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

Bài 2 :

a) \(3^2.4^3-3^2+333=3^2\left(4^3-1\right)+9.37=9.63+9.37=9\left(63+37\right)=9.100=900\)

b) \(5.4^3+24.5+41^0=20.4^2+20.6+1=20\left(16+6\right)+1=20.22+1=441\)

c) \(2^3.4^2+3^2.5-40.1^{2023}=8.16+9.5-40.1=128+45-40=128+5=133\)

A = |\(x\) + 5| + 2023

|\(x\) + 5| ≥ 0 ⇒| \(x\) + 5| + 2023 ≥ 2023⇒ A(min) = 2023 xảy ra khi \(x\) = -5

B = (\(x+2\))² - 2023

(\(x\) + 2)2 ≥ 0 ⇒ (\(x\) + 2)2 ≥ - 2023 ⇒ A(min) = -2023  xảy ra khi \(x\) = -2

C = \(x^2\) - 6\(x\) + 20

C = (\(x^2\) - 3\(x\)) - ( 3\(x\) - 9) + 11

C = \(x\)(\(x-3\)) - 3(\(x\) -3) + 11

C = (\(x-3\))(\(x\)-3) + 11

C = (\(x-3\))² + 11

(\(x\) -3)² ≥ 0 ⇒ (\(x\) - 3)² + 11 ≥ 11 vậy C(min) = 11 xảy ra khi \(x=3\)

D = \(x^2\) + 10\(x\) - 25

D = \(x^2\) + 5\(x\) + 5\(x\) + 25 - 55

D = (\(x^2\) + 5\(x\)) + (5\(x\) + 25) - 50

D = \(x\)(\(x\) + 5) + 5(\(x\) + 5)  - 50

D = (\(x\) +5)(\(x\) + 5) - 50

D = ( \(x\) + 5)² - 50

(\(x+5\))² ≥ 0 ⇒ (\(x\) + 5)² - 50 ≥ -50 ⇒ D(min) = -50 xảy ra khi \(x\) = -5

`(2^x+1)^2 =25`

`=> (2^x+1)^2 = (+-5)^2`

\(\Rightarrow\left[{}\begin{matrix}2^x+1=5\\2^x+1=-5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2^x=4\\2^x=-6\end{matrix}\right.\)

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

\(\left(x+6\right)\left(5^x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+6=0\\5^x-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-6\\5^x=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=0\end{matrix}\right.\)

\(\left(x-3\right)^{2023}=x-3\)

\(\Rightarrow\left(x-3\right)^{2023}-\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left[\left(x-3\right)^{2022}-1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\\left(x-3\right)^{2022}-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\\left(x-3\right)^{2022}=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x-3=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

`#3107.101107`

1.

`(2^x + 1)^2 = 25`

`=> (2^x + 1)^2 = (+-5)^2`

`=>`\(\left[{}\begin{matrix}2^x+1=5\\2^x+1=-5\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2^x=4\\2^x=-6\left(\text{vô lý}\right)\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=2\\x\in\varnothing\end{matrix}\right.\)

Vậy, `x =2.`

2.

`(x + 6)(5x - 1) = 0`

`=>`\(\left[{}\begin{matrix}x+6=0\\5x-1=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=-6\\5x=1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=-6\\x=\dfrac{1}{5}\end{matrix}\right.\)

Vậy, `x \in {-6; 1/5}`

3.

`2*3^(x + 3) + 3^(2 + x) = 891`

`=> 2* 3^x * 3^3 + 3^2 * 3^x = 891`

`=> 54*3^x + 9*3^x = 891`

`=> 3^x * (54 + 9) = 891`

`=> 3^x * 63 = 891`

`=> 3^x = 891 \div 63`

`=> 3^x = 891/63`

Bạn xem lại đề.

4.

`(x - 3)^2023 = x - 3`

`=> (x - 3)^2023 - (x - 3) = 0`

`=> (x - 3) * [ (x - 3)^2022 - 1] = 0`

`=>`\(\left[{}\begin{matrix}x-3=0\\\left(x-3\right)^{2022}-1=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=3\\\left(x-3\right)^{2022}=1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=3\\\left(x-3\right)^{2022}=\left(\pm1\right)^{2022}\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=3\\x-3=1\\x-3=-1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=3\\x=4\\x=2\end{matrix}\right.\)

Vậy, `x \in {2; 3; 4}.`

a) \(\dfrac{10^{12}+5^{11}.2^9-5^{13}.2^8}{4.5^5.10^6}\)

\(=\dfrac{2^{12}.5^{12}+5^{11}.2^9-5^{13}.2^8}{2^2.5^5.2^6.5^6}\)

\(=\dfrac{2^{12}.5^{12}+5^{11}.2^9-5^{13}.2^8}{2^8.5^{11}}\)

\(=\dfrac{\left(2^8.5^{11}\right)\left(2^4.5+2-5^2\right)}{2^8.5^{11}}\)

\(=2^4.5+2-5^2\)

\(=57\)

b) \(\dfrac{\left[5\left(x-y\right)^4-3\left(x-y\right)^3+4\left(x-y\right)^2\right]}{\left(y-x\right)^2}\)

\(=\dfrac{\left(x-y\right)^2\left[5\left(x-y\right)^2-3\left(x-y\right)+4\right]}{\left(y-x\right)^2}\)

\(=\dfrac{\left(x^2+y^2-2xy\right)\left[5\left(x-y\right)^2-3\left(x-y\right)+4\right]}{\left(y^2+x^2-2xy\right)}\)

\(=5\left(x-y\right)^2-3\left(x-y\right)+4\)

c) \(\dfrac{\left(x+y\right)^5-2\left(x+y\right)^4+3\left(x+y\right)^3}{-5\left(x+y\right)^3}\)

\(=\dfrac{\left(x+y\right)^3\left[5\left(x+y\right)^2-2\left(x+y\right)+3\right]}{-5\left(x+y\right)^3}\)

\(=\dfrac{5\left(x+y\right)^2-2\left(x+y\right)+3}{-5}\)

Bài 1 :

\(M=\dfrac{30-2^{20}}{2^{18}}=\dfrac{2.15-2^{20}}{2^{18}}=\dfrac{15}{2^{17}}-2^2=\dfrac{15}{2^{17}}-4< 0\left(\dfrac{15}{2^{17}}< 1\right)\)

\(N=\dfrac{3^5}{1^{2021}+2^3}=\dfrac{3^5}{9}=\dfrac{3^5}{3^2}=3^3=27\)

\(\Rightarrow M< N\)

Bài 3 :

a) \(t^2+5t-8\) khi \(t=2\)

\(=5^2+2.5-8\)

\(=25+10-8\)

\(=27\)

b) \(\left(a+b\right)^2-\left(b-a\right)^3+2021\left(1\right)\)

\(\left\{{}\begin{matrix}a=5\\b=a+1=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=11\\b-a=1\end{matrix}\right.\)

\(\left(1\right)=11^2-1^3+2021=121-1+2021=2141\)

c) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)

\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)

\(\left(1\right)=1^3=1\)

Bài 2 :

a) \(...=2^3\left(19-14\right)+1=8.5+1=41\)

b) \(...=100-\left[60:\left(5^2-15\right)\right]=100-\left[60:10\right]=100-6=94\)

c) \(...=160:\left[17+\left(9.5-\left(14+2^3\right)\right)\right]=160:\left[17+\left(45-22\right)\right]=160:\left[17+23\right]=160:40=4\)

d) \(...=798+100\left[16-2\left(25-22\right)\right]=798+100\left[16-2.3\right]=798+100.10=798+1000=1798\)

bạn có thể check lại đề bài câu a được không ạ

là x^2 = x^5 á bạn

a) A=x^3 + 3x^2*5 + 3x*5^2 + 5^3

       =(x+5)^3

Thay x = -10 vào biểu thức A ta được:

A  = (-10+5)^3

    =(-5)^3

    =-75

Làm tương tự nhé

bài vách ngọc ngà và bài cà phê ko đường