Những lần sau cố gắng đặt câu hỏi dưới dạng công thức nhé!

Lời giải:

ĐK: \(x\ne-10;x\ne-9\)

Phương trình tương đương:

\(9\left(x+10\right)\left(x+9\right)^2+10\left(x+9\right)\left(x+10\right)^2=810\left(x+9\right)+900\left(x+10\right)\\ \Leftrightarrow\left(x+10\right)\left(x+9\right)\left[9\left(x+9\right)+10\left(x+10\right)\right]=90\left[9\left(x+9\right)+10\left(x+10\right)\right]\\ \Leftrightarrow\left[\left(x+10\right)\left(x+9\right)-90\right].\left[9\left(x+9\right)+10\left(x+10\right)\right]=0\\ \Leftrightarrow\left(x^2+19x\right)\left(19x+181\right)=0\)

\( \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = - 19\\ x = - \dfrac{{181}}{{19}} \end{array} \right.\left( {tm} \right)\)

cộng 1 vào từng phân thức để tử là x+19

nên x=-19

\(\Leftrightarrow\dfrac{x+9}{10}+1+\dfrac{x+10}{9}+1=\dfrac{9}{x+10}+1+\dfrac{10}{x+9}+1\)

\(\Leftrightarrow\dfrac{x+19}{10}+\dfrac{x+19}{9}=\dfrac{x+19}{x+10}+\dfrac{x+19}{x+9}\)

\(\Leftrightarrow\left(x+9\right)\left(\dfrac{1}{10}+\dfrac{1}{9}-\dfrac{1}{x+10}-\dfrac{1}{x+9}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+9=0\\\dfrac{1}{10}+\dfrac{1}{9}-\dfrac{1}{x+10}-\dfrac{1}{x+9}=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-9\\x=0\end{matrix}\right.\)

Bạn nhớ ĐKXĐ

phải đặt x+19 lầm nhân tử chung chứ sao lại là x+9

9 x 2 = 18

9 x 5 = 45

9 x 8 = 72

9 x 10 = 90

2 x 9 = 18

5 x 9 = 45

8 x 9 = 72

10 x 9 = 90

\(\Leftrightarrow\frac{9\left(X+9\right)\left(X+9\right)\left(X+10\right)+10\left(X+10\right)\left(X+10\right)\left(X+9\right)}{90\left(X+10\right)\left(X+9\right)}=\frac{9.90\left(X+9\right)+10.90\left(X+10\right)}{90\left(X+10\right)\left(X+9\right)}\)

\(\Rightarrow9\left(X+9\right)^2\left(X+10\right)+10\left(X+10\right)^2\left(X+9\right)=810\left(X+9\right)+900\left(X+10\right)\)

\(\Leftrightarrow\left(9X+90\right)\left(X^2+18X+81\right)+\left(10X+90\right)\left(X^2+20X+100\right)=810X+7290+900X+9000\)

\(\Leftrightarrow\)9X³+162X²+729X+90X²+1620X+7290+10X³+200X²+1000X+90X²+1800X+9000=1710X+16290

\(\Leftrightarrow\)19X³+542X²+5149X+16290=1710X+16290

\(\Leftrightarrow\)19X³+542X²=16290-16290+1710X-5149X

\(\Leftrightarrow\)19X³+542X²=-3439X

\(\Leftrightarrow\)19X³+542X²+3439X=0

RỒI GIẢI TIẾP

nốt đi bạn

ĐKXĐ: x≠-10; x≠-9

Ta có: \(\frac{x+9}{10}+\frac{x+10}{9}=\frac{10}{x+9}+\frac{9}{x+10}\)

Vậy: x=0

a) \(\dfrac{5}{7}\times\dfrac{5}{9}+\dfrac{4}{9}\times\dfrac{5}{7}\)

\(=\dfrac{5}{7}\times\left(\dfrac{4}{9}+\dfrac{5}{9}\right)\)

\(=\dfrac{5}{7}\times1\)

\(=\dfrac{5}{7}\)

b) \(\dfrac{1}{10}+\dfrac{5}{9}+\dfrac{4}{9}+\dfrac{9}{10}-1\)

\(=\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\left(\dfrac{1}{10}+\dfrac{9}{10}-1\right)\)

\(=1+0\)

\(=1\)

c) \(\dfrac{5}{7}\times\dfrac{5}{9}+\dfrac{4}{9}\times\dfrac{5}{7}+\dfrac{2}{7}\)

\(=\dfrac{5}{7}\times\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{2}{7}\)

\(=\dfrac{5}{7}+\dfrac{2}{7}\)

\(=1\)

d) \(\dfrac{2}{7}+\dfrac{2}{8}+\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{4}{7}\)

\(=\left(\dfrac{2}{8}+\dfrac{1}{4}\right)+\left(\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{4}{7}\right)\)

\(=\left(\dfrac{1}{4}+\dfrac{1}{4}\right)+1\)

\(=\dfrac{1}{2}+1\)

\(=\dfrac{3}{2}\)

e) \(\dfrac{4}{5}+\dfrac{3}{10}+\dfrac{2}{10}+0,7\)

\(=\dfrac{4}{5}+\dfrac{5}{10}+\dfrac{7}{10}\)

\(=\dfrac{4}{5}+\dfrac{12}{10}\)

\(=\dfrac{4}{5}+\dfrac{6}{5}\)

\(=\dfrac{10}{5}\)

\(=2\)

g) \(362\times728+326\times272\)

\(=326\times\left(728+272\right)\)

\(=326\times1000\)

\(=326000\)