\(\dfrac{x}{0,9}=\dfrac{5}{6}\)
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GIÚP MÌNH VỚI MN ƠIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII
\(A=\dfrac{\left(20.5\right)^5.5^5}{100^5}=\dfrac{100^5.3125}{100^5}=3125\)
\(B=\dfrac{\left(0,3.3\right)^5}{\left(0,3\right)^5.0,3}=\dfrac{\left(0,3\right)^5.3^5}{\left(0,3\right)^5.0,3}=\dfrac{3^5}{0,3}=810\)
\(x+0,9=20,17\Leftrightarrow x=20,17-0,9=19,24\)
\(47,5-x=4,64\Leftrightarrow x=47,5-4,64=42,86\)
\(3x=\dfrac{5}{4}-\dfrac{1}{2}\Leftrightarrow3x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{4}\)
\(a.x=20,17-0,9=19,27\)
\(b.47,5-x=4,56\Leftrightarrow x=42,86\)
\(c.3.x=\dfrac{5}{4}-\dfrac{1}{2}\Leftrightarrow3x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{4}\)
a, \(4\times\left(-\dfrac{1}{2}\right)^3-2\times\left(-\dfrac{1}{2}\right)^2+3\times\left(-\dfrac{1}{2}\right)+1\)
\(=\left(-\dfrac{1}{2}\right)\left[\left(4\times-\dfrac{1}{2}\right)-\left(2\times-\dfrac{1}{2}\right)+3\right]+1\)
\(=\left(-\dfrac{1}{2}\right)\left(-2+1+3\right)+1\)
\(=\left(-\dfrac{1}{2}\right)2+1\)
\(=-1+1\)
\(=0\)
@Trịnh Thị Thảo Nhi
a, 4×(−12)3−2×(−12)2+3×(−12)+14×(−12)3−2×(−12)2+3×(−12)+1
=(−12)[(4×−12)−(2×−12)+3]+1=(−12)[(4×−12)−(2×−12)+3]+1
=(−12)(−2+1+3)+1=(−12)(−2+1+3)+1
=(−12)2+1=(−12)2+1
=−1+1=−1+1
=0=0
\(A=\dfrac{115-\dfrac{10}{7}-\dfrac{5}{11}+\dfrac{5}{23}}{403-\dfrac{26}{7}-\dfrac{13}{11}+\dfrac{13}{23}}+\dfrac{\dfrac{3}{5}+\dfrac{3}{13}-0,9}{\dfrac{7}{91}+0,2-\dfrac{3}{30}}\)
\(=\dfrac{5\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}{13\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}+\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{13}-\dfrac{3}{10}\right)}{\dfrac{1}{13}+\dfrac{1}{5}-\dfrac{3}{10}}\)
\(=\dfrac{5}{13}+3\)
\(=\dfrac{5}{13}+\dfrac{39}{13}\)
\(=\dfrac{44}{13}\)
\(=3\dfrac{5}{13}\)
CHÚC BN HC TỐT
\(A=\dfrac{5\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}{13\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}+\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{13}-\dfrac{3}{10}\right)}{\dfrac{1}{13}+\dfrac{1}{5}-\dfrac{3}{10}}\)
=5/13+3
=39/13+5/13=44/13
a) \(\left\{{}\begin{matrix}5x+3y=-7\\2x-4y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x+3y=-7\\x-2y=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}5x+3y=-7\\x=3+2y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5.\left(3+2y\right)+3y=-7\\x=3+2y\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}13y=-22\\x=3+2y\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-22}{13}\\x=3+2.\dfrac{-22}{13}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-22}{13}\\x=\dfrac{-5}{13}\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm là: \(\left\{{}\begin{matrix}y=\dfrac{-22}{13}\\x=\dfrac{-5}{13}\end{matrix}\right.\).
b)\(\left\{{}\begin{matrix}7x+14y=17\\2x+4y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}14x+28y=34\\14x+28y=35\end{matrix}\right.\) (vô nghiệm)
Vậy hệ phương trình vô nghiệm.
Mấy bài dễ tự làm nhé:D
1)
Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{bk}{b\left(k+1\right)}=\dfrac{k}{k+1}\\\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{dk}{d\left(k+1\right)}=\dfrac{k}{k+1}\end{matrix}\right.\)
Ta có điều phải chứng minh
\(\left\{{}\begin{matrix}\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\\\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\end{matrix}\right.\)
Ta có điều phải chứng minh
0.75
\(\dfrac{x}{0,9}=\dfrac{5}{6}\)
\(\Rightarrow x=\dfrac{0,9.5}{6}\)
\(\Rightarrow x=0,75\)