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a, Ta có: \(142-\left(12x+30\right)=10^{5-3}\)
\(=>142-12x-30=10^2\)
\(112-12x=100\)
\(=>12x=112-100=12=>x=1\)
Vậy số cần tìm là 1;
b, Ta có: \(=>\left(5x+3^4\right)=6^9:6^8.3^4\)
\(=>5x+3^4=6.3^4=>5x=6.3^4-3^4\)
\(=>5x=5.3^4=>x=3^4=81\)
Vậy x=81;
CHÚC BẠN HỌC TỐT.......
Ta có: \(71.2-6(2x+5)=10^5:10^3\)
\(\Rightarrow142-12x-30=10^2\)
\(\Rightarrow142-30-10^2=12x\)
\(\Rightarrow142-30-100=12x\)
\(\Rightarrow12=12x\)
\(\Rightarrow x=\dfrac{12}{12}\)
\(\Rightarrow x=1\)
Vậy \(x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2^x=2\Rightarrow x=1\)
\(2^{2x+2}=8^2\Rightarrow2^{2x+2}=2^6\Rightarrow2x+2=6\)\(2x=6-2=3\Rightarrow x=3:2=\frac{3}{2}\)
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Giải:
\(\left(2x-5\right)^3=8\)
\(\Leftrightarrow\left(2x-5\right)^3=2^3\)
Vì \(3=3\)
Nên \(2x-5=2\)
\(\Leftrightarrow2x=2+5\)
\(\Leftrightarrow2x=7\)
\(\Leftrightarrow x=\dfrac{7}{2}=3,5\)
Vậy \(x=3,5\).
Chúc bạn học tốt!
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c/ 2x - 1 = \(5^{98}:5^{96}\)
2x - 1 = \(5^2\) = 25
2x = 25 + 1 = 26
x = 26 : 2
x = 13
d/ 7x + 3 = \(3^5.2^3.9\)
7x + 3 = \(3^5.3^2.8=3^7.8=2187.8\)
7x + 3 = \(17496\)
7x = 17496 - 3 = 17493
x = 17493 : 7
x = 2499
e/\(2^{2x+6}=1\)
\(2^{2x+6}=2^0\)
2x + 6 = 0
2x = 0 - 6 = - 6
x = - 6 : 2
x = - 3
j/ \(2^x=8\)
\(2^x=2^3\)
x = 3
g/ \(2^x:2^3=16\)
\(2^{x-3}=2^4\)
x - 3 = 4
x = 4 + 3
x = 7
h/ \(2^x+2^{x+1}+2^{x+2}=56\)
\(2^x\left(1+2+2^2\right)\) = 56
\(2^x.7=56\)
\(2^x=56:7\)
\(2^x=8\)
\(2^x=2^3\)
x = 3
Bài a, b thiên phong giải r, mk chỉ làm những bài còn lại thôi. Chúc bạn học tốt!!!
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\(a)2x^2-98=0\)
\(2x^2=0+98\)
\(2x^2=98\)
\(x^2=98:2\)
\(x^2=49\)
\(\rightarrow x^2=7^2\)
\(\rightarrow x=7\)
Vậy x = 7
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=1+5+5^2+..+5^{49}+5^{50}\)
\(5A=5+5^2+5^3+...+5^{50}+5^{51}\)
\(5A-A=\left(5+5^2+5^3+...+5^{51}\right)-\left(1+5+5^2+...+5^{50}\right)\)
\(4A=\left(5-5\right)+\left(5^2-5^2\right)+...+\left(5^{50}+5^{50}\right)+5^{51}-1\)
\(4A=0+0+...+0+5^{51}-1\)
\(4A=5^{51}-1\)
\(A=\frac{5^{51}-1}{4}\)
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Ta có " (x - 5)7 = (x - 5)4
=> (x - 5)7 - (x - 5)4 = 0
<=> (x - 5)4[(x - 5)3 - 1] = 0
\(\Leftrightarrow\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^3-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\\left(x-5\right)^3=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x-5=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=6\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 71.2-6(2x+5)=105:103
<=> 71.2-6(2x+5)=102
142-6(2x+5)=100
6(2x+5)=142-100
6(2x+5)=42
2x+5=7
2x=7-5
2x=2
x=2:2
x=1
\(\left(2x-5\right)^3=8\)
\(\left(2x-5\right)^3=2^3\)
\(\Rightarrow2x-5=2\)
\(2x=7\)
\(x=\frac{7}{2}\)
Vây ...
\(\left(2x-5\right)^3=8\)
\(\Leftrightarrow2x-5=2\)
\(2x=2+5\)
\(2x=7\)
\(x=7:2\)
\(x=3,5\)
@Bảo
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