\(a)5-\left(x-6\right)=4\left(3-2x\right)\)

\(\Leftrightarrow5-x+6=12-8x\)

\(\Leftrightarrow-x+8x=12-5-6\)

\(\Leftrightarrow7x=1\Leftrightarrow x=\frac{1}{7}\)

a) 5-(x-6)=4(3-2x)

<=>5-x-6=12-8x

<=>-x+8x=2-5-6

<=>7x=1

<=>x=1/7

\(b)3-4x\left(25-2x\right)=8x^2+x-300\)

\(\Leftrightarrow3-100x+8x^2=8x^2+x-300\)

\(\Leftrightarrow101x-303=0\)

\(\Leftrightarrow101\left(x-3\right)=0\)

\(\Leftrightarrow x-3=0\Leftrightarrow x=3\)

\(\left(\frac{x+14}{200}+1\right)+\left(\frac{x+27}{187}+1\right)+\left(\frac{x+105}{109}+1\right)=\left(\frac{x+200}{14}+1\right)\)

\(+\left(\frac{x+187}{27}+1\right)+\left(\frac{x+109}{105}+1\right)\)

\(\Rightarrow\frac{x+214}{200}+\frac{x+214}{187}+\frac{x+214}{109}-\frac{x+214}{14}-\frac{x+214}{27}-\frac{x+214}{105}=0\)

\(\Rightarrow\left(x+214\right)\left(\frac{1}{200}+\frac{1}{187}+\frac{1}{109}-\frac{1}{14}-\frac{1}{27}-\frac{1}{105}\right)=0\)

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

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 \(S=\left\{-214\right\}\)

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:

\(\frac{x+214}{200}+\frac{x+214}{187}+\frac{x+214}{109}-\frac{x+214}{14}-\frac{x+214}{27}-\frac{x+214}{105}=0\)

\(\Leftrightarrow x+214=0\Rightarrow x=-214\)

Ta có :\(\frac{x+14}{200}+\frac{x+27}{187}+\frac{x+105}{109}=\frac{x+200}{14}+\frac{x+187}{27}+\frac{x+109}{105}\)

=> \(\left(\frac{x+14}{200}+1\right)+\left(\frac{x+27}{187}+1\right)+\left(\frac{x+105}{109}+1\right)=\left(\frac{x+200}{14}+1\right)+\left(x+\frac{287}{27}\right)+\)

\(+\left(\frac{x+109}{105}+1\right)\)

=> \(\frac{x+214}{200}+\frac{x+214}{187}+\frac{x+214}{109}=\frac{x+214}{14}+\frac{x+214}{27}+\frac{x+214}{105}\)

=> \(\frac{x+214}{200}+\frac{x+214}{187}+\frac{x+214}{109}-\frac{x+214}{14}-\frac{x+214}{27}-\frac{x+214}{105}=0\)

=> \(\left(x+214\right)\left(\frac{1}{200}+\frac{1}{187}+\frac{1}{109}-\frac{1}{14}-\frac{1}{27}-\frac{1}{105}\right)=0\)

=> x + 214 = 0 (Vì \(\frac{1}{200}+\frac{1}{187}+\frac{1}{109}-\frac{1}{14}-\frac{1}{27}-\frac{1}{105}\ne0\))

=> x = -214

Vậy x = -214

\(\frac{x+1}{99}+\frac{x+2}{98}=\frac{x+3}{97}+\frac{x+4}{96}\)

\(\Rightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1=\frac{x+3}{97}+1+\frac{x+4}{96}+1\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}=\frac{x+100}{97}+\frac{x+100}{96}\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}-\frac{x+100}{97}-\frac{x+100}{96}=0\)

\(\Rightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)=0\)

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

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

a, \(2\left(\frac{11x}{12}+\frac{1}{3}\right)=2-\frac{x}{6}\)

\(2\left(\frac{11x+4}{12}\right)-2+\frac{x}{6}=0\)

\(\frac{44x+8}{12}-2+\frac{x}{6}=0\)

\(\frac{44x+8}{12}-\frac{24}{12}+\frac{2x}{12}=0\)

\(\frac{44x+8-24+2x}{12}=\frac{46x-16}{12}=0\)

\(\Leftrightarrow46x-16=0\)

\(\Leftrightarrow46x=16\Rightarrow x=\frac{8}{23}\)

Vậy nghiệm của pt là x=8/23

k mk

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.

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.

Phương trình 4:
\(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15\)
\(\Rightarrow\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}-15=0\)
\(\Rightarrow\left(\frac{x-90}{10}-1\right)+\left(\frac{x-76}{12}-2\right)+\left(\frac{x-58}{14}-3\right)+\left(\frac{x-36}{16}-4\right)+\left(\frac{x-15}{17}-5\right)=0\)
\(\Rightarrow\frac{x-90-10}{10}+\frac{x-76-24}{12}+\frac{x-58-42}{14}+\frac{x-36-64}{16}+\frac{x-15-85}{17}=0\)
\(\Rightarrow\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)
Do \(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\)
\(\Rightarrow x-100=0\)
\(\Rightarrow x=100\)
Vậy x = 100.