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Nguyễn Trần Thành ĐạtXuân Tuấn TrịnhHung nguyenHoang HungQuan Ace Legona giúp với
Sửa đề: \(\dfrac{100+\dfrac{99}{2}+\dfrac{98}{3}+...+\dfrac{1}{100}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{101}}-2\)
\(=\dfrac{\left(\dfrac{99}{2}+1\right)+\left(\dfrac{98}{3}+1\right)+...+\left(\dfrac{1}{100}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{101}}-2\)
\(=\dfrac{\dfrac{101}{2}+\dfrac{101}{3}+...+\dfrac{101}{100}+\dfrac{101}{101}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}+\dfrac{1}{101}}-2\)
\(=\dfrac{101\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}+\dfrac{1}{101}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}+\dfrac{1}{101}}-2\)
\(=101-2=99\)
Vậy...
Giải:
1) \(\dfrac{-1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)\)
\(=\dfrac{-1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)\)
\(=\dfrac{-1}{12}-\dfrac{55}{24}\)
\(=\dfrac{-19}{8}\)
2) \(-1,75-\left(\dfrac{-1}{9}-2\dfrac{1}{18}\right)\)
\(=-\dfrac{7}{4}+\dfrac{1}{9}+2\dfrac{1}{18}\)
\(=-\dfrac{7}{4}+\dfrac{1}{9}+\dfrac{37}{18}\)
\(=\dfrac{5}{12}\)
3) \(-\dfrac{5}{6}-\left(-\dfrac{3}{8}+\dfrac{1}{10}\right)\)
\(=-\dfrac{5}{6}+\dfrac{3}{8}-\dfrac{1}{10}\)
\(=-\dfrac{67}{120}\)
4) \(\dfrac{2}{5}+\left(-\dfrac{4}{3}\right)+\left(-\dfrac{1}{2}\right)\)
\(=\dfrac{2}{5}-\dfrac{4}{3}-\dfrac{1}{2}\)
\(=-\dfrac{43}{30}\)
5) \(\dfrac{3}{12}-\left(\dfrac{6}{15}-\dfrac{3}{10}\right)\)
\(=\dfrac{3}{12}-\dfrac{6}{15}+\dfrac{3}{10}\)
\(=\dfrac{3}{20}\)
6) \(\left(8\dfrac{5}{11}+3\dfrac{5}{8}\right)-3\dfrac{5}{11}\)
\(=8\dfrac{5}{11}+3\dfrac{5}{8}-3\dfrac{5}{11}\)
\(=8+\dfrac{5}{11}+3+\dfrac{5}{8}-3-\dfrac{5}{11}\)
\(=8+\dfrac{5}{8}\)
\(=\dfrac{69}{8}\)
7) \(-\dfrac{1}{4}.13\dfrac{9}{11}-0,25.6\dfrac{2}{11}\)
\(=-\dfrac{1}{4}.13\dfrac{9}{11}-\dfrac{1}{4}.6\dfrac{2}{11}\)
\(=-\dfrac{1}{4}\left(13\dfrac{9}{11}+6\dfrac{2}{11}\right)\)
\(=-\dfrac{1}{4}\left(13+\dfrac{9}{11}+6+\dfrac{2}{11}\right)\)
\(=-\dfrac{1}{4}\left(13+6+1\right)\)
\(=-\dfrac{1}{4}.20=-5\)
8) \(\dfrac{4}{9}:\left(-\dfrac{1}{7}\right)+6\dfrac{5}{9}:\left(-\dfrac{1}{7}\right)\)
\(=\dfrac{4}{9}\left(-7\right)+6\dfrac{5}{9}\left(-7\right)\)
\(=-7\left(\dfrac{4}{9}+6\dfrac{5}{9}\right)\)
\(=-7\left(\dfrac{4}{9}+6+\dfrac{5}{9}\right)\)
\(=-7\left(6+1\right)\)
\(=-7.7=-49\)
Vậy ...
5. phân tích ra : \(1+\dfrac{a}{b}+\dfrac{b}{a}+1\)
áp dụng bđ cosy
\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}=2\)
=> đpcm
6. \(x^2-x+1=x^2-2.\dfrac{1}{2}.x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
hay với mọi x thuộc R đều là nghiệm của bpt
7.áp dụng bđt cosy
\(a^4+b^4+c^4+d^4\ge2\sqrt{a^2.b^2.c^2.d^2}=4abcd\left(đpcm\right)\)
Đặt \(B=\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{n}\)
Đặt \(A=\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{n-\left(n-2\right)}{n-2}+\dfrac{n-\left(n-1\right)}{n-1}\)
\(=\dfrac{n}{1}+\dfrac{n}{2}+...+\dfrac{n}{n-2}+\dfrac{n}{n-1}-1-1-...-1\)
\(=n+\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}-\left(n-1\right)\)
\(=\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}+\dfrac{n}{n}=n\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2}\right)=n.B\)
\(A:B=n\)
Giải các phương trình
\(a,3x-2=2x-3\)
\(\Leftrightarrow3x-2x=-3+2\)
\(\Leftrightarrow x=-1\)
Vậy pt có tập nghiệm S = { - 1 }
\(b,2x+3=5x+9\)
\(\Leftrightarrow2x-5x=9-3\)
\(\Leftrightarrow-3x=6\)
\(\Leftrightarrow x=-2\)
Vậy pt có tập nghiệm S = { - 2 }
\(c,11x+42-2x=100-9x-22\)
\(\Leftrightarrow11x-2x+9x=100-22-42\)
\(\Leftrightarrow18x=36\)
\(\Leftrightarrow x=2\)
Vậy pt có tập nghiệm S = { - 2 }
\(d,2x-\left(3-5x\right)=4\left(x+3\right)\)
\(\Leftrightarrow2x-3+5x=4x+12\)
\(\Leftrightarrow2x+5x-4x=12+3\)
\(\Leftrightarrow3x=15\)
\(\Leftrightarrow x=5\)
Vậy pt có tập nghiệm S = { - 5 }
\(e,\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=\dfrac{5}{3}+2x\)
\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{5.2}{6}+\dfrac{2x.6}{6}\)
\(\Leftrightarrow9x+6-3x-1=10+12x\)
\(\Leftrightarrow9x-3x-12x=10-6+1\)
\(\Leftrightarrow-6x=5\)
\(\Leftrightarrow x=-\dfrac{5}{6}\)
Vậy pt có tập nghiệm S = { - \(\dfrac{5}{6}\) }
f,\(\dfrac{x+4}{5}-x+4=\dfrac{x}{3}-\dfrac{x-2}{2}\)
\(\Leftrightarrow\dfrac{6\left(x+4\right)}{30}-\dfrac{30x}{30}+\dfrac{4.30}{30}=\dfrac{10x}{30}-\dfrac{15\left(x-2\right)}{30}\)
\(\Leftrightarrow6x+24-30x+120=10x-15x+30\)
\(\Leftrightarrow6x-30x-10x+15x=30-24-120\)
\(\Leftrightarrow-19x=-114\)
\(\Leftrightarrow x=6\)
Vậy pt có tập nghiệm S = { - 6 }
\(g,\left(2x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=1\end{matrix}\right.\)
Vậy pt có tập nghiệm S = { \(1;-\dfrac{1}{2}\) }
\(h,\left(x+\dfrac{2}{3}\right)\left(x-\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{3}=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy pt có tập nghiệm S = { \(-\dfrac{2}{3};\dfrac{1}{2}\) }
\(i,\left(3x-1\right)\left(2x-3\right)\left(2x-3\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(2x-3\right)^2\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\2x-3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=\dfrac{3}{2}\\x=-5\end{matrix}\right.\)
Vậy pt có tập nghiệm S = { \(\dfrac{1}{3};\dfrac{3}{2};-5\) }
\(k,3x-15=2x\left(x-5\right)\)
\(\Leftrightarrow3x-15=2x^2-10x\)
\(\Leftrightarrow-2x^2+3x+10x=15\)
\(\Leftrightarrow-2x^2+13x-15=0\)
\(\Leftrightarrow-2x^2+10x+3x-15=0\)
\(\Leftrightarrow\left(x-5\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy pt có tập nghiệm S = { \(5;\dfrac{3}{2}\) }
\(m,\left|x-2\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
Vậy pt có tập nghiệm S = { -1; 5 }
\(n,\left|x+1\right|=\left|2x+3\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=2x+3\\x+1=-2x-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy pt có tập nghiệm S = { \(-2;-\dfrac{4}{3}\) }
\(j,\dfrac{7x-3}{x-1}=\dfrac{2}{3}\) ĐKXĐ : x≠ 1
\(\Leftrightarrow3\left(7x-3\right)=2\left(x-1\right)\)
\(\Leftrightarrow21x-9=2x-2\)
\(\Leftrightarrow x=\dfrac{7}{19}\) ( t/m )
Vậy pt có tập nghiệm S = { \(\dfrac{7}{19}\) }
đ, ĐKXĐ : x ≠ - 1
\(\dfrac{2\left(3-7x\right)}{1+x}=\dfrac{1}{2}\)
\(\Leftrightarrow4\left(3-7x\right)=1+x\)
\(\Leftrightarrow12-28x=1+x\)
\(\Leftrightarrow-29x=-11\)
\(\Leftrightarrow x=\dfrac{11}{29}\) ( t/m)
Vậy pt có tập nghiệm S = { \(\dfrac{11}{29}\) }
\(y,\dfrac{x+5}{x-5}-\dfrac{x-5}{x+5}=\dfrac{20}{x^2-25}\) ĐKXĐ : \(\left\{{}\begin{matrix}x\ne5\\x\ne-5\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{\left(x+5\right)^2-\left(x-5\right)^2}{\left(x-5\right)\left(x+5\right)}=\dfrac{20}{\left(x-5\right)\left(x+5\right)}\)
\(\Rightarrow20x=20\)
\(\Leftrightarrow x=1\) ( t/m )
Vậy pt có tập nghiệm S = { 1 }
\(\dfrac{1}{x-1}+\dfrac{2}{x+1}=\dfrac{x}{x^2-1}\) ĐKXĐ : \(\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{x+1+2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x}{\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow3x-1=x\)
\(\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\)( t/m)
Vậy pt có tập nghiệm S = { \(\dfrac{1}{2}\) }
Ta có: \(\dfrac{n^3-1}{n^3+1}=\dfrac{\left(n-1\right)\left(n^2+n+1\right)}{\left(n+1\right)\left(n^2-n+1\right)}=\dfrac{\left(n-1\right)[\left(n+0,5\right)^2+0,75]}{\left(n+1\right)[\left(n-0,5\right)^2+0,75]}\)
Thay vào M ta có:
\(M=\dfrac{2,5^2+0.75}{3.\left(1,5^2+0,75\right)}.\dfrac{2.\left(3,5^2+0,75\right)}{4.\left(2,5^2+0,75\right)}...\dfrac{99[\left(100,5\right)^2+0,75]}{101.[\left(99,5\right)^2+0,75}\)
\(=\dfrac{1.2.3...99}{3.4.5...101}.\dfrac{\left(2,5^2+0,75\right).\left(3,5^2+0,75\right)...[\left(100,5\right)^2+0,75]}{\left(1,5^2+0,75\right).\left(2,5^2+0,75\right)...[\left(99,5\right)^2+0,75]}\)\(=\dfrac{1.2}{100.\left(101\right)}.\dfrac{\left(100,5\right)^2+0,75}{1,5^2+0,75}=\dfrac{2}{3}.\dfrac{\left(100^2+100+1\right)}{3.100.101}>\dfrac{2}{3}\left(đpcm\right)\)