Tìm x biết:
\(\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\right)=\frac{0,33.x}{2009}\)
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a) \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)
\(\Leftrightarrow2^x\left(1+2^1+2^2+2^2\right)=15.2^x\)
\(\Leftrightarrow15.2^x=480\)
\(\Leftrightarrow2^x=480:15\)
\(\Leftrightarrow2^x=32\)
\(\Leftrightarrow2^x=2^5\)
=> x = 5
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{1}{3}\left(1-\frac{1}{100}\right)=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{1}{3}.\frac{99}{100}=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{1.33}{1.100}=\frac{0,33.x}{2009}\)
\(\Leftrightarrow\frac{33}{100}=\frac{0,33.x}{2009}\)
\(\Leftrightarrow33.x=66297\)
\(\Leftrightarrow x=22099\)
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(\Rightarrow\frac{99}{100}=\frac{0.33.x}{2009}\)
\(\Rightarrow100.0.33.x=99.2009\)
\(\Rightarrow0x=198891\Rightarrow\)không có GT x thỏa mãn
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\)
\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{97}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
Còn lại thì dễ rồi bạn nhé
bạn ơi như là cô giáo cho đề sai rồi kết quả phải là \(\frac{375}{376}\)thì mới giải được
Ta có:
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{125}{376}\)
\(\Rightarrow\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x.\left(x+3\right)}\right)=\frac{125}{376}\)
\(\Rightarrow\frac{1}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{125}{376}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{x+3}=\frac{125}{376}:\frac{1}{3}=\frac{375}{376}\)
\(\Rightarrow\frac{1}{x+3}=1-\frac{375}{376}=\frac{1}{376}\Leftrightarrow x+3=376\Leftrightarrow x=373\)
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{125}{376}\)
\(3.\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}\right)=3.\frac{125}{376}\)
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}=\frac{375}{376}\)
\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{375}{376}\)
\(1-\frac{1}{x+3}=\frac{375}{376}\)
\(\frac{x+2}{x+3}=\frac{375}{376}\)
=> x + 2 = 375
=> x = 375 - 2
=> x = 373
1/3.(1-1/4+1/4-1/7+......+1/x-1/(x+3)=6/19
1/3.(1-1/x+3)=6/19
1-1/x+3=6/19:1/3
1-1/x+3=18/19
1/x+3=1-18/19
1/x+3=1/19
=> x+3=19
=>x=19-3
x=16
Đặt biểu thức là A, ta có:
3A=\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.11}+...+\frac{3}{x\left(x+3\right)}\)
3A=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}\)
3A=1-\(\frac{1}{x+3}\)
A=\(\frac{1}{3}-\frac{3}{x+3}\)
=>\(\frac{1}{3}-\frac{3}{x+3}\) =\(\frac{6}{19}\) =>x=168
\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{97\cdot100}=\frac{0,33\cdot x}{2009}\cdot3\)
\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}=\frac{0,99\cdot x}{2009}\)
\(\frac{100}{100}-\frac{1}{100}=\frac{0,99x}{2009}\)
\(\frac{99}{100}=\frac{0,99x}{2009}\)
=>0,99x*100=2009*99
99x=2009*99
=>x=2009
Vậy x=2009
\(0,33\cdot\frac{x}{2009}\) hay \(\frac{0,33\cdot x}{2009}\)
\(\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{97.100}\right)=\frac{0,33x}{2009}\)
\(\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\right)=\frac{0,33x}{2009}\)
\(\left(1-\frac{1}{4}+\frac{1}{4}-...-\frac{1}{100}\right)=\frac{0,33x}{2009}\)
\(1-\frac{1}{100}=\frac{0,33x}{2009}\)
\(\frac{99}{100}=\frac{0,33x}{2009}\Rightarrow2009x99=0,33x\times100\)
198891:100:0,33=6027=x