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1.
a. Ta có: \(A=2^{300}=2^{3.100}=\left(2^3\right)^{100}=8^{100}\)
\(B=3^{200}=3^{2.100}=\left(3^2\right)^{100}=9^{100}\)
Mà \(8^{100}< 9^{100}\)
\(\Rightarrow A< B\)
b. Ta có: \(A=2^{332}< 2^{333}=2^{3.111}=\left(2^3\right)^{111}=8^{111}\)
\(B=3^{223}>3^{222}=3^{2.111}=\left(3^2\right)^{111}=9^{111}\)
Mà \(8^{111}< 9^{111}\)
\(\Rightarrow A< B\)
c. Ta có: \(A=2^{91}=2^{13.7}=\left(2^{13}\right)^7=8192^7\)
\(B=5^{35}=5^{5.7}=\left(5^5\right)^7=3125^7\)
Mà \(8192^7>3125^7\)
\(\Rightarrow A>B\)
Câu 2:
a: =>(x-6)(x-7)=0
=>x=6 hoặc x=7
b: =>\(x^8\left(x^2-25\right)=0\)
\(\Leftrightarrow x^8\left(x-5\right)\left(x+5\right)=0\)
hay \(x\in\left\{0;5;-5\right\}\)
Lời giải:
a. $(x+5)-20=123$
$x+5=123+20=143$
$x=143-5=138$
b.
$-18+(3-x)=25$
$3-x=25+18=43$
$x=3-43=-404
c. $4(x-2)+11=-13$
$4(x-2)=-13-11$
$4(x-2)=-24$
$x-2=-24:4=-6$
$x=-6+2=-4$
d.
$(x+10):(-20)=6$
$x+10=6(-20)=-120$
$x=-120-10=-130$
a)(x+5)-20=123
(x+5)=123+20
x+5=143
x=143-5
x=138
b)-18 + (3-x)=25
(3-x)=25-(-18)
3-X=43
x=3-43
x=-40
c)4.(x-2):(-20)=6
4.(x-2)=6.(-20)
4.(x-2)=-120
x-2=(-120):4
x-2=-30
x=(-30)+2
x=-28
d)
( x + 10 ) : ( - 20 ) = 6
x+10=6.(-20)
x+10=-120
x=(-120)-10
x=-130
1. 11(x-9)=77
x-9=7
x=16
2. 4(x-3)=48
x-3=12
x=15
3.3(x-8)=81
(x-8)=27
x=35
Đề bài : \(\dfrac{97}{20}\times\left(\dfrac{25}{8}-\dfrac{221}{200}\right)< x< \dfrac{91}{10}\times\left(\dfrac{137}{20}+\dfrac{11}{4}\right)\)
\(\dfrac{97}{20}\times\left(\dfrac{25}{8}-\dfrac{221}{200}\right)=\dfrac{97}{20}\times\dfrac{101}{50}=\dfrac{9797}{1000}\)
\(\dfrac{91}{10}\times\left(\dfrac{137}{20}+\dfrac{11}{4}\right)=\dfrac{91}{10}\times\dfrac{48}{5}=\dfrac{2184}{25}=\dfrac{2184\times40}{25\times40}=\dfrac{87360}{1000}\)
\(\Rightarrow\dfrac{9797}{1000}< x< \dfrac{87360}{1000}\)
Vậy \(x\in\left(\dfrac{9797}{1000};\dfrac{87360}{1000}\right)\)