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a) \(2.5^2.3^2+\left\{\left[2.5^3-\left(5x+4\right).5\right]:\left(2^2.3.5\right)\right\}=453\)
\(2.25.9+\left\{\left[2.125-\left(5x+4\right).5\right]:\left(4.3.5\right)\right\}=453\)
\(50.9+\left\{\left[250-\left(5x+4\right).5\right]:60\right\}=453\)
\(450+\left\{\left[250-\left(5x+4\right).5\right]:60\right\}=453\)
\(\left[250-\left(5x+4\right).5\right]:60=453-450\)
\(\left[250-\left(5x+4\right).5\right]:60=3\)
\(250-\left(5x+4\right).5=3.60\)
\(250-\left(5x+4\right).5=180\)
\(\left(5x+4\right).5=250-180\)
\(\left(5x+4\right).5=70\)
\(5x+4=70:5\)
\(5x+4=14\)
\(5x=14-4\)
\(5x=10\)
\(x=10:5\)
\(x=2\)
Vậy \(x=2\)
b) \(\left|x-\frac{1}{3}\right|=2\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=-2\\x-\frac{1}{3}=2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\left(-2\right)+\frac{1}{3}\\x=2+\frac{1}{3}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-5}{3}\\x=\frac{7}{3}\end{cases}}\)
Vậy \(x\in\left\{\frac{-5}{7};\frac{7}{3}\right\}\)
c, \(5^{x+4}-3\cdot5^{x+3}=2\cdot5^{11}\)
\(\Leftrightarrow5^{x+3}\cdot5-3\cdot5^{x+3}=2\cdot5^{11}\)
\(\Leftrightarrow5^{x+3}\left(5-3\right)=2\cdot5^{11}\)
\(\Leftrightarrow5^{x+3}\cdot2=2\cdot5^{11}\)
\(\Leftrightarrow5^{x+3}=5^{11}\)
\(\Leftrightarrow x+3=11\)
\(\Leftrightarrow x=8\)
Vậy x = 8
d, \(2^x+2^{x+1}+2^{x+2}+2^{x+3}+2^{x+4}+2^{x+5}=480\)
\(\Leftrightarrow2^x\left(1+2+2^2+2^3+2^4+2^5\right)=480\)
\(\Leftrightarrow2^x\cdot63=480\)
\(\Leftrightarrow2^x=\frac{160}{21}\)
\(\Leftrightarrow x\approx2,93\)
c) \(3^2+2^4-\left(6^8:6^6-6^2\right)< 5^x< 125\)
\(=9+16-\left(6^{8-6}-36\right)< 5^x< 5^3\)
\(=25-\left(6^2-36\right)< 5^x< 5^3\)
\(=25-\left(36-36\right)< 5^x< 5^3\)
\(=25-0< 5^x< 5^3\)
\(=25< 5^x< 5^3\)
\(=5^2< 5^x< 5^3\)
Vì \(5^2=25\) và \(5^3=125\) nên \(x\) không thể thỏa mãn đề bài
⇒ \(x\) không thỏa mãn đề bài
mk chưa hc đến cái toán đó nếu mk hc rồi mk sẽ giúp bn nha
\(.5^{x-3}-2.5^2=3.5^2\)
\(5^{x-3}=25\)
Ta có :
\(5^2=25\)
=> x-3 = 2
=> x=5
5x-3-2.52=3.52
5x-3=3.52+2.52
5x-3=52.(3+2)
5x-3=52.5
5x-3=53
=>x-3=3
x=6
\(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{5}{11}.\)
\(\frac{1}{2}\left(1-\frac{1}{x+2}\right)=\frac{5}{11}\)
\(1-\frac{1}{x+2}=\frac{5}{11}:\frac{1}{2}\)
\(1-\frac{1}{x+2}=\frac{5}{11}\cdot2=\frac{10}{11}\)
\(\frac{1}{x+2}=1-\frac{10}{11}\)
\(\frac{1}{x+2}=\frac{1}{11}\)
\(\Rightarrow x+2=11\)
\(\Rightarrow x=11-2=9\)
vậy x=9
ko viết đề bài nha
\(5^{2x-3}-2.5^2=5^2.3\)
\(\Rightarrow5^{2x-3}=5^2.3+5^2.2\)
\(\Rightarrow5^{2x-3}=5^2.\left(2+3\right)=5^3\)
\(\Rightarrow2x-3=3\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=3\)
\(5^{2x-3}\)-2.\(5^2\)= \(5^2\).3
\(5^{2x-3}\) _2.25=25.3
\(5^{2x-3_{ }}\)_50=75
\(5^{2x-3}\)= 75_50
\(5^{2x-3}\)= \(5^2\)
suy ra : 2x-3=2
2x=2.3
2x=6
x=6:2
x=3
Làm từng bài nha bạn
\(a)\) \(\left|x\right|=3,5\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=3,5\\x=-3,5\end{cases}}\)
Vậy \(x=3,5\) hoặc \(x=-3,5\)
Chúc bạn học tốt ~
5x-11 - 2.52 = 3.53
5x-11 - 2.25 = 3.125
5x-11 - 50 = 375
5x-11 = 375 + 50
5x-11 = 425
.....
=>x thuộc rỗng nhé
5x-11-2.25=3.125
5x-11-50=375
5x-11=375+50
5x-11=425
đề sai rồi vì 5 mũ mấy cũng ko bằng 425 đâu