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Tìm x thỏa mãna, $\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{97.100}=|\frac{x}{3}|$b,  $\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{97.101}=|\frac{5x-4}{101}|$c,\(\left(1-\frac{1}{2}\right)\left(1-\fra...

Loan~♡๖ۣۜTεαм ๖ۣۜCà ๖ۣۜK... · Accepted Answer

a, $\frac{1}{1.4}$+$\frac{1}{4.7}$+......+$\frac{1}{97.100}$= |$\frac{x}{3}$|$\Rightarrow$$\frac{1}{3}$ ( $\frac{3}{1.4}$+$\frac{3}{4.7}$+.......+$\frac{3}{97.100}$)= |$\frac{x}{3}$|$\Rightarrow$$\frac{1}{3}$ ( 1  - $\frac{1}{4}$+ $\frac{1}{4}$-$\frac{1}{7}$+......+$\frac{1}{97}$-$\frac{1}{100}$) = |$\frac{x}{3}$|$\Rightarrow$$\frac{1}{3}$ ( 1-$\frac{1}{100}$) = |$\frac{x}{3}$|$\Rightarrow$$\frac{1}{3}$ . $\frac{99}{100}$ = |$\frac{x}{3}$|$\Rightarrow$$\frac{33}{100}$ = |$\frac{x}{3}$|$\Rightarrow$$\frac{x}{3}$= $\orbr{\begin{cases}\frac{33}{100}\\frac{-33}{100}\end{cases}}$Với $\frac{x}{3}$ = $\frac{33}{100}$$\Rightarrow$100x= 33.3 $\Rightarrow$100x=99$\Rightarrow$x=$\frac{99}{100}$Với $\frac{x}{3}$=$\frac{-33}{100}$$\Rightarrow$100x=-33.3$\Rightarrow$100x=-99$\Rightarrow$x=$\frac{-99}{100}$Vậy x=$\orbr{\begin{cases}\frac{99}{100}\\frac{-99}{100}\end{cases}}$Câu trả lời: a, $\frac{1}{1.4}$+$\frac{1}{4.7}$+......+$\frac{1}{97.100}$= |$\frac{x}{3}$|$\Rightarrow$$\frac{1}{3}$ ( $\frac{3}{1.4}$+$\frac{3}{4.7}$+.......+$\frac{3}{97.100}$)= |$\frac{x}{3}$|$\Rightarrow$$\frac{1}{3}$ ( 1  - $\frac{1}{4}$+ $\frac{1}{4}$-$\frac{1}{7}$+......+$\frac{1}{97}$-$\frac{1}{100}$) = |$\frac{x}{3}$|$\Rightarrow$$\frac{1}{3}$ ( 1-$\frac{1}{100}$) = |$\frac{x}{3}$|$\Rightarrow$$\frac{1}{3}$ . $\frac{99}{100}$ = |$\frac{x}{3}$|$\Rightarrow$$\frac{33}{100}$ = |$\frac{x}{3}$|$\Rightarrow$$\frac{x}{3}$= $\orbr{\begin{cases}\frac{33}{100}\\frac{-33}{100}\end{cases}}$Với $\frac{x}{3}$ = $\frac{33}{100}$$\Rightarrow$100x= 33.3 $\Rightarrow$100x=99$\Rightarrow$x=$\frac{99}{100}$Với $\frac{x}{3}$=$\frac{-33}{100}$$\Rightarrow$100x=-33.3$\Rightarrow$100x=-99$\Rightarrow$x=$\frac{-99}{100}$Vậy x=$\orbr{\begin{cases}\frac{99}{100}\\frac{-99}{100}\end{cases}}$

Loan~♡๖ۣۜTεαм ๖ۣۜCà ๖ۣۜK... · Answer

b, $\frac{4}{1.5}$+ $\frac{4}{5.9}$+......+ $\frac{4}{97.101}$= |$\frac{5x-4}{101}$|$\Rightarrow$1-$\frac{1}{5}$+$\frac{1}{5}$-$\frac{1}{9}$+......+$\frac{1}{97}$-$\frac{1}{101}$= |$\frac{5x-4}{101}$|$\Rightarrow$1-$\frac{1}{101}$= |$\frac{5x-4}{101}$$\Rightarrow$ $\frac{100}{101}$= |$\frac{5x-4}{101}$|$\Rightarrow$$\frac{5x-4}{101}$ =$\orbr{\begin{cases}\frac{100}{101}\\frac{-100}{101}\end{cases}}$Với $\frac{5x-4}{101}$ =$\frac{100}{101}$$\Rightarrow$(5x-4).101=100.101$\Rightarrow$505x-404=10100$\Rightarrow$505x=10504$\Rightarrow$x=$\frac{104}{5}$Với $\frac{5x-4}{101}$=$\frac{-100}{101}$$\Rightarrow$(5x-4). 101=-100.101$\Rightarrow$505x-404=-10100$\Rightarrow$505x=-9696$\Rightarrow$x=$\frac{-96}{5}$Vậy x=$\orbr{\begin{cases}\frac{104}{5}\\frac{-96}{5}\end{cases}}$Câu trả lời: b, $\frac{4}{1.5}$+ $\frac{4}{5.9}$+......+ $\frac{4}{97.101}$= |$\frac{5x-4}{101}$|$\Rightarrow$1-$\frac{1}{5}$+$\frac{1}{5}$-$\frac{1}{9}$+......+$\frac{1}{97}$-$\frac{1}{101}$= |$\frac{5x-4}{101}$|$\Rightarrow$1-$\frac{1}{101}$= |$\frac{5x-4}{101}$$\Rightarrow$ $\frac{100}{101}$= |$\frac{5x-4}{101}$|$\Rightarrow$$\frac{5x-4}{101}$ =$\orbr{\begin{cases}\frac{100}{101}\\frac{-100}{101}\end{cases}}$Với $\frac{5x-4}{101}$ =$\frac{100}{101}$$\Rightarrow$(5x-4).101=100.101$\Rightarrow$505x-404=10100$\Rightarrow$505x=10504$\Rightarrow$x=$\frac{104}{5}$Với $\frac{5x-4}{101}$=$\frac{-100}{101}$$\Rightarrow$(5x-4). 101=-100.101$\Rightarrow$505x-404=-10100$\Rightarrow$505x=-9696$\Rightarrow$x=$\frac{-96}{5}$Vậy x=$\orbr{\begin{cases}\frac{104}{5}\\frac{-96}{5}\end{cases}}$

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