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bài1
a, M= -[-(a+b)-(-a+b-a-b)]
= -(-a-b+a-b+a+b)
= -(a-b)
b,thay a=-5, b=-3 vào biểu thức M đã thu gọn ta đc: M= -[-5-(-3)] = -(-5+3) = -(-2) =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) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)
\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)
\(\left(1\right)=1^3=1\)
a) 1619 và 825
Ta có :
1619 = ( 24 )19 = 276
825 = ( 23 )25 = 275
Vì 276 > 275 Nên 1619 > 825
b) 536 và 1124
Ta có :
536 = ( 53 )12 = 12512
1124 = ( 112 )12 = 12112
Vì 12512 > 12112 Nên 536 > 1124
1.
\(M=3^0+3^1+......+3^{50}.\)
\(\Rightarrow3M=3+3^2+.......+3^{51}\)
\(\Rightarrow3M-M=\left(3+3^2+.......+3^{51}\right)-\left(3^0+3+.....+3^{50}\right)\)
\(\Rightarrow2M=3^{51}-1\)
\(\Rightarrow M=\frac{3^{51}-1}{2}\)
2.
\(a,\)Ta có : \(16^{19}=\left(2^4\right)^{19}=2^{76}\)
\(8^{25}=\left(2^3\right)^5=2^{75}\)
Vì \(2^{76}>2^{75}\Rightarrow16^{19}>8^{25}\)
\(b,\)Ta có : \(5^{36}=\left(5^3\right)^{12}=125^{12}\)
\(11^{24}=\left(11^2\right)^{12}=121^{12}\)
Vì \(125^{12}>121^{12}\Rightarrow5^{36}>11^{24}\)
\(1,\\ \left(a+1\right)\left(b+2\right)=5\\Vậy:\left(a+1\right);\left(b+2\right)\inƯ\left(5\right)=\left\{1;5\right\}\\ TH1:a+1=1\Rightarrow a=0;b+2=5\Rightarrow b=3\left(Loại,vì:a< b\right)\\ TH2:a+1=5\Rightarrow a=4;b+2=1\Rightarrow b=-1\left(Nhận,vì:a>b\right)\\ \Rightarrow\left(a;b\right)=\left(4;-1\right)\)
\(2,\\ \left(a+1\right).\left(b+3\right)=6\\ \Rightarrow\left(a+1\right);\left(b+3\right)\inƯ\left(6\right)=\left\{1;2;3;6\right\}\\ \Rightarrow TH1:a+1=1\Rightarrow a=0;b+3=6\Rightarrow b=3\left(Loại,vì:a< b\right)\\ TH2:a+1=2\Rightarrow a=1;b+3=3\Rightarrow b=0\left(Nhận,vì:a>b\right)\\ TH3:a+1=3\Rightarrow a=2;b+3=2\Rightarrow b=-1\left(Nhận,vì:a>b\right)\\ TH4:a+1=6\Rightarrow a=5;b+3=1\Rightarrow b=-2\left(Nhận,vì:a>b\right)\\ Vậy:\left(a;b\right)=\left(1;0\right).hoặc\left(a;b\right)=\left(2;-1\right).hoặc\left(a;b\right)=\left(5;-2\right)\)
