141 + 130 x 3 = \(\dfrac{x+280}{x}\) + 390
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`@` `\text {Ans}`
`\downarrow`
`a,`
`x - 280 \div 35 = 554`
`=> x - 8 = 554`
`=> x = 554 + 8`
`=> x = 562`
Vậy, `x = 562`
`b,`
`x + (5 \div 3) = 500`
`=> x + 5/3 = 500`
`=> x = 500 - 5/3`
`=> x = 1495/3`
Vậy, `x = 1495/3`
`c,`
`390 \div (5x - 5) = 39`
`=> 5x - 5 = 390 \div 39`
`=> 5x - 5 = 10`
`=> 5x = 10 + 5`
`=> 5x = 15`
`=> x = 15 \div 5`
`=> x = 3`
Vậy, `x = 3`
`d,`
`34x - 14x = 200`
`=> (34 - 14)x = 200`
`=> 20x = 200`
`=> x = 200 \div 20`
`=> x = 10`
Vậy, `x = 10.`
`@` `\text {Kaizuu lv uuu}`
a) \(x-280:35=554\)
\(x-8=554\)
\(x=562\)
b) \(x+\left(5:3\right)=500\)
\(x+\dfrac{5}{3}=500\)
\(x=500-\dfrac{5}{3}=\dfrac{1500}{3}-\dfrac{5}{3}=\dfrac{1495}{3}\)
c) \(390:\left(5x-5\right)=39\)
\(5x-5=\dfrac{39}{390}=\dfrac{1}{10}\)
\(5x=5+\dfrac{1}{10}\)
\(5x=\dfrac{51}{10}\)
\(x=\dfrac{51}{10}.\dfrac{1}{5}=\dfrac{51}{50}\)
d) \(34x-14x=200\)
\(20x=200\)
\(x=200:20=10\)
`3 xx (x/4 + x/28 + x/70 + x/130) =60/13`
`x/4 + x/28 + x/70 + x/130 = 20/13`
`x xx (1/4+1/28+1/70+1/130)=20/13`
`x xx 4/13= 20/13`
`x=5`
`3 xx (x/4 + x/28 + +x/70 +x/130) =60/13`
`x xx (1/4 + 1/28 + +1/70 +1/130)=20/13
`x xx 4/13 = 20/13`
`x=5`
ta có : \(\dfrac{392-x}{32}+\dfrac{390-x}{34}+\dfrac{388-x}{36}+\dfrac{386-x}{38}\)+\(\dfrac{384-x}{40}=-5\)
\(\Leftrightarrow\)\(\dfrac{392-x}{32}+1+\dfrac{390-x}{34}+1+\dfrac{388-x}{36}+1\)+\(\dfrac{384-x}{40}+1=0\)
\(\Leftrightarrow\)\(\dfrac{424-x}{32}+\dfrac{424-x}{34}+\dfrac{424-x}{36}+\dfrac{424-x}{38}+\dfrac{424-x}{40}=0\)\(\Leftrightarrow\left(424-x\right)\left(\dfrac{1}{32}+\dfrac{1}{34}+\dfrac{1}{36}+\dfrac{1}{38}+\dfrac{1}{40}\right)=0\)
\(\Leftrightarrow x=424\)(vì \(\dfrac{1}{32}+\dfrac{1}{34}+\dfrac{1}{36}+\dfrac{1}{38}+\dfrac{1}{40}\ne0\))
Vậy tập nghiệm của phương trình là s=\(\left\{424\right\}\)
Giải:
a) \(\dfrac{120^3}{40^3}=\left(\dfrac{120}{40}\right)^3=30^3=2700\)
b) \(\dfrac{390^4}{130^4}=\left(\dfrac{390}{130}\right)^4=30^4=810000\)
c) \(\dfrac{3^2}{\left(0,375\right)^2}=\left(\dfrac{3}{0,375}\right)^2=8^2=64\)
Đáp số: a) 2700; b) 810000; c) 64.
Chúc bạn học tốt!!!
a, \(\left\{{}\begin{matrix}x+y=4\\\left(x^2+y^2\right)\left(x^3+y^3\right)=280\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\\left(x^2+y^2\right)\left(x^2+y^2-xy\right)=70\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\\left(16-2xy\right)\left(16-3xy\right)=70\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\3x^2y^2-40xy+93=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\\left[{}\begin{matrix}xy=\dfrac{31}{3}\\xy=3\end{matrix}\right.\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x+y=4\\xy=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\\\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x+y=4\\xy=\dfrac{31}{3}\end{matrix}\right.\)
Phương trình này vô nghiệm
Vậy hệ đã cho có nghiệm \(\left(x;y\right)\in\left\{\left(1;3\right);\left(3;1\right)\right\}\)
b, ĐK: \(xy>0\)
\(\left\{{}\begin{matrix}\sqrt{\dfrac{2x}{y}}+\sqrt{\dfrac{2y}{x}}=3\\x-y+xy=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2x}{y}+\dfrac{2y}{x}+4=9\\x-y+xy=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x^2+y^2\right)=5xy\\x-y+xy=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-y\right)\left(x-2y\right)=0\\x-y+xy=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}2x=y\\x=2y\end{matrix}\right.\\x-y+xy=3\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}y=2x\\x-y+xy=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=2x\\2x^2-x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=2x\\\left(x+1\right)\left(2x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=-2\\x=-1\end{matrix}\right.\\\left\{{}\begin{matrix}y=3\\x=\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x=2y\\x-y+xy=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2y\\2y^2+y-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=3\\y=\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\)
Vậy ...
Bài làm :
\(a,\left[\left(250-25\right):15\right]:x=\left(450-60\right):130\)
\(\Rightarrow\left(225:15\right):x=390:130\)
\(\Rightarrow15:x=3\)
\(\Rightarrow x=15:3\)
\(\Rightarrow x=5\)
\(b,390-\left(x-7\right)=169:13\)
\(\Rightarrow390-\left(x-7\right)=13\)
\(\Rightarrow x-7=390-13\)
\(\Rightarrow x-7=377\)
\(\Rightarrow x=377+7\)
\(\Rightarrow x=384\)
\(c,x-6:2-\left(48-24\right):2:6-3=0\)
\(\Rightarrow x-6:2-24:2:6=3\)
\(\Rightarrow x-3-2=3\)
\(\Rightarrow x=3+2+3\)
\(\Rightarrow x=8\)
Học tốt nhé
a) ((250-25):15):x=(450-60):130
(225:15):x=390:130
15:x=3
x=15:3=5
b) 390-(x-7)=169:13
390-x+7=13
x+7=390-13=377
x=377-7=370
c)x-6:2-(48-24):2:6-3=0
x-3-24:2:6-3=0
x-3-2-3=0
x-8=0
x=8
tk cho mk :>>
9) Theo bài, ta có : 5x = 4y
\(\Rightarrow\dfrac{x}{4}=\dfrac{y}{5}\)
Mà y - x = 7
Theo tính chất dãy tỉ số bằng nhau, ta có :
\(\dfrac{x}{4}=\dfrac{y}{5}=\dfrac{y-x}{5-4}=\dfrac{7}{1}=7\)
Do đó : \(\dfrac{x}{4}=7\Rightarrow x=7.4=28\)
\(\dfrac{y}{5}=7\Rightarrow y=7.5=35\)
Vậy x = 28 ; y = 35
=>280/x+1+390=141+390
=>280/x+391=140+391
=>280/x=140
hay x=2