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Bai 1: \(\left(-2\right)^3.\left(\dfrac{3}{4}-0,25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)
\(=\left(-8\right).\left(\dfrac{3}{4}-\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=\left(-8\right).\dfrac{1}{2}:\left(\dfrac{27}{12}-\dfrac{14}{12}\right)\)
\(=\left(-4\right):\dfrac{13}{12}\)
\(=\left(-4\right).\dfrac{12}{13}\)
\(=\dfrac{-48}{13}\)
Bai 2:
\(a,4\dfrac{1}{3}:\dfrac{x}{4}=6:0,3\)
\(\dfrac{13}{3}:\dfrac{x}{4}=20\)
\(\dfrac{x}{4}=\dfrac{13}{3}:20\)
\(\dfrac{x}{4}=\dfrac{13}{60}\)
➩ \(x.60=4.13\) ➩ \(x.60=52\) ➩ \(x=\dfrac{13}{15}\)
Vay \(x=\dfrac{13}{15}\)
\(b,\left(2^3:4\right).2^{\left(x+1\right)}=64\)
\(\left(8:4\right).2^{x+1}=64\)
\(2.2^{x+1}=64\)
\(2^{x+1}=32\)
➩ \(2^{x+1}=2^5\) ➩ \(x+1=5\) ➩ \(x=4\)
Vay \(x=4\)
\(B=\dfrac{\dfrac{2}{10}-\dfrac{3}{8}+\dfrac{5}{11}}{\dfrac{-3}{10}+\dfrac{9}{16}-\dfrac{15}{22}}\)\(-\dfrac{1}{3}\)
\(B=\dfrac{\dfrac{2}{10}-\dfrac{6}{16}+\dfrac{10}{22}}{\dfrac{-3}{10}+\dfrac{9}{16}-\dfrac{15}{22}}\)\(-\dfrac{1}{3}\)
\(B=\dfrac{2.\left(\dfrac{1}{10}-\dfrac{3}{16}+\dfrac{5}{22}\right)}{-3.\left(\dfrac{1}{10}-\dfrac{3}{16}+\dfrac{5}{22}\right)}\)\(-\dfrac{1}{3}\)
\(B=\dfrac{-2}{3}-\dfrac{1}{3}=-1\)
6)a) \(\left|\dfrac{5}{3}:x\right|=\left|\dfrac{-1}{6}\right|\)
⇒ \(\left|\dfrac{5}{3}:x\right|=\dfrac{1}{6}\)
⇒ \(\dfrac{5}{3}:x=\dfrac{1}{6}\) hoặc \(\dfrac{5}{3}:x=\dfrac{-1}{6}\)
*TH1 : \(\dfrac{5}{3}:x=\dfrac{1}{6}\)
⇒ \(x=\dfrac{5}{3}:\dfrac{1}{6}=10\)
*TH2 : \(\dfrac{5}{3}:x=\dfrac{-1}{6}\)
⇒ \(x=\dfrac{5}{3}:\dfrac{-1}{6}=-10\)
Vậy \(x\) ∈ \(\left\{10;-10\right\}\)
\(b,\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|-\dfrac{3}{4}=\left|\dfrac{-3}{4}\right|\)
⇒ \(\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|-\dfrac{3}{4}=\dfrac{3}{4}\)
⇒\(\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|=\dfrac{3}{4}+\dfrac{3}{4}=\dfrac{3}{2}\)
⇒ \(\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{3}{2}\) hoặc \(\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{-3}{2}\)
TH1 : \(\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{3}{2}\)
⇒ \(\dfrac{3}{4}x=\dfrac{3}{2}+\dfrac{3}{4}=\dfrac{9}{4}\)
⇒\(x=\dfrac{9}{4}:\dfrac{3}{4}=3\)
TH2 : \(\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{-3}{2}\)
⇒ \(\dfrac{3}{4}x=\dfrac{-3}{2}+\dfrac{3}{4}=\dfrac{-3}{4}\)
⇒ \(x=\dfrac{-3}{4}:\dfrac{3}{4}=-1\)
Vậy \(x\) ∈ \(\left\{3;1\right\}\)
\(\dfrac{8^{14}}{4^4.64^5}=\dfrac{\left(2^3\right)^{14}}{\left(2^2\right)^4.\left(2^5\right)^5}=\dfrac{2^{42}}{2^8.2^{25}}=2^{42-\left(8+25\right)}=2^9\)
\(\dfrac{9^{10}.27^7}{81^7.3^{15}}=\dfrac{\left(3^2\right)^{10}.\left(3^3\right)^7}{\left(3^4\right)^7.3^{15}}=\dfrac{3^{20}.3^{21}}{3^{28}.3^{15}}=\dfrac{3^{20+21}}{3^{28+15}}=\dfrac{3^{41}}{3^{41}.3^2}=\dfrac{1}{3^2}=\dfrac{1}{9}\)
1/
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{x-y}{5-6}=\dfrac{36}{-1}=-36\)
\(\Rightarrow\left\{{}\begin{matrix}x=-36\cdot5=-180\\y=-36\cdot6=-216\\z=-36\cdot4=-144\end{matrix}\right.\)
2/
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{y+z}{3+4}=\dfrac{28}{7}=4\)
\(\Rightarrow\left\{{}\begin{matrix}x=4\cdot7=28\\y=4\cdot3=12\\z=4\cdot4=16\end{matrix}\right.\)
3/
\(\dfrac{x}{1,2}=\dfrac{y}{1,3}\Leftrightarrow\dfrac{2x}{2,4}=\dfrac{y}{1,3}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x}{2,4}=\dfrac{y}{1,3}=\dfrac{2x-y}{2,4-1,3}=\dfrac{5,5}{1,1}=5\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5\cdot2,4}{2}=6\\y=5\cdot1,3=6,5\\z=5\cdot1,4=7\end{matrix}\right.\)
4/
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{0,5}=\dfrac{y}{0,3}=\dfrac{x-y}{0,5-0,3}=\dfrac{1}{0,2}=5\)
\(\Rightarrow\left\{{}\begin{matrix}x=5\cdot0,5=2,5\\y=5\cdot0,3=1,5\\z=5\cdot0,2=1\end{matrix}\right.\)
5/
\(z=\dfrac{x}{0,3}\Leftrightarrow z=\dfrac{3x}{0,9}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(z=\dfrac{3x}{0,9}=\dfrac{z-3x}{1-0,9}=\dfrac{1}{0,1}=10\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{10\cdot0,9}{3}=3\\y=10\cdot0,7=7\\z=10\end{matrix}\right.\)
a) \(4\dfrac{1}{3}:\dfrac{x}{4}=6:0,3\)
\(\dfrac{13}{3}:\dfrac{x}{4}=20\)
\(\dfrac{x}{4}=\dfrac{13}{3}:20\)
\(\dfrac{x}{4}=\dfrac{13}{3}\cdot\dfrac{1}{20}\)
\(\dfrac{x}{4}=\dfrac{13}{60}\)
\(x=\dfrac{13}{60}\cdot4\)
Vậy \(x=\dfrac{13}{15}\)
b)\(2^3:4.2^{\left(x+1\right)}=64\)
\(8:4.2^{\left(x+1\right)}=64\)
\(2.2^{\left(x+1\right)}=64\)
\(2\cdot2^x.2=64\)
\(4.2^x=64\)
\(2^x=64:4\)
\(2^x=16\)
\(2^x=2^4\)
Vậy \(x=4\)
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a/ \(4\dfrac{1}{3}:\dfrac{x}{4}=6:0,3\)
\(\Leftrightarrow\dfrac{13}{3}:\dfrac{x}{4}=20\)
\(\Leftrightarrow\dfrac{52}{3x}=20\)
\(\Leftrightarrow x=\dfrac{13}{15}\)
Vậy..
b/ \(\left(x-1\right)^5=-32\)
\(\Leftrightarrow\left(x-1\right)^5=\left(-2\right)^5\)
\(\Leftrightarrow x-1=-2\)
\(\Leftrightarrow x=-1\)
Vậy..
c/ \(\left(2^3:4\right).2^{x+1}=64\)
\(\Leftrightarrow2.2^{x+1}=64\)
\(\Leftrightarrow2^{x+2}=2^6\)
\(\Leftrightarrow x+2=6\)
\(\Leftrightarrow x=4\)
Vậy..
d/ \(\left|3-2x\right|-3=-3\)
\(\Leftrightarrow\left|3-2x\right|=0\)
\(\Leftrightarrow3-2x=0\)
\(\Leftrightarrow x=\dfrac{3}{2}\)
Vậy..
e/ \(\left|x+\dfrac{4}{5}\right|-\dfrac{1}{7}=0\)
\(\Leftrightarrow\left|x+\dfrac{4}{5}\right|=\dfrac{1}{7}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{4}{5}=\dfrac{1}{7}\\x+\dfrac{4}{5}=-\dfrac{1}{7}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{23}{35}\\x=-\dfrac{33}{35}\end{matrix}\right.\)
Vậy..
4\(\dfrac{1}{3}\):\(\dfrac{x}{4}\)=6:0,3
=>\(\dfrac{13}{3}\):\(\dfrac{x}{4}\)=20
=>\(\dfrac{x}{4}\)=\(\dfrac{13}{3}\):20=\(\dfrac{13}{60}\)
=>x\(\in\varnothing\)
\(4\dfrac{1}{3}:\dfrac{x}{4}=6:0,3\)
\(\dfrac{13}{3}:\dfrac{x}{4}=20\)
\(\dfrac{x}{4}=20.\dfrac{13}{3}\)
\(\dfrac{x}{4}=\dfrac{260}{3}\)
\(\dfrac{3x}{12}=\dfrac{1040}{12}\)
\(3x=1040\)
\(x=1040:3\)
\(x=\dfrac{1040}{3}\)
\(1\dfrac{2}{3}:\dfrac{x}{4}=6:0,3\)
\(\dfrac{5}{3}:\dfrac{x}{4}=20\)
\(\dfrac{x}{4}=\dfrac{5}{3}:20\)
\(\dfrac{x}{4}=\dfrac{1}{12}\)
\(x=\dfrac{1}{3}\)