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

Ta có: \(M=1\cdot2+3\cdot4+\cdots+57\cdot58\)

\(=1\left(1+1\right)+3\left(3+1\right)+\cdots+57\left(57+1\right)\)

\(=\left(1^2+3^2+\cdots+57^2\right)+\left(1+3+\cdots+57\right)\)

Đặt \(A=1^2+3^2+\cdots+57^2\)

\(=1^2+2^2+3^2+4^2+\cdots+57^2+58^2-\left(2^2+4^2+\cdots+58^2\right)\)

\(=\frac{58\left(58+1\right)\left(2\cdot58+1\right)}{6}-2^2\left(1^2+2^2+\cdots+29^2\right)\)

\(=\frac{58\cdot59\cdot117}{6}-4\cdot\frac{29\left(29+1\right)\left(2\cdot29+1\right)}{6}\)

\(=29\cdot59\cdot39-4\cdot\frac{29\cdot30\cdot59}{6}=29\cdot59\cdot39-4\cdot29\cdot5\cdot59\)

=32509

Đặt B=1+3+...+57

Số số hạng của dãy số là:

(57-1):2+1=56:2+1=28+1=29(số)

Tổng của dãy số là: \(B=\left(57+1\right)\cdot\frac{29}{2}=58\cdot\frac{29}{2}=29\cdot29=841\)

M=A+B

=32509+841

=33350

\(N=1^2+3^2+\cdots+57^2\)

=>N=A

=32509

M-N

=33350-32509

=841

\(-22x^3-\left(-21x^3+19x^2+23^0\right)-\left(-x^3-18x^2\right)+\left(x^2-23^1\right)\)

\(=-22x^3+21x^3-19x^2-1+x^3+18x^2+x^2-23\)

\(=\left(-22x^3+21x^3+x^3\right)+\left(-19x^2+18x^2+x^2\right)+\left(-1-23\right)\)

\(=0x^3+0x^2-24\)

\(=-24\)

Vậy biểu thức trên có giá trị không phụ thuộc vào biến.

bằng 2¹⁰¹ + 2 trên 3

a)1

b)0

\(A=\frac{x^2+y^2-z^2+2xy}{x^2-y^2+z^2+2xz}\)

       \(=\frac{\left(x^2+2xy+y^2\right)-z^2}{\left(x^2+2xz+z^2\right)-y^2}\)

         \(=\frac{\left(x+y\right)^2-z^2}{\left(x+z\right)^2-y^2}\)

           \(=\frac{\left(x+y+z\right)\left(x+y+z\right)}{\left(x+y+z\right)\left(x-y+z\right)}\)

               \(=\frac{x+y-z}{x-y+z}\)

Ta thay : \(x=0;y=2009;z=2010\) ta được :

\(A=\frac{0+2009-2010}{0-2009+2010}=-\frac{1}{1}=-1\)

Chúc bạn học tốt !!!

\(A=\frac{x^2+y^2-z^2+2xy}{x^2-y^2+z^2+2xz}=\frac{\left(x^2+2xy+y^2\right)-z^2}{\left(x^2+2xz+z^2\right)-y^2}=\frac{\left(x+y\right)^2-z^2}{\left(x+z\right)^2-y^2}\)

\(=\frac{\left(x+y+z\right)\left(x+y-z\right)}{\left(x+y+z\right)\left(x-y+z\right)}=\frac{x+y-z}{x-y+z}\)

Thay \(\hept{\begin{cases}x=0\\y=2009\\z=2010\end{cases}}\) vào biểu thức :

\(\Rightarrow A=\frac{0+2009-2010}{0-2009+2010}=-1\)

\(=3^{2+3}+2^{3+2}\)

\(=3^5+2^5\)

\(=243+32=275\)

\(3^2.3^3+2^3.2^2=3^{2+3}+2^{3+2}=3^5+2^5=243+32=275\)

ko có biết

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

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

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