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a: Sửa đề: \(\dfrac{12}{-6}=\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{2}{-z}=\dfrac{-t}{-9}\)
=>\(\dfrac{x}{5}=\dfrac{y}{-3}=\dfrac{-2}{z}=\dfrac{t}{9}=-2\)
=>\(x=-2\cdot5=-10;y=-2\cdot\left(-3\right)=6;z=\dfrac{-2}{-2}=1;t=9\cdot\left(-2\right)=-18\)
b: \(\dfrac{-24}{-6}=\dfrac{x}{3}=\dfrac{4}{y^2}=\dfrac{z^3}{-2}\)
=>\(\dfrac{x}{3}=\dfrac{4}{y^2}=\dfrac{z^3}{-2}=4\)
=>\(\left\{{}\begin{matrix}x=4\cdot3=12\\y^2=\dfrac{4}{4}=1\\z^3=-2\cdot4=-8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=12\\y\in\left\{1;-1\right\}\\z=-2\end{matrix}\right.\)
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a) \(58+7x=100\)
\(=>7x=100-58\)
\(=>7x=42\)
\(=>x=42:7\)
\(=>x=6\)
b) \(3x-7=28\)
\(=>3x=28+7\)
\(=>3x=35\)
\(=>x=35:3\)
\(=>x=\dfrac{35}{3}\)
c) \(x-56:4=16\)
\(=>x-14=16\)
\(=>x=16+14\)
\(=>x=30\)
d) \(101+\left(36-4x\right)=105\)
\(=>36-4x=105-101\)
\(=>36-4x=4\)
\(=>4x=36-4\)
\(=>4x=32\)
\(=>x=32:4\)
\(=>x=8\)
e) \(\left(x-12\right):12=12\)
\(=>x-12=12.12\)
\(=>x-12=144\)
\(=>x=144-12\)
\(=>x=132\)
f) \(\left(3x-2^4\right).7^3=2.7^4\)
\(=>3x-2^4=2.7^4:7^3\)
\(=>3x-16=2.7=14\)
\(=>3x=14+16\)
\(=>3x=30\)
\(=>x=30:3\)
\(=>x=10\)
i) \(\left(10+2x\right).4^{2011}=4^{2013}\)
\(=>10+2x=4^{2013}:4^{2011}\)
\(=>10+2x=4^2=16\)
\(=>2x=16-10\)
\(=>2x=6\)
\(=>x=6:2\)
\(=>x=3\)
\(#WendyDang\)
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a: x-56:4=16
nên x-14=16
hay x=30
b: \(101+\left(36-4x\right)=105\)
\(\Leftrightarrow36-4x=4\)
hay x=8
Vì a: x-56:4=16
=>x-14=16
Hay:x=30
Vì b: 101+(36−4x)=105101+(36−4x)=105
⇔36−4x=4⇔36−4x=4
Hay:x=8
Vậy x={30:8}
Nếu sai thì thui nhá,đừng chửi mình!
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1: \(\Leftrightarrow x=UCLN\left(24;36;150\right)=6\)
2: \(\Leftrightarrow x\in\left\{24;48;72;...\right\}\)
mà 16<=x<=50
nên \(x\in\left\{24;48\right\}\)
3: \(\Leftrightarrow x\inƯ\left(6\right)\)
mà x>-10
nên \(x\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
4: \(\Leftrightarrow x\in BC\left(4;5;8\right)\)
\(\Leftrightarrow x\in\left\{...;-40;0;40;80;120;160;200;...\right\}\)
mà -20<x<180
nên \(x\in\left\{0;40;80;120;160\right\}\)
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Câu 3:
a) \(\dfrac{12}{36}=\dfrac{12:12}{36:12}=\dfrac{1}{3}\)
\(\dfrac{-16}{20}=\dfrac{-16:4}{20:4}=\dfrac{-4}{5}\)
b) \(\dfrac{21}{105}=\dfrac{21:21}{105:21}=\dfrac{1}{5}\)
\(\dfrac{35}{150}=\dfrac{35:5}{150:5}=\dfrac{7}{30}\)
Câu 4:
a) \(\dfrac{3}{10}+\dfrac{5}{10}=\dfrac{3+5}{10}=\dfrac{8}{10}=\dfrac{4}{5}\)
b) Ta có: \(\left(-27\right)\cdot36+64\cdot\left(-27\right)+23\cdot\left(-100\right)\)
\(=\left(-27\right)\cdot\left(64+36\right)+23\cdot\left(-100\right)\)
\(=-27\cdot100-23\cdot100\)
\(=100\left(-27-23\right)\)
\(=-50\cdot100=-5000\)
c) \(\dfrac{5}{8}+\dfrac{3}{12}=\dfrac{15}{24}+\dfrac{6}{24}=\dfrac{21}{24}=\dfrac{7}{8}\)
d) Ta có: \(\dfrac{-2}{17}+\dfrac{3}{19}+\dfrac{-15}{17}+\dfrac{16}{19}+\dfrac{5}{6}\)
\(=\left(-\dfrac{2}{17}+\dfrac{-15}{17}\right)+\left(\dfrac{3}{19}+\dfrac{16}{19}\right)+\dfrac{5}{6}\)
\(=-1+1+\dfrac{5}{6}\)
\(=\dfrac{5}{6}\)
\(\Leftrightarrow\dfrac{x}{4}=\dfrac{36}{y^2}=\dfrac{z^3}{16}=4\)
=>x=16; \(y\in\left\{3;-3\right\}\); z=4