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Những câu hỏi liên quan
LV
4
TQ
1
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 1
\(\dfrac{x-90}{10}+\dfrac{x-76}{12}+\dfrac{x-58}{14}+\dfrac{x-36}{16}+\dfrac{x-15}{17}=15\)
\(\Leftrightarrow\dfrac{x-90}{10}-1+\dfrac{x-76}{12}-2+\dfrac{x-58}{14}-3+\dfrac{x-36}{16}-4+\dfrac{x-15}{17}-5=0\)
\(\Leftrightarrow\dfrac{x-100}{10}+\dfrac{x-100}{12}+\dfrac{x-100}{14}+\dfrac{x-100}{16}+\dfrac{x-100}{17}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\right)=0\)
\(\Leftrightarrow x-100=0\) (do \(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\ne0\))
\(\Leftrightarrow x=100\)
VQ
3
NH
29 tháng 3 2023
\(\dfrac{x-187}{13}+\dfrac{x-170}{15}+\dfrac{x-149}{17}+\dfrac{x-124}{19}=10\)
`<=>(x-187)/13+(x-170)/15+(x-149)/17+(x-124)/19-10=0`
`<=>(x-187)/13-1+(x-170)/15-2+(x-149)/17-3+(x-124)/19-4=0`
`<=>(x-200)/13+(x-200)/15+(x-200)/17+(x-200)/19=0`
`<=>(x-200)(1/13+1/15+1/17+1/19)=0`
`<=>x-200=0(1/13+1/15+1/17+1/19>0)`
`<=>x=200`
29 tháng 3 2023
\(=>\left(\dfrac{x-187}{13}-1\right)+\left(\dfrac{x-170}{15}-2\right)+\left(\dfrac{x-149}{17}-3\right)+\left(\dfrac{x-124}{19}-4\right)=0\)\(< =>\left(\dfrac{x-187}{13}-\dfrac{13}{13}\right)+\left(\dfrac{x-170}{15}-\dfrac{30}{15}\right)+\left(\dfrac{x-149}{17}-\dfrac{51}{17}\right)+\left(\dfrac{x-124}{19}-\dfrac{76}{19}\right)=0\)
\(< =>\left(\dfrac{x-200}{13}\right)+\left(\dfrac{x-200}{15}\right)+\left(\dfrac{x-200}{17}\right)+\left(\dfrac{x-200}{19}\right)=0\)
\(< =>\left(x-200\right)\left(\dfrac{1}{13}+\dfrac{1}{15}+\dfrac{1}{17}+\dfrac{1}{19}\right)=0\)
\(< =>x-200=0\)
<=>x=200
M
Giải Phương Trình Sau:
\(\dfrac{x-342}{15}+\dfrac{x-323}{17}+\dfrac{x-300}{19}+\dfrac{x-273}{21}=10\)
1
AH
Akai Haruma
Giáo viên
19 tháng 4 2021
Lời giải:
PT $\Leftrightarrow \frac{x-342}{15}-1+\frac{x-323}{17}-2+\frac{x-300}{19}-3+\frac{x-273}{21}-4=0$
$\Leftrightarrow \frac{x-357}{15}+\frac{x-357}{17}+\frac{x-357}{19}+\frac{x-357}{21}=0$
$(x-357)\left(\frac{1}{15}+\frac{1}{17}+\frac{1}{19}+\frac{1}{21}\right)=0$
Dễ thấy: $\frac{1}{15}+\frac{1}{17}+\frac{1}{19}+\frac{1}{21}\neq 0$
$\Rightarrow x-357=0$
$\Rightarrow x=357$
15 tháng 3 2020
\(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15\)
\(\Leftrightarrow\frac{x-90}{10}-1+\frac{x-76}{12}-2+\frac{x-58}{14}-3+\frac{x-36}{16}-4+\frac{x-15}{17}-5=0\)
\(\Leftrightarrow\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)
có : \(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\)
\(\Leftrightarrow x-100=0\)
\(\Leftrightarrow x=100\)
15 tháng 3 2020
\(pt\)\(\Leftrightarrow\)\(({x-90\over10}-1)+({x-76\over12}-2)+\)\(+({x-58\over14}-3)+({x-36\over16}-4)+({x-15\over17}-5)=0\)
\(\Leftrightarrow\)\(({x-100\over10})+({x-100\over12})+({x-100\over14})+({x-100\over16})\)
\(+({x-100\over17})=0\)
\(\Leftrightarrow\)\((x-100)({1\over10}+{1\over12}+{1\over14}+{1\over16}+{1\over17})=0\)
\(\Rightarrow\)\(x-100=0\)
\(\Rightarrow\)\(x=100\)
26 tháng 7 2023
a: \(=\dfrac{3}{5}:\dfrac{7}{5}=\dfrac{3}{5}\cdot\dfrac{5}{7}=\dfrac{3}{7}\)
b: \(=\dfrac{9}{17}\left(\dfrac{8}{5}-\dfrac{3}{5}\right)+\dfrac{8}{17}\)
=9/17+8/17=1
c: =>x-3/10=7/15*1/5=7/75
=>x=7/75+3/10=59/150
NC
1
5 tháng 1 2020
Bài làm
\(\frac{x+19}{27}-\frac{x+17}{29}=\frac{x+15}{31}-\frac{x+13}{33}\)
\(\Leftrightarrow\left(\frac{x+19}{27}+1\right)-\left(\frac{x+17}{29}+1\right)=\left(\frac{x+15}{31}+1\right)-\left(\frac{x+13}{33}+1\right)\)
\(\Leftrightarrow\frac{x+46}{27}-\frac{x+46}{29}=\frac{x+46}{31}-\frac{x+46}{33}\)
\(\Leftrightarrow\left(x+46\right).\frac{1}{27}-\left(x+46\right).\frac{1}{29}=\left(x+46\right).\frac{1}{31}-\left(x+46\right).\frac{1}{33}\)
\(\Leftrightarrow\left(x+46\right).\frac{1}{27}-\left(x+46\right).\frac{1}{29}-\left(x+46\right).\frac{1}{31}+\left(x+46\right).\frac{1}{33}=0\)
\(\Leftrightarrow\left(x+46\right)\left(\frac{1}{27}-\frac{1}{29}-\frac{1}{31}\right)=0\)
Mà \(\left(\frac{1}{27}-\frac{1}{29}-\frac{1}{31}\right)>0\forall x\)
\(\Leftrightarrow x+46=0\)
\(\Leftrightarrow x=-46\)
Vậy phương trình trên có tập nghiệm S = { -46 }
# Học tốt #
ĐKXĐ: \(x\ge15\)
Đặt \(\sqrt{x-15}=t\ge0\Rightarrow x=t^2+15\)
Pt trở thành:
\(t^2+15-t=17\Leftrightarrow t^2-t-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=-1< 0\left(loại\right)\\t=2\end{matrix}\right.\)
\(\Rightarrow\sqrt{x-15}=2\Rightarrow x=19\)