\(\frac{1927-x}{91}+\frac{1925-x}{93}+\frac{1923-x}{95}+\frac{1921-x}{97}+4=0\)

\(\Leftrightarrow\left(\frac{1927-x}{91}+1\right)+\left(\frac{1925-x}{93}+1\right)+\left(\frac{1923-x}{95}+1\right)+\left(\frac{1921-x}{97}+1\right)=0\)

\(\Leftrightarrow\frac{1927-x+91}{91}+\frac{1925-x+93}{93}+\frac{1923-x+95}{95}+\frac{1921-x+97}{97}=0\)

\(\Leftrightarrow\frac{2018-x}{91}+\frac{2018-x}{93}+\frac{2018-x}{95}+\frac{2018-x}{97}=0\)

\(\Leftrightarrow\left(2018-x\right)\left(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\right)=0\)

Vì \(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\ne0\)\(\Rightarrow2018-x=0\)\(\Leftrightarrow x=2018\)

Vậy tập nghiệm của phương trình là \(S=\left\{2018\right\}\)

<=> (1927-x/91 + 1)+(1925-x/93 + 1) + (1923-x/95 + 1) + (1921-x/97 + 1) = 0

<=> 2018-x/91 + 2018-x/93 + 2018-x/95 + 2018-x/97 = 0

<=> (2018-x).(1/91 + 1/93 + 1/95 + 1/97) = 0

<=> 2018-x = 0 ( vì 1/91 + 1/93 + 1/95 + 1/97 > 0 )

<=> x = 2018

Vậy x = 2018

Tk mk nha

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 \(200-x=0\Rightarrow x=200\)

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 \(x+100=0\Rightarrow x=-100\)

1/

99-x/101+97-x/103+95-x/105+93-x/107

\(\Leftrightarrow\)(99-x/101+1)+(97-x/103+1)+(95-x/105+1)+(93-x/107+1)=0

\(\Leftrightarrow\)100-x/101+100-x/103+100-x/105+100-x/107=0

\(\Leftrightarrow\)(100-x)(1/101+1/103+1/105+1/107)=0

\(\Leftrightarrow\)100-x=0\(\Leftrightarrow\)

x=-100

Giải:

\(\dfrac{1909-x}{91}+\dfrac{1907-x}{93}+\dfrac{1905-x}{95}+\dfrac{1903-x}{97}+4=0\)

\(\Leftrightarrow\dfrac{1909-x}{91}+1+\dfrac{1907-x}{93}+1+\dfrac{1905-x}{95}+1+\dfrac{1903-x}{97}+1=0\)

\(\Leftrightarrow\dfrac{1909-x+91}{91}+\dfrac{1907-x+93}{93}+\dfrac{1905-x+95}{95}+\dfrac{1903-x+97}{97}=0\)

\(\Leftrightarrow\dfrac{2000-x}{91}+\dfrac{2000-x}{93}+\dfrac{2000-x}{95}+\dfrac{2000-x}{97}=0\)

\(\Leftrightarrow\left(2000-x\right)\left(\dfrac{1}{91}+\dfrac{1}{93}+\dfrac{1}{95}+\dfrac{1}{97}\right)=0\)

Vì \(\left(\dfrac{1}{91}+\dfrac{1}{93}+\dfrac{1}{95}+\dfrac{1}{97}\right)\ne0\)

Nên \(2000-x=0\)

\(\Leftrightarrow x=2000\)

Vậy \(x=2000\)

1) PT \(\Leftrightarrow\left(\dfrac{x+1}{35}+1\right)+\left(\dfrac{x+3}{33}+1\right)=\left(\dfrac{x+5}{31}+1\right)+\left(\dfrac{x+7}{29}+1\right)\)

\(\Leftrightarrow\dfrac{x+36}{35}+\dfrac{x+36}{33}=\dfrac{x+36}{31}+\dfrac{x+36}{29}\)

\(\Leftrightarrow\left(x+36\right)\left(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}\right)=0\)

\(\Leftrightarrow x+36=0\) (Do \(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}>0\))

\(\Leftrightarrow x=-36\).

Vậy nghiệm của pt là x = -36.

4.

\(\dfrac{x+1}{99}+\dfrac{x+3}{97}+\dfrac{x+5}{95}=\dfrac{x+7}{93}+\dfrac{x+9}{91}+\dfrac{x+11}{89}\\ \Rightarrow\left(\dfrac{x+1}{99}+1\right)+\left(\dfrac{x+3}{97}+1\right)+\left(\dfrac{x+5}{95}+1\right)=\left(\dfrac{x+7}{93}+1\right)+\left(\dfrac{x+9}{91}+1\right)+\left(\dfrac{x+11}{89}+1\right)\\ \Rightarrow\dfrac{x+100}{99}+\dfrac{x+100}{97}++\dfrac{x+100}{95}=\dfrac{x+100}{93}+\dfrac{x+100}{91}+\dfrac{x+100}{89}\\ \Rightarrow\left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}-\dfrac{1}{93}-\dfrac{1}{91}-\dfrac{1}{89}\right)=0\\ \Leftrightarrow x+100=0\Leftrightarrow x=-100\)

\(\text{1) }\dfrac{\left(2x-3\right)\left(2x+3\right)}{8}=\dfrac{\left(x-4\right)^2}{6}+\dfrac{\left(x-2\right)^2}{3}\\ \Leftrightarrow\dfrac{\left(2x-3\right)\left(2x+3\right)}{8}\cdot24=\left[\dfrac{\left(x-4\right)^2}{6}+\dfrac{\left(x-2\right)^2}{3}\right]24\\ \Leftrightarrow3\left(4x^2-9\right)=4\left(x^2-8x+16\right)+8\left(x^2-4x+4\right)\\ \Leftrightarrow12x^2-27=4x^2-32x+64+8x^2-32x+32\\ \Leftrightarrow12x^2-27=12x^2-64x+96\\ \Leftrightarrow12x^2-12x^2+64x=96+27\\ \Leftrightarrow64x=123\\ \Leftrightarrow x=\dfrac{123}{64}\\ \text{Vậy }S=\left\{\dfrac{123}{64}\right\}\\ \)

\(\text{2) }x+2-\dfrac{2x-\dfrac{2x-5}{6}}{15}=\dfrac{7x-\dfrac{x-3}{2}}{5}\\ \Leftrightarrow\left(x+2-\dfrac{2x-\dfrac{2x-5}{6}}{15}\right)15=\dfrac{7x-\dfrac{x-3}{2}}{5}\cdot15\\ \Leftrightarrow15x+30-2x-\dfrac{2x-5}{6}=21x-\dfrac{3x-9}{2}\\ \Leftrightarrow15x-2x-\dfrac{2x-5}{6}-21x+\dfrac{3x-9}{2}=-30\\ \Leftrightarrow-8x-\dfrac{2x-5}{6}+\dfrac{3x-9}{2}=-30\\ \Leftrightarrow\left(-8x-\dfrac{2x-5}{6}+\dfrac{3x-9}{2}\right)6=-30\cdot6\\ \Leftrightarrow-48x-2x+5+9x-27=-180\\ \Leftrightarrow-41x==-158\\ \Leftrightarrow x=\dfrac{158}{41}\\ \text{Vậy }S=\left\{\dfrac{158}{41}\right\}\)

\(\text{3) }1-\dfrac{x-\dfrac{1+x}{3}}{3}=\dfrac{x}{2}-\dfrac{2x-\dfrac{10-7}{3}}{2}\\ \Leftrightarrow\left(1-\dfrac{x-1-x}{3}\right)6=\left(\dfrac{x}{2}-\dfrac{2x-1}{2}\right)6\\ \Leftrightarrow6+2=-3x+3\\ \Leftrightarrow-3x=8-3\\ \Leftrightarrow-3x=5\\ \Leftrightarrow x=-\dfrac{5}{3}\\ \\ \text{Vậy }S=\left\{-\dfrac{5}{3}\right\}\)

\(\dfrac{1909-x}{91}+\dfrac{1907-x}{93}+\dfrac{1905-x}{95}+\dfrac{1903}{97}+4=0\) ( Sửa đề)

⇔ \(\dfrac{1909-x}{91}+1+\dfrac{1907-x}{93}+1+\dfrac{1905-x}{95}+1+\dfrac{1903}{97}+1=0\) ⇔ \(\dfrac{2000-x}{91}+\dfrac{2000-x}{93}+\dfrac{2000-x}{95}+\dfrac{2000-x}{97}=0\)

⇔ \(\left(2000-x\right)\left(\dfrac{1}{91}+\dfrac{1}{93}+\dfrac{1}{95}+\dfrac{1}{97}\right)=0\)

Do : \(\dfrac{1}{91}+\dfrac{1}{93}+\dfrac{1}{95}+\dfrac{1}{97}>0\)

\(\text{⇔}2000-x=0\)

\(\text{⇔}x=2000\)

Vậy ,....