Giải bất phương trình:
\(\frac{x-15}{173}+\frac{x-13}{171}\le\frac{x-11}{169}+\frac{x-9}{167}\)
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20 tháng 9 2020
1) \(\frac{x-1}{x+3}-\frac{x}{x-3}=\frac{4x+15}{9-x^2}\)
ĐKXĐ : \(x\ne\pm3\)
\(\Leftrightarrow\frac{x-1}{x+3}-\frac{x}{x-3}=\frac{-4x-15}{x^2-9}\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{-4x-15}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow\frac{x^2-4x+3}{\left(x-3\right)\left(x+3\right)}-\frac{x^2+3x}{\left(x-3\right)\left(x+3\right)}=\frac{-4x-15}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow\frac{x^2-4x+3-x^2-3x}{\left(x-3\right)\left(x+3\right)}=\frac{-4x-15}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow-7x+3=-4x-15\)
\(\Leftrightarrow-7x+4x=-15-3\)
\(\Leftrightarrow-3x=-18\)
\(\Leftrightarrow x=6\)( tmđk )
Vậy x = 6 là nghiệm của phương trình
2) 2x + 3 < 6 - ( 3 - 4x )
<=> 2x + 3 < 6 - 3 + 4x
<=> 2x - 4x < 6 - 3 - 3
<=> -2x < 0
<=> x > 0
Vậy nghiệm của bất phương trình là x > 0
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.
7 tháng 4 2019
ta có:
\(\frac{x+2}{2013}+\frac{x+5}{2010}>\frac{x+8}{2007}+\frac{x+11}{2004}\)
\(\Leftrightarrow\left(\frac{x+2}{2013}+1\right)+\left(\frac{x+5}{2010}+1\right)>\left(\frac{x+8}{2007}+1\right)+\left(\frac{x+11}{2004}+1\right)\)
\(\Leftrightarrow\frac{x+2015}{2013}+\frac{x+2015}{2010}>\frac{x+2015}{2007}+\frac{x+2015}{2004}\)
\(\Leftrightarrow\frac{x+2015}{2013}+\frac{x+2015}{2010}-\frac{x+2015}{2007}-\frac{x+2015}{2004}>0\)
\(\Leftrightarrow\left(x+2015\right)\left(\frac{1}{2013}+\frac{1}{2010}-\frac{1}{2007}-\frac{1}{2004}\right)>0\)
\(\Rightarrow\orbr{\begin{cases}\hept{\begin{cases}x+2015>0\\\frac{1}{2013}+\frac{1}{2010}-\frac{1}{2007}-\frac{1}{2004}>0\end{cases}}\\\hept{\begin{cases}x+2015< 0\\\frac{1}{2013}+\frac{1}{2010}-\frac{1}{2007}-\frac{1}{2004}< 0\end{cases}}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}\hept{\begin{cases}x+2015>0\\\frac{1}{2013}+\frac{1}{2010}-\frac{1}{2007}-\frac{1}{2004}>0\end{cases}}\\\hept{\begin{cases}x+2015< 0\\\frac{1}{2013}+\frac{1}{2010}-\frac{1}{2007}-\frac{1}{2004}< 0\end{cases}}\end{cases}}\)
DD
8 tháng 5 2017
Cái bài đầu giải BPT bn ghi cái dj ak ,mik cx k hỉu nữa
V mik giải bài 2 nghen, sửa lại đề bài đầu rồi mik giải cho
\(3x-3=|2x+1|\)
Điều kiện: \(3x-3\ge0\Leftrightarrow3x\ge3\Leftrightarrow x\ge1\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=3x-3\\2x+1=-3x+3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3x=-1-3\\2x+3x=-1+3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-x=-3\\5x=2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\left(n\right)\\x=\frac{2}{5}\left(l\right)\end{cases}}}\)
Vậy S={3}
Cài đề câu b ,bn xem lại nhé!
8 tháng 5 2017
\(\frac{2x-3}{35}+\frac{x\left(x-2\right)}{7}>\frac{x^2}{7}-\frac{2x-3}{5}\)
\(\Leftrightarrow\frac{2x-3}{35}+\frac{5x\left(x-2\right)}{35}-\frac{5x^2}{35}+\frac{7\left(2x-3\right)}{35}>0\)
\(\Leftrightarrow2x-3+5x\left(x-2\right)-5x^2+7\left(2x-3\right)>0\)
\(\Leftrightarrow2x-3+5x^2-10x-5x^2+14x-21>0\)
\(\Leftrightarrow6x-24>0\)
\(\Leftrightarrow x>4\)
VẬY TẬP NGHIỆM CỦA BẤT PHƯƠNG TRÌNH LÀ : S = { \(x\text{\x}>4\)}
\(\frac{6x+1}{18}+\frac{x+3}{12}\le\frac{5x+3}{6}+\frac{12-5x}{9}\)
\(\Leftrightarrow\frac{6\left(6x+1\right)}{108}+\frac{9\left(x+3\right)}{108}\le\frac{18\left(5x+3\right)}{108}+\frac{12\left(12-5x\right)}{108}\)
\(\Leftrightarrow36x+6+9x+27\le90x+54+144-60x\)
\(\Leftrightarrow36x+6+9x+27-90x-54-144+60x\le0\)
\(\Leftrightarrow15x-165\le0\)
\(\Leftrightarrow x\le11\)
VẬY TẬP NGHIỆM CỦA BẤT PHƯƠNG trình ..........
tk mk nka !!! chúc bạn học tốt !!!
5 tháng 5 2020
Câu 1:
PT <=> \(\left(\frac{x-85}{15}-1\right)+\left(\frac{x-74}{13}-2\right)+\left(\frac{x-67}{11}-3\right)+\left(\frac{x-64}{9}-4\right)=0\)
<=> \(\frac{x-100}{15}+\frac{x-100}{13}+\frac{x-100}{11}+\frac{x-100}{9}=0\)
<=> \(\left(x-100\right)\left(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\right)=0\)
Mà \(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\ne0\)
<=> x - 100 = 0
<=> x = 100
Câu 2
PT <=> \(\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)=\left(\frac{x+6}{94}+1\right)+\left(\frac{x+8}{92}+1\right)\)
<=> \(\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)
<=> \(\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)
Mà \(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0\)
<=> x + 100 = 0
<=> x = -100
Câu 3:
PT <=> \(\left(\frac{x+2}{13}+1\right)+\left(\frac{2x+45}{15}-1\right)=\left(\frac{3x+8}{37}+1\right)+\left(\frac{4x+69}{9}-1\right)\)
<=> \(\frac{x+15}{13}+\frac{2\left(x+15\right)}{15}=\frac{3\left(x+15\right)}{37}+\frac{4\left(x+15\right)}{9}\)
<=> \(\left(x+15\right)\left(\frac{1}{13}+\frac{2}{15}-\frac{3}{37}-\frac{4}{9}\right)=0\)
mà \(\frac{1}{13}+\frac{2}{15}-\frac{3}{37}-\frac{4}{9}\ne0\)
<=> x+15 = 0
<=> x = -15
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 5 2020
1/
\(\Leftrightarrow\frac{x-85}{15}-1+\frac{x-74}{13}-2+\frac{x-67}{11}-3+\frac{x-64}{9}-4=0\)
\(\Leftrightarrow\frac{x-100}{15}+\frac{x-100}{13}+\frac{x-100}{11}+\frac{x-100}{9}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\right)=0\)
\(\Rightarrow x=100\)
2/
\(\frac{x+2}{98}+1+\frac{x+4}{96}+1-1-\frac{x+6}{94}-1-\frac{x+8}{92}=0\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)
\(\Rightarrow x=-100\)
3/
\(\frac{x+2}{13}+1+\frac{2x+45}{15}-1-1-\frac{3x+8}{37}+1-\frac{4x+69}{9}=0\)
\(\Leftrightarrow\frac{x+15}{13}+\frac{2\left(x+15\right)}{15}-\frac{3\left(x+15\right)}{37}-\frac{4\left(x+15\right)}{9}=0\)
\(\Leftrightarrow\left(x+15\right)\left(\frac{1}{13}+\frac{2}{15}-\frac{3}{37}-\frac{4}{9}\right)=0\)
\(\Rightarrow x=-15\)
N
4 tháng 4 2019
a)\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\right)=0\)(Trừ từng số hạng cho 1;2;3;4 rồi nhóm)
Vậy x=100.
b)\(\Leftrightarrow\left(x-14\right)\left(\frac{1}{13}-\frac{1}{15}-\frac{1}{27}+\frac{1}{29}\right)=0\)(Trừ từng số cho 1)
Vậy x=14.
7 tháng 8 2020
Bài làm:
PT:
đkxđ: \(x\ne0;x\ne2\)
Ta có: \(\frac{x+2}{x-2}=\frac{2}{x^2-2x}+\frac{1}{x}\)
\(\Leftrightarrow\frac{x\left(x+2\right)}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}+\frac{x-2}{x\left(x-2\right)}\)
\(\Rightarrow x^2+2x=2+x-2\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\left(vl\right)\\x+1=0\end{cases}}\Rightarrow x=-1\)
BPT:
Ta có: \(\frac{x+1}{2}-x\le\frac{1}{2}\)
\(\Leftrightarrow\frac{x+1}{2}-x-\frac{1}{2}\le0\)
\(\Leftrightarrow\frac{x+1-2x-1}{2}\le0\)
\(\Leftrightarrow\frac{-x}{2}\le0\)
\(\Rightarrow-x\le0\)
\(\Rightarrow x\ge0\)
MN
7 tháng 8 2020
a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne2\end{cases}}\)
\(\frac{x+2}{x-2}=\frac{2}{x^2-2x}+\frac{1}{x}\)
\(\Leftrightarrow\frac{2}{x\left(x-2\right)}+\frac{1}{x}-\frac{x+2}{x-2}=0\)
\(\Leftrightarrow\frac{2+x-2-x^2-2x}{x\left(x-2\right)}=0\)
\(\Leftrightarrow-x^2-x=0\)
\(\Leftrightarrow-x\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{cases}}}\)
Vậy \(S=\left\{-1\right\}\)
b) \(\frac{x+1}{2}-x\le\frac{1}{2}\)
\(\Leftrightarrow x+1-2x-1\le0\)
\(\Leftrightarrow-x\le0\)
\(\Leftrightarrow x\ge0\)
Vậy \(x\ge0\)
\(\frac{x-15}{173}+\frac{x-13}{171}\le\frac{x-11}{169}+\frac{x-9}{167}\)
\(\Leftrightarrow\frac{x-15}{173}+1+\frac{x-13}{171}+1\le\frac{x-11}{169}+1+\frac{x-9}{167}+1\)
\(\Leftrightarrow\frac{x-15+173}{173}+\frac{x-13+171}{171}\le\frac{x-11+169}{169}+\frac{x-9+167}{167}\)
\(\Leftrightarrow\frac{x+158}{173}+\frac{x+158}{171}-\frac{x+158}{169}-\frac{x+158}{167}\le0\)
\(\Leftrightarrow\left(x+158\right)\left(\frac{1}{173}+\frac{1}{171}-\frac{1}{169}-\frac{1}{167}\right)\le0\Rightarrow x+158\le0\)
\(\Leftrightarrow x\le-158\)Vậy tập nghiệm bfp là \(\left\{x|x\le-158\right\}\)