\(M=343-8x^3-64+8x^3=279\\ N=8x^3-1-1+8x^3=16x^3=16\cdot1000=16000\)

\(M=343-8x^3-\left(64-8x^3\right)=343-64=279\)

(biểu thức này ko phụ thuộc vào biến x nên ko cần thay)

\(N=8x^3-1-1+8x^3=16x^3\)

Thay \(x=10\)

\(N=16\cdot10^3=16\cdot1000=16000\)

`|2x+1|-3=x+4`

`<=>|2x+1|=x+4+3=x+7(x>=-7)`

`**2x+1=x+7`

`<=>x=7-1=6(tm)`

`**2x+1=-x-7`

`<=>3x=-6`

`<=>x=-2(tm)`

`|3x-5|=1-3x(x<=1/3)`

`**3x-5=1-3x`

`<=>6x=6`

`<=>x=1(l)`

`**3x-5=3x-1`

`<=>-5=-1` vô lý

`|2x+2|+|x-1|=10`

Nếu `x>=1`

`pt<=>2x+2+x-1=10`

`<=>3x+1=10`

`<=>3x=9`

`<=>x=3(tm)`

Nếu `x<=-1`

`pt<=>-2x-2+1-x=10`

`<=>-1-3x=10`

`<=>-11=3x`

`<=>x=-11/3(tm)`

Nếu `-1<=x<=1`

`pt<=>2x+2+1-x=10`

`<=>x+3=10`

`<=>x=7(l)`

Vậy `S={3,-11/3}`

d) 

+) Với \(x< -4\), PT \(\Rightarrow3-x-x-4-2x-6=10\) \(\Leftrightarrow x=-\dfrac{17}{4}\)  (Nhận)

+) Với \(-4\le x\le-3\), PT \(\Rightarrow3-x+x+4-2x-6=10\) \(\Leftrightarrow x=-\dfrac{9}{2}\) (Loại)

+) Với \(-3< x\le3\), PT \(\Rightarrow3-x+x+4+2x+6=10\) \(\Leftrightarrow x=-\dfrac{3}{2}\) (Nhận)

+) Với \(x>3\), PT \(\Rightarrow x+3+x+4+2x+6=10\) \(\Leftrightarrow x=-\dfrac{3}{4}\) (Loại)

  Vậy \(x\in\left\{-\dfrac{3}{2};-\dfrac{17}{4}\right\}\) 

\(3^{200}=9^{100}>4^{100}\\ 5^{200}=25^{100}< 64^{100}=4^{300}\\ 6^{50}=36^{25}>7^{25}\\ 8^{40}=64^{20}>10^{20}\\ 16^{20}=256^{10}>32^{10}\)

tick mik nha!!

3200=9100>41005200=25100<64100=4300650=3625>725840=6420>10201620=25610>3210

a: \(3^{200}=\left(3^2\right)^{100}=9^{100}>4^{100}\)

b: \(5^{200}=\left(5^2\right)^{100}=25^{100}\)

\(4^{300}=\left(4^3\right)^{100}=64^{100}\)

mà 25<64

nên \(5^{200}< 4^{300}\)

c: \(6^{50}=\left(6^2\right)^{25}=36^{25}>7^{25}\)

1: =>\(5^{x-2}-9=2^4-\left(6^2-6^2\right)\)

=>\(5^{x-2}=16+9=25\)

=>x-2=2

=>x=4

2: \(\Leftrightarrow3^x+16=19^6:19^5-3=19-3=16\)

=>3^x=0

=>x=0

3: \(\Leftrightarrow2^x+2^x\cdot16=272\)

=>2^x*17=272

=>2^x=16

=>x=4

4: \(\Leftrightarrow2^{x-1}+3=24-\left(4^2-2^2+1\right)=24-\left(16-4+1\right)\)

=>\(2^{x-1}+3=24-16+4-1=8+4-1=12-1=11\)

=>2^x-1=8

=>x-1=3

=>x=4

\(2,\)

\(a,20-\left[4^2+\left(x-6\right)\right]=90\)

\(\Rightarrow20-16-x+6=90\)

\(\Rightarrow10-x=90\)

\(\Rightarrow x=-80\)

Vậy: \(x=-80\)

\(b,\left(x+3\right)\left(2x-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+3=0\\2x-4=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

Vậy: \(x\in\left\{-3;2\right\}\)

\(c,1000:\left[30+\left(2^x-6\right)\right]=3^2+4^2\left(x\in N\right)\)

\(\Rightarrow1000:\left(30+2^x-6\right)=25\)

\(\Rightarrow24+2^x=40\)

\(\Rightarrow2^x=16\)

\(\Rightarrow x=4\)

Vậy: \(x=4\)

:))))

\(2,\)

\(a,20-\left[42+\left(x-6\right)\right]=90\)

\(\Rightarrow20-42-x+6-90=0\)

\(\Rightarrow x=-106\)

Vậy: \(x=-106\)

\(b,\left(x+3\right)\left(2x-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+3=0\\2x-4=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

Vậy: \(x\in\left\{-3;2\right\}\)

\(c,1000:\left[30+\left(2x-6\right)\right]=32+42\left(x\in N\right)\)

\(\Rightarrow1000:\left(30+2x-6\right)=74\)

\(\Rightarrow1000:\left(24+2x\right)=74\)

\(\Rightarrow24+2x=\dfrac{500}{37}\)

\(\Rightarrow2x=-\dfrac{388}{37}\)

\(\Rightarrow x=-\dfrac{194}{37}\)

Mà \(x\in N\)

\(\Rightarrow x\in\varnothing\)

Vậy: \(x\in\varnothing\)

Ta có:

97 x 32 ⋮ 8

8 ⋮ 8

⇒ 97 x 32 + 8 ⋮ 8

Ta có:

2020 x 30 ⋮ 10

8 x 5 = 40 ⋮ 10

⇒ 2020 x 30 + 8 x 5 ⋮ 10