Tìm x, biết:
2/(x+2).(x+4) + 4/(x+4).(x+8) + 8/(x+8).(x+16) = x/(x+2).(x+14)
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Tìm x biết:
2/(x+2)(x+4)+ 4/(x+4)(x+8)+6/(x+8)(x+14)=x/(x+2)(x+14)
2. Tìm x biết:a)2(x+2)(x+4)dfrac{2}{left(x+2right)left(x+4right)} + 4(x+4)(x+8)dfrac{4}{left(x+4right)left(x+8right)} + 6(x+8)(x+14)dfrac{6}{left(x+8right)left(x+14right)} x(x+2)(x+14)dfrac{x}{left(x+2right)left(x+14right)}b)x2023dfrac{x}{2023} + x+12022dfrac{x+1}{2022}+ x+22021dfrac{x+2}{2021} +...+ x+20221dfrac{x+2022}{1} + 2023 0.Gíup mình giải 2 bài này với!Cảm ơn các bạn rất nhiều!!!
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2. Tìm x biết:
a)2(x+2)(x+4)\dfrac{2}{\left(x+2\right)\left(x+4\right)} + 4(x+4)(x+8)\dfrac{4}{\left(x+4\right)\left(x+8\right)} + 6(x+8)(x+14)\dfrac{6}{\left(x+8\right)\left(x+14\right)} = x(x+2)(x+14)\dfrac{x}{\left(x+2\right)\left(x+14\right)}
b)x2023\dfrac{x}{2023} + x+12022\dfrac{x+1}{2022} x+22021\dfrac{x+2}{2021} +...+ x+20221\dfrac{x+2022}{1} + 2023 = 0.
Gíup mình giải 2 bài này với!
Cảm ơn các bạn rất nhiều!!!
tìm x:
8.(x-6)+4.(x-3)-2.(+7)=-x+4
-7.(8-x)-6.(9+x)=20-x
9.(x-7)-20.(4-3x)=-7x+15
-16.(6+x)-30.(2-x)=-40x-16
17.(x-6)-14.(x+2)=4.(x-6)-2.(x-2)
mn giải giúp mình vs ạ.Mình đang cần gấp!!!!
\(8x-48+4x-12-14=-x+4\)
\(\Leftrightarrow12x-75=-x+4\Leftrightarrow13x=79\Leftrightarrow x=\dfrac{79}{13}\)
\(-7\left(8-x\right)-6\left(x+9\right)=20-x\Leftrightarrow-56+7x-6x-54=20-x\)
\(\Leftrightarrow2x=130\Leftrightarrow x=65\)
\(9x-63-80+60x=-7x+15\Leftrightarrow76x=158\Leftrightarrow x=\dfrac{79}{38}\)
\(-96-16x-60+30x=-40x-16\Leftrightarrow54x=140\Leftrightarrow x=\dfrac{70}{27}\)
\(17x-102-14x-28=4x-24-2x+4\Leftrightarrow x=110\)
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Tìm x, biết:
a) x : 4 = 8
b) 4 × x = 12 + 8
c) 30 : x = 14 + 16
a) x : 4 = 8
x = 8 × 4
x = 32
b) 4 × x = 12 + 8
4 × x = 20
x = 20 : 4
x = 5
c) 30 : x = 14 + 16
30 : x = 30
x = 30 : 30
x = 1
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Tìm x: 2/(x+2)(x+4)+4/(x+4)(x+8)+6/(x+8)(x+14)=x/(x+2)(x+14)
\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
<=>\(\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
<=>\(\frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
<=>\(\frac{12}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
<=>x = 12
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\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Leftrightarrow\frac{12}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Leftrightarrow x=12\)
Vậy \(x=12\)
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nè cấc bạn sao đó lại bằng 12 khi ko có cái gì để suy ra
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DB
14 tháng 8 2023 lúc 10:52
2. Tìm x biết:
a)dfrac{2}{left(x+2right)left(x+4right)} + dfrac{4}{left(x+4right)left(x+8right)} + dfrac{6}{left(x+8right)left(x+14right)} dfrac{x}{left(x+2right)left(x+14right)}.
b)dfrac{x}{2023} + dfrac{x+1}{2022} + dfrac{x+2}{2021} +...+ dfrac{x+2022}{1} + 2023 0.
Gíup mình giải 2 bài này với!
Cảm ơn các bạn rất nhiều!!!
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2. Tìm x biết:
a)\(\dfrac{2}{\left(x+2\right)\left(x+4\right)}\) + \(\dfrac{4}{\left(x+4\right)\left(x+8\right)}\) + \(\dfrac{6}{\left(x+8\right)\left(x+14\right)}\) = \(\dfrac{x}{\left(x+2\right)\left(x+14\right)}\).
b)\(\dfrac{x}{2023}\) + \(\dfrac{x+1}{2022}\) + \(\dfrac{x+2}{2021}\) +...+ \(\dfrac{x+2022}{1}\) + 2023 = 0.
Gíup mình giải 2 bài này với!
Cảm ơn các bạn rất nhiều!!!
a/
\(VT=\dfrac{\left(x+4\right)-\left(x+2\right)}{\left(x+2\right)\left(x+4\right)}+\dfrac{\left(x+8\right)-\left(x+4\right)}{\left(x+4\right)\left(x+8\right)}+\dfrac{\left(x+14\right)-\left(x+8\right)}{\left(x+8\right)\left(x+14\right)}=\)
\(=\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+8}+\dfrac{1}{x+8}-\dfrac{1}{x+14}=\)
\(=\dfrac{1}{x+2}-\dfrac{1}{x+14}=\dfrac{12}{\left(x+2\right)\left(x+14\right)}\)
\(\Rightarrow\dfrac{12}{\left(x+2\right)\left(x+14\right)}=\dfrac{x}{\left(x+2\right)\left(x+14\right)}\left(x\ne-2;x\ne-14\right)\)
\(\Rightarrow x=12\)
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\(\dfrac{x}{2023}+\dfrac{x+1}{2022}+...+\dfrac{x+2022}{1}+2023=0\)
\(\dfrac{1}{2023}x+\dfrac{1}{2022}x+\dfrac{1}{2022}\cdot1+...+\dfrac{1}{1}x+\dfrac{1}{1}\cdot2022+2023=0\)
\(x\left(\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}\right)+\left(\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023\right)=0\)
\(x\left(\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}\right)=\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023\)
\(x=\dfrac{\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023}{\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}}\)
\(x=\dfrac{\dfrac{1}{2022}+\dfrac{2022}{2022}+\dfrac{2}{2021}+\dfrac{2021}{2021}+...+\dfrac{2022}{1}+\dfrac{1}{1}}{\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}}\)
\(x=\dfrac{\dfrac{2023}{2022}+\dfrac{2023}{2021}+...+\dfrac{2023}{1}}{\dfrac{1}{2022}+\dfrac{1}{2021}+...+\dfrac{1}{1}}=2023\)
Vậy x = 2023
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2 phần ( x + 2 ). ( x + 4 ) + 4 phần ( x + 4 ) .( x + 8 ) + 6 phần( x + 8 ) .( x + 14 ) = x phần( x + 2 ) ( x + 14 ) với ( x ∉ { − 2 , − 4 , − 8 , − 14 } ) .
Bài 3. Tìm x biết:
1) |x| = 10 2) |x - 8| = 0 3) 7 + |x| = 12
4) |x + 1| = 3 5) 15 - x = 16 - (14 - 42) 6) 210 - (x - 12) = 168
1 , 10
2 , 8
3 , 5
4 , 2
5 , 1
6 ,
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1/ `|x|=10<=> x=\pm 10`
2/ `|x-8|=0<=>x-8=0<=>x=8`
3/ `7+|x|=12<=>|x|=5<=>x=\pm 5`
4/ `|x+1|=3`
$\Leftrightarrow\left[\begin{array}{1}x+1=3\\x+1=-3\end{array}\right.\\\Leftrightarrow\left[\begin{array}{1}x=3\\x=-4\end{array}\right.$
5/ `15-x=16-(14-42)`
`<=>15-x=16+28`
`<=>15-x=44`
`<=>x=-29`
6/ `210-(x-12)=168`
`<=>210-x+12=168`
`<=>222-x=168`
`<=>x=54`
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1.
\(\left|x\right|=10\Leftrightarrow x=\pm10\)
2.
\(\left|x-8\right|=0\Leftrightarrow x-8=0\Leftrightarrow x=8\)
3.
\(7+\left|x\right|=12\Leftrightarrow\left|x\right|=5\Leftrightarrow x=\pm5\)
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Tinh gia tri cua phan so: A= 2 x 3 + 2 x 4 x 8 + 4 x 8 x 16 + 8 x 16 x 32/ 3x 4 + 2 x 6 x 8 + 4 x 12 x 16 + 8 x 24 x 32