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1 cách cực ngu
\(\frac{x+107}{7}+\frac{x+6}{47}+\frac{x+1}{33}+\frac{x+184}{21}=0\)
\(\Leftrightarrow1551\left(x+107\right)+231\left(x+6\right)+329\left(x+1\right)+517\left(x+184\right)=0\)
\(\Leftrightarrow1551x+165957+231x+1386+329x+517x+95128=0\)
\(\Leftrightarrow2628x+262800=0\)
\(\Leftrightarrow2628x=-262800\)
\(\Leftrightarrow x=-100\)
Ta có : y=205x-1023
mà x=2015y-2031
\(\Rightarrow y=205.\left(2015y-2031\right)-1023\)
\(\Rightarrow y=413075y-416355-1023=413075y-417378\)
\(\Rightarrow417378=413075y-y=413074y\)
\(\Rightarrow y=\frac{417378}{413074}=1,010419441\approx1,01\)
Thay y= 1,010419441 vào x=2015-y-2031
\(\Rightarrow x=2015.1,010419441-2031=4,995172778\approx5\)
Vậy x= 5 ; y= 1,01
Chọn mình nha camr ơn, chúc bạn học tốt nha
x = 2015y - 2031 = 2015(205x - 1023) - 2031 = 413075x - 2061345 - 2031 = x + 413074x - 2063376
=> x = 2063376 : 413074 = \(4\frac{205540}{206537}\) => y = 205.\(4\frac{205540}{206537}\)- 1023 = \(1024\frac{2152}{206537}\)- 1023 = \(1\frac{2152}{206537}\)
Câu B đây;vừa bị lag
B, \(\frac{x+1}{35}\)+\(\frac{x+3}{33}\)=\(\frac{x+5}{31}\)+\(\frac{x+7}{29}\)
⇔ \(\frac{x+1}{35}\)+1+\(\frac{x+3}{33}\)+1=\(\frac{x+5}{31}\)+1+\(\frac{x+7}{29}\)+1
⇔ \(\frac{x+36}{35}\)+\(\frac{x+36}{33}\)-\(\frac{x+36}{31}\)-\(\frac{x+36}{29}\)=0
⇔ (x+36)(\(\frac{1}{35}\)+\(\frac{1}{33}\)-\(\frac{1}{31}\)-\(\frac{1}{29}\))=0
Mà \(\frac{1}{35}\)+\(\frac{1}{33}\)-\(\frac{1}{31}\)-\(\frac{1}{29}\)<0
⇔ x+36=0
⇔ x=-36
Vậy tập nghiệm của phương trình đã cho là:S={-36}
câu C tương tự nhé
\(\frac{x-1986-1987}{1985}+\frac{x-1985-1987}{1986}+\frac{x-1985-1986}{1987}=3\)
=> \(\left(\frac{x-1986-1987}{1985}-1\right)+\left(\frac{x-1985-1987}{1986}-1\right)+\left(\frac{x-1985-1986}{1987}-1\right)=3-3\)
=> \(\frac{x-1985-1986-1987}{1985}+\frac{x-1985-1986-1987}{1986}+\frac{x-1985-1986-1987}{1987}=0\)
=> \(\left(x-1985-1986-1987\right).\left(\frac{1}{1985}+\frac{1}{1986}+\frac{1}{1987}\right)=0\)
=> \(\left(x-5958\right).\left(\frac{1}{1985}+\frac{1}{1986}+\frac{1}{1987}\right)=0\)
Mà \(\frac{1}{1985}+\frac{1}{1986}+\frac{1}{1987}\ne0\)
=> x - 5958 = 0
=> x = 5958
\(\frac{x-21}{1999}+\frac{x-33}{1987}\le\frac{x+6}{2026}+\frac{x+11}{2031}\)
<=> \(\frac{x-21}{1999}-1+\frac{x-33}{1987}-1\le\frac{x+6}{2026}-1+\frac{x+11}{2031}-1\)
<,=>. \(\frac{x-2020}{1999}+\frac{x-2020}{1987}\le\frac{x-2020}{2026}+\frac{x-2020}{2031}\)
<=> \(\left(x-2020\right)\left(\frac{1}{1999}+\frac{1}{1987}-\frac{1}{2026}-\frac{1}{2031}\right)\le0\) (1)
Vì \(\frac{1}{1999}+\frac{1}{1987}-\frac{1}{2026}-\frac{1}{2031}\ge0\)
Nên (1) \(x-2020\le0\Leftrightarrow x\le2020\)