Tìm x biết 66 -5.3x = 21
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Trả lời:
\(3x+1+5\times3x+2=144\)
\(\Leftrightarrow3x+1+15x+2=144\)
\(\Leftrightarrow18x=141\)
\(\Leftrightarrow x=\frac{47}{8}\)
Vậy \(x=\frac{47}{8}\)
(x/6+x/10+...+x/66+x/78)*2=220/39*2
x/12+x/20+...x/132+x/156=440/39
x/2-x/3+x/3-x/4+...+x/11-x/12+x/12-x/13=440/39
x/2-x/13=440/39
11x/26=440/39
33x/78=880/78
33x=880
x=880/33
x=80/3=26, (6)
d: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{y-x}{3-4}=\dfrac{5}{-1}=-5\)
Do đó: x=-20; y=-15
\(a,5^x+5^{x+2}=650\\ \Rightarrow a,5^x+5^x.25=650\\ \Rightarrow26.5^x=650\\ \Rightarrow5^x=25\\ \Rightarrow5^x=5^2\\ \Rightarrow x=2\)
\(b,3^{x.1}+5.3^{x.1}=162\\ \Rightarrow3^x+5.3^x=162\\ \Rightarrow6.3^x=162\\ \Rightarrow3^x=27\\ \Rightarrow3^x=3^3\\ \Rightarrow x=3\)
: x+53=66=>x=66-35=31
x-5+21= 447=>x=447-21+5=431
x3-3x = 0=>x(x^2-3)=0
=>x={0;\(\sqrt{3}\);\(-\sqrt{3}\)}
\(\frac{x}{6}+\frac{x}{10}+\frac{x}{15}+\frac{x}{21}+\frac{x}{28}+\frac{x}{36}+\frac{x}{45}+\frac{x}{55}+\frac{x}{66}+\frac{x}{78}+\frac{x}{78}=\frac{220}{39}\)
\(\Leftrightarrow x\left(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}+\frac{1}{55}+\frac{1}{66}+\frac{1}{78}+\frac{1}{78}\right)=\frac{220}{39}\)
\(\Leftrightarrow x\cdot\frac{20}{39}=\frac{220}{39}\Rightarrow x=11\)
\(\frac{x}{6}+\frac{x}{10}+\frac{x}{15}+\frac{x}{21}+\frac{x}{28}+\frac{x}{36}+\frac{x}{45}+\frac{x}{55}+\frac{x}{66}+\frac{x}{78}+\frac{x}{78}=\frac{220}{39}\)
\(=>x=\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}+\frac{1}{55}+\frac{1}{66}+\frac{1}{78}+\frac{1}{78}=\frac{220}{39}\)
\(x\cdot\frac{20}{39}=\frac{220}{39}\)
\(x=\frac{220}{39}:\frac{20}{39}=11\)
66 - 5.3x = 21
5.3x = 66 - 21
5.3x = 45
3x = 45 : 5
3x = 9
3x = 32
x = 2
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