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a; (2y+1).7^4=3.7^5
=> (2y+1).2401=50421
=> 2y+1=21
=> 2y=20
=> y= 10 . Vậy y= 10
b; 4.2y+1-3=125
=> 4. 2y.2-3=125
=> 4.2y.2=128
=> 2y.8=128
=> 2y=16
=> 2y=2^4
=> y=4 .Vậy y=4
c; 15-2.|y-3|=-5
=> 2.|y-3|=10
=> y-3=5
=> y=8 . Vậy y=8
d; 2y+5 chia hết cho y+1
=> 2(y+1)+3 chia hết cho y+1
=> 3 chia hết cho y+1
=> y+1E{ 1;3;-1;-3}
=> yE{ 0; 2;-2;-4} .Vậy y= 0;2;-2;-4
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1.
a. 9 < \(^{3^y}\)< 81
9=3\(^{^2}\)
81=3\(^{^4}\)
\(\Rightarrow\)y=3 vì \(3^2< 3^3< 3^4\)
b.25\(\le\)5\(^{^y}\)\(\le\)125
25=5\(^{^2}\)
125=5\(^{^3}\)
\(\Rightarrow\)y=2 và 3 vì 5\(^{^2}\)\(\le\)5\(^{^2}\)(5\(^{^3}\))\(\le\)5\(^{^3}\)
c.
16\(\ge\)4\(^{^y}\)\(\ge\)1024
16=4\(^{^2}\)
1024=4\(^{^5}\)
\(\Rightarrow\)y=2,3,4,5 vì 4\(^{^2}\)\(\ge\)4\(^{^2\left(4^3;4^4;4^5\right)}\)
\(^{^2}\)
\(\Rightarrow\)
2.
a.3x\(^{2^y}\)=48
\(^{2^y}\)=48:3
\(^{2^y}\)=16
\(\Rightarrow\)y=4 vì \(^{2^4}\)=16
b.5x\(y^7\)=640
\(y^7\)=640:5
\(y^7\)=128
\(\Rightarrow\)y=2 vì \(2^7\)=128
c.\(y^{100}\)=\(y^2\)
\(\Rightarrow\)y=1 vì:
\(1^{100}\)=1
\(1^2\)=1
d.(y-3)\(^{^5}\)=243
\(\Rightarrow\)y-3=3 vì 3\(^{^5}\)=243
y=3+3
y=6
e.(2y+1)\(^{^3}\)=125
\(\Rightarrow\)2y+1=5 vì 5\(^{^3}\)=125
2y=5-1=4
y=4:2=2
i.2\(^{^{y+3}}\)+2\(^{^y}\)=288
2\(^{^y}\).2\(^{^3}\)+2\(^{^y}\)=288
2\(^{^y}\).2\(^{^3}\)=288-2\(^{^y}\)
2\(^{^y}\).8=288-2\(^{^y}\)
8=(288-2\(^{^y}\)):2\(^{^y}\)
8=288:2\(^{^y}\)-2\(^{^y}\):2\(^{^y}\)
8=288:2\(^{^y}\)-1
288:2\(^{^y}\)=8+1=9
2\(^{^y}\)=288:9=32
\(\Rightarrow\)y=5 vì 2\(^{^5}\)=32
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Phần trong ngoặc.........phân tích cơ số ra thừa số nguyên tố:
\(125^3\cdot7^5-175^5:5=\left(5^3\right)^3\cdot7^5-\left(5^2\cdot7\right)^5:5=5^9\cdot7^5-5^9\cdot7^5=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,4^{2x-6}=1\)\(\Leftrightarrow2x-6=0\Leftrightarrow2x=6\Rightarrow x=3\)
\(b,2^{2x-1}=16\Rightarrow2^{2x-1}=2^4\)
\(\Rightarrow2x-1=4\Rightarrow2x=5\Rightarrow x=\frac{5}{2}\)
\(c,5< 5^x< 125\Rightarrow5^1< 5^x< 5^3\)\(\Rightarrow1< x< 3\)
\(d,5^{x+1}=125\Rightarrow5^{x+1}=5^3\Rightarrow x+1=3\Rightarrow x=2\)
\(a,4^{2x-6}=1\)
\(4^{2x-6}=4^0\)
\(\Rightarrow2x-6=0\)
\(\Rightarrow2x=6\Leftrightarrow x=3\)
\(b,2^{x-1}=16\)
\(2^{x-1}=2^4\)
\(\Rightarrow x-1=4\)
\(\Rightarrow x=5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
125(28+72)-25(3^2.4+64)
=125.100-25(9.4+64)
=125.100-25.(36+64)
=125.100-25.100
=12500-2500
=10000
5x=125y-2
=>5x=53y-6
=>x=3y-6=>4x=12y-24 (1)
và 22y-5=16x+1
=>22y-5=24x+4
=>2y-5=4x+4 =>2y-9=4x (2)
Lấy (2) trừ đi (1) ta có:
4x-4x=2y-9-12y+24
=>-10y+15=0
-10y=-15
y=1,5
=>x=3.1,5-6-1,5
=>x+y=1,5-1,5=0