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\(^{3^2}\).\(^{3^3}\)+\(2^3\).\(2^2\)
(\(^{2^3}\).\(^{3^3}\))+(\(2^2\).\(^{3^2}\)
=275
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 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) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)
\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)
\(\left(1\right)=1^3=1\)
\(x^3-3x^2y+3xy^2-y^3\)
\(=x^3-3\cdot x^2\cdot y+3\cdot x\cdot y^2-y^3\)
\(=\left(x-y\right)^3\)
Thay x=3 và y=2 vào ta có:
\(\left(3-2\right)^3=1^3=1\)
3A-A= 3^2+3^3+....+3^101-3 -3^2-3^3-....-3^100
A= (3^101-3 ) :2
A = 3 + 32 + 33 + .... + 3100
3A = 32 + 33 + 34 + ... + 3101
3A - A = (32 + 33 + 34 + ... + 3101) - (3 + 32 + 33 + ... + 3100)
2A = 3101 - 3
=> A = \(\frac{3^{101}-3}{2}\)
Ủng hộ mk nha !!! ^_^
\(a,x^2y-8x+xy-8=xy\left(x+1\right)-8\left(x+1\right)=\left(xy-8\right)\left(x+1\right)\\ b,=\left(x+3y\right)^2-9=\left(x+3y-3\right)\left(x+3y+3\right)\)
\(A=3x^2\left(2x^2-7x-2\right)-6x^2\left(x^2-4x-1\right)-3x^3+15\\ A=6x^4-21x^3-6x^2-6x^4+24x^3+6x^2-3x^3+15\\ A=15\left(đpcm\right)\)
\(Sửa:\left(6x^3-7x^2+2x\right):\left(2x+1\right)\\ =\left(6x^3+3x^2-10x^2-5x\right):\left(2x+1\right)\\ =\left[3x^2\left(2x+1\right)-5x\left(2x+1\right)\right]:\left(2x+1\right)\\ =3x^2-5x\)
\(=3^{2+3}+2^{3+2}\)
\(=3^5+2^5\)
\(=243+32=275\)
\(3^2.3^3+2^3.2^2=3^{2+3}+2^{3+2}=3^5+2^5=243+32=275\)