Tính hợp lý nếu có thể:
A, 2/13 nhân (-5/3)+11/13 nhân (-5/3)
B,(-1/3)^2+(-1/3)^3 nhân 27+(-2017/2018)^0
C, (1,2-√1/4):1và 1/20+|3/4-1,25|-(-3/2)^2
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Vì \(a^2+b^2+c^2=1\)
\(\Rightarrow-1\le a,b,c\le1\)
\(\Rightarrow a-1\le0;b-1\le0;c-1\le0\)
Lây cai xau trừ cai trươc được
\(\left(a^3+b^3+c^3\right)-\left(a^2+b^2+c^2\right)=0\)
\(\Leftrightarrow a^2\left(a-1\right)+b^2\left(b-1\right)+c^2\left(c-1\right)=0\)
Ta co \(VT\le0\)
Dâu = xảy ra khi: \(\left(a,b,c\right)=\left\{0,0,1;0,1,0;1,0,0\right\}\)
\(\Rightarrow S=1\)
=(1+4+42) +(43+44+45)+....+(42017+42018+42019)
=(1+4+42)+43(1+4+42)+.....+42017(1+4+42)
=(1+4+42)(1+43+46+....+42017)
=(1+4+16)(1+43+46+.....+42017)
=21(1+43+46+...+42017)
Vậy 21(1+43+46+.....+42017) chia hết cho 21
\(1+4+4^2+4^3+4^4+....+4^{2019}\)
\(=\left(1+4+4^2\right)+\left(4^3+4^4+4^5\right)+......+\left(4^{2017}+4^{2018}+4^{2019}\right)\)
\(=\left(1+4+4^2\right)+4^3\left(1+4+4^2\right)+.....+4^{2017}\left(1+4+4^2\right)\)
\(=\left(1+4+4^2\right)\left(1+4^3+.....+4^{2017}\right)\)
\(=21\left(1+4^3+....+4^{2017}\right)\)
Mà \(21⋮21\Rightarrow21\left(1+4^3+.....+4^{2017}\right)⋮21\)
Vậy biểu thức trên chia hết cho 21(đpcm)
a. x + \(\dfrac{3}{7}\)= \(\dfrac{2}{5}:\dfrac{18}{25}=>x+\dfrac{3}{7}=\dfrac{2}{5}\)x\(\dfrac{35}{18}=>x+\dfrac{3}{7}=\dfrac{7}{9}\)
=> x = \(\dfrac{7}{9}-\dfrac{3}{7}=\dfrac{49}{63}-\dfrac{27}{63}=\dfrac{22}{63}\)
b. \(x\) x \(\dfrac{5}{9}\)= \(\dfrac{4}{5}-\dfrac{1}{3}\)
=> \(x\) x \(\dfrac{5}{9}\)= \(\dfrac{12}{15}-\dfrac{5}{15}=>x\) x \(\dfrac{5}{9}\)= \(\dfrac{7}{15}\)
=> x = \(\dfrac{7}{15}:\dfrac{5}{9}\)
=> x = \(\dfrac{21}{25}\)
\(a.x+\dfrac{3}{7}=\dfrac{2}{5}:\dfrac{18}{35}\\x+\dfrac{3}{7}=\dfrac{2}{5}\times\dfrac{35}{18} \\ x+\dfrac{3}{7}=\dfrac{7}{9}\\ x=\dfrac{7}{9}-\dfrac{3}{7}\\ x=\dfrac{22}{63}\)
\(b.x\times\dfrac{5}{9}=\dfrac{4}{5}-\dfrac{1}{3}\\x\times\dfrac{5}{9}=\dfrac{7}{15}\\ x=\dfrac{7}{15}:\dfrac{5}{9}\\ x= \dfrac{21}{25}\)
\(\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+....+\frac{1}{99^2}\)
\(< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+.....+\frac{1}{98\cdot99}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{98}-\frac{1}{99}\)
\(=1-\frac{1}{99}\)
\(< 1\)
\(< \frac{5}{4}\)
\(\frac{1}{2}.2^n+2^{2+n}=9.2^5\)
\(\frac{1}{2}.2^n+4.2^n=9.2^5\)
\(2^n.\left(\frac{1}{2}+4\right)=9.2^5\)
\(2^n.\frac{9}{2}=9.2^5\)
\(2^n=9.\frac{2}{9}.2^5\)
\(2^n=2.2^5\)
\(2^n=2^6\)
\(\Rightarrow n=6\)
\(\frac{1}{2}\)\(\times\)\(2^n\)\(+\)\(2^{2+n}\)\(=\)\(9\)\(\times\)\(2^5\)
\(\frac{1}{2}\)\(\times\)\(2^n\)\(\times\)\((\)\(1\)\(+\)\(2^2\)\()\)\(=\)\(9\)\(\times\)\(2^5\)
\(\frac{1}{2}\)\(\times\)\(2^n\)\(\times\)\(5\)\(=\)\(9\)\(\times\)\(2^5\)
\(2^n\)\(=\)\(9\)\(\times\)\(32\)\(\div\)\(5\)\(\times\)\(2\)
\(2^n\)\(=\)115,2
2, \(=>9A=3^3+3^5+3^7+......+3^{39}+3^{41}\)
\(=>9A-A=3^{41}-3\)
\(=>A=\dfrac{3^{41}-3}{8}\)
CHÚC BẠN HỌC TỐT........