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\(3\times\left(\frac{x}{4}+\frac{x}{28}+\frac{x}{70}+\frac{x}{130}\right)=\frac{60}{13}\)
=> \(\frac{x}{4}+\frac{x}{28}+\frac{x}{70}+\frac{x}{130}=\frac{20}{13}\)
=> \(\frac{x}{1\cdot4}+\frac{x}{4\cdot7}+\frac{x}{7\cdot10}+\frac{x}{10\cdot13}=\frac{20}{13}\)
=> \(\frac{x}{3}\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}\right)=\frac{20}{13}\)
=> \(\frac{x}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{10}-\frac{1}{13}\right)=\frac{20}{13}\)
=> \(\frac{x}{3}\left(1-\frac{1}{13}\right)=\frac{20}{13}\)
=> \(\frac{x}{3}\cdot\frac{12}{13}=\frac{20}{13}\)
=> \(\frac{x}{3}=\frac{20}{13}:\frac{12}{13}=\frac{20}{13}\cdot\frac{13}{12}=\frac{5}{3}\)
=> x = 5
\(3\cdot\left(\frac{x}{4}+\frac{x}{28}+\frac{x}{70}+\frac{x}{130}\right)=\frac{60}{13}\)
\(3\cdot\left(\frac{x}{1\cdot4}+\frac{x}{4\cdot7}+\frac{x}{7\cdot10}+\frac{x}{10\cdot13}\right)=\frac{60}{13}\)
\(3\left(x-3\right)\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}\right)=\frac{60}{13}\)
\(\left(3x-9\right)\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}\right)=\frac{60}{13}\)
\(\left(3x-9\right)\left(1-\frac{1}{13}\right)=\frac{60}{13}\)
\(\left(3x-9\right)\cdot\frac{12}{13}=\frac{60}{13}\)
\(3x-9=\frac{\frac{60}{13}}{\frac{12}{13}}\)
\(3x-9=5\)
\(3x=5+9\)
\(3x=14\)
\(x=\frac{14}{3}\approx4,667\)
CÁCH LÀM NHƯ SAU :
(7/28 + 1/28) + 1/70 + 1/130 + 1/x.(x+3)
8/28 + 1/70 +1/130 +1/x.(x+3)
2/7+1/70+1/130+1/x.(x+3)
(20/70 +1/70)+1/130+1/x.(x+3)
3/10+1/130+1/x.(x+3)
39/130+1/130+1/x.(x+3)
4/13+1/x.(x+3)
Đến đây bn tự làm hộ mình vớ. chúc hok tốt k cho mình nhé
\(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+\frac{1}{x\left(x+3\right)}\)
\(=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{x\left(x+3\right)}\)
\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{x}-\frac{1}{x+3}\right)\)
\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{13}+\frac{1}{x}-\frac{1}{x+3}\right)\)
\(=\frac{1}{3}\left(\frac{12}{13}+\frac{1}{x}-\frac{1}{x+3}\right)\)
\(=\frac{1}{3}.\frac{12}{13}+\frac{1}{3}.\frac{1}{x}-\frac{1}{3}.\frac{1}{x+3}\)
\(=\frac{4}{13}+\frac{1}{3x}-\frac{1}{3x+3}\)
\(=\frac{4}{13}+\frac{1}{3x}-\frac{1}{3x+3}\)
\(=\frac{4}{13}+\frac{1}{3x}=\frac{1}{3x+3}\)
\(=\frac{4}{13}+\frac{1}{3x}=\frac{1}{3x+3}\)
\(=\frac{4}{13}+\frac{1}{3x}=\frac{1}{3}.\frac{1}{x+3}\)
\(=\frac{4}{13}=\frac{1}{3}.\frac{1}{x+3}-\frac{1}{3x}\)
\(=\frac{4}{13}=\frac{1}{3}.\frac{1}{x+3}-\frac{1}{3}.\frac{1}{x}\)
\(=\frac{4}{13}=\frac{1}{3}\left(\frac{1}{x+3}-\frac{1}{x}\right)\)
\(=\frac{4}{13}:\frac{1}{3}=\frac{1}{x+1}-\frac{1}{x}\)
\(=\frac{12}{13}=\frac{1}{x+1}-\frac{1}{x}\)
\(=\frac{12}{13}=\frac{x-\left(x+1\right)}{\left(x+1\right)x}\)
\(=\frac{12}{13}=-\frac{1}{x^2+x}\)
\(\Leftrightarrow=12\left(x^2+x\right)=13.\left(-1\right)\)
\(=12\left(x^2+x\right)=-13\)
\(=x^2+x=-\frac{13}{12}\)
\(=x\left(x+1\right)=-\frac{13}{12}\)
.... Chiụ
\(\frac{x+9}{13-x}=\frac{6}{5}\)
\(\Rightarrow\orbr{\begin{cases}x+9=6\\13-x=5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=8\end{cases}}\)
\(\frac{120-x}{30}=3\frac{1}{2}\)
\(\Rightarrow\frac{120-x}{30}=\frac{7}{2}\)
\(\Rightarrow\frac{120-x}{30}=\frac{105}{30}\)
\(\Rightarrow120-x=105\)
\(\Rightarrow x=120-105\)
\(\Rightarrow x=15\)
(x + 5 ) x \(\frac{19}{13}=57\)
x + 5 = \(57\div\frac{19}{13}=39\)
x = 39 -5 = 34
b) 12 . 15 - x = 120 \(\cdot\frac{1}{4}\div\frac{1}{4}=120\)
180 - x = 120
x = 180 - 120 = 60
\(\frac{3}{4}x\frac{8}{9}x...x\frac{120}{121}=\frac{1x3}{2x2}x\frac{2x4}{3x3}x...x\frac{10x12}{11x11}=\frac{1x3x2x4x...x10x12}{2x2x3x3x...x11x11}\)
\(\frac{\left(1x2x3x4x...x10\right)x\left(3x4x5x6x...x12\right)}{\left(2x3x4x5x...x11\right)x\left(2x3x4x5x...x11\right)}=\frac{12}{11x2}=\frac{6}{11}\)
X x\(\frac{1}{2}\)+ X x\(\frac{3}{4}\)=\(\frac{6}{7}\)
=>X x(\(\frac{1}{2}\)+ \(\frac{3}{4}\)) =\(\frac{6}{7}\)
=>X x \(\frac{5}{4}\)=\(\frac{6}{7}\)
=>X=\(\frac{6}{7}\): \(\frac{5}{4}\)
=>X=\(\frac{24}{35}\)
a)\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}......\frac{99}{100}\)
\(=\frac{1.2.3.4.....99}{2.3.4.5.6.....100}\)
\(=\frac{1}{100}\)
b) Tương tự như câu a
\(61+32,5\times4=130+\frac{120+x}{x}\)
\(61+130=130+\frac{x}{x}+\frac{120}{x}\)
\(\Rightarrow1+\frac{120}{x}=61\)
\(\Rightarrow\frac{120}{x}=61-1\)
\(\Rightarrow\frac{120}{x}=\frac{60}{1}\)
\(\Rightarrow x=1\times120\div60\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
61+32,5x4=130+\(\frac{x+120}{x}\)
61+130=130+\(\frac{x+120}{x}\)
61=\(\frac{x+120}{x}\)
61(x+120)=x
61x+7320=x
61x-x=7320
60x=7320
x=7320:60
x=122
vậy x= 122