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\(\left(\frac{x+14}{200}+1\right)+\left(\frac{x+27}{187}+1\right)+\left(\frac{x+105}{109}+1\right)=\left(\frac{x+200}{14}+1\right)\)
\(+\left(\frac{x+187}{27}+1\right)+\left(\frac{x+109}{105}+1\right)\)
\(\Rightarrow\frac{x+214}{200}+\frac{x+214}{187}+\frac{x+214}{109}-\frac{x+214}{14}-\frac{x+214}{27}-\frac{x+214}{105}=0\)
\(\Rightarrow\left(x+214\right)\left(\frac{1}{200}+\frac{1}{187}+\frac{1}{109}-\frac{1}{14}-\frac{1}{27}-\frac{1}{105}\right)=0\)
Mà \(\frac{1}{200}+\frac{1}{187}+\frac{1}{109}-\frac{1}{14}-\frac{1}{27}-\frac{1}{105}\ne0\)
\(\Rightarrow x+214=0\)
\(\Rightarrow x=-214\)
Vậy x = -214
\(\Leftrightarrow\frac{x+14}{200}+\frac{x+27}{187}+\frac{x+105}{109}-\frac{x+200}{14}-\frac{x+187}{27}-\frac{x+109}{105}=0\)
\(\Leftrightarrow\left(\frac{x+14}{200}+1\right)+\left(\frac{x+27}{187}+1\right)+\left(\frac{x+105}{109}+1\right)-\left(\frac{x+200}{14}+1\right)-\left(\frac{x+187}{27}+1\right)-\left(\frac{x+109}{105}+1\right)=0\)\(\Leftrightarrow\frac{x+214}{200}+\frac{x+214}{187}+\frac{x+214}{109}-\frac{x+214}{14}-\frac{x+214}{27}-\frac{x+214}{105}=0\)
\(\Leftrightarrow\left(x+214\right)\left(\frac{1}{200}+\frac{1}{187}+\frac{1}{109}-\frac{1}{14}-\frac{1}{27}-\frac{1}{105}\right)=0\)
Mà \(\frac{1}{200}+\frac{1}{187}+\frac{1}{109}< \frac{1}{14}+\frac{1}{27}+\frac{1}{105}\Rightarrow\frac{1}{200}+\frac{1}{187}+\frac{1}{109}-\frac{1}{14}-\frac{1}{27}-\frac{1}{105}\ne0\)
\(\Rightarrow x+214=0\)
\(\Rightarrow x=-214\)
Vậy x=-214
\(\frac{x+14}{200}+\frac{x+27}{187}+\frac{x+105}{109}=\frac{x+200}{14}+\frac{x+187}{27}+\frac{x+109}{105}\\\Leftrightarrow \frac{x+14}{200}+1+\frac{x+27}{187}+1+\frac{x+105}{109}+1=\frac{x+200}{14}+1+\frac{x+187}{27}+1+\frac{x+109}{105}+1\\\Leftrightarrow \frac{x+214}{200}+\frac{x+214}{187}+\frac{x+214}{109}-\frac{x+214}{14}-\frac{x+214}{27}-\frac{x+214}{105}=0\\\Leftrightarrow \left(x+214\right)\left(\frac{1}{200}+\frac{1}{187}+\frac{1}{109}-\frac{1}{14}-\frac{1}{27}-\frac{1}{105}\right)=0\)
\(\Leftrightarrow x+214=0\left(vi\frac{1}{200}+\frac{1}{187}+\frac{1}{109}-\frac{1}{14}-\frac{1}{27}-\frac{1}{105}\ne0\right)\\\Leftrightarrow x=-214 \)
Vậy tập nghiệp của phương trình trên là \(S=\left\{-214\right\}\)
Giải:
Ta có:
\(\frac{x+1}{15}+\frac{x+2}{7}+\frac{x+4}{4}+6=0\)
\(\Leftrightarrow\frac{x}{15}+\frac{1}{15}+\frac{x}{7}+\frac{2}{7}+\frac{x}{4}+\frac{4}{4}+6=0\)
\(\Leftrightarrow\frac{x}{15}+\frac{x}{7}+\frac{x}{4}=-\frac{772}{105}\)
\(\Leftrightarrow x\left(\frac{1}{15}+\frac{1}{7}+\frac{1}{4}\right)=-\frac{772}{105}\)
\(\Leftrightarrow x=-16\)
Vậy phương trình trên có nghiệm là x = -16.
b. Cách làm tương tự.
Chúc bạn học tốt@@
\(a)5-\left(x-6\right)=4\left(3-2x\right)\)
\(\Leftrightarrow5-x+6=12-8x\)
\(\Leftrightarrow-x+8x=12-5-6\)
\(\Leftrightarrow7x=1\Leftrightarrow x=\frac{1}{7}\)
a) 5-(x-6)=4(3-2x)
<=>5-x-6=12-8x
<=>-x+8x=2-5-6
<=>7x=1
<=>x=1/7
\(\frac{x+1}{99}+\frac{x+2}{98}=\frac{x+3}{97}+\frac{x+4}{96}\)
\(\Rightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1=\frac{x+3}{97}+1+\frac{x+4}{96}+1\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}=\frac{x+100}{97}+\frac{x+100}{96}\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}-\frac{x+100}{97}-\frac{x+100}{96}=0\)
\(\Rightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)=0\)
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
(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
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
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
Ta có : \(\frac{149-x}{25}+\frac{170-x}{23}+\frac{187-x}{21}+\frac{200-x}{19}=10\)
\(\Leftrightarrow\frac{149-x}{25}-1+\frac{170-x}{23}-2+\frac{187-x}{21}-3+\frac{200-x}{19}-4=0\)
\(\Leftrightarrow\frac{124-x}{25}+\frac{124-x}{23}+\frac{124-x}{21}+\frac{124-x}{19}=0\)
\(\Leftrightarrow\left(124-x\right)\left(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\right)=0\)
Vì \(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\ne0\)
Nên : 124 - x = 0
<=> x = 124
Vậy x = 124
\(\Leftrightarrow\frac{108-x}{92}+1+\frac{107-x}{93}+1+\frac{106-x}{94}+1+\frac{105-x}{95}=0.\)
\(\Leftrightarrow\frac{108+92-x}{92}+\frac{107+93-x}{93}+\frac{106+94-x}{94}+\frac{105+95-x}{95}=0\)
\(\Leftrightarrow\frac{200-x}{92}+\frac{200-x}{93}+\frac{200-x}{94}+\frac{200-x}{95}=0\)
\(\Leftrightarrow\left(200-x\right)\left(\frac{1}{92}+\frac{1}{93}+\frac{1}{94}+\frac{1}{95}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}200-x=0\Leftrightarrow x=200\\\frac{1}{92}+\frac{1}{93}+\frac{1}{94}+\frac{1}{95}\ne0\end{cases}}\)
Vậy nghiệm của phương trình là 200
Ta có :
\(\frac{108-x}{92}+\frac{107-x}{93}+\frac{106-x}{94}+\frac{105-x}{95}+4=0\)
\(\Leftrightarrow\)\(\left(\frac{108-x}{92}+1\right)+\left(\frac{107-x}{93}+1\right)+\left(\frac{106-x}{94}+1\right)+\left(\frac{105-x}{95}+1\right)+\left(4-4\right)=0\)
\(\Leftrightarrow\)\(\frac{200-x}{92}+\frac{200-x}{93}+\frac{200-x}{94}+\frac{200-x}{95}=0\)
\(\Leftrightarrow\)\(\left(200-x\right)\left(\frac{1}{92}+\frac{1}{93}+\frac{1}{94}+\frac{1}{95}\right)=0\)
Vì \(\frac{1}{92}+\frac{1}{93}+\frac{1}{94}+\frac{1}{95}\ne0\)
\(\Rightarrow\)\(200-x=0\)
\(\Rightarrow\)\(x=200\)
Vậy \(x=200\)
Chúc bạn học tốt ~
Pt <=> \(\left(\frac{x+14}{200}+1\right)+\left(\frac{x+27}{187}+1\right)+\left(\frac{x+105}{109}+1\right)=\left(\frac{x+200}{14}+1\right)+\left(\frac{x+187}{27}+1\right)+\left(\frac{x+109}{105}+1\right)\)<=> \(\frac{x+14+200}{200}+\frac{x+27+187}{187}+\frac{x+105+109}{109}=\frac{x+200+14}{14}+\frac{x+187+27}{27}+\frac{x+109+105}{105}\)<=> \(\frac{x+214}{200}+\frac{x+214}{187}+\frac{x+214}{109}=\frac{x+214}{14}+\frac{x+214}{27}+\frac{x+214}{105}\)
<=> \(\frac{x+214}{200}+\frac{x+214}{187}+\frac{x+214}{109}-\frac{x+214}{14}-\frac{x+214}{27}-\frac{x+214}{105}=0\)
<=> \(\left(x+214\right)\left(\frac{1}{200}+\frac{1}{187}+\frac{1}{109}-\frac{1}{14}-\frac{1}{27}-\frac{1}{105}\right)=0\)
Vì \(\left(\frac{1}{200}+\frac{1}{187}+\frac{1}{109}-\frac{1}{14}-\frac{1}{27}-\frac{1}{105}\right)\ne0\)
<=> \(x+214=0\)
<=> \(x=-214\)
Ta có:
\(\frac{x+14}{200}+\frac{x+27}{187}+\frac{x+105}{109}=\frac{x+200}{14}+\frac{x+187}{27}+\frac{x+109}{105}\)
Cộng thêm mỗi phân thức 1 ta được:
\(\frac{x+214}{200}+\frac{x+214}{187}+\frac{x+214}{109}-\frac{x+214}{14}-\frac{x+214}{27}-\frac{x+214}{105}=0\)
\(\Leftrightarrow x+214=0\Rightarrow x=-214\)