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\(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)
\(\Rightarrow\dfrac{59-x}{41}+1+\dfrac{57-x}{43}+1+\dfrac{55-x}{45}+1+\dfrac{53-x}{47}+1+\dfrac{51-x}{49}+1=0\)\(\Rightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}+\dfrac{100-x}{49}=0\)
\(\Rightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\right)=0\)
\(\Rightarrow100-x=0\Rightarrow x=100\)
\(\frac{2^5.15^4}{5^4.6^3}=\frac{2^5.\left(3.5\right)^4}{5^4.\left(2.3^3\right)}=\frac{2^5.3^4.5^4}{5^4.2^3.3^3}=\frac{2^5.3^4.5^4}{2^3.3^3.5^4}=2^2.3=4.3=12\)
k cho mik nha
a, (42.43 / 21.0) .2003
= (42.43/0) .2003
= 0.2003
= 0
b, 1203/403
= (120/40)3
= 33
= 9
c, [(-2,5) .0,38 .0,4] - [(-1,25).0,62.(-0,8)]
= {[(-2,5) .0,4] .0,38} - {[(-1,25) .(-0,8)] .0,62}
= [(-1) .0,38] - (1 .0,62)
= (-0.38) - 0,62
= -1
MK chỉnh dấu ngoặc hơi loằng ngoằng
Biến đổi :
\(\frac{125\cdot8^2}{30^6\cdot\left(-15\right)^2}=\frac{5^3\cdot\left(2^3\right)^2}{2^6\cdot3^6\cdot5^6\cdot3^2\cdot5^2}\)
\(=\frac{5^3\cdot2^6}{2^6\cdot3^8\cdot5^8}=\frac{1}{3^8\cdot5^5}\)
\(\Rightarrow\left(\frac{-2}{5}\right)^5:\left(\frac{125\cdot8^2}{30^6\cdot\left(-15\right)^2}\right)^2\)
\(=\frac{\left(-2\right)^5}{5^5}:\frac{1^2}{\left(3^8\cdot5^5\right)^2}\)
\(=\frac{\left(-2\right)^5\cdot3^{16}\cdot5^{10}}{5^5}\)
\(=\left(-2\right)^5\cdot3^{16}\cdot5^5\)
\(\frac{11}{15}+\frac{23}{16}-\left(5+\frac{27}{16}-\frac{19}{15}\right)+\left(\frac{-1}{2}\right)^2\)
\(=\frac{11}{15}+\frac{23}{16}-5-\frac{27}{16}+\frac{19}{15}+\frac{1}{4}\)
\(=\left(\frac{11}{15}+\frac{19}{15}\right)+\left(\frac{23}{16}-\frac{27}{16}\right)-5+\frac{1}{4}\)
\(=\frac{30}{15}-\frac{4}{16}-5+\frac{1}{4}\)
\(=2-\frac{1}{4}-5+\frac{1}{4}\)
\(=-3\)
học tốt ngôlãmtân
\(\frac{11}{15}+\frac{23}{16}-\left(5+\frac{27}{16}-\frac{19}{15}\right)+\left(-\frac{1}{2}\right)^2\) = \(\frac{11}{15}+\frac{23}{16}-5-\frac{27}{16}+\frac{19}{15}+\frac{1}{4}\)
\(=2-\frac{1}{4}-5+\frac{1}{4}=2-5=-3\)
Kb với mình nha!
Tử số: 2^19 x (3^3)^3 x 5+15 x 4^9 x(3^2)^4
=2^19 x3^9x5 + 15 x(2^2)^9 x 3^8
= 2^19 x 3^9 x 5 +3 x 5 x 2^18 x 3^8
= 2^19 x 3^9 x 5+ 3^9 x 5 x 2^18
= 5 x 3^9 x 2^18 (2+1)
=5 x 3^10 x 2^18
Mẫu số
= (2 x 3)^9 x 2^10 -12^10
= 2^9 x 3^9 x 2^10 - (2^2x3)^10
= 2^9 x 3^9 x 2^10 -2^20 x 3^10
= 2^19 x 3^9 - 2^20 x 3^10
= 2^19 x 3^9 (1-2 x 3)
= 2^19 x 3^9 x(-5)
Chia cả tử và mẫu ta có
(5 x 3^10 x 2^18) / (2^19 x 3^9 x (-5)) = -3/2
\(H=\frac{2^{19}.27^3.5-15.\left(-4\right)^9.9^4}{6^9.2^{10}-\left(-12\right)^{10}}\)
\(\Rightarrow\)\(H=\frac{2^{19}.3^9.5-3.5-1.2^{18}.3^8}{2^9.3^9.2^{10}-6^{10}.2^{10}}=\frac{2^{19}.3^9.5-3^9.5-2^{18}}{2^{19}.3^9-3^{10}.2^{10}.2^{10}}=\frac{2^{19}.3^9.5-3^9.5-2^{18}}{2^{19}.3^9-3^{10}.2^{20}}\)
\(\Rightarrow H=\frac{2^{18}.3^9.5\left(2-1\right)}{2^{19}.3^9.\left(1-3.2\right)}=\frac{5}{2.\left(-5\right)}=\frac{-1}{2}\)
Vậy \(H=\frac{-1}{2}\)
\(\dfrac{40^{30}.3^{43}}{2^{57}.27^{15}}=\dfrac{\left(2^3.5\right)^{30}.3^{43}}{2^{57}.\left(3^3\right)^{15}}=\dfrac{2^{90}.5^{30}.3^{43}}{2^{57}.3^{45}}\)
\(=\dfrac{2^{33}.2^{57}.5^{30}.3^{43}}{2^{57}.3^{43}.3^2}=\dfrac{2^{33}.5^{30}}{3^2}=\dfrac{2^{33}.5^{30}}{9}\)
bạn gì kết quả to lắm nên mình chỉ để như vậy thôi