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NT
0
8 tháng 8 2017
Câu a :
Ta có :
\(\dfrac{x}{y}=\dfrac{7}{3}\Leftrightarrow\) \(\dfrac{x}{7}=\dfrac{y}{3}\) .
Áp dụng dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{7}=\dfrac{y}{3}=\dfrac{5x}{35}=\dfrac{2y}{6}=\dfrac{5x-2y}{35-6}=\dfrac{87}{29}=3\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{7}=3\Rightarrow x=21\\\dfrac{y}{3}=3\Rightarrow y=9\end{matrix}\right.\)
Vậy ......................
Câu b :
Áp dụng dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{19}=\dfrac{y}{21}=\dfrac{2x}{38}=\dfrac{y}{21}=\dfrac{2x-y}{38-21}=\dfrac{34}{17}=2\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{19}=2\Rightarrow x=38\\\dfrac{y}{21}=2\Rightarrow y=42\end{matrix}\right.\)
Vậy ....................
8 tháng 8 2017
Làm mấy câu bạn kia chưa làm:v
\(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\)
\(\Rightarrow\dfrac{x^3}{2^3}=\dfrac{y^3}{4^3}=\dfrac{z^3}{6^3}\)
\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}\)
\(\Rightarrow\left(\dfrac{x}{2}\right)^2=\left(\dfrac{y}{4}\right)^2=\left(\dfrac{z}{6}\right)^2\)
\(\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{36}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{36}\)
\(=\dfrac{x^2+y^2+z^2}{4+16+36}=\dfrac{14}{56}=\dfrac{1}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=\dfrac{1}{4}.4=1\Rightarrow x=\pm1\\y^2=\dfrac{1}{4}.16=4\Rightarrow y=\pm2\\z=\dfrac{1}{4}.36=9\Rightarrow z=\pm3\end{matrix}\right.\)
SS
25 tháng 6 2019
\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8-6^8.20}\)
\(A=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8-\left(2.3\right)^8.2^2.5}\)
\(A=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8-2^{10}.3^8.5}\)
\(A=\frac{2^{10}.\left(3^8-3^9\right)}{2^{10}.3^8.\left(1-5\right)}=\frac{3^8-3^9}{3^8.\left(-4\right)}=\frac{3^8.\left(1-3\right)}{3^8.\left(-4\right)}=\frac{-2}{-4}=\frac{1}{2}\)
Vậy A = \(\frac{1}{2}\)
\(B=\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
\(B=\frac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)
\(B=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)
\(B=\frac{2^{19}.3^9+3^9.2^{18}.5}{2^{19}.3^9+2^{20}.3^{10}}\)
\(B=\frac{2^{18}.3^9.\left(2+5\right)}{2^{19}.3^9\left(1+2.3\right)}=\frac{7}{2.7}=\frac{1}{2}\)
Vậy B = \(\frac{1}{2}\)
HN
28 tháng 8 2021
a,
\(5^{x+4}-3.5^{x+3}=2.5^{11}\)
\(\Rightarrow5^{x+3}\left(5-3\right)=2.5^{11}\)
\(\Rightarrow5^{x+3}2=2.5^{11}\)
\(\Rightarrow5^{x+3}=5^{11}\)
\(\Rightarrow x+3=11\)
\(\Rightarrow x=8\)
HN
28 tháng 8 2021
b, (Check lai xem de sai o dau khong nhe)
\(3.5^{x+2}+4.5^{x+3}=19.5^{10}\)
Dat 5x ra ben ngoai
\(\Rightarrow5^x.5^23+5^x:5^{-3}.4\)
\(\Rightarrow5^x\left(5^2.3+5^{-3}.4\right)\)
\(\Rightarrow5^x\left(5^{-3}.5^5.3+5^{-3}.4\right)\)
\(\Rightarrow5^x[5^{-3}\left(5^53+4\right)\)
\(\Rightarrow5^x[5^{-3}\left(3125.3+4\right)\)
\(\Rightarrow5^x\left(5^{-3}\right).9379\)
=> Khong tim duoc gia tri cua x \(\Rightarrow x\in\varnothing\)
YN
9 tháng 1 2022
Answer:
\(\frac{15}{34}+\frac{7}{21}+\frac{19}{34}-\frac{20}{15}+\frac{3}{7}\) (Mình đã sửa đề nhé.)
\(=\left(\frac{15}{34}+\frac{19}{34}\right)+\left(\frac{7}{21}+\frac{3}{7}\right)-\frac{20}{15}\)
\(=1+\left(-1\right)+\frac{3}{7}\)
\(=\frac{3}{7}\)
\(\frac{3}{8}.27\frac{1}{5}-51\frac{1}{5}.38+19\)
\(=\frac{3}{8}\left(-24\right).38+19\)
\(=-342+19\)
\(=-323\)
\(-\frac{2}{7}.\frac{21}{8}.-\frac{7}{2}\)
\(=\left(\frac{-2}{7}.\frac{-7}{2}\right)\frac{21}{8}\)
\(=\frac{21}{8}\)
KK
0
`#3107`
\(\left(2x-\dfrac{1}{2}\right)^2+\dfrac{3}{7}=\dfrac{19}{8}\\ \Rightarrow\left(2x-\dfrac{1}{2}\right)^2=\dfrac{19}{8}-\dfrac{3}{7}\\ \Rightarrow\left(2x-\dfrac{1}{2}\right)^2=\dfrac{109}{56}\\ \Rightarrow\left(2x-\dfrac{1}{2}\right)^2=\left(\sqrt{\dfrac{109}{56}}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{1}{2}=\sqrt{\dfrac{109}{56}}\\2x-\dfrac{1}{2}=-\sqrt{\dfrac{109}{56}}\end{matrix}\right.\)
Bạn xem lại đề, số lớn quá ;-;.