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giúp e với em cần gấp ạ

\(A=3^2\cdot4^3-3^2+333\)

\(=3^2\cdot63+333\)

\(=3^2\cdot\left(63+37\right)\)

\(=30^2\)

\(B=5\cdot3^2+4\cdot3^2=3^2\cdot9=9^2\)

d: Ta có: \(D=5^3+6^3+59\)

\(=125+216+59\)

\(=400=20^2\)

Ông ơi

thằng  trung anh ơiiiiiiiiiiiii

c: \(C=5\cdot4^3+2^4\cdot5+41\)

\(=5\cdot64+16\cdot5+41\)

\(=320+80+41\)

\(=441=21^2\)

\(B=5.3^2+4.3^2=3^2\left(5+4\right)=9.9=9^2\)

\(C=5^3+6^3+59=125+216+59=400=20^2\)

\(D=5.4^3+2^4.5+41=320+80+41=441=21^2\)

b: \(5\cdot3^2+4\cdot3^2=3^2\cdot9=3^4\)

c: \(5^3+6^3+59=20^2\)

\(5\cdot4^3+2^4\cdot5+41=21^2\)

a: \(2^5\cdot5^2-8^2-7=27^2\)

b: \(2^3+3^2\cdot3^2-40=7^2\)

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

a) \(2^3.3^2=8.9=72\)

b) \(5^{10}:5^7=5^2=25\)

c) \(2^6:2=2^5=32\)

d) \(7^4:7^4=7^0=1\)

e) \(9^5:9^5=9^0=1\)

a) 23.32=8.9=7223.32=8.9=72

b) 510:57=52=25510:57=52=25

c) 5

32

d) $1$

e) $1$

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

Ta có: \(1^3+2^3+3^3+...+n^3=\left(1+2+3+...+n\right)^2\)

a, \(1^3+2^3+3^3+4^3=\left(1+2+3+4\right)^2=10^2\)

b, \(1^3+2^3+3^3+4^3+5^3=\left(1+2+3+4+5\right)^2=15^2\)

c,

\(3^6:3^2+2^3\cdot2^2-2\\ =3^{6-2}+2^{3+2-1}\\ =3^4+2^4\)

\(a,1^3+2^3+3^3+4^3.\)

\(=\left(1+2+3+4\right)^2.\)

\(=10^2=100.\)

\(b,1^3+2^3+3^3+4^3+5^3.\)

\(=\left(1+2+3+4+5\right)^2.\)

\(=15^2=225.\)

(2 phần a, b thì áp dụng công thức: \(1^3+2^3+3^3+...+n^3=\left(1+2+3+...+n\right)^2.\))

\(c,3^6:3^2+2^3.2^2-2.\)

\(=3^{6-2}+2^{3+2}+2.\)

\(=3^4+2^5-2.\)(đến đoạn này mới thấy hình như đề có vấn đề).