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

\(\Leftrightarrow\frac{1}{x\left(x+4\right)+5\left(x+4\right)}+\frac{1}{x\left(x+5\right)+6\left(x+5\right)}+\frac{1}{x\left(x+6\right)+7\left(x+6\right)}=\frac{1}{18}\)(điều kiện: \(x\ne\left\{-4;-5;-6;-7\right\}\) )

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

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

\(\Leftrightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Rightarrow54=\left(x+4\right)\left(x+7\right)\)

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

\(\Leftrightarrow x\left(x+13\right)-2\left(x+13\right)=0\Leftrightarrow\left(x+13\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x=-13\\x=2\end{cases}}\)(thỏa mãn ĐKXĐ)

Vậy tập nghiệm của pt là: \(S=\left\{-13;2\right\}\)

Lâu lắm không làm nhể

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

\(\Rightarrow\frac{1}{x^2+4x+5x+20}+\frac{1}{x^2+5x+6x+30}+\frac{1}{x^2+6x+7x+42}=\frac{1}{18}\)

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

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

Dùng công thứ \(\frac{1}{x.\left(x+1\right)}=\frac{1}{x}-\frac{1}{x+1}\)

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

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

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

\(\Rightarrow\frac{3}{\left(x+4\right).\left(x+7\right)}=\frac{1}{18}\Rightarrow\left(x+4\right).\left(x+7\right)=54\)

\(\Rightarrow\hept{\begin{cases}x+4=6\\x+7=9\end{cases}}\)hoặc \(\hept{\begin{cases}x+4=-6\\x+7=-9\end{cases}}\)

Suy ra \(x=3\)hoặc \(x=-3\)

Thêm đk: \(x\in Z\)

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

\(\frac{1}{\left(x^2+4x\right)+\left(4x+20\right)}+\frac{1}{\left(x^2+5x\right)+\left(6x+30\right)}+\frac{1}{\left(x^2+6x\right)+\left(7x+42\right)}=\frac{1}{18}\)

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

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

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

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

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

\(\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Rightarrow\left(x+4\right)\left(x+7\right)=54\)

đến đây b tự lập bảng nhé~

Ta có : \(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+\left|x+\frac{1}{12}\right|+...+\left|x+\frac{1}{110}\right|\ge0\forall x\)

=> 11x \(\ge\)0

=> x  \(\ge\)0 

Khi đó \(\orbr{\begin{cases}x+\frac{1}{2}+x+\frac{1}{6}+x+\frac{1}{12}+...+x+\frac{1}{110}=11x\left(10\text{ số hạng x }\right)\\x+\frac{1}{2}+x+\frac{1}{6}+x+\frac{1}{12}+...+x+\frac{1}{110}=-11x\left(10\text{ số hạng x}\right)\end{cases}}\)

=> \(\orbr{\begin{cases}10x+\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)=11x\\10x+\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)=-11x\end{cases}}\)

=> \(\orbr{\begin{cases}10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)=11x\\10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)=-11x\end{cases}}\)

=> \(\orbr{\begin{cases}10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)=11x\\10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)=-11x\end{cases}}\)

=> \(\orbr{\begin{cases}10x+\left(1-\frac{1}{11}\right)=11x\\10x+\left(1-\frac{1}{11}\right)=-11x\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{10}{11}\\21x=-\frac{10}{11}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{10}{11}\left(\text{tm}\right)\\x=-\frac{10}{231}\left(\text{loại}\right)\end{cases}}}\)

Vậy \(x=\frac{10}{11}\)

ai làm cho mik đi

\(a)\)Áp dụng tính chất của dãy tỉ số bằng nhau , ta có :

\(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}=\frac{2\cdot(2x+3)-(4x+5)}{2\cdot(5x+2)-(10x+2)}=\frac{4x+6-4x-5}{10x+4-10x-2}=\frac{1}{2}\)

Suy ra :

\(\frac{2x+3}{5x+2}=\frac{1}{2}\Rightarrow1\cdot(5x+2)=2\cdot(2x+3)\)

\(5x+2=4x+6\)

\(5x-4x=6-2\)

\(x=4\)

\(b)\)Ta có : \(\frac{4}{x-3}=\frac{8}{y-6}=\frac{20}{z-15}\)

\(\Rightarrow\frac{x-3}{4}=\frac{y-6}{8}=\frac{z-15}{20}\)

\(\Rightarrow\frac{x}{4}-\frac{3}{4}=\frac{y}{8}-\frac{6}{8}=\frac{z}{20}-\frac{15}{20}\)

\(\Rightarrow\frac{x}{4}-\frac{3}{4}=\frac{y}{8}-\frac{3}{4}=\frac{z}{20}-\frac{3}{4}\)

\(\Rightarrow\frac{x}{4}=\frac{y}{8}=\frac{z}{20}\)

Đặt : \(\frac{x}{4}=\frac{y}{8}=\frac{z}{20}=k\Rightarrow x=4k;y=8k;z=20k\)

Thay vào đề , ta có : xyz = 640

\(\Rightarrow4k\cdot8k\cdot20k=640\)

\(\Rightarrow640k^3=640\)

\(\Rightarrow k^3=1\)

\(\Rightarrow k=1\)

\(\Rightarrow x=4;y=8;z=20\)

Vậy

cam on nha

\(2.x=\frac{1+2+3+...+9}{1-2+3-4+5-6+7-8+9}+\frac{25.150-60.5+20.75}{1+2+3+...+99}\)

\(2.x=\frac{\left(9+1\right).9:2}{\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+\left(7-8\right)+9}+\frac{2.3.5^2.\left(5^2-2+2.5\right)}{\left(1+99\right).99:2}\)

\(2.x=\frac{45}{\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)+9}+\frac{2.3.5^2.33}{100.99.\frac{1}{2}}\)

\(2x=\frac{45}{5}+\frac{50.99}{50.2.99.\frac{1}{2}}=9+\frac{1}{2.\frac{1}{2}}=9+1=10\)

=> 2x = 10

x = 5

Bài 1:

a, \(\frac{1}{-16}-\frac{3}{45}=\frac{-1}{16}-\frac{1}{15}\)

\(=\frac{-15}{240}-\frac{16}{240}\)

\(=\frac{-31}{240}\)

b, \(=\frac{-10}{12}-\frac{-12}{12}\)

\(=\frac{2}{12}=\frac{1}{6}\)

c, \(=\frac{-30}{6}-\frac{1}{6}\)

\(=\frac{-31}{6}\)

Bài 2:

a, \(x=-\frac{1}{2}-\frac{3}{4}\)

\(x=-\frac{1}{4}\)

b,   \(\frac{1}{2}+x=-\frac{11}{2}\)

\(x=-\frac{11}{2}-\frac{1}{2}\)

\(x=-6\)

Bạn nhớ k đúng và chọn câu trả lời này nhé!!!! Mình giải đúng và chính xác hết ^_^

a, 10+15+20+....+295+x.300+x=67

10+15+20+...+295+x(300+1)=67

10+15+20+...+295+x.301=67

8845+x.301=67

67-8845=x.301

-8878=x.301

x=-29/149/301

b, 

\(\frac{1}{7.6}+\frac{1}{6.5}+\frac{1}{5.4}+\frac{1}{4.3}+\frac{1}{3.2}+\frac{1}{2.1}-\frac{1}{x+1}=\frac{59}{77}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}-\frac{1}{x+1}=\frac{59}{77}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}-\frac{1}{x+1}=\frac{59}{77}\)\(1-\frac{1}{7}-\frac{1}{x+1}=\frac{59}{77}\)

\(\frac{6}{7}-\frac{1}{x+1}=\frac{59}{77}\)

\(\frac{1}{x+1}=\frac{6}{7}-\frac{59}{77}\)

\(\frac{1}{x+1}=\frac{1}{11}\)

suy ra x+1=11

suy ra x=10

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}=\frac{2001}{2003}\)

\(\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}\right)=\frac{1}{2}\cdot\frac{2001}{2003}\)

\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)

\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)

\(\frac{1}{x+1}=\frac{1}{2}-\frac{2001}{4006}\)

\(\frac{1}{x+1}=\frac{1}{2003}\)

\(\Rightarrow x+1=2003\)

\(x=2002\)

Vậy x = 2002

Bài này lớp 6 thật à bạn. 

Mik mới lớp 5 thui à.  Sorry nha. 

\(\frac{x}{5}=\frac{20}{x}\)

Ta có :

x . x = 20 . 5

x² = 100

=> x = 10 hoặc x = -10

XxX=20x5

X=100

10

tương đương vỡi^2=100. tương đương với x=10