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\(\Leftrightarrow\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{x\left(x+3\right)}=\dfrac{375}{376}\)
\(\Leftrightarrow1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{375}{376}\)
\(\Leftrightarrow1-\dfrac{1}{x+3}=\dfrac{375}{376}\)
\(\Leftrightarrow\dfrac{1}{x+3}=1-\dfrac{375}{376}=\dfrac{1}{376}\)
\(\Rightarrow x+3=376\)
\(\Rightarrow x=373\)
\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{125}{376}\)
\(\Leftrightarrow\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{125}{376}\)
\(\Leftrightarrow\dfrac{1}{3}\left(1-\dfrac{1}{x+3}\right)=\dfrac{125}{376}\left(x\ne0;x\ne-3\right)\)
\(\Leftrightarrow\dfrac{x+3-1}{x+3}=\dfrac{3.125}{376}\Leftrightarrow\dfrac{x+2}{x+3}=\dfrac{3.125.}{376}.\dfrac{\left(x+3\right)}{x+3}\)
\(\Leftrightarrow376\left(x+2\right)=3.125.\left(x+3\right)\)
\(\Leftrightarrow376x+752=375x+1125\)
\(\Leftrightarrow376x-375x=1125-752\Leftrightarrow x=373\left(x\in N^{\cdot}\right)\)
a) Ta có: \(A=\dfrac{4}{1\cdot4}+\dfrac{4}{4\cdot7}+\dfrac{4}{7\cdot10}+...+\dfrac{4}{31\cdot34}\)
\(=\dfrac{4}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{31\cdot34}\right)\)
\(=\dfrac{4}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{31}-\dfrac{1}{34}\right)\)
\(=\dfrac{4}{3}\left(1-\dfrac{1}{34}\right)\)
\(=\dfrac{4}{3}\cdot\dfrac{33}{34}=\dfrac{22}{17}\)
\(\frac{2\frac{1}{2}x-1}{\frac{2}{3}}=\frac{\frac{-2}{3}}{1-2\frac{1}{2}x}\) ĐKXĐ \(x\ne\frac{2}{5}\)
\(\Leftrightarrow\)\(\frac{\frac{5}{2}x-1}{\frac{2}{3}}=\frac{\frac{2}{3}}{\frac{5}{2}x-1}\)\(\Leftrightarrow\)\(\left(\frac{5}{2}x-1\right)^2=\frac{4}{9}\)\(\Leftrightarrow\)\(\frac{25}{4}x^2-5x+1=\frac{4}{9}\)
\(\Leftrightarrow\)\(\frac{25}{4}x^2-5x+\frac{5}{9}=0\)\(\Leftrightarrow\)\(\frac{25}{4}x^2-\frac{25}{6}x-\frac{5}{6}x+\frac{5}{9}=0\)
\(\Leftrightarrow\)\(\left(\frac{25}{4}x^2-\frac{25}{6}x\right)-\left(\frac{5}{6}x-\frac{5}{9}\right)=0\)\(\Leftrightarrow\)\(\frac{25}{2}x\left(\frac{1}{2}x-\frac{1}{3}\right)-\frac{5}{3}\left(\frac{1}{2}x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\)\(\left(\frac{25}{2}x-\frac{5}{3}\right)\left(\frac{1}{2}x-\frac{1}{3}\right)=0\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{2}{15}\end{cases}}\)
a: \(A=6\left(x+\dfrac{1}{3}\right)^2-7>=-7>-8\forall x\)
\(B=-8-\left(3.75-x\right)^2\le-8\)
Do đó: A>B
b: \(A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}=\dfrac{15}{16}\)
\(B=\left(\dfrac{1}{2}\right)^4=\dfrac{1}{16}\)
Do đó: A>B
Bài 1:
1: \(M=\left|x-1\right|+x+2\)
Trường hợp 1: x>=1
M=x-1+x+2=2x+1
Trường hợp 2: x<1
M=1-x+x+2=3
2: \(N=x-3+\left|x-3\right|\)
Trường hợp 1: x>=3
\(N=x-3+x-3=2x-6\)
Trường hợp 2: x<3
\(N=x-3+3-x=0\)
3: \(P=2x-1-\left|x-2\right|\)
Trường hợp 1: x<2
\(P=2x-1-\left(2-x\right)=2x-1-2+x=3x-3\)
TRường hợp 2: x>=2
\(P=2x-1-x+2=x+1\)
\(a,\Leftrightarrow\left[{}\begin{matrix}-\dfrac{4}{3}x+\dfrac{1}{2}=\dfrac{1}{2}\\-\dfrac{4}{3}x+\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{4}\end{matrix}\right.\\ c,\Leftrightarrow\left(\dfrac{1}{2}\right)^x\left(1+\dfrac{1}{4}\right)=\dfrac{5}{4}\\ \Leftrightarrow\left(\dfrac{1}{2}\right)^x=1\Leftrightarrow x=0\)
b: Ta có: \(3^x+3^{x+2}=20\)
\(\Leftrightarrow3^x\cdot10=20\)
\(\Leftrightarrow3^x=2\left(loại\right)\)
\(\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{3}=\dfrac{23}{12}\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{23}{12}+\dfrac{1}{3}=\dfrac{9}{4}\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{3}{2}\\x-\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}+\dfrac{1}{2}=2\\x=-\dfrac{3}{2}+\dfrac{1}{2}=-1\end{matrix}\right.\)
\(\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{3}=\dfrac{23}{12}\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{23}{12}+\dfrac{1}{3}\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{9}{4}=\left(\dfrac{3}{2}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{3}{2}\\x-\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
a) \(\left(x+\dfrac{1}{2}\right)+\left(x+\dfrac{1}{6}\right)+\left(x+\dfrac{1}{12}\right)+....+\left(x+\dfrac{1}{9900}\right)\)
\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\right)=1\)
\(\Leftrightarrow50x+\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)=1\)
\(\Leftrightarrow50x+\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=1\)
\(\Leftrightarrow50x+\left(1-\dfrac{1}{100}\right)=1\)
\(\Leftrightarrow50x+\dfrac{99}{100}=1\)
\(\Leftrightarrow50x=\dfrac{1}{100}\Rightarrow x=\dfrac{1}{5000}\)
b) \(A=\dfrac{3^2}{1.4}+\dfrac{3^2}{4.7}+\dfrac{3^2}{7.10}+...+\dfrac{3^2}{202.205}\)
\(A=\dfrac{3^2}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{202}-\dfrac{1}{205}\right)\)
\(A=\dfrac{9}{3}\cdot\left(1-\dfrac{1}{205}\right)\)
\(A=\dfrac{9}{3}\cdot\dfrac{204}{205}=\dfrac{615}{205}\)
a) \(\left(x+\dfrac{1}{2}\right)+\left(x+\dfrac{1}{6}\right)+\left(x+\dfrac{1}{12}\right)+....+\left(x+\dfrac{1}{9900}\right)=1\)
\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\right)=1\)
\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)=1\)
Có tất cả : (99 - 1) : 1 + 1 = 99 (số x)
\(\Rightarrow99x+\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=1\)
\(\Rightarrow99x+\left(1-\dfrac{1}{100}\right)=1\)
\(\Rightarrow99x+\dfrac{99}{100}=1\Rightarrow99x=1-\dfrac{99}{100}\)
\(\Rightarrow99x=\dfrac{1}{100}\Rightarrow x=\dfrac{1}{100.99}=\dfrac{1}{9900}\)
b) \(A=\dfrac{3^2}{1.4}+\dfrac{3^2}{4.7}+\dfrac{3^2}{7.10}+....+\dfrac{3^2}{202.205}\)
\(A=\dfrac{3^2}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{202}-\dfrac{1}{205}\right)\)
\(A=\dfrac{9}{3}\cdot\left(1-\dfrac{1}{205}\right)\)
\(A=3\cdot\dfrac{204}{205}=\dfrac{615}{205}\)