n) 558 - (15 : x + 29).17 = 14
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AL
1
25 tháng 6 2023
558 - (15 : x + 29) x 17 = 14
\(\Rightarrow\left(15:x+29\right)\times17=558-14\)
\(\Rightarrow\left(15:x+29\right)\times17=544\)
\(\Rightarrow15:x+29=544:17\)
\(\Rightarrow15:x+29=32\)
\(\Rightarrow15:x=32-29\)
\(\Rightarrow15:x=3\)
\(\Rightarrow x=15:3\)
\(\Rightarrow x=5\)
Vậy x = 5
\(\left(x+40\right)\times15=75\times24\)
\(\Rightarrow\left(x+40\right)\times15=1800\)
\(\Rightarrow x+40=1800:15\)
\(\Rightarrow x+40=120\)
\(\Rightarrow x=120-40\)
\(\Rightarrow x=80\)
Vậy x = 80
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 \(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\ne0\))
\(\Leftrightarrow x=100\)
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 #
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
\(pt\)\(\Leftrightarrow\)\(({x-90\over10}-1)+({x-76\over12}-2)+\)\(+({x-58\over14}-3)+({x-36\over16}-4)+({x-15\over17}-5)=0\)
\(\Leftrightarrow\)\(({x-100\over10})+({x-100\over12})+({x-100\over14})+({x-100\over16})\)
\(+({x-100\over17})=0\)
\(\Leftrightarrow\)\((x-100)({1\over10}+{1\over12}+{1\over14}+{1\over16}+{1\over17})=0\)
\(\Rightarrow\)\(x-100=0\)
\(\Rightarrow\)\(x=100\)
4 tháng 3 2020
\(\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+17}{83}+\frac{x+116}{4}=0\)
\(\Leftrightarrow\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+17}{83}+\frac{x+100}{4}+4=0\)
\(\Leftrightarrow\left(\frac{x+14}{86}+1\right)+\left(\frac{x+15}{85}+1\right)+\left(\frac{x+14}{86}+1\right)+\left(\frac{x+13}{87}+1\right)+\frac{x+100}{4}=0\)
\(\Leftrightarrow\frac{x+100}{86}+\frac{x+100}{85}+\frac{x+100}{84}+\frac{x+100}{83}+\frac{x+100}{4}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow x+100=0\left(vì\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\ne0\right)\)
\(\Leftrightarrow x=-100\)
vậy.............................
1 tháng 7 2017
Ta có : 17 - 14(x + 1) = 13 - 4(x + 1) - 5(x - 3)
<=> 17 - 14x - 14 = 13 - 4x - 4 - 5x + 15
<=> -14x + 3 = -9x + 24
<=> -14x + 9x = 24 - 3
<=> -5x = 21
=> x = -4,2
1 tháng 7 2017
Ta có : 5x + 3,5 + (3x - 4) = 7x - 3(x - 0,5)
<=> 5x + 3,5 + 3x - 4 = 7x - 3x + 1,5
<=> 8x - 0,5 = 4x + 1,5
=> 8x - 4x = 1,5 + 0,5
=> 4x = 2
=> x = \(\frac{1}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 12 2020
ĐKXĐ: \(x\ge15\)
Đặt \(\sqrt{x-15}=t\ge0\Rightarrow x=t^2+15\)
Pt trở thành:
\(t^2+15-t=17\Leftrightarrow t^2-t-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=-1< 0\left(loại\right)\\t=2\end{matrix}\right.\)
\(\Rightarrow\sqrt{x-15}=2\Rightarrow x=19\)
XO
12 tháng 2 2020
Ta có : \(\frac{x+14}{186}+\frac{x+15}{185}+\frac{x+16}{184}+\frac{x+17}{183}+\frac{x+216}{4}=0\)
=> \(\frac{x+14}{186}+\frac{x+15}{185}+\frac{x+16}{184}+\frac{x+17}{183}+\frac{x+200+16}{4}=0\)
=> \(\frac{x+14}{186}+\frac{x+15}{185}+\frac{x+16}{184}+\frac{x+17}{183}+\frac{x+200}{4}+4=0\)
=> \(\left(\frac{x+14}{186}+1\right)+\left(\frac{x+15}{185}+1\right)+\left(\frac{x+16}{184}+1\right)+\left(\frac{x+17}{183}\right)+\frac{x+200}{4}=0\)
=> \(\frac{x+200}{186}+\frac{x+200}{185}+\frac{x+200}{184}+\frac{x+200}{183}+\frac{x+200}{4}=0\)
=> \(\left(x+200\right)\left(\frac{1}{186}+\frac{1}{185}+\frac{1}{184}+\frac{1}{183}+\frac{1}{4}\right)=0\)
Vì \(\frac{1}{186}+\frac{1}{185}+\frac{1}{184}+\frac{1}{4}\ne0\)
nên x + 200 = 0
=> x = - 200
Vậy x = - 200
W
12 tháng 2 2020
Từ đề bài, ta có:
\(1+\frac{x+14}{186}+1+\frac{x+15}{185}+1+\frac{x+16}{184}+1+\frac{x+17}{183}+1+\frac{x+216}{4}=5\)
\(\Leftrightarrow\frac{200+x}{186}+\frac{200+x}{185}+\frac{200+x}{184}+\frac{200+x}{183}+\frac{200+x}{4}=5\)
\(\Leftrightarrow\left(200+x\right)\left(\frac{1}{186}+\frac{1}{185}+\frac{1}{184}+\frac{1}{183}+\frac{1}{4}\right)=5\)
Bạn xem có sai đề bài không ạ :D Thiết nghĩ vế phải phải là 5 chứ. Nếu đề bài đúng thì đến bước trên bạn tự tính nhé. Lười tính :)
Chúc bạn học tốt!
4 tháng 2 2017
Phương trình 1:
\(\frac{x-85}{15}+\frac{x-74}{13}+\frac{x-67}{11}+\frac{x-64}{9}=10\)
\(\Rightarrow\frac{x-85}{15}+\frac{x-74}{13}+\frac{x-67}{11}+\frac{x-64}{9}-10=0\)
\(\Rightarrow\left(\frac{x-85}{15}-1\right)+\left(\frac{x-74}{13}-2\right)+\left(\frac{x-67}{11}-3\right)+\left(\frac{x-64}{9}-4\right)=0\)
\(\Rightarrow\frac{x-85-15}{15}+\frac{x-74-26}{13}+\frac{x-67-33}{11}+\frac{x-64-36}{9}=0\)
\(\Rightarrow\frac{x-100}{15}+\frac{x-100}{13}+\frac{x-100}{11}+\frac{x-100}{9}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\right)=0\)
Do \(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\ne0\)
\(\Rightarrow x-100=0\)
\(\Rightarrow x=100\)
Vậy x = 100.
4 tháng 2 2017
Phương trình 3:
\(\frac{1909-x}{91}+\frac{1907-x}{93}+\frac{1905-x}{95}+\frac{1903-x}{97}+4=0\)
\(\Rightarrow\left(\frac{1909-x}{91}+1\right)+\left(\frac{1907-x}{93}+1\right)+\left(\frac{1905-x}{95}+1\right)+\left(\frac{1903-x}{97}+1\right)=0\)
\(\Rightarrow\frac{1909-x+91}{91}+\frac{1907-x+93}{93}+\frac{1905-x+95}{95}+\frac{1903-x+97}{97}=0\)
\(\Rightarrow\frac{2000-x}{91}+\frac{2000-x}{93}+\frac{2000-x}{95}+\frac{2000-x}{97}=0\)
\(\Rightarrow\left(2000-x\right)\left(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\right)=0\)
Do \(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\ne0\)
\(\Rightarrow2000-x=0\)
\(\Rightarrow x=2000\)
Vậy x = 2000.
15 tháng 4 2021
10/29+11/29+12/29+...+19/29
=10+11+12+13+14+15+=16+17+18+19/29
=145/29
\(558-\left(15:x+29\right)\cdot17=14\)
=>\(\left(15:x+29\right)\cdot17=558-14=544\)
=>\(15:x+29=544:17=32\)
=>15:x=32-29=3
=>\(x=\dfrac{15}{3}=5\)