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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
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
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
AH
Akai Haruma
Giáo viên
5 tháng 1 2019
Bài 1:
\(\frac{99-x}{101}+\frac{97-x}{103}+\frac{95-x}{105}+\frac{93-x}{107}=-4\)
\(\Leftrightarrow \frac{99-x}{101}+1+\frac{97-x}{103}+1+\frac{95-x}{105}+1+\frac{93-x}{107}+1=0\)
\(\Leftrightarrow \frac{99-x+101}{101}+\frac{97-x+103}{103}+\frac{95-x+105}{105}+\frac{93-x+107}{107}=0\)
\(\Leftrightarrow \frac{200-x}{101}+\frac{200-x}{103}+\frac{200-x}{105}+\frac{200-x}{107}=0\)
\(\Leftrightarrow (200-x)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)
Vì \(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\neq 0\) nên suy ra \(200-x=0\Rightarrow x=200\)
AH
Akai Haruma
Giáo viên
5 tháng 1 2019
Bài 2:
\(\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+116}{4}=0\)
\(\Leftrightarrow \frac{x+14}{86}+1+\frac{x+15}{85}+1+\frac{x+16}{84}+1+\frac{x+17}{83}+1+\frac{x+116}{4}-4=0\)
\(\Leftrightarrow \frac{x+100}{86}+\frac{x+100}{85}+\frac{x+100}{84}+\frac{x+100}{83}+\frac{x+100}{4}=0\)
\(\Leftrightarrow (x+100)\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)=0\)
Vì \(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\neq 0\). Do đó \(x+100=0\Rightarrow x=-100\)
13 tháng 10 2021
3: \(\left(x+5\right)\left(x^2-5x+25\right)-x\left(x-4\right)^2+16x\)
\(=x^3+125-x^3+8x^2-16x+16x\)
\(=8x^2+125\)
\(\left(2x-6\right)\left(x^2+2\right)=\left(2x-6\right)\left(8x-10\right)\)
\(\Leftrightarrow\left(2x-6\right)\left(x^2+2\right)-\left(2x-6\right)\left(8x-10\right)=0\)
\(\Leftrightarrow\left(2x-6\right)\left(x^2+2-8x+10\right)=0\)
\(\Leftrightarrow2\left(x-3\right)\left(x^2-6x-2x-12\right)=0\)
\(\Leftrightarrow2\left(x-3\right)\left(x-6\right)\left(x-2\right)=0\)
\(\Rightarrow x\in\left\{3;6;2\right\}\)
\(\left(5x-1\right)^2=\left(3x+5\right)^2\)
\(\Leftrightarrow\left(5x-1\right)^2-\left(3x+5\right)^2=0\)
\(\Leftrightarrow\left(5x-1-3x-5\right)\left(5x-1+3x+5\right)=0\)
\(\Leftrightarrow\left(2x-6\right)\left(8x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-6=0\\8x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{-1}{2}\end{cases}}}\)