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\(9^{x+1}+5.3^{2x}=324\)
\(9^x.9+5.\left(3^2\right)^x=324\)
\(9^x.9+5.9^x=324\)
\(9^x.\left(5+9\right)=324\)
\(9^x.14=324\)
\(9^x=\frac{324}{14}\)
\(\Rightarrow x\in\varnothing\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 5x + 1 - 2.5x = 75
<=> 5x.5 - 2.5x = 75
<=> 5x.3 = 75
<=> 5x = 25
<=> 5x = 52
<=> x = 2
Vậy x = 2
b) 9x + 1 - 5.32x = 324
<=> (32)x + 1 - 5.32x = 324
<=> 32x + 2 - 5.32x = 324
<=> 32x.32 - 5.32x = 324
<=> 32x . 4 = 324
<=> 32x = 81
<=> 32x = 34
<=> 2x = 4
<=> x = 2
Vậy x = 2
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(9^{x+1}-5.3^{2x}=324=>9^x.9-5.\left(3^2\right)^x=324=>\left(3^2\right)^x.9-5.\left(3^2\right)^x=324\)
\(=>3^{2x}.\left(9-5\right)=324=>3^{2x}=\frac{324}{4}=81=3^4=>2x=4=>x=2\)
vậy x=2
tick nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
\(9^{x+1}-5.3^{2x}=324\)
\(\Rightarrow9^x.9-5.9^x=324\)
\(\Rightarrow9^x\left(9-5\right)=324\)
\(\Rightarrow9^x.4=324\)
\(\Rightarrow9^x=81\)
\(\Rightarrow9^x=9^2\)
vậy x=2
\(\Rightarrow x=2\)
\(9^{x+1}-5.3^{2x}=324\)
\(9^x.9-5.9^x=324\)
\(9^x.\left(9-5\right)=324\)
\(9^x.4=324\)
\(9^x=81=9^2\)
\(\Rightarrow x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(5^{x+1}-2.5^x=375\)
\(\Rightarrow5^x\left(5-2\right)=375\)
\(\Rightarrow5^x.3=375\)
\(\Rightarrow5^x=125=5^3\)
\(\Rightarrow x=3\)
b) \(9^{x+1}-5.3^{2x}=324\)
\(\Rightarrow3^{2\left(x+1\right)}-5.3^{2x}=324\)
\(\Rightarrow3^2\left(3^{x+1}-5.3^x\right)=324\)
\(\Rightarrow9.3^x\left(3-5\right)=324\)
\(\Rightarrow3^x.\left(-2\right)=36\)
\(\Rightarrow3^x=-18=3^2.\left(-2\right)\)(vô lí vì 3x không chia hết cho 2)
c) \(\left(1-x\right)^5=32=2^5\)
\(\Rightarrow1-x=2\)
\(\Rightarrow x=-1\)
d) \(3.5^{2x+1}-3.25^x=300\)
\(\Rightarrow3\left(5^{2x}.5-5^{2x}\right)=300\)
\(\Rightarrow5^{2x}\left(5-1\right)=100\)
\(\Rightarrow5^{2x}.4=100\)
\(\Rightarrow5^{2x}=25=5^2\)
\(\Rightarrow2x=2\)
\(\Rightarrow x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta cố bdt \(|a|+|b|\ge|a+b|\), dễ dàng chứng mình bằng bình phương 2 vế. Dấu = sảy ra <=>IaI.IbI=a.b <=> a.b>=0
áp dụng vào từng câu
a)A=Ix+1I+Ix+2I+Ix+3I+I-x-4I+I-x-5I ( vì Ix+4I=I-x=4I, Ix+5I=I-x-5I
A>=I(x+1)+(-x-5)I+I(x+2)+(-x-4)I +Ix+3I=4+2+Ix+3I=6+Ix+3I>=6
Dấu bằng khi (x+1)(-x-5)>=0;(x+2)(-x-4)>=0;Ix+3I=0 =>x=-3
b) LÀm tương tự MinB=18
Dấu = khi (2x+1)(-2x-11)>=0;(2x+3)(-2x-9)>=0;(2x+5)(-2x-7)>=0 <=>-7/2<=x<=-5/2
Ta có :
9x + 1 - 5 . 32x = 324
(32)x + 1 - 5 . 32x = 324
32x + 2 - 5 . 32x = 324
9 . 32x - 5 . 32x = 324
32x (9 - 5) = 324
⇒ 32x = 324 : 4 = 81 ⇒ x = 2