tính hợp lí ( nếu có thể )
1000 + ( - 670 ) + 297 + ( - 330 )
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a/ ( -500 ) + (-174) + 1999 + (-266) + ( - 1499 )
=[(-174)+(-266)]+[1999+(-1499)+(-500)]
=(-440)+0
=-440
b/ 1000 + ( - 670 ) + 2004 + ( -300 )
=2004+1000-970
=2004+30
=2034
\(a,A=\left(2^{17}+17^2\right)\left(9^{15}-3^{30}\right).\left(2^4-32^2\right)\)
\(\Rightarrow A=\left(2^{17}+17^2\right)\left(3^{30}-3^{30}\right).\left(2^4-32^2\right)\)
\(\Rightarrow A=\left(2^{17}+17^2\right).0.\left(2^4-32^2\right)\)
\(\Rightarrow A=0.\)
\(\Rightarrow B=\left(2^8.8^3\right):\left(2^5.2^3\right)\)
\(\Rightarrow B=\left(2^8.2^9\right):\left(2^5.2^3\right)\)
\(\Rightarrow B=2^{17}:2^8\)
\(\Rightarrow B=2^9\)
\(\Rightarrow B=512\)
\(c,C=64^4.16^5:4^{20}\)
\(\Rightarrow C=2^{24}.2^{20}:2^{40}\)
\(\Rightarrow C=2^4\)
\(\Rightarrow C=16\)
a: \(A=\left(2^{17}+17^2\right)\cdot\left(2^4-32^2\right)\cdot\left(3^{30}-3^{30}\right)=0\)
b: \(B=\dfrac{2^8\cdot2^9}{2^5\cdot2^3}=2^9\)
c: \(C=2^{24}\cdot2^{20}:2^{40}=2^4=16\)
110+220+330+440+560+670+780+890+500
= (110+890)+(220+780)+(330+670)+(440+560)+500
= 1000+1000+1000+1000+500
=4500
\(41\dfrac{8}{23}-\left(6\dfrac{7}{32}+15\dfrac{8}{23}\right)\)
\(=41\dfrac{8}{23}-6\dfrac{7}{32}-15\dfrac{8}{23}\)
\(=26-6\dfrac{7}{32}\)
\(=20-\dfrac{7}{32}\)
\(=\dfrac{633}{32}\)
\(41\dfrac{8}{23}-\left(6\dfrac{7}{32}+15\dfrac{8}{23}\right)\)
\(=41\dfrac{8}{23}-6\dfrac{7}{32}-15\dfrac{8}{23}\)
\(=26-6\dfrac{7}{32}\)
\(=20-\dfrac{7}{32}\)
\(=\dfrac{633}{32}\)
Lời giải:
Gọi tổng trên là $A$
$A=2(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.11}+...+\frac{1}{100.103})$
$A=\frac{2}{3}(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.11}+...+\frac{3}{100.103})$
$=\frac{2}{3}(\frac{4-1}{1.4}+\frac{7-4}{4.7}+...+\frac{103-100}{100.103})$
$=\frac{2}{3}(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{100}-\frac{1}{103})$
$=\frac{2}{3}(1-\frac{1}{103})$
$=\frac{2}{3}.\frac{102}{103}=\frac{68}{103}$
Bạn Akai Haruma đáp án của bạn đúng khi phân số 1/7*11 là 1/7*10
=(1000+297)-(670+330)
=1000+297-1000
=297