\(\dfrac{2}{(x + 2)(x + 4)}+\dfrac{4}{(x + 4)(x + 8)}+\dfrac{6}{(x + 8)(x + 14)}=\dfrac{x}{(x + 2)(x + 14)}\)
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AH
Akai Haruma
Giáo viên
16 tháng 9 2023
Lời giải:
Điều kiện: $x\neq -2; x\neq -2; x\neq -8; x\neq -14$
Đề bài
$\Rightarrow \frac{(x+4)-(x+2)}{(x+2)(x+4)}+\frac{(x+8)-(x+4)}{(x+4)(x+8)}+\frac{(x+14)-(x+8)}{(x+8)(x+14)}=\frac{x}{(x+2)(x+14)}$
$\Rightarrow \frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{(x+2)(x+14)}$
$\Rightarrow \frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{(x+2)(x+14)}$
$\Rightarrow \frac{12}{(x+2)(x+14)}=\frac{x}{(x+2)(x+14)}$
$\Rightarrow 12=x$ (thỏa mãn)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
14 tháng 8 2023
a/
\(VT=\dfrac{\left(x+4\right)-\left(x+2\right)}{\left(x+2\right)\left(x+4\right)}+\dfrac{\left(x+8\right)-\left(x+4\right)}{\left(x+4\right)\left(x+8\right)}+\dfrac{\left(x+14\right)-\left(x+8\right)}{\left(x+8\right)\left(x+14\right)}=\)
\(=\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+8}+\dfrac{1}{x+8}-\dfrac{1}{x+14}=\)
\(=\dfrac{1}{x+2}-\dfrac{1}{x+14}=\dfrac{12}{\left(x+2\right)\left(x+14\right)}\)
\(\Rightarrow\dfrac{12}{\left(x+2\right)\left(x+14\right)}=\dfrac{x}{\left(x+2\right)\left(x+14\right)}\left(x\ne-2;x\ne-14\right)\)
\(\Rightarrow x=12\)
14 tháng 8 2023
\(\dfrac{x}{2023}+\dfrac{x+1}{2022}+...+\dfrac{x+2022}{1}+2023=0\)
\(\dfrac{1}{2023}x+\dfrac{1}{2022}x+\dfrac{1}{2022}\cdot1+...+\dfrac{1}{1}x+\dfrac{1}{1}\cdot2022+2023=0\)
\(x\left(\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}\right)+\left(\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023\right)=0\)
\(x\left(\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}\right)=\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023\)
\(x=\dfrac{\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023}{\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}}\)
\(x=\dfrac{\dfrac{1}{2022}+\dfrac{2022}{2022}+\dfrac{2}{2021}+\dfrac{2021}{2021}+...+\dfrac{2022}{1}+\dfrac{1}{1}}{\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}}\)
\(x=\dfrac{\dfrac{2023}{2022}+\dfrac{2023}{2021}+...+\dfrac{2023}{1}}{\dfrac{1}{2022}+\dfrac{1}{2021}+...+\dfrac{1}{1}}=2023\)
Vậy x = 2023
AH
Akai Haruma
Giáo viên
13 tháng 7 2018
Lời giải:
PT \(\Leftrightarrow \frac{(x+4)-(x+2)}{(x+2)(x+4)}+\frac{(x+8)-(x+4)}{(x+4)(x+8)}+\frac{(x+14)-(x+8)}{(x+8)(x+14)}=\frac{x}{(x+2)(x+14)}\)
\(\Leftrightarrow \frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{(x+2)(x+14)}\)
\(\Leftrightarrow \frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{(x+2)(x+14)}\)
\(\Leftrightarrow \frac{12}{(x+2)(x+14)}=\frac{x}{(x+2)(x+14)}\)
\(\Rightarrow x=12\) (thỏa mãn)
Vậy......
11 tháng 3 2023
f: =>\(\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)
=>28-6(x+2)=-9-5(x-4)
=>28-6x-12=-9-5x+20
=>-6x+16=-5x+11
=>-x=-5
=>x=5
12 tháng 3 2023
d: =>\(\dfrac{12x+1}{11x-4}=\dfrac{20x+17-20x+8}{18}=\dfrac{25}{18}\)
=>25(11x-4)=18(12x+1)
=>275x-100=216x+18
=>59x=118
=>x=2
f: =>\(\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)
=>28-6(x+2)=-9-5(x-4)
=>28-6x-12=-9-5x+20
=>-6x+16=-5x+11
=>-x=-5
=>x=5
Y
17 tháng 1 2023
\(1,\left(dk:x\ne0,-1,4\right)\)
\(\Leftrightarrow\dfrac{9}{x+1}+\dfrac{2}{x-4}-\dfrac{11}{x}=0\)
\(\Leftrightarrow\dfrac{9x\left(x-4\right)+2x\left(x+1\right)-11\left(x+1\right)\left(x-4\right)}{x\left(x+1\right)\left(x-4\right)}=0\)
\(\Leftrightarrow9x^2-36x+2x^2+2x-11x^2+44x-11x+44=0\)
\(\Leftrightarrow-x=-44\)
\(\Leftrightarrow x=44\left(tm\right)\)
\(2,\left(đk:x\ne4\right)\)
\(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{2+x}{x-4}-\dfrac{3}{2\left(x-4\right)}+\dfrac{5}{6}=0\)
\(\Leftrightarrow\dfrac{14.2-6\left(2+x\right)-3.3+5\left(x-4\right)}{6\left(x-4\right)}=0\)
\(\Leftrightarrow28-12-6x-9+5x-20=0\)
\(\Leftrightarrow-x=13\)
\(\Leftrightarrow x=-13\left(tm\right)\)
HN
1
5 tháng 2 2021
ĐKXĐ: x\(\ne\)4
\(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{3}{2\left(4-x\right)}-\dfrac{5}{6}\)
\(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}+\dfrac{3}{2\left(x-4\right)}=-\dfrac{5}{6}\)
\(\Rightarrow14\cdot2-6\cdot\left(x+2\right)+3\cdot3=-5\left(x-4\right)\)
\(\Leftrightarrow28-6x-12+9=-5x+20\)
\(\Leftrightarrow25-6x=-5x+20\Leftrightarrow-6x+5x=20-25\Leftrightarrow-x=-5\Leftrightarrow x=5\left(TM\right)\)Vậy...
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+8}+\dfrac{1}{x+8}-\dfrac{1}{x+14}=\dfrac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Leftrightarrow\dfrac{x}{\left(x+2\right)\left(x+14\right)}=\dfrac{x+14-x-2}{\left(x+2\right)\left(x+14\right)}=\dfrac{12}{\left(x+2\right)\left(x+14\right)}\)
=>x=12