Tính: 2x + 2x+1= 96
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Xét cấp số cộng 1, 6, 11, ..., 96.
Ta có: 96 = 1 + 5(n − 1) ⇒ n = 20
Suy ra
Và 2x.20 + 970 = 1010
Từ đó x = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
2x + 2x + 1 = 96
2x + 2x . 2 = 96
2x . (1 + 2) = 96
2x . 3 = 96
2x = 96 : 3
2x = 32
2x = 25
x = 5
![](https://rs.olm.vn/images/avt/0.png?1311)
2x - 4(5 - x) = 2x + 16
=> 4(5 - x) = 2x - (2x + 16)
=> 4(5 - x) = 2x - 2x - 16
=> 4(5 - x) = -16
=> 5 - x = -4
=> x = 9
Vậy...
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a)16.4x=48
\(4^2.4^x=4^{8^{ }}\)
\(4^x=4^{8^{ }}:4^2\)
\(4^x=4^{6^{ }}\)
\(x=6\)
b)(X-2)(X-5)=0
\(\Rightarrow x-2=0\rightarrow x=2\)
\(x-5=0\rightarrow x=5\)
Vậy x∈ {2;5}
c)2x+2x+1=96
\(2^x.1+2^{x^{ }}.2\)
\(2^x.\left(1+2\right)=96\)
\(2^x.3=96\)
\(2^x=96:3\)
\(2^x=32\)
\(2^x=2^5\)
\(x=5\)
\(Zzz\) 😪
chị giúp em đc mấy bài trong trang của em mới đăng đc ko ạ?
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a: =>12x-64=32
=>12x=96
=>x=8
b: =>x-1=5
=>x=6
c: =>2^x*3=96
=>2^x=32
=>x=5
![](https://rs.olm.vn/images/avt/0.png?1311)
9) Ta có: \(\dfrac{2x+5}{x+3}+1=\dfrac{4}{x^2+2x-3}-\dfrac{3x-1}{1-x}\)
\(\Leftrightarrow\left(2x+5\right)\left(x-1\right)+x^2+2x-3=4+\left(3x-1\right)\left(x+3\right)\)
\(\Leftrightarrow2x^2-2x+5x-5+x^2+2x-3-4-3x^2-10x+x+3=0\)
\(\Leftrightarrow-4x=9\)
hay \(x=-\dfrac{9}{4}\)
10) Ta có: \(\dfrac{x-1}{x+3}-\dfrac{x}{x-3}=\dfrac{7x-3}{9-x^2}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3-7x}{\left(x-3\right)\left(x+3\right)}\)
Suy ra: \(x^2-4x+3-x^2-3x-3+7x=0\)
\(\Leftrightarrow0x=0\)(luôn đúng)
Vậy: S={x|\(x\notin\left\{3;-3\right\}\)}
11) Ta có: \(\dfrac{5+9x}{x^2-16}=\dfrac{2x-1}{x+4}+\dfrac{3x-1}{x-4}\)
\(\Leftrightarrow\dfrac{\left(2x-1\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}+\dfrac{\left(3x-1\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{9x+5}{\left(x-4\right)\left(x+5\right)}\)
Suy ra: \(2x^2-9x+4+3x^2+12x-x-4-9x-5=0\)
\(\Leftrightarrow5x^2-7x=0\)
\(\Leftrightarrow x\left(5x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{7}{5}\end{matrix}\right.\)
12) Ta có: \(\dfrac{2x}{2x-1}+\dfrac{x}{2x+1}=1+\dfrac{4}{\left(2x-1\right)\left(2x+1\right)}\)
\(\Leftrightarrow\dfrac{2x\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}+\dfrac{x\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}=\dfrac{4x^2-1+4}{\left(2x-1\right)\left(2x+1\right)}\)
Suy ra: \(4x^2+2x+2x^2-x-4x^2-3=0\)
\(\Leftrightarrow2x^2+x-3=0\)
\(\Leftrightarrow2x^2+3x-2x-3=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2^{x+2}.2^x=96\)
\(2^x.2^2-2^x=96\)
\(2^x.\left(2^2-1\right)=96\)
\(2^x.\left(4-1\right)=96\)
\(2^x.3=96\)
\(2^x=96:3\)
\(2^x=32=2^5\)
\(\Rightarrow x=5\)
\(#WendyDang\)
( Mong bạn lần sau để đúng khối, vì lớp 4 chưa học số mũ đâu nhé. )
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{8}=\dfrac{2x+3y-3z}{2\cdot15+3\cdot10-3\cdot8}=\dfrac{96}{36}=\dfrac{8}{3}\)
Do đó: \(\left\{{}\begin{matrix}x=24\\y=\dfrac{80}{3}\\z=\dfrac{64}{3}\end{matrix}\right.\)
2x + 2x + 1 = 96
<=> 2x + 2x.2 = 96
<=> 3.2x = 96
<=> 2x = 32
<=> 2x = 25
<=> x = 5
2x + 2x + 1 = 96
<=> 2x + 2x.2 = 96
<=> 3.2x = 96
<=> 2x = 32
<=> 2x = 25
<=> x = 5