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\(A=\left(3+1\right)\left(3^2+1\right)...\left(3^{64}+1\right)\)
\(\Leftrightarrow\left(3-1\right)A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)...\left(3^{64}+1\right)\)
\(\Leftrightarrow2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(\Leftrightarrow2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)
\(\Leftrightarrow2A=3^{128}-1\)
\(\Leftrightarrow A=\frac{3^{128}-1}{2}\)
a) \(\left(5xy^3\right)^2-2.5xy^3.6yz^2+\left(6yz^2\right)^2\)=\(\left(5xy^3-6yz^2\right)^2\)
b) \(\left(\frac{1}{3}u^2v^3\right)^2-2.\frac{1}{3}u^2v^3.\frac{1}{2}u^3v+\left(\frac{1}{2}u^3v\right)^2\)=\(\left(\frac{1}{3}u^2v^3-\frac{1}{2}u^3v\right)^2\)
\(a.\) Với \(a+b+c=0\) thì \(\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{\left(-c\right).\left(-a\right).\left(-b\right)}{abc}=\frac{-abc}{abc}=-1\)
\(b.\) Công thức tổng quát: \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
Ta có:
\(\frac{1}{x\left(x+1\right)}=\frac{1}{x}-\frac{1}{x+1}\)
\(\frac{1}{\left(x+1\right)\left(x+2\right)}=\frac{1}{x+1}-\frac{1}{x+2}\)
\(\frac{1}{\left(x+2\right)\left(x+3\right)}=\frac{1}{x+2}-\frac{1}{x+3}\)
\(\frac{1}{\left(x+3\right)\left(x+4\right)}=\frac{1}{x+3}-\frac{1}{x-4}\)
\(\frac{1}{\left(x+4\right)\left(x+5\right)}=\frac{1}{x+4}-\frac{1}{x+5}\)
Do đó, suy ra được: \(A=\frac{1}{x}-\frac{1}{x+5}=\frac{x+5-x}{x\left(x+5\right)}=\frac{5}{x\left(x+5\right)}\)
2/ Áp dụng BĐT Bunhiacopxki \(\left(ax+by\right)^2\le\left(a^2+b^2\right)\left(x^2+y^2\right)\)
\(\Leftrightarrow a^2x^2+b^2y^2+2abxy\le a^2x^2+a^2y^2+b^2x^2+b^2y^2\)
\(\Leftrightarrow bx^2+ay^2-2abxy\ge0\)
\(\Leftrightarrow\left(bx-ay\right)^2\ge0\)(đúng) Dấu "=" xảy ra khi x/a=y/b
Ta có: \(\left(x+4y\right)^2\le\left(1^2+2^2\right)\left(x^2+4y^2\right)=5\left(x^2+4y^2\right)\)
Mà a + 4b = 1
\(\Rightarrow x^2+4y^2\ge\frac{1}{5}\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}\frac{1}{x}=\frac{2}{2y}=\frac{1}{y}\\x+4y=1\end{cases}}\Rightarrow x=y=\frac{1}{5}\)
\(a,\left(x+2\right)\left(x-2\right)+4x=4+x^2\)
\(\Leftrightarrow x^2-4+4x-4-x^2=0\)
\(\Leftrightarrow4x=16\)
\(\Rightarrow x=4\)
\(b,2\left(x+5\right)\left(x-5\right)+5x=8+x^2\)
\(\Leftrightarrow2x^2-50+5x-8-x^2=0\)
\(\Leftrightarrow x^2-5x-42=0\Rightarrow\left(x^2-5x+\dfrac{25}{4}\right)-\dfrac{193}{4}=0\Rightarrow\left(x-\dfrac{5}{2}\right)^2=\dfrac{193}{4}\Rightarrow\left[{}\begin{matrix}x-\dfrac{5}{2}=\sqrt{\dfrac{193}{4}}\\x-\dfrac{5}{2}=-\sqrt{\dfrac{193}{4}}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{193}{4}}+\dfrac{5}{2}\\x=-\sqrt{\dfrac{193}{4}}+\dfrac{5}{2}\end{matrix}\right.\)
\(c,\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)
\(\Leftrightarrow x^2+8x+16-x^2-1-16=0\)
\(\Leftrightarrow8x-1=0\Leftrightarrow8x=1\Rightarrow x=\dfrac{1}{8}\)
\(d,\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5x^2-245=0\)\(\Leftrightarrow2x=235\Leftrightarrow x=117,5\)
a) 3 u 2 − 8 u + 3 ( u 2 + 1 ) ( u − 1 ) b) 1 − 4 u 4 ( 4 u + 1 )