Bài 3: Giải các phương trình sau.
\(\frac{x+1}{15}+\frac{x+2}{7}+\frac{x+4}{4}+6=0\)
\(\frac{x+14}{200}+\frac{x+27}{187}+\frac{x+105}{109}=\frac{x+200}{14}+\frac{x+187}{27}+\frac{x+109}{105}\)
K
Khách
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6 tháng 2 2019
\(a)5-\left(x-6\right)=4\left(3-2x\right)\)
\(\Leftrightarrow5-x+6=12-8x\)
\(\Leftrightarrow-x+8x=12-5-6\)
\(\Leftrightarrow7x=1\Leftrightarrow x=\frac{1}{7}\)
6 tháng 2 2019
a) 5-(x-6)=4(3-2x)
<=>5-x-6=12-8x
<=>-x+8x=2-5-6
<=>7x=1
<=>x=1/7
1 tháng 3 2017
\(\left(\frac{x+14}{200}+1\right)+\left(\frac{x+27}{187}+1\right)+\left(\frac{x+105}{109}+1\right)=\left(\frac{x+200}{14}+1\right)\)
\(+\left(\frac{x+187}{27}+1\right)+\left(\frac{x+109}{105}+1\right)\)
\(\Rightarrow\frac{x+214}{200}+\frac{x+214}{187}+\frac{x+214}{109}-\frac{x+214}{14}-\frac{x+214}{27}-\frac{x+214}{105}=0\)
\(\Rightarrow\left(x+214\right)\left(\frac{1}{200}+\frac{1}{187}+\frac{1}{109}-\frac{1}{14}-\frac{1}{27}-\frac{1}{105}\right)=0\)
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
18 tháng 2 2020
\(\Leftrightarrow\frac{x+14}{200}+\frac{x+27}{187}+\frac{x+105}{109}-\frac{x+200}{14}-\frac{x+187}{27}-\frac{x+109}{105}=0\)
\(\Leftrightarrow\left(\frac{x+14}{200}+1\right)+\left(\frac{x+27}{187}+1\right)+\left(\frac{x+105}{109}+1\right)-\left(\frac{x+200}{14}+1\right)-\left(\frac{x+187}{27}+1\right)-\left(\frac{x+109}{105}+1\right)=0\)\(\Leftrightarrow\frac{x+214}{200}+\frac{x+214}{187}+\frac{x+214}{109}-\frac{x+214}{14}-\frac{x+214}{27}-\frac{x+214}{105}=0\)
\(\Leftrightarrow\left(x+214\right)\left(\frac{1}{200}+\frac{1}{187}+\frac{1}{109}-\frac{1}{14}-\frac{1}{27}-\frac{1}{105}\right)=0\)
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
HY
18 tháng 2 2020
\(\frac{x+14}{200}+\frac{x+27}{187}+\frac{x+105}{109}=\frac{x+200}{14}+\frac{x+187}{27}+\frac{x+109}{105}\\\Leftrightarrow \frac{x+14}{200}+1+\frac{x+27}{187}+1+\frac{x+105}{109}+1=\frac{x+200}{14}+1+\frac{x+187}{27}+1+\frac{x+109}{105}+1\\\Leftrightarrow \frac{x+214}{200}+\frac{x+214}{187}+\frac{x+214}{109}-\frac{x+214}{14}-\frac{x+214}{27}-\frac{x+214}{105}=0\\\Leftrightarrow \left(x+214\right)\left(\frac{1}{200}+\frac{1}{187}+\frac{1}{109}-\frac{1}{14}-\frac{1}{27}-\frac{1}{105}\right)=0\)
\(\Leftrightarrow x+214=0\left(vi\frac{1}{200}+\frac{1}{187}+\frac{1}{109}-\frac{1}{14}-\frac{1}{27}-\frac{1}{105}\ne0\right)\\\Leftrightarrow x=-214 \)
Vậy tập nghiệp của phương trình trên là \(S=\left\{-214\right\}\)
24 tháng 8 2020
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 \(\left(\frac{1}{200}+\frac{1}{187}+\frac{1}{109}-\frac{1}{14}-\frac{1}{27}-\frac{1}{105}\right)\ne0\)
<=> \(x+214=0\)
<=> \(x=-214\)
NC
24 tháng 8 2020
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:
\(\frac{x+214}{200}+\frac{x+214}{187}+\frac{x+214}{109}-\frac{x+214}{14}-\frac{x+214}{27}-\frac{x+214}{105}=0\)
\(\Leftrightarrow x+214=0\Rightarrow x=-214\)
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
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.
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.
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