P=33.3−1 + 2−3.24 /10−3:10−2+(0,1)0

a: Ta có: 10−{[(x:3+17):10+3·24]:10}=5

⇔[(x:3+17):10+48]:10=5

⇔(x:3+17):10+48=50

⇔(x:3+17):10=2

⇔x:3+17=20

⇔x:3=3

\(\Leftrightarrow\) X=3.3

\(\Leftrightarrow\) X= 9

Vậy x= 9 

Hoktot~

a: Ta có: \(81^{125}=3^{500}\)

\(27^{130}=3^{390}\)

mà 500>390

nên \(81^{125}>27^{130}\)

a: Ta có: \(10-\left\{\left[\left(x:3+17\right):10+3\cdot2^4\right]:10\right\}=5\)

\(\Leftrightarrow\left[\left(x:3+17\right):10+48\right]:10=5\)

\(\Leftrightarrow\left(x:3+17\right):10+48=50\)

\(\Leftrightarrow\left(x:3+17\right):10=2\)

\(\Leftrightarrow x:3+17=20\)

\(\Leftrightarrow x:3=3\)

hay x=9

b: Ta có: \(2448:\left[119-\left(x-6\right)\right]=24\)

\(\Leftrightarrow119-\left(x-6\right)=102\)

\(\Leftrightarrow x-6=17\)

hay x=23

c: Ta có: \(165-\left(35:x+3\right)\cdot19=13\)

\(\Leftrightarrow\left(35:x+3\right)\cdot19=152\)

\(\Leftrightarrow\left(35:x+3\right)=8\)

\(\Leftrightarrow35:x=5\)

hay x=7

Lời giải:

$4^{30}=(4^3)^{10}=64^{10}> 48^{10}=(2.24)^{10}=2^{10}.24^{10}> 3.24^{10}$

4^30=2^30*2^30

=2^30*4^15

3*24^10=3*3^10*8^10=3^11*2^30

mà 4^30>3^11

nên 2^30+3^30+4^30>3*24^10

Ta có: 4^30=2^30.2^30=2^30.4^15

3.24^10=3.(3.2^3)^10=2^30.3^11

Ta thấy: 3^11<3^15<4^15 => 4^15>3^11

Vì 4^15>3^11 nên 2^30.4^15>2^30.3^11

=>2^30+3^30+4^30>3.24^10

Bài 3:

a: \(S=1+5^2+5^4+\cdots+5^{200}\)

=>25S=\(5^2+5^4+5^6+\cdots+5^{202}\)

=>25S-S=\(5^2+5^4+\cdots+5^{202}-1-5^2-\cdots-5^{200}\)

=>24S=\(5^{202}-1\)

=>\(S=\frac{5^{202}-1}{24}\)

b: \(4^{30}=\left(2^2\right)^{30}=2^{60}=2^{30}\cdot2^{30}=8^{10}\cdot4^{15}\)

\(3\cdot24^{10}=3\cdot3^{10}\cdot8^{10}=8^{10}\cdot3^{11}\)

mà \(4^{15}>3^{11}\)

nên \(4^{30}>3\cdot24^{10}\)

=>\(2^{30}+3^{30}+4^{30}>3\cdot24^{10}\)

Bài 2:

a: |2x-3|>5

=>\(\left[\begin{array}{l}2x-3>5\\ 2x-3<-5\end{array}\right.\Rightarrow\left[\begin{array}{l}2x>8\\ 2x<-2\end{array}\right.\Rightarrow\left[\begin{array}{l}x>4\\ x<-1\end{array}\right.\)

c: |3x-1|<=7

=>-7<=3x-1<=7

=>-6<=3x<=8

=>\(-2\le x\le\frac83\)

d: \(\left|3x-5\right|+\left|2x+3\right|=7\) (1)

TH1: \(x<-\frac32\)

=>2x+3<0; 3x-5<0

(1) sẽ trở thành: -2x-3-3x+5=7

=>-5x+2=7

=>-5x=5

=>x=-1(loại)

TH2: -3/2<=x<5/3

=>2x+3>=0; 3x-5<0

(1) sẽ trở thành: 2x+3-3x+5=7

=>-x+8=7

=>-x=-1

=>x=-1(nhận)

TH3: x>=5/3

=>2x+3>0; 3x-5>=0

(1) sẽ trở thành: 2x+3+3x-5=7

=>5x-2=7

=>5x=9

=>x=9/5(nhận)

\(3\times24^{10}\)

\(=3\times\left(2^3\times3\right)^{10}\)

\(=3\times3^{10}\times\left(2^3\right)^{10}\)

\(=3^{11}\times2^{30}\)

\(=3^{11}\times\left(2^2\right)^{15}\)

\(=3^{11}\times4^{15}\)

Vì \(3^{11}\)<\(4^{15}\left(3;4;11;15\inℕ\right)\)

Nên \(3^{11}\times4^{15}\)< \(4^{15}\times4^{15}=4^{30}\)

Do đó : \(3\times24^{10}\)< \(4^{30}\)

Vậy \(2^{30}+3^{30}+4^{30}\)> \(3\times24^{10}\)