102 x 2 + 102 x 4 + 102 x 7 +102 x 8
nhớ diễn giải ra mình sẽ tick cho
^_^ >_<
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x=(1+2+3-4-5-6)+...+(97+98+99-100-101-102)
x=-9+...+-9
x=-9.17
x=-153
Ta có :\(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)
<=> \(\left(\frac{x-5}{100}-1\right)+\left(\frac{x-4}{101}-1\right)+\left(\frac{x-3}{102}-1\right)=\left(\frac{x-100}{5}-1\right)+\left(\frac{x-101}{4}-1\right)+\left(\frac{x-102}{3}-1\right)\)
<=> \(\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}=\frac{x-105}{5}+\frac{x-105}{4}+\frac{x-105}{3}\)
<=> \(\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}\right)=\left(x-105\right)\left(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}\right)\)
<=> \(\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)
<=> x - 105 = 0 (Vì \(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\ne0\))
<=> x = 105
Vậy nghiệm phương trình là x = 105
\(\frac{x-2016}{100}+\frac{x-2014}{102}+\frac{x-2016}{104}+...+\frac{x-2}{2114}=1008\)
\(\Rightarrow\frac{x-2016}{100}-1+\frac{x-2014}{102}-1+...+\frac{x-2}{2114}-1=0\)
\(\Rightarrow\frac{x-2116}{100}+\frac{x-2116}{102}+...+\frac{x-2116}{2114}=0\)
\(\Rightarrow\left(x-2116\right)\left(\frac{1}{100}+\frac{1}{102}+...+\frac{1}{2114}\right)=0\)
mà \(\frac{1}{100}+\frac{1}{102}+...+\frac{1}{2114}\ne0\)
\(\Rightarrow x-2116=0\)
\(\Rightarrow x=2116\)
P/s màu mè ghê ha =))
\(\frac{x-2016}{100}+\frac{x-2014}{102}+...+\frac{x-2}{2114}=1008\)
\(=>\frac{x-2016}{100}+\frac{x-2014}{102}+...+\frac{x-2}{2114}-1008=0\)
\(=>\frac{x-2016}{100}-1+\frac{x-2014}{102}-1+...+\frac{x-2}{2114}-1=0\)
\(=>\frac{x-2116}{100}+\frac{x-2116}{102}+...+\frac{x-2116}{2114}=0\)
\(=>\left(x-2116\right).\left(\frac{1}{100}+\frac{1}{102}+...+\frac{1}{2114}\right)=0\)
Do \(\frac{1}{100}+\frac{1}{102}+...+\frac{1}{2114}\ne0\)
\(=>x-2116=0\)
\(=>x=2116\)
\(\orbr{\begin{cases}x>3\\x< -4\end{cases}}\)Thiếu chỗ khai căn ra là :
Giá trị tuyệt đối của \(x+\frac{1}{2}\)\(>3,5\)
TH1 Khi \(x>0\)thì
\(x+\frac{1}{2}>3,5\Leftrightarrow x>3\)
TH2 Khi \(x< 0\)thì
\(-\left(x+\frac{1}{2}\right)>3,5\)
\(\Leftrightarrow-x-\frac{1}{2}>3,5\Leftrightarrow-x>4\)
\(\Leftrightarrow x< -4\)
Đó như vậy có hai cái nha :
\(\orbr{\begin{cases}x>3\\x< -4\end{cases}}\)
\(\left(x-3\right)\left(x+4\right)>0\)
\(\Leftrightarrow x^2+x-12>0\)
\(\Leftrightarrow x^2+2.\frac{1}{2}x+\frac{1}{4}-12,35>0\)
\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2-12,35>0\)
Bất đẳng thức lớn hơn 0 khi và chỉ khi
\(\left(x+\frac{1}{2}\right)^2>12,35\)
Khai căn hai vế ra tức là căn hai vế ý
\(x+\frac{1}{2}>3,5\)
\(\Leftrightarrow x>3\)
A={ 13;15;17;19;21;23;25;27;29;31;33;35;37;39;41;43;45;47;49;51;53;55;57;59;61;63;65;67;69;71;73;75;77;79;81;83;85;87;89;91;93;95;97;99;101}
102 x 2 + 102 x 4 + 102 x 7 +102 x 8
= 120 x ( 2 x 4 x 7 x 8 )
= 120 x 448
= 53760
102x(2+4+7+8)=102x21=2142