\(\frac{x-18}{74}+\frac{x-20}{72}+\frac{x-22}{70}=3\)

\(\Leftrightarrow\)\(\left(\frac{x-18}{74}-1\right)+\left(\frac{x-20}{72}-1\right)+\left(\frac{x-22}{70}-1\right)=0\)

\(\Leftrightarrow\)\(\frac{x-92}{74}+\frac{x-92}{72}+\frac{x-92}{70}=0\)

\(\Leftrightarrow\)\(\left(x-92\right)\left(\frac{1}{74}+\frac{1}{72}+\frac{1}{70}\right)=0\)

Mà \(\frac{1}{74}+\frac{1}{72}+\frac{1}{70}\ne0\)

\(\Leftrightarrow\)\(x+92=0\)

\(\Leftrightarrow\)\(x=-92\)

Vậy S = { - 92 }

Ta có :

\(\frac{x-18}{74}+\frac{x-20}{72}+\frac{x-22}{70}=3\)

\(\Leftrightarrow\)\(\left(\frac{x-18}{74}-1\right)+\left(\frac{x-20}{72}-1\right)+\left(\frac{x-22}{70}-1\right)=3-3\)

\(\Leftrightarrow\)\(\frac{x-92}{74}+\frac{x-92}{72}+\frac{x-92}{70}=0\)

\(\Leftrightarrow\)\(\left(x-92\right)\left(\frac{1}{74}+\frac{1}{72}+\frac{1}{70}\right)=0\)

Vì \(\frac{1}{74}+\frac{1}{72}+\frac{1}{70}\ne0\)

\(\Rightarrow\)\(x-92=0\)

\(x=92\)

Vậy \(x=92\)

Chúc bạn học tốt 

bài 1+2: phân tích mẫu thành nhân tử r` áp dụng 

1/ab=1/a-1/b 

bài 3+4: quy đồng rút gọn blah...

\(\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}\)

\(\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)

\(\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)

dit me may

a: \(\Leftrightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{9}-\dfrac{1}{10}\right)\cdot\left(x-1\right)+\dfrac{1}{10}x-x=-\dfrac{9}{10}\)

\(\Leftrightarrow\dfrac{9}{10}x-\dfrac{9}{10}-\dfrac{9}{10}x=-\dfrac{9}{10}\)

=>-9/10=-9/10(luôn đúng)

b: \(\Leftrightarrow\dfrac{195x+195+130x+195+117x+195+100x+195}{195}=\dfrac{22\cdot39+4\cdot65+6\cdot39+40\cdot5}{195}\)

=>347x+780=1552

=>347x=772

hay x=772/347

\(\Leftrightarrow\frac{4x^2}{5}\times\frac{2x-3}{6}-\frac{3x-10}{15}\times\frac{4x^2+3}{3}=\frac{22x^2}{45}\)

\(\Leftrightarrow\frac{4x^2\left(2x-3\right)}{30}-\frac{\left(3x-10\right)\left(4x^2+3\right)}{45}=\frac{22x^2}{45}\)

\(\Leftrightarrow\frac{12x^2\left(2x-3\right)}{90}-\frac{2\left(3x-10\right)\left(4x^2+3\right)}{90}=\frac{44x^2}{90}\)

\(\Leftrightarrow12x^2\left(2x-3\right)-2\left(3x-10\right)\left(4x^2+3\right)=44x^2\)

\(\Leftrightarrow24x^2-36x^2-2\left(12x^3+9x-40x^2-30\right)=44x^2\)

\(\Leftrightarrow24x^2-36x^2-24x^3-18x+80x^2+60=44x^2\)

\(\Leftrightarrow24x^3-36x^2-24x^3-18x+80x^2-44x^2=-60\)

\(\Leftrightarrow\left(24x^3-24x^3\right)+\left(-36x^2+80x^2-44x^2\right)-18x=-60\)

\(\Leftrightarrow-18x=-60\)

\(\Leftrightarrow x=\frac{-60}{-18}\)

\(\Leftrightarrow x=\frac{10}{3}\)

\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)

<=> \(\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

<=>\(\frac{\left(x+6\right)\left(x+7\right)+\left(x+4\right)\left(x+7\right)+\left(x+4\right)\left(x+5\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

Từ đó, bạn tính ra nhá! Hơi dài, ai có cách nào ngắn hơn thì nói với mình nha!

minh ko biet

Đk:\(\left(x\ne-4;x\ne-5;x\ne-6;x\ne-7\right)\)

\(\Rightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Rightarrow\frac{3}{x^2+11x+28}=\frac{1}{18}\)

\(\Leftrightarrow x^2+11x+28=54\)

\(\Rightarrow x^2+11x-26=0\)

\(\Rightarrow\left(x-2\right)\left(x+13\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=2\\x=-13\end{array}\right.\)

Vậy pt có tập nghiệm là S={2,-13}

Đk:

Vậy pt có tập nghiệm là S={2,-13}