\(\left(x-5\right)^{2022}=\left(x-5\right)^{2024}\\ \Rightarrow\left(x-5\right)^{2022}-\left(x-5\right)^{2024}=0\\ \Rightarrow\left(x-5\right)^{2022}\left[1-\left(x-5\right)^2\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(x-5\right)^{2022}=0\\1-\left(x-5\right)^2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^2=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\\left(x-5\right)^2=\left(\pm1\right)^2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=6\\x=4\end{matrix}\right.\)

`#3107.101107`

\(\left(x-5\right)^{2022}=\left(x-5\right)^{2024}\)

\(\Rightarrow\left(x-5\right)^{2022}-\left(x-5\right)^{2024}=0\\ \Rightarrow\left(x-5\right)^{2022}\cdot\left[1-\left(x-5\right)^2\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(x-5\right)^{2022}=0\\1-\left(x-5\right)^2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^2=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\\left(x-5\right)^2=\left(\pm1\right)^2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x-5=1\\x-5=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=6\\x=4\end{matrix}\right.\)

Vậy, \(x\in\left\{4;5;6\right\}.\)

Bài 9,

6²x73+36x3³=36x73+36x27=36(73+27)=36x100=3600.

197-\([\)6x(5-1)²+2022⁰\(]\):5=197-\([\)6x16+1\(]\):5=197-97:5=197-97/5=888/5.

Bài 10,

21-4x=13

=>4x=21-13=8

=>x=8:4=2.

30:(x-3)+1=4⁵:4³=4²=16

=>30:(x-3)=16-1=15

=>x-3=30:15=2

=>x=2+3=5.

(x-1)³+5x6=38

=>(x-1)³+30=38

=>(x-1)³=38-30=8=2³

=>x-1=2

=>x=3.

Bài 1: 

a) 0²⁰⁰² < 0²⁰²³

b) 2022⁰ = 2023⁰

c) 54⁹ < 55¹⁰

d) ( 4 + 5 )³ > 4² + 5²

đ) 9² - 3² > ( 9 - 3 )²

Bài 2:

a) 3² x 4³ - 3² + 333

= 9 x 64 - 9 + 333

= 576 - 9 + 333

= 567 + 333

= 900

b) 5 x 4³ + 24 x 5 + 41⁰

= 5 x 64 + 24 x 5 + 1

= 5 x ( 64 + 24 ) + 1

= 5 x 88 + 1

= 440 + 1

= 441

c) 2³ x 4² + 3² x 5 - 40 x 1²⁰²³

= 8 x 16 + 9 x 5 - 40 x 1

= 128 + 45 - 40

= 133

Bài 1 :

a) \(0^{2002}=0;0^{2023}=0\Rightarrow0^{2002}=0^{2023}\)

b) \(2022^0=1;2023^0=1\Rightarrow2022^0=2023^0\)

c) \(54^9< 55^9;55^9< 55^{10}\Rightarrow54^9< 55^{10}\)

d) \(\left(4+5\right)^3>\left(4+5\right)^2;\left(4+5\right)^2>4^2+5^2\Rightarrow\left(4+5\right)^3>4^2+5^2\)

đ) \(9^2-3^2=81-9=82;\left(9-3\right)^2=6^2=36\Rightarrow9^2-3^2>\left(9-3\right)^2\)

1. \(6x^3-8=40\\ 6x^3=48\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2

2. \(4x^5+15=47\\ 4x^5=32\\ x^5=8\\ \Rightarrow x\in\varnothing\left(\text{vì }x\in N\right)\)Vậy x ∈ ∅

3. \(2x^3-4=12\\ 2x^3=16\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2

4. \(5x^3-5=0\\ 5x^3=5\\ x^3=1\\ \Rightarrow x=1\)Vậy x = 1

5. \(\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)Vậy \(x\in\left\{5;6\right\}\)

6. \(\left(3x-2\right)^{20}=\left(3x-1\right)^{20}\\ \Rightarrow3x-2=3x-1\\ 3x-3x=2-1\\ 0=1\left(\text{vô lí}\right)\)Vậy x ∈ ∅

7. \(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\\ \left(3x-1\right)^{10}=\left[\left(3x-1\right)^2\right]^{10}\\ \Rightarrow\left(3x-1\right)^2=3x-1\\ \left(3x-1\right)^2-\left(3x-1\right)=0\\ \left(3x-1\right)\left[\left(3x-1\right)-1\right]=0\\ \left(3x-1\right)\left(3x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x-1=0\\3x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=1\\3x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\left(\text{loại vì }x\in N\right)\\x=\frac{2}{3}\left(\text{loại vì }x\in N\right)\end{matrix}\right.\)Vậy x ∈ ∅

8. \(\left(2x-1\right)^{50}=2x-1\\ \left(2x-1\right)^{50}-\left(2x-1\right)=0\\ \left(2x-1\right)\left[\left(2x-1\right)^{49}-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\\left(2x-1\right)^{49}=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=1\\2x-1=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\left(\text{loại vì }x\in N\right)\\x=1\left(t/m\right)\end{matrix}\right.\)Vậy x = 1

9. \(\left(\frac{x}{3}-5\right)^{2000}=\left(\frac{x}{3}-5\right)^{2008}\\ \left(\frac{x}{3}-5\right)^{2008}-\left(\frac{x}{3}-5\right)^{2000}=0\\ \left(\frac{x}{3}-5\right)^{2000}\left[\left(\frac{x}{3}-5\right)^8-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(\frac{x}{3}-5\right)^{2000}=0\\\left(\frac{x}{3}-5\right)^8=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}-5=0\\\frac{x}{3}-5=1\\\frac{x}{3}-5=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}=5\\\frac{x}{3}=6\\\frac{x}{3}=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\cdot3=15\\x=6\cdot3=18\\x=4\cdot3=12\end{matrix}\right.\)Vậy \(x\in\left\{15;18;12\right\}\)

\(1.6x^3-8=40\\ \Leftrightarrow6x^3=48\\ \Leftrightarrow x^3=8\Leftrightarrow x^3=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{2;-2\right\}\)

\(2.4x^3+15=47\) (T nghĩ đề là mũ 3)

\(\Leftrightarrow4x^3=32\Leftrightarrow x^3=8=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{2;-2\right\}\)

Câu 3, 4 tương tự nhé.

a,\(\left(x-1\right)^2+\left(y-3\right)^{10}+\left(z+4\right)^{100}=0\)0(1)

Có \(\hept{\begin{cases}\left(x-1\right)^2\ge0\\\left(y-3\right)^{10}\ge0\\\left(z+4\right)^{100}\ge0\end{cases}}\)(2)

Từ (1) và (2)\(\Rightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y-3\right)^{10}=0\\\left(z+4\right)^{100}=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x-1=0\Rightarrow x=1\\y-3=0\Rightarrow y=3\\z+4=0\Rightarrow z=-4\end{cases}}\)

Em làm tương tự với câu b, không hiểu gì thì hỏi anh

Ta có: \(\frac{2z-4x}{3}=\frac{3x-2y}{4}=\frac{4y-3z}{2}\)

=>\(\frac{6z-12x}{9}=\frac{12x-8y}{16}=\frac{8y-6z}{4}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{6z-12x}{9}=\frac{12x-8y}{16}=\frac{8y-6z}{4}=\frac{6x-12x+12x-8y+8y-6z}{9+16+4}=0\)

=>12x=8y=6z

=>\(\frac{12x}{24}=\frac{8y}{24}=\frac{6z}{24}\)

=>\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)

Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=k\)

=>x=2k; y=3k; z=4k

Vì x;y;z là các số nguyên dương nên k là số nguyên dương

\(200

=>\(200<\left(3k\right)^2+\left(4k\right)^2<450\)

=>\(200<25k^2<450\)

=>\(8

mà k là số nguyên dương

nên k∈{3;4}

TH1: k=3

=>\(\begin{cases}x=2\cdot3=6\\ y=3\cdot3=9\\ z=4\cdot3=12\end{cases}\)

TH2: k=4

=>\(\begin{cases}x=2\cdot4=8\\ y=3\cdot4=12\\ z=4\cdot4=16\end{cases}\)

mà k

\(\left(2x+4\right)^{2024}+\left(\left|3y-9\right|\right)^{2023}=0\) (*) 

Ta có: \(\left(2x+4\right)^{2024}\ge0\forall x\) (vì có số mũ chẵn) (1)

\(\left(\left|3y-9\right|\right)^{2023}\ge0\forall y\) (vì giá trị tuyệt đối luôn ≥0) (2) 

Từ (1) và (2) ta có: 

\(\Rightarrow\left\{{}\begin{matrix}2x+4=0\\3y-9=0\end{matrix}\right.\)

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

Vậy: ... 

Áp dụng tính chất của dãy tỉ số bằng nhau,ta có:
\(\dfrac{x}{y}=\dfrac{y}{z}=\dfrac{z}{x}=\dfrac{x+y+z}{y+z+x}=\dfrac{x+y+z}{x+y+z}=1\)
\(\Rightarrow\left\{{}\begin{matrix}x=y\\y=z\\z=x\end{matrix}\right.\)
Do đó \(\left\{{}\begin{matrix}x-y=0\\y-z=0\\z-x=0\end{matrix}\right.\)
Thay vào biểu thức \(P=\left(x-y\right)^{2022}+\left(y-z\right)^{2023}+\left(x-z-1\right)^{202}\),ta có:
\(P=0^{2022}+0^{2023}+\left(-1\right)^{202}\)
\(=0+0+1\)
\(=1\)

a: \(8x\left(2023+x\right)-8x\left(x+2024\right)=56\)

=>\(8x\left(x+2023-x-2024\right)=56\)

=>-8x=56

=>\(x=\frac{56}{-8}=-7\) (nhận)

b: \(5-2025x=9-2026\left(x-1\right)\)

=>-2025x+5=9-2026x+2026

=>-2025x+5=-2026x+2035

=>x=2035-5=2030(nhận)

c: \(\left(-12\right)^2\cdot x=56+10\cdot13x\)

=>144x=56+130x

=>14x=56

=>x=4(nhận)

d: -(x-32+11)=(21-33-x+7)

=>-(x-21)=(-x-5)

=>-x+21=-x-5

=>21=-5(vô lý)

=>x∈∅

e: \(-2\left(x+6\right)+6\left(x+10\right)=8\)

=>-2x-12+6x+60=8

=>4x+52=8

=>4x=8-52=-44

=>\(x=-\frac{44}{4}=-11\) (nhận)