^3=1^3-1+2^3-2+3^3-3+...+n^3-n+\left(1+2+3+...+n\right)\)\(=0+1.2.3+2.3.4+...+\left(n-1\right)n\left(n+1\right)+\dfrac{n\left(n+1\right)}{2}\)(*)
Xét \(B=1.2.3+2.3.4+...+\left(n-1\right)n\left(n+1\right)\)
\(4B=1.2.3.4+2.3.4.4+...+\left(n-1\right)n\left(n+1\right).4=1.2.3.4+2.3.4.5-1.2.3.4+...+\left(n-1\right)n\left(n+1\right)\left(n+2\right)-\left(n-2\right)\left(n-1\right)n\left(n+1\right)\)
\(=\left(n-1\right)n\left(n+1\right)\left(n+2\right)\)
\(\Rightarrow B=\dfrac{\left(n-1\right)n\left(n+1\right)\left(n+2\right)}{4}\)
từ (*): \(1^3+2^3+...+n^3=\dfrac{\left(n-1\right)n\left(n+1\right)\left(n+2\right)}{4}+\dfrac{n\left(n+1\right)}{2}\)
\(=\dfrac{n\left(n+1\right)}{2}\left[\dfrac{\left(n-1\right)\left(n+2\right)}{2}+1\right]=\dfrac{n\left(n+1\right)}{2}.\dfrac{n^2+n-2+2}{2}=\left[\dfrac{n\left(n+1\right)}{2}\right]^2\)
do đó \(A=\sqrt{\left[\dfrac{n\left(n+1\right)}{2}\right]^2}=\dfrac{n\left(n+1\right)}{2}=1+2+...+n\)(đpcm)