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a)Nhận xét
\(\dfrac{n^3+1}{n^3-1}=\dfrac{\left(n+1\right)\left(n^2-n+1\right)}{\left(n-1\right)\left(n^2+n+1\right)}=\dfrac{\left(n+1\right)\left[\left(n-0,5\right)^2+0;75\right]}{\left(n-1\right)\left[\left(n+0,5\right)^2+0,75\right]}\)
Áp dụng công thức trên:
\(A=\dfrac{2^3+1}{2^3-1}.\dfrac{3^3+1}{3^3-1}....\dfrac{9^3+1}{9^3-1}\)
\(=\dfrac{\left(2+1\right)\left[\left(2-0,5\right)^2+0,75\right]}{\left(2-1\right)\left[\left(2+0,5\right)^2+0,75\right]}.\dfrac{\left(3+1\right)\left[\left(3-0,5\right)^2+0,75\right]}{\left(3-1\right)\left[\left(3+0,5\right)^2+0,75\right]}...\dfrac{\left(9+1\right)\left[\left(9-0,5\right)^2+0,75\right]}{\left(9-1\right)\left[\left(9+0,5\right)^2+0,75\right]}\)
\(=\dfrac{3\left(1,5^2+0,75\right)}{\left(2,5^2+0,75\right)}.\dfrac{4\left(2,5^2+0,75\right)}{2\left(3,5^2+0,75\right)}...\dfrac{10\left(8,5^2+0,75\right)}{8\left(9,5^2+0,75\right)}\)
\(=\dfrac{3.4....10}{1.2.....8}.\dfrac{1,5^2+0,75}{9,5^2+0,75}\)
\(=\dfrac{9.10}{2}.\dfrac{3}{91}\)
\(=\dfrac{3}{2}.\dfrac{90}{91}< \dfrac{3}{2}\)
\(\Rightarrowđpcm\)
b) Làm tương tự
\(=\dfrac{\dfrac{2^3\cdot3^2}{3^3\cdot4^2}\cdot\left(-1\right)}{\dfrac{2^2\cdot\left(-5\right)^3}{5^2\cdot2^6\cdot3^3}}=\dfrac{-\dfrac{1}{2}\cdot\dfrac{1}{3}}{-\dfrac{1}{2^4}\cdot5\cdot\dfrac{1}{3^3}}=\dfrac{1}{6}:\dfrac{5}{2^4\cdot3^3}\)
\(=\dfrac{1}{6}\cdot\dfrac{2^4\cdot3^3}{5}=\dfrac{2^3\cdot3^2}{5}=\dfrac{72}{5}\)
Theo AM-GM ta có:
\(\left\{{}\begin{matrix}b^2+1\ge2\sqrt{b^2}=2b\\a^2+b^2\ge2\sqrt{a^2b^2}=2ab\end{matrix}\right.\)
\(\Rightarrow a^2+2b^2+1\ge2ab+2b\Rightarrow a^2+2b^2+3\ge2ab+2b+2\)
\(=2\left(ab+b+1\right)\Rightarrow\dfrac{1}{a^2+2b^2+3}\le\dfrac{1}{2\left(ab+b+1\right)}\)
Tương tự cho 2 BĐT còn lại ta có:
\(\dfrac{1}{b^2+2c^2+3}\le\dfrac{1}{2\left(bc+c+1\right)};\dfrac{1}{c^2+2a^2+3}\le\dfrac{1}{2\left(ca+a+1\right)}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\le\dfrac{1}{2}\left(\dfrac{1}{ab+b+1}+\dfrac{1}{bc+c+1}+\dfrac{1}{ca+a+1}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{ab}{ab+b+1}+\dfrac{b}{ab+b+1}+\dfrac{1}{ab+b+1}\right)\left(abc=1\right)\)
\(=\dfrac{1}{2}\left(\dfrac{ab+b+1}{ab+b+1}\right)=\dfrac{1}{2}=VP\)
\(\dfrac{1.2}{1.1}.\dfrac{2.3}{2.2}.\dfrac{3.4}{3.3}.\dfrac{4.5}{4.4}...\dfrac{10.11}{10.10}\left(x-2\right)=-20x+40\)
\(\Leftrightarrow\dfrac{2.3.4...11}{1.2.3...10}\left(x-2\right)=-20x+40\)
\(\Leftrightarrow11\left(x-2\right)=-20x+40\)
\(\Leftrightarrow11x-22=-20x+40\)
\(\Leftrightarrow31x=62\)
\(\Rightarrow x=2\)
\(=>\dfrac{2\cdot1}{1\cdot1}\cdot\dfrac{2\cdot3}{2\cdot2}\cdot\dfrac{3\cdot4}{3\cdot3}\cdot......\cdot\dfrac{10\cdot11}{10\cdot10}\cdot\left(x-2\right)=-20\left(x+1\right)+60\)=>11*(x-2)=-20*(x+1)+60
=>11x-22=-20x-20+60
=>31x=62
=>x=2
\(C=\left(\dfrac{1}{\left(a^2+1\right)\left(a+1\right)^2}+\dfrac{2}{\left(a+1\right)^3}\cdot\dfrac{a+1}{a}\right):\dfrac{a-1}{a^3}\)
\(=\left(\dfrac{1}{\left(a^2+1\right)\left(a+1\right)^2}+\dfrac{2}{a\left(a+1\right)^2}\right):\dfrac{a-1}{a^3}\)
\(=\dfrac{a+2\cdot\left(a^2+1\right)}{a\left(a^2+1\right)\left(a+1\right)^2}\cdot\dfrac{a^3}{a-1}\)
\(=\dfrac{2a\left(a+1\right)}{\left(a^2+1\right)\cdot\left(a+1\right)^3}\cdot\dfrac{a^2}{a-1}\)
\(=\dfrac{2a^3}{\left(a^2+1\right)\left(a+1\right)^2\cdot\left(a-1\right)}\)
\(a,2\left(5x+1\right)-7\left(3x-2\right)=4\left(2x-1\right)+3\left(2-x\right)\)
\(\Leftrightarrow10x+2-21x+14=8x-4+6-3x\)
\(\Leftrightarrow-16x=-14\)
\(\Rightarrow x=\dfrac{7}{8}\)
\(b,-4\left(\dfrac{1}{2}x-3\right)+\dfrac{7}{2}\left(2x-1\right)+x=5x\left(1-x\right)\)
\(\Leftrightarrow-2x+12+7x-\dfrac{7}{2}+x=5x-5x^2\)
\(\Leftrightarrow5x^2+x+\dfrac{17}{2}=0\)
Cái này không biết tách kiểu gì cho vừa nên bạn nhấn máy tính nhé
Mode 5 3 rồi lần lượt điền vào theo thứ tự trên thì
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{10}+\dfrac{13i}{10}\\x=-\dfrac{1}{10}-\dfrac{13i}{10}\end{matrix}\right.\)
\(a,\dfrac{8y}{3x^2}.\dfrac{9x^2}{4y^2}=\dfrac{72x^2y}{12x^2y^2}=\dfrac{6}{y}\\b,\dfrac{3x+x^2}{x^2+x+1}.\dfrac{3x^3-3}{x+3}=\dfrac{x\left(x+3\right)3\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x+3\right)}=3x\left(x-1\right)=3x^2-3x \)
\(c,\dfrac{2x^2+4}{x-3}.\dfrac{3x+1}{x-1}.\dfrac{6-2x}{x^2+2}=\dfrac{2\left(x^2+2\right)\left(3x+1\right)2\left(3-x\right)}{\left(x-3\right)\left(x-1\right)\left(x^2+2\right)}=\dfrac{-4\left(3x+1\right)}{x-1}=\dfrac{-12x-4}{x-1}\)
\(d,\dfrac{2x^2}{3y^3}:\left(-\dfrac{4x^3}{21y^2}\right)=\dfrac{-2x^2.21y^2}{3y^3.4x^3}=\dfrac{-42x^2y^2}{12x^3y^3}=\dfrac{-7}{2xy}\)
\(e,\dfrac{2x+10}{x^3-64}:\dfrac{\left(x+5\right)^2}{2x-8}=\dfrac{2\left(x+5\right)}{\left(x-4\right)\left(x^2+4x+16\right)}.\dfrac{2\left(x-4\right)}{\left(x+5\right)^2}=\dfrac{4}{\left(x+5\right)\left(x^2+4x+16\right)}=\dfrac{4}{x^3+9x^2+16x+80}\)
\(f,\dfrac{1}{x+y}\left(\dfrac{x+y}{xy}-x-y\right)-\dfrac{1}{x^2}:\dfrac{y}{x}=\dfrac{1}{x+y}\left(\dfrac{\left(x+y\right)\left(1-xy\right)}{xy}\right)-\dfrac{x}{x^2y}=\dfrac{1-xy}{xy}-\dfrac{x}{x^2y}=\dfrac{-x^2y}{x^2y}=-1\)
Cách làm tương tự bài dưới đây.
Câu hỏi của Nguyễn Bạch Gia Chí - Toán lớp 8 | Học trực tuyến