3/x=y/28=-39/91
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Theo đề ta có 3/x=-39/91 => 3.13/13x=39/91=>39/13x=39/91=> x= 91:13=7(*)
Ta lại có từ (*) x=7 => 3/7=y/28=> 3.4/7.4=y/28=>12/28=y/28=> y=12
Ta có: \(\dfrac{3}{x}=\dfrac{y}{8}=\dfrac{-39}{91}\)
\(\Rightarrow\) \(\dfrac{3}{x}=\dfrac{y}{8}=\dfrac{-3}{7}\)
\(\Rightarrow\) \(\dfrac{-3}{-x}=\dfrac{\dfrac{y}{4}}{7}=\dfrac{-3}{7}\)
\(\Rightarrow\) \(\left\{{}\begin{matrix}-x=7\\\dfrac{y}{4}=-3\end{matrix}\right.\)
\(\Rightarrow\) \(\left\{{}\begin{matrix}x=-7\\y=-12\end{matrix}\right.\) (TM)
Vậy x = -7; y = -12
Chúc bn học tốt!
\(\dfrac{3}{x-5}=\dfrac{-4}{x-2}\left(x\notin\left\{5;2\right\}\right)\)
\(\Rightarrow3\left(x-2\right)=-4\left(x-5\right)\)
\(\Rightarrow3x-6=-4x+20\)
\(\Rightarrow3x+4x=20+6\)
\(\Rightarrow7x=26\)
\(\Rightarrow x=\dfrac{26}{7}\) (thỏa)
_________
\(\dfrac{3}{x}=\dfrac{y}{28}=\dfrac{-39}{91}\left(x\ne0\right)\)
\(\Rightarrow\dfrac{3}{x}=\dfrac{y}{28}=\dfrac{-3}{7}\)
+) \(\dfrac{3}{x}=-\dfrac{3}{7}\)
\(\Rightarrow x=\dfrac{3\cdot-7}{3}=-7\) (thỏa)
+) \(\dfrac{y}{28}=-\dfrac{3}{7}\)
\(\Rightarrow y=\dfrac{28\cdot-3}{7}=-12\)
Vì x⋮39, x⋮65, x⋮91 nên x ϵ B(39,65,91) 39=3.13 65=5.13 91=7.13 BCNN(39,65,91)=13^3=2197 BC(39,65,91)=B(2197)=(0,2197,4394...) mà 400<x<2600 nên xϵ 2197 vì không dùng được dấu ngoặc nhọn nên dùng ngoặc tròn
a, \(x\) ⋮ 39; \(x\) ⋮ 65; \(x\) ⋮ 91; ⇒ \(x\) \(\in\)B(39; 65; 91)
39 = 3.13; 65 = 5.13; 91 = 7.13
⇒ BCNN(39; 65; 91) = 3.5.7.13 = 1365
⇒ \(x\) \(\in\)BC(39; 65; 91) = {0; 1365; 2730;...;}
mà 400 < \(x\) < 2600
⇒ \(x\) = 1365
b, \(x\) ⋮ 12; \(x\)⋮ 21; \(x\) ⋮ 28 ⇒\(x\) \(\in\) BC(12; 21; 28)
12 = 22.3; 21 = 3.7; 28 = 22.7 ⇒ BCNN(12; 21; 28) = 22.3.7=84
\(x\) \(\in\) BC(12; 21; 28) = {0; 84; 168; 252;336; 420; 504;...}
Mà \(x\) < 500 nên \(x\) \(\in\) {0; 84; 168; 252; 336; 420}
Ta có:
\(y'=\left(3^{x+1}\right)'\)
\(=3^{x+1}ln3\)
\(\Rightarrow A\)
-Chúc bạn học tốt-
Lời giải chi tiết:
84 = 80 + 4 | 42 = 40 + 2 | 55 = 50 + 5 |
77 = 70 + 7 | 91 = 90 + 1 | 39 = 30 + 9 |
28 = 20 + 8 | 63 = 60 + 3 | 99 = 90 + 9 |
84=80+4 42=40+2 55=50+5
77=70+7 91=90+1 39=30+9
28=20+8 63=60+3 99=90+9
\(\dfrac{x+3}{97}+\dfrac{x+5}{95}+\dfrac{x+9}{91}=\dfrac{x+91}{9}+\dfrac{x+92}{8}+\dfrac{x+61}{39}\)
=> \(\dfrac{x+3}{97}+1+\dfrac{x+5}{95}+1+\dfrac{x+9}{91}+1=\dfrac{x+91}{9}+1+\dfrac{x+92}{8}+1+\dfrac{x+61}{39}+1\)
=> \(\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}=\dfrac{x+100}{9}+\dfrac{x+100}{8}+\dfrac{x+100}{39}\)
=> \(\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}-\dfrac{x+100}{9}-\dfrac{x+100}{8}-\dfrac{x+100}{39}=0\)
=> \(\left(x+100\right).\left(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\right)=0\)
=> x = - 100 (do \(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\ne0\)
Ta có: \(\dfrac{x+3}{97}+\dfrac{x+5}{95}+\dfrac{x+9}{91}=\dfrac{x+91}{9}+\dfrac{x+92}{8}+\dfrac{x+61}{39}\)
\(\Leftrightarrow\dfrac{x+3}{97}+1+\dfrac{x+5}{95}+1+\dfrac{x+9}{91}+1=\dfrac{x+91}{9}+1+\dfrac{x+92}{8}+1+\dfrac{x+61}{39}+1\)
\(\Leftrightarrow\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}=\dfrac{x+100}{9}+\dfrac{x+100}{8}+\dfrac{x+100}{39}\)
\(\Leftrightarrow\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}-\dfrac{x+100}{9}-\dfrac{x+100}{8}-\dfrac{x+100}{39}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\right)=0\)
mà \(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\ne0\)
nên x+100=0
hay x=-100
Vậy: S={-100}