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\(a,2\cdot3^3-3\cdot2^2+7^2-5^2\)
\(=2\cdot27-3\cdot4+49-25\)
\(=54-12+49-25\)
\(=42+24\)
\(=66\)
Sao k có ai giúp mk hết vậy >:((, thôi để mk tự giúp mk vậy :>. E mới nghĩ ra cách này có gì sai anh giúp đỡ.
Cách 1 - Ta có :
\(A=\frac{1}{1.2}+\frac{1}{1.3}+\frac{1}{1.4}+...+\frac{1}{3.2}+\frac{1}{3.3}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{6}+\frac{1}{9}\)
\(\Rightarrow A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{6}+\frac{1}{9}\)
\(\Rightarrow A=\frac{5}{6}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{6}+\frac{1}{9}\)
Mà \(\frac{5}{6}>\frac{2}{3}\Rightarrow\frac{5}{6}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{6}+\frac{1}{9}>\frac{2}{3}\)
\(\Leftrightarrowđpcm\)
\(\frac{1}{1.1}+\frac{1}{1.3}+\frac{1}{3.2}+...+\frac{1}{49.25}\)
\(=\frac{2}{2}.\left(\frac{1}{1.1}+\frac{1}{1.3}+\frac{1}{3.2}+...+\frac{1}{49.25}\right)\)
\(=\frac{2}{1.1.2}+\frac{2}{1.3.2}+\frac{2}{3.2.2}+...+\frac{2}{49.25.2}\)
\(=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{49.50}\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(=2.\left(1-\frac{1}{50}\right)\)
\(=2.\frac{49}{50}\)
\(=\frac{49}{25}\)
Chúc bạn học tốt !!!
G=\(\frac{8.9+8.9.10+8.9+11+8.9+2}{4.6+4.6.6+4.6.4+4.6.8}\)
G=\(\frac{8.9.\left(1+10+1+1\right)+11+2}{4.6.\left(1+6+4+8\right)}\)
G=\(\frac{72.13+13}{24.19}\)
G=\(\frac{13.\left(72+1\right)}{24.19}=\frac{13.73}{24.19}\)
G=\(\frac{949}{456}\)
\(2011^{2011}.\left(7^{10}:7^8-3.2^4-2^{2011}:2^{2011}\right)\)
\(=2011^{2011}.\left(7^{10-8}-3.2^4-2^{2011-2011}\right)\)
\(=2011^{2011}.\left(7^2-3.2^4-2^0\right)\)
\(=2011^{2011}.\left(49-3.16-1\right)\)
\(=2011^{2011}.\left(49-48-1\right)\)
\(=2011^{2011}.\left(1-1\right)\)
\(=2011^{2011}.0\)
\(=0\)
bó tay hết cách