= 10/3.x+ 67/4=-53/4

   10/3.x=-53/4-67/4

   10/3.x=-30

   x=-30:10/3

   x=-9

nhân chéo là đc:

3(x+2)=-4(x-5)

3x+6=-4x+20

3x+4x=20-6

7x     =14

 x      =2

Vậy x=2

\(\frac{x+3}{-4}=-\frac{9}{x+3}\)

\(\Leftrightarrow\left(x+3\right)\left(x+3\right)=-4\cdot\left(-9\right)\)

\(\Leftrightarrow\left(x+3\right)^2=36\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+3\right)^2=6^2\\\left(x+3\right)^2=\left(-6\right)^2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=6\\x+3=-6\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-9\end{cases}}\)

Vậy ....

quy đồng

\(\left(x+3\right)^2=36\)

\(\left(x+3\right)^2-6^2=0\)

áp dụng định lí "  \(a^2-b^2=\left(a+b\right)\left(a-b\right)\) ta được

\(\left(x+3-6\right)\left(x+3+6\right)=0\)

\(x=3,x=-9\)

nhân chéo: 

(x+3)(x+3)=-4.(-9)

(x+3)^2   =36

Hai TH: 

TH1: x+3=6

=>   x=3

TH2: x+3=-6

=>  x=-9

Vậy x=3 hoặc x=-9

\(3\frac{1}{3}x+16\frac{3}{4}=-13,25\)

\(\frac{10}{3}x+\frac{67}{4}=\frac{-53}{4}\)

\(\frac{10}{3}x=-30\)

\(x=-9\)

\(3\frac{1}{3}x+16\frac{3}{4}=-13,25\)

\(\frac{10}{3}x+\frac{67}{4}=\frac{-53}{4}\)

\(\frac{10}{3}x=-30\)

\(x=-9\)

\(3\frac{1}{3}x+16\frac{3}{4}=-13,25\)

\(\frac{10}{3}x=-13,25-16\frac{3}{4}\)

\(\frac{10}{3}x=-30\)

\(x=-9\)

Bài 2 : 

a) (1) \(CaO+CO_2\rightarrow CaCO_3\)

     (2) \(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)

     (3) \(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\)

     (4) \(CaCO_3\underrightarrow{t^o}CaO+CO_2\)

     (5) \(CaO+2HCl\rightarrow CaCl_2+H_2O\)

 Chúc bạn học tốt

b)(1) \(S+O_2\underrightarrow{t^o}SO_2\)

    (2) \(2SO_2+O_2\underrightarrow{t^o,V_2O_5}2SO_3\)

    (3) \(SO_3+H_2O\rightarrow H_2SO_4\)

    (4) \(H_2SO_4+CuO\rightarrow CuSO_4+H_2O\)

    (5) \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)

    (6) \(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)

    (7) \(SO_2+2NaOH\rightarrow Na_2SO_3+H_2O\)

 Chúc bạn học tốt

11)\(\dfrac{3x+1}{x-5}+\dfrac{2x}{x-5}=\dfrac{3x+2x+1}{x-5}=\dfrac{5x+1}{x-5}\)

12)\(\dfrac{4-x^2}{x-3}+\dfrac{2}{x^2-9}=\dfrac{4-x^2}{x-3}+\dfrac{2}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(4-x^2\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{2}{\left(x-3\right)\left(x+3\right)}=\dfrac{2+\left(2-x\right)\left(2+x\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)

13)

\(\dfrac{3}{4x-2}+\dfrac{2x}{4x^2-1}=\dfrac{3}{2\left(2x-1\right)}+\dfrac{2x}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{3\left(2x+1\right)}{2\left(2x-1\right)\left(2x+1\right)}+\dfrac{2.2x}{2\left(2x-1\right)\left(2x+1\right)}=\dfrac{6x+3+4x}{2\left(2x-1\right)\left(2x+1\right)}=\dfrac{10x+3}{2\left(2x-1\right)\left(2x+1\right)}\)

14)

\(\dfrac{2x+1}{2x-4}+\dfrac{5}{x^2-4}=\dfrac{2x+1}{2\left(x-2\right)}+\dfrac{5}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(2x+1\right)\left(x+2\right)}{2\left(x-2\right)\left(x+2\right)}+\dfrac{5.2}{2\left(x-2\right)\left(x+2\right)}=\dfrac{2x^2+5x+12}{2\left(x-2\right)\left(x+2\right)}\)

Phân tích đa thức thành nhân tử :

\(52.143-52.39-8.26\)

\(=52.143-52.39-52.4\)

\(=52.\left(143-39-4\right)\)

#~~ Hết~~#