a)\(\left(5.4^{11}-3.16^5\right):4^{10}\)
b)\(\dfrac{7256.4375-725}{3650+4375.7255}\)
c)100+98+96+...+4+2-97-95-...-3-1
giúp mình nhé
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a: \(=347\cdot4\cdot9\cdot400:8=347\cdot36\cdot50=624600\)
c: \(=16:\left\{400:\left[200-37-138\right]\right\}\)
\(=16:\left\{400:25\right\}=16:16=1\)
e: \(=46-\left[300:15\right]-2=46-20-2=24\)
Bài 1
4.|347|+(-40)+3150-|-307|
=347+-40+3150+-307
=(347+-40+-307)+3150
=0+3150
=3150
\(\frac{7256.4375-725}{4375.7255+3650}=\frac{\left(7255+1\right).4375-725}{4375.7255+3650}=\frac{7255.4375+4375-725}{7255.4375+3650}=\frac{7255.4375+3650}{7255.4375+3650}=1\)
\(\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^{10}\left(11+5\right)}{3^9.2^4}=\frac{3.16}{16}=3\)
\(\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}\left(13+65\right)}{2^8.104}=\frac{2^2.78}{26.2^2}=\frac{78}{26}=3\)
\(\left(125^3.7^5-175^5.5\right):2001^{2002}\) ( bạn xem lại đề xem sai đâu ko nhé )
Để Thiên giải câu 3 cho:
(1253.75 -1755;5):20012001
\(=\left[\left(5^3\right)^3.7^5-175^5:5\right]:2001^{2002}\)
\(=\left(5^9.7^5-175:5\right):2001^{2002}\)
\(=\left(5^5.5^4.7^4.7-175^4.175:5\right):2001^{2002}\)
\(=\left(5^5.35^4.7-175^4.35\right):2001^{2002}\)
\(=\left(5^4.35^4.5.7-175^4.35\right):2001^{2002}\)
\(=\left(175^4.35-175^4.35\right):2001^{2002}\)
\(=0:2001^{2002}\)
\(=0\)
Ta có:
\( \dfrac{7256.4375-725}{3650+4375.7255}\\ =\dfrac{7255.4375+4375-7255}{3650+4375.7255}\\ =\dfrac{7255.4375+\left(4375-725\right)}{3650+4375.7255}\\ =\dfrac{7255.4375+3650}{3650+4375.7255}\\ =1\)
Vậy giá trị của \(\dfrac{7256.4375-725}{3650+4375.7255}=1\)
\(\dfrac{7256\cdot4375-725}{3650+4375\cdot7255}\\ =\dfrac{\left(7255+1\right)\cdot4375-725}{7255\cdot4375+3650}\\ =\dfrac{7255\cdot4375+4375-725}{7255\cdot4375+3650}\\ =\dfrac{7255\cdot4375+3650}{7255\cdot4375+3650}\\ =1\)
Ta có: \(7256.4375-\dfrac{725}{3650}+4375.7255\)
\(=7256.4375+4375.7255-\dfrac{29}{146}\)
\(=4375\left(7256-7255\right)-\dfrac{29}{146}\)
\(=4375.1-\dfrac{29}{146}\)
\(=\dfrac{638750}{126}-\dfrac{29}{126}\)
\(=\dfrac{638721}{126}=\dfrac{70969}{14}\)
Chúc bn học tốt!!!
\(\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)=\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+5}{95}+1\right)\)
\(\Leftrightarrow\dfrac{x+2+98}{98}+\dfrac{x+3+97}{97}-\dfrac{x+4+96}{96}-\dfrac{x+5+95}{95}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\right)=0\)
\(\Rightarrow x+100=0\)
\(\Leftrightarrow x=-100\)
Vậy...
\(100 + 98 + 96 + ... + 2 - 97 - 95 - ... -1\)
\(= 100 + (98 - 97) + (96-95) +... + (2 - 1)\)
\(= 100 + 1 + 1 + 1 + ... +1\)
\(=100+1\cdot49\)
\(= 100 + 49 \)
\(= 149\)