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Giải:
a) \(1\dfrac{1}{2}.2\dfrac{1}{3}+1\dfrac{1}{3}.\dfrac{1}{2}\)
\(=\dfrac{3}{2}.\dfrac{7}{3}+\dfrac{4}{3}.\dfrac{1}{2}\)
\(=\dfrac{21}{6}+\dfrac{4}{6}\)
\(=\dfrac{1}{6}\left(21+4\right)\)
\(=\dfrac{1}{6}.25=\dfrac{25}{6}\)
b) \(\dfrac{1}{9}.\dfrac{2}{145}-4\dfrac{1}{3}.\dfrac{2}{145}+\dfrac{2}{145}\)
\(=\dfrac{1}{9}.\dfrac{2}{145}-\dfrac{13}{3}.\dfrac{2}{145}+\dfrac{2}{145}\)
\(=\dfrac{2}{145}\left(\dfrac{1}{9}-\dfrac{13}{3}+1\right)\)
\(=\dfrac{2}{145}\left(-\dfrac{29}{9}\right)\)
\(=-\dfrac{2}{45}\)
Vậy ...
\(a,=\dfrac{\left(x-2\right)^2-\left(x+2\right)^2}{\left(x-2\right)^2\left(x+2\right)^2}:\dfrac{x-2+x+2}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{-8x}{\left(x-2\right)^2\left(x+2\right)^2}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{2x}=\dfrac{-4}{\left(x-2\right)\left(x+2\right)}\)
\(b,=\dfrac{5x^2+26xy+5y^2+5x^2-26xy+5y^2}{x\left(x-5y\right)\left(x+5y\right)}\cdot\dfrac{\left(x-5y\right)\left(x+5y\right)}{x^2+y^2}\\ =\dfrac{10\left(x^2+y^2\right)}{x\left(x^2+y^2\right)}=\dfrac{10}{x}\)
\(\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\)
\(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\)
\(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{1}{x+1}+\dfrac{-\left(x+3\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{\left(3x+1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}+\dfrac{-\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{\left(3x+1\right)\left(x+1\right)-\left(x-1\right)^2-\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{3x^2+4x+1-\left(x^2-2x+1\right)-\left(x^2+2x+3\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x^2+4x+3}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x^2+x+3x+3}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x\left(x+1\right)+3\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x+3}{\left(x-1\right)^2}\)
\(\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\)
\(=\dfrac{\left(3x+1\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}+\dfrac{-\left(x+3\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{3x^2+4x+1-x^2+2x-1-x^2-2x+3}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x^2+4x+3}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{x^2+3x+x+3}{\left(x-1\right)^2\left(x+1\right)}\)
\(=\dfrac{\left(x+3\right)\left(x+1\right)}{\left(x-1\right)^2\left(x+1\right)}=\dfrac{x+3}{\left(x-1\right)^2}\)
a: \(=\dfrac{4x^2+4x+1-\left(4x^2-4x+1\right)}{\left(2x-1\right)\left(2x+1\right)}\cdot\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{8x}{2x+1}\cdot\dfrac{5}{4x}=\dfrac{10}{2x+1}\)
c: \(=\dfrac{1}{x-1}-\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\left(\dfrac{x+1-x+1}{\left(x-1\right)^2\cdot\left(x+1\right)}\right)\)
\(=\dfrac{1}{x-1}-\dfrac{x}{x^2+1}\cdot\dfrac{2}{\left(x-1\right)}=\dfrac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}=\dfrac{x-1}{x^2+1}\)
\(a,A=\dfrac{5-3}{5+2}=\dfrac{2}{7}\\ b,B=\dfrac{3x-9+2x+6-3x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ c,C=AB=\dfrac{x-3}{x+2}\cdot\dfrac{2}{x-3}=\dfrac{2}{x+2}\\ C=-\dfrac{1}{3}\Leftrightarrow x+2=-6\Leftrightarrow x=-8\left(tm\right)\)
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
1: Ta có: \(\dfrac{x-4}{3}+2x=\dfrac{4x-2}{6}\)
\(\Leftrightarrow2x-8+12x=4x-2\)
\(\Leftrightarrow10x=6\)
hay \(x=\dfrac{3}{5}\)
2: Ta có: \(\dfrac{5x-2}{5}-2=\dfrac{1-2x}{3}\)
\(\Leftrightarrow15x-6-30=10-20x\)
\(\Leftrightarrow35x=46\)
hay \(x=\dfrac{46}{35}\)
3: Ta có: \(\dfrac{x-2}{2}-\dfrac{2}{3}=x-1\)
\(\Leftrightarrow3x-6-4=6x-6\)
\(\Leftrightarrow-3x=4\)
hay \(x=-\dfrac{4}{3}\)
Giải:
a) \(1\dfrac{1}{2}.2\dfrac{1}{3}+1\dfrac{1}{3}.\dfrac{1}{2}\)
\(=\dfrac{3}{2}.\dfrac{7}{3}+\dfrac{4}{3}.\dfrac{1}{2}\)
\(=\dfrac{21}{6}+\dfrac{4}{6}\)
\(=\dfrac{1}{6}\left(21+4\right)\)
\(=\dfrac{25}{6}\)
b) \(\dfrac{1}{9}.\dfrac{2}{145}-4\dfrac{1}{3}.2\dfrac{2}{145}+\dfrac{2}{145}\)
\(=\dfrac{1}{9}.\dfrac{2}{145}-\dfrac{13}{3}.\dfrac{292}{145}+\dfrac{2}{145}\)
\(=\dfrac{2}{145}\left(\dfrac{1}{9}-\dfrac{13}{3}.146+1\right)\)
\(=\dfrac{2}{145}\left(-\dfrac{5684}{9}\right)\)
\(=-\dfrac{392}{45}\)
Vậy ...