Cho A=\(\frac{109-x}{91}+\frac{107-x}{93}+\frac{105-x}{95}+\frac{103-x}{97}\)và B=-2
Tìm x để A=B
K
Khách
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2 tháng 2 2017
(109-x)/91+(107-x)/93+(105-x)/95+(103-x)/97=-4
[(109-x)/91 +1]+[(107-x)/93 +1]+[(105-x)/95 +1]+[(103-x)/97 +1]-4=-4
(109+91-x)/91+(107+93-x)/93+(105+95-x)/95+(103+97-x)/97=-4+4
(200-x)/91+(200-x)/93+(200-x)/95+(200-x)/97=0
(200-x)(1/91+1/93+1/95+1/97)=0
Ma : 1/91+1/93+1/95+1/97\(\ne\)0
=>200-x=0
=>x=200
KN
11 tháng 2 2020
\(\frac{x+1}{99}+\frac{x+2}{98}=\frac{x+3}{97}+\frac{x+4}{96}\)
\(\Rightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1=\frac{x+3}{97}+1+\frac{x+4}{96}+1\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}=\frac{x+100}{97}+\frac{x+100}{96}\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}-\frac{x+100}{97}-\frac{x+100}{96}=0\)
\(\Rightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)=0\)
Dễ thấy \(\left(\frac{1}{99}< \frac{1}{98}< \frac{1}{97}< \frac{1}{96}\right)\)nên \(\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)\ne0\)
\(\Rightarrow x+100=0\Rightarrow x=-100\)
Vậy x = -100
\(\frac{109-x}{91}+\frac{107-x}{93}+\frac{105-x}{95}+\frac{103-x}{97}+4=0\)
\(\Rightarrow\frac{109-x}{91}+1+\frac{107-x}{93}+1+\frac{105-x}{95}+1+\frac{103-x}{97}+1=0\)
\(\Rightarrow\frac{200-x}{91}+\frac{200-x}{93}+\frac{200-x}{95}+\frac{200-x}{97}=0\)
\(\Rightarrow\left(200-x\right)\left(\frac{1}{91}+\frac{1}{93}-\frac{1}{95}-\frac{1}{97}\right)=0\)
Dễ thấy \(\left(\frac{1}{91}>\frac{1}{93}>\frac{1}{95}>\frac{1}{97}\right)\)nên \(\left(\frac{1}{91}+\frac{1}{93}-\frac{1}{95}-\frac{1}{97}\right)\ne0\)
\(\Rightarrow200-x=0\Rightarrow x=200\)
Vậy x = 200
11 tháng 2 2020
b, \(\frac{x+1}{99}+1+\frac{x+2}{98}+1=\frac{x+3}{97}+1+\frac{x+4}{96}+1\)
\(\frac{x+200}{99}+\frac{x+200}{98}=\frac{x+200}{97}+\frac{x+200}{96}\)
\(\frac{x+200}{99}+\frac{x+200}{98}-\frac{x+200}{97}-\frac{x+200}{96}=0\)
\(\left(x+200\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)=0\)
mà\(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\ne0\)
==> x+200=0
<=>x=-200
Vậy nghiệm của phương trình là x=-200
11 tháng 2 2020
c, \(\frac{109-x}{91}+1+\frac{107-x}{93}+1+\frac{105-x}{95}+1+\frac{103-x}{97}+1=0\)
\(\frac{200-x}{91}+\frac{200-x}{93}+\frac{200-x}{95}+\frac{200-x}{97}=0\)
\(\left(200-x\right)\left(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\right)=0\)
mà \(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\ne0\)
==>200-x=0
<=>x=200
vậy nghiệm của pt là x=200
22 tháng 2 2018
Ta có: \(\frac{99-x}{101}+\frac{97-x}{103}+\frac{95-x}{105}+\frac{93-x}{107}=-4\)
\(\Leftrightarrow\frac{99-x}{101}-1+\frac{97-x}{103}-1+\frac{95-x}{105}-1+\frac{93-x}{107}-1=-4+4\)
\(\Leftrightarrow\frac{200-x}{101}+\frac{200-x}{103}+\frac{200-x}{105}+\frac{200-x}{107}=0\)
\(\Leftrightarrow\left(200-x\right).\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)
Vì \(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\ne0\)
=> 200 - x = 0
=> x = 200
Vậy x = 200
M
4
26 tháng 3 2019
\(\left(\frac{99-x}{101}+1\right)+\left(\frac{97-x}{103}+1\right)+\left(\frac{95-x}{105}+1\right)+\left(\frac{93-x}{107}+1\right)=-4+4\)
\(\frac{200-x}{101}+\frac{200-x}{103}+\frac{200-x}{105}+\frac{200-x}{107}=0\)
\(\left(200-x\right)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\) mà \(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\ne0\)
\(\Rightarrow200-x=0\Rightarrow x=200\)
k nha
26 tháng 3 2019
\(\left(\frac{99-x}{101}+1\right)+\left(\frac{97-x}{103}+1\right)+\left(\frac{95-x}{105}+1\right)+\left(\frac{93-x}{107}+1\right)=-4+4\)
\(\frac{110-x}{101}+\frac{110-x}{103}+\frac{110-x}{105}+\frac{110-x}{107}=0\)
\(\left(110-x\right).\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)
\(\Rightarrow110-x=0\)( vì \(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\ne0\) )
\(\Rightarrow x=110\)
vậy x=110
DN
5
16 tháng 2 2019
\(\left(2x-6\right)\left(x^2+2\right)=\left(2x-6\right)\left(8x-10\right)\)
\(\Leftrightarrow\left(2x-6\right)\left(x^2+2\right)-\left(2x-6\right)\left(8x-10\right)=0\)
\(\Leftrightarrow\left(2x-6\right)\left(x^2+2-8x+10\right)=0\)
\(\Leftrightarrow2\left(x-3\right)\left(x^2-6x-2x-12\right)=0\)
\(\Leftrightarrow2\left(x-3\right)\left(x-6\right)\left(x-2\right)=0\)
\(\Rightarrow x\in\left\{3;6;2\right\}\)
16 tháng 2 2019
\(\left(5x-1\right)^2=\left(3x+5\right)^2\)
\(\Leftrightarrow\left(5x-1\right)^2-\left(3x+5\right)^2=0\)
\(\Leftrightarrow\left(5x-1-3x-5\right)\left(5x-1+3x+5\right)=0\)
\(\Leftrightarrow\left(2x-6\right)\left(8x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-6=0\\8x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{-1}{2}\end{cases}}}\)
23 tháng 4 2020
Sửa đề: \(\frac{x-1}{99}+\frac{x-3}{97}+\frac{x-5}{95}+\frac{x-7}{93}+\frac{x-95}{5}+x=105\)
Ta có: \(\frac{x-1}{99}+\frac{x-3}{97}+\frac{x-5}{95}+\frac{x-7}{93}+\frac{x-95}{5}+x=105\)
\(\Leftrightarrow\frac{x-1}{99}+\frac{x-3}{97}+\frac{x-5}{95}+\frac{x-7}{93}+\frac{x-95}{5}+x-105=0\)
\(\Leftrightarrow\frac{x-1}{99}-1+\frac{x-3}{97}-1+\frac{x-5}{95}-1+\frac{x-7}{93}-1+\frac{x-95}{5}-1+x-100=0\)
\(\Leftrightarrow\frac{x-100}{99}+\frac{x-100}{97}+\frac{x-100}{95}+\frac{x-100}{93}+\frac{x-100}{5}+\frac{x-100}{1}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}+\frac{1}{93}+\frac{1}{5}+1\right)=0\)
mà \(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}+\frac{1}{93}+\frac{1}{5}+1\ne0\)
nên x-100=0
hay x=100
Vậy: x=100
7 tháng 3 2018
Bạn tham khảo nhé
\(a)\) \(\frac{x-1}{2003}+\frac{x-2}{2002}+\frac{x-3}{2001}-3=0\)
\(\Leftrightarrow\)\(\left(\frac{x-1}{2003}-1\right)+\left(\frac{x-2}{2002}-1\right)+\left(\frac{x-3}{2001}-1\right)+\left(-3+3\right)=0\)
\(\Leftrightarrow\)\(\frac{x-2004}{2003}+\frac{x-2004}{2002}+\frac{x-2004}{2001}=0\)
\(\Leftrightarrow\)\(\left(x-2004\right)\left(\frac{1}{2003}+\frac{1}{2002}+\frac{1}{2001}\right)=0\)
Vì \(\frac{1}{2003}+\frac{1}{2002}+\frac{1}{2001}\ne0\)
\(\Rightarrow\)\(x-2004=0\)
\(\Rightarrow\)\(x=2004\)
Vậy \(x=2004\)
Chúc bạn học tốt ~
7 tháng 3 2018
\(b)\) \(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}=-4\)
\(\Leftrightarrow\)\(\left(\frac{315-x}{101}+1\right)+\left(\frac{313-x}{103}+1\right)+\left(\frac{311-x}{105}+1\right)+\left(\frac{309-x}{107}+1\right)=-4+4\)
\(\Leftrightarrow\)\(\frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=0\)
\(\Leftrightarrow\)\(\left(416-x\right)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)
Vì \(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\ne0\)
\(\Rightarrow\)\(416-x=0\)
\(\Rightarrow\)\(x=416\)
Vậy \(x=416\)
Chúc bạn học tốt ~
\(\frac{109-x}{91}+\frac{107-x}{93}+\frac{105-x}{95}+\frac{103-x}{97}=-2\)
\(\Leftrightarrow\frac{109-x}{91}+1+\frac{107-x}{93}+1+\frac{105-x}{95}+1+\frac{103-x}{97}+1=2\)
\(\Leftrightarrow\left(200-x\right)\left(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\right)=2\)
\(\Rightarrow x=200-\frac{2}{\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}}\)
Bạn tự bấm máy, kết quả chẳng đẹp gì
\(A=2B\) thì còn có lý
Cảm ơn bạn.