simple equation
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Kudo Shinichi 21/03/2017 at 22:40
Ta có :
\(\dfrac{2}{3}+\dfrac{m}{4}+\dfrac{n}{6}=6\Rightarrow\dfrac{m}{4}+\dfrac{n}{6}=5\dfrac{1}{3}\)
=> m/4 = 2/6 = 2/3
m/4 = 16/3 : ( 3+2 ) x 2 = 32/15
n/6 = 16/3-32/15 = 16/5
=> m x n = 32 x 16 = 512
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FA KAKALOTS 03/02/2018 at 12:39
Ta có :
23+m4+n6=6⇒m4+n6=513
=> m/4 = 2/6 = 2/3
m/4 = 16/3 : ( 3+2 ) x 2 = 32/15
n/6 = 16/3-32/15 = 16/5
=> m x n = 32 x 16 = 512
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Tạ Đức Duy 23/06/2017 at 23:12
why m/4 =2/6=2/3 what what ?? i think you wrong ;(
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Hồ Thu Giang 19/03/2017 at 09:36
- If a + b + c \(\ne\)0 then
\(\dfrac{2a}{b+c}=\dfrac{2b}{a+c}=\dfrac{2c}{a+b}=\dfrac{2a+2b+2c}{b+c+a+c+a+b}=\dfrac{2a+2b+2c}{2a+2b+2c}=1\)
=> \(m=1\)
- If a + b + c = 0 then
b + c = -a
a + c = -b
a + b = -c
=> \(\dfrac{2a}{b+c}=\dfrac{2b}{a+b}=\dfrac{2c}{a+b}=\dfrac{2a}{-a}=\dfrac{2b}{-b}=\dfrac{2c}{-c}=-2\)
=> \(m=-2\)
Jeff Bezos selected this answer. -
FA KAKALOTS 03/02/2018 at 12:39
- If a + b + c ≠
0 then
2ab+c=2ba+c=2ca+b=2a+2b+2cb+c+a+c+a+b=2a+2b+2c2a+2b+2c=1
=> m=1
- If a + b + c = 0 then
b + c = -a
a + c = -b
a + b = -c
=> 2ab+c=2ba+b=2ca+b=2a−a=2b−b=2c−c=−2
=> m=−2
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\(\dfrac{3-x}{5}+\dfrac{x+y}{10}=\dfrac{x-y}{4}\)
\(\Leftrightarrow\dfrac{12-4x}{20}+\dfrac{2x+2y}{20}=\dfrac{5x-5y}{20}\)
\(\Leftrightarrow12-4x+2x+2y=5x-5y\)
\(\Leftrightarrow2y+5y=5x+4x-2x-12\)
\(\Leftrightarrow7y=7x-12\Leftrightarrow y=x-\dfrac{12}{7}\)
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FA KAKALOTS 03/02/2018 at 12:40
3−x5+x+y10=x−y4
⇔12−4x20+2x+2y20=5x−5y20
⇔12−4x+2x+2y=5x−5y
⇔2y+5y=5x+4x−2x−12
⇔7y=7x−12⇔y=x−127
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Hồ Thu Giang 19/03/2017 at 09:42
If x = 3
=> \(\dfrac{a-3}{3}=\dfrac{3b-5}{5}\)
=> \(5\left(a-3\right)=3\left(3b-5\right)\)
=> 5a - 15 = 9b - 15
=> 5a = 9b
=> \(\dfrac{a}{b}=\dfrac{9}{5}\)
=> \(\dfrac{b}{a}=\dfrac{5}{9}\)
=> \(\dfrac{a}{b}-\dfrac{b}{a}=\dfrac{9}{5}-\dfrac{5}{9}=\dfrac{56}{45}\)
Jeff Bezos selected this answer. -
FA KAKALOTS 03/02/2018 at 12:40
If x = 3
=> a−33=3b−55
=> 5(a−3)=3(3b−5)
=> 5a - 15 = 9b - 15
=> 5a = 9b
=> ab=95
=> ba=59
=> ab−ba=95−59=5645
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FA KAKALOTS 03/02/2018 at 12:40
i will you to make this topic :)
replace x=1 into the polution above , we have
p+4q=161
=> q= (161-p)/4
=> q=40-(p-1)/4 => q<40(1)
because the p and q are primes so (p-1) divisible for 4 =>p-1 >=4 => p>=5 (2)
Merged (1) with (2) we have q ={2;3;5;7;11;13;17;19;23;29;31}
then you try to the each of case : it will have 2 cases (correct satisfies)
this case p=37 <=> q=31 and p=133 <=> q=7 -
Tạ Đức Duy 23/06/2017 at 23:10
i will you to make this topic :)
replace x=1 into the polution above , we have
p+4q=161
=> q= (161-p)/4
=> q=40-(p-1)/4 => q<40(1)
because the p and q are primes so (p-1) divisible for 4 =>p-1 >=4 => p>=5 (2)
Merged (1) with (2) we have q ={2;3;5;7;11;13;17;19;23;29;31}
then you try to the each of case : it will have 2 cases (correct satisfies)
this case p=37 <=> q=31 and p=133 <=> q=7
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We have : 7*(x*4) = 7*(3x - 20) = 21 - 5(3x - 20) = 121 - 15x
So 121 - 15x = 1 => 15x = 120 => x = 8
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Hồ Thu Giang 19/03/2017 at 09:46
(x + 2) : 2 = (5x - 8) : 4
=> \(\dfrac{x+2}{2}=\dfrac{5x-8}{4}\)
<=> 4(x + 2) = 2(5x - 8)
<=> 4x + 8 = 10x - 16
=> 24 = 6x
=> x = 4
Jeff Bezos selected this answer. -
FA KAKALOTS 03/02/2018 at 12:41
(x + 2) : 2 = (5x - 8) : 4
=> x+22=5x−84
<=> 4(x + 2) = 2(5x - 8)
<=> 4x + 8 = 10x - 16
=> 24 = 6x
=> x = 4
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17x - 85 = 85 - 17x
<=> 17x + 17x = 85 + 85
<=> 34x = 170
<=> x = 5
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We have : x*(1*2) = x*(3 - 2) = x*1 = 3x - 1
So 3x - 1 = 2 => 3x = 3 => x = 1
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\(\dfrac{x}{1.2}+\dfrac{x}{2.3}+\dfrac{x}{3.4}+...+\dfrac{x}{1999.2000}=1\)
\(x\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{1999.2000}\right)=1\)
\(x\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{1999}-\dfrac{1}{2000}\right)=1\)
\(x\left(1-\dfrac{1}{2000}\right)=1\)
\(x.\dfrac{1999}{2000}=1\)
\(x=\dfrac{2000}{1999}\)
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FA KAKALOTS 03/02/2018 at 12:41
x1.2+x2.3+x3.4+...+x1999.2000=1
x(11.2+12.3+13.4+...+11999.2000)=1
x(1−12+12−13+13−14+...+11999−12000)=1
x(1−12000)=1
x.19992000=1
x=20001999
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\(\dfrac{a-b}{b}=\dfrac{a}{b}-1\).So\(\dfrac{a}{b}-1=\dfrac{3}{5}\Rightarrow\dfrac{a}{b}=\dfrac{8}{5}\)
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Replace x = 2 into the given equations,we have :
\(\left\{{}\begin{matrix}2p+q=91\\p-q=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2p+p-2=91\\q=p-2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3p=93\\q=p-2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}p=31\\q=29\end{matrix}\right.\)=> pq = 899
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FA KAKALOTS 03/02/2018 at 12:42
Replace x = 2 into the given equations,we have :
{2p+q=91p−q=2⇒{2p+p−2=91q=p−2
⇒{3p=93q=p−2⇒{p=31q=29
=> pq = 899
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mathlove 17/03/2017 at 10:52
It is very easy. the done equation is writend as \(\left(x-3\right)\left(\dfrac{13}{6}-\dfrac{4}{3}-\dfrac{7}{6}\right)=8\)
\(\Leftrightarrow-\dfrac{1}{3}\left(x-3\right)=8\Leftrightarrow x=-21\) .
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FA KAKALOTS 03/02/2018 at 12:42
It is very easy. the done equation is writend as (x−3)(136−43−76)=8
⇔−13(x−3)=8⇔x=−21
.
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Ace Legona 17/03/2017 at 11:53
@mathlove: \(3-x\ne x-3\) ???
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mathlove 17/03/2017 at 11:02
Because (a+1) satisfies 2(x+1)=3(x-1), so that 2(a+1+1)=3a , a = 4.
With a = 4, the equation to solve become 2[3(2+x) - 2(4-x)] = 16 , then x = 2.
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FA KAKALOTS 03/02/2018 at 12:42
Because (a+1) satisfies 2(x+1)=3(x-1), so that 2(a+1+1)=3a , a = 4.
With a = 4, the equation to solve become 2[3(2+x) - 2(4-x)] = 16 , then x = 2.
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FA KAKALOTS 03/02/2018 at 12:42
Replace x = 3 into the equation,we have :
m−33+34=1712⇔m3−1=23⇔m3=53⇔m=5
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Replace x = 3 into the equation,we have :
\(\dfrac{m-3}{3}+\dfrac{3}{4}=\dfrac{17}{12}\Leftrightarrow\dfrac{m}{3}-1=\dfrac{2}{3}\Leftrightarrow\dfrac{m}{3}=\dfrac{5}{3}\Leftrightarrow m=5\)
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3 - a = 4 - b = 2 => a = 1 ; b = 2 => a + b + c = 502
=> m =\(\dfrac{502}{2008}=\dfrac{1}{4}\)
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»ﻲ†hïếu๖ۣۜGïลﻲ« 26/03/2017 at 10:25
Replacing x = 1/2 we have: 4x - 3a = 0
Have 4.1/2 - 3a = 0
Have 2 - 3a = 0
Have 3a = 2
Has a = 2/3
Dress a = 2/3 good
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Love people Name Jiang 17/03/2017 at 18:51
Replacing x = 1/2 we have: 4x - 3a = 0
<=> 4.1/2 - 3a = 0
=> 2 - 3a = 0
=> 3a = 2
=> a = 2/3
Dress a = 2/3 good