`P=((3+x)/(3-x)-(3-x)/(3+x)+(4x^2)/(x^2-9)):((2x+1)/(x+3)-1)`

`=((4x^2-(3-x)^2-(3+x)^2)/(x^2-9)):((2x+1-x-3)/(x+3))`

`=((4x^2-x^2+6x-9-x^2-6x-9)/(x^2-9)):((x-2)/(x+3))`

`=((2x^2-18)/(x^2-9))*(x+3)/(x-2)`

`=((2(x^2-9))/(x^2-9))*(x+3)/(x-2)`

`=(2x+6)/(x-2)`

ĐKXĐ: \(x\ne\pm3;x\ne-\dfrac{1}{2};x\ne2\)

\(P=\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{4x^2}{\left(3-x\right)\left(3+x\right)}\right):\dfrac{2x+1-x-3}{x+3}\)

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

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

\(=\dfrac{12x-4x^2}{3-x}\cdot\dfrac{1}{x-2}\)

\(=\dfrac{4x\left(3-x\right)}{3-x}\cdot\dfrac{1}{x-2}\) \(=\dfrac{4x}{x-2}\)

Chắc mình làm tắt quá để mình làm lại bước biến đổi.

`P=((3+x)/(3-x)-(3-x)/(3+x)+(4x^2)/(x^2-9)):((2x+1)/(x+3)-1)`

`=((x-3)/(x+3)+(4x^2)/(x^2-9)-(x+3)/(x-3)):((2x+1-x-3)/(x+3))`

`=((x-3)^2/(x^2-9)+(4x^2)/(x^2-9)-(x+3)^2/(x^2-9)):((x-2)/(x+3))`

`=(((x-3)^2+4x^2-(x+3)^2)/(x^2-9))*(x+3)/(x-2)`

`=(x^2-6x+9+4x^2-x^2-6x-9)/(x^2-9)*(x+3)/(x-2)`

`=(4x^2-12x)/(x^2-9)*(x+3)/(x-2)`

`=(4x(x-3))/((x-3)(x+3))*(x+3)/(x-2)`

`=(4x)/(x+3)*(x+3)/(x-2)`

`=(4x)/(x-2)`

a) \(A\left(x\right)=3x^3-4x^4-2x^3+4x^4-5x+3\)

\(\Rightarrow A\left(x\right)=-4x^4+4x^4+3x^3-2x^3-5x+3\)

\(\Rightarrow A\left(x\right)=x^3-5x+3\)

\(B\left(x\right)=5x^3-4x^2-5x^3-4x^2-5x-3\)

\(\Rightarrow B\left(x\right)=5x^3-5x^3-4x^2-4x^2-5x-3\)

\(\Rightarrow B\left(x\right)=-8x^2-5x-3\)

b) \(A\left(x\right)+B\left(x\right)=x^3-5x+3+\left(-8x^2-5x-3\right)\)

\(\Rightarrow A\left(x\right)+B\left(x\right)=x^3-5x+3-8x^2-5x-3\)

\(\Rightarrow A\left(x\right)+B\left(x\right)=x^3-8x^2-5x-5x+3-3\)

\(\Rightarrow A\left(x\right)+B\left(x\right)=x^3-8x^2-10x\)

\(A\left(x\right)-B\left(x\right)=x^3-5x+3-\left(-8x^2-5x-3\right)\)

\(\Rightarrow A\left(x\right)-B\left(x\right)=x^3-5x+3+8x^2+5x+3\)

\(\Rightarrow A\left(x\right)-B\left(x\right)=x^3+8x^2-5x+5x+3+3\)

\(\Rightarrow A\left(x\right)-B\left(x\right)=x^3+8x^2+6\)

a, x= -3

b, x= -3, x= 3/2

Sao khó vậy mày

nhầm b, x= -3 hoặ x= 9/4

lúc nãy đánh nhầm

\(\left(-1\frac{1}{2}\right)\left(-1\frac{1}{3}\right)\left(-1\frac{1}{4}\right)...\left(-1\frac{1}{2003}\right)\left(-1\frac{1}{2004}\right)\)

\(=-\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{2004}{2003}.\frac{2005}{2004}\)

\(=-\frac{3.4.5.....2004.2005}{2.3.4.....2003.2004}=\frac{-2005}{2}\)

\(11-\left(3x-1\right)=\frac{9}{2}-\left(5-3,5x\right)\)

\(=>11-3x+1=\frac{9}{2}-5+3,5x\)

\(=>-3x+12=3,5x-\frac{1}{2}\)

\(=>-3x-3,5x=-\frac{1}{2}-12\)

\(=>-6,5x=-12,5\)

\(=>x=\frac{-12,5}{-6,5}=\frac{25}{13}\)

Ủng hộ nha

\(11-\left(3x-1\right)=\frac{9}{2}-\left(5-3,5x\right)\)

\(11-3x+1=\frac{9}{2}-5+3,5x\)

\(12-3x=-\left(0,5\right)+3,5x\)

\(12,5-3x=3,5x\)

\(12,5=6,5x\)

\(x=12,5:6,5=\frac{25}{13}\)

này mà lớp 1 á

Đùa nhau à

đây không phải bài lớp 1

a) \(2\left|\frac{3}{2}x-\frac{1}{4}\right|=\left|-\frac{5}{4}\right|\)

\(\Leftrightarrow\orbr{\begin{cases}2\left|\frac{3}{2}x-\frac{1}{4}\right|=\frac{5}{4}\\2\left|\frac{3}{2}x-\frac{1}{4}\right|=-\frac{5}{4}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left|\frac{3}{2}x-\frac{1}{4}\right|=\frac{5}{4}:2\\\left|\frac{3}{2}x-\frac{1}{4}\right|=-\frac{5}{4}:2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left|\frac{3}{2}x-\frac{1}{4}\right|=\frac{5}{8}\\\left|\frac{3}{2}x-\frac{1}{4}\right|=-\frac{5}{8}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{3}{2}x-\frac{1}{4}=\frac{5}{8}\\\frac{3}{2}x-\frac{1}{4}=-\frac{5}{8}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{3}{2}x=\frac{5}{8}+\frac{1}{4}\\\frac{3}{2}x=-\frac{5}{8}+\frac{1}{4}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{3}{2}x=\frac{7}{8}\\\frac{3}{2}x=-\frac{3}{8}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{8}:\frac{3}{2}\\x=-\frac{3}{8}:\frac{3}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{8}.\frac{2}{3}\\x=-\frac{3}{8}.\frac{2}{3}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{12}\\x=-\frac{1}{4}\end{cases}}\)

Vậy \(x\in\left\{\frac{7}{12};-\frac{1}{4}\right\}\)

x^2 - 4x + 4 = 5 ( x - 2 ) 

x^2 - 4x + 4 - 5 ( x - 2 ) = 0 

( x - 2 ) ^2 - 5 ( x - 2 ) = 0 

( x - 2 ) ( x - 2 - 5 ) = 0

( x - 2 ) ( x - 7 ) = 0 

x  - 2 = 0 hoặc x - 7 = 0 

x = 2 hoặc x = 7 

c) (x-2)^2=5(x-2)

=> x-2=5 hoặc x-2 =0

=> x=7 hoặc x=2

d) (2x-3)^2=(5-x)^2

=> 2x-3=5-x

=> x=8/3

4x^2 - 12x + 9 = ( 5 - x ) ^2 

( 2x - 3 ) ^2 = ( 5 - x ) ^2 

( 2x - 3 ) ^2 - ( 5 - x ) ^2 = 0 

( 2x - 3 - 5 + x ) ( 2x - 3 + 5 - x ) = 0 

( 3x - 8 ) ( x + 2 ) = 0 

3x - 8 = 0 hoặc x + 2 = 0 

x = 8/3 hoặc x = -2 

1)

a)

\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)

\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)

\(\frac{-20}{5}< x< \frac{-3}{10}\)

\(\frac{-40}{10}< x< \frac{-3}{10}\)

\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)

\(\left(\frac{-5}{3}\right)^3< x< \frac{-24}{35}.\frac{-5}{6}\)

\(\frac{25}{3}< x< \frac{-4}{7}.\frac{1}{1}\)

\(\frac{-25}{3}< x< \frac{-4}{7}\)

\(\frac{-175}{21}< x< \frac{-12}{21}\)

\(\Rightarrow Z\in\left\{-13;-14;-15;-16;...;-174\right\}\)

2)

a)

\(\left(\frac{9}{10}-\frac{15}{16}\right).\left(\frac{5}{12}-\frac{11}{15}-\frac{7}{20}\right)\)

\(=\left(\frac{72}{80}-\frac{75}{80}\right).\left(\frac{25}{60}-\frac{44}{60}-\frac{21}{60}\right)\)

\(=\frac{-3}{80}.\frac{-40}{60}\)

\(=\frac{-1}{-2}.\frac{-1}{-20}\)

\(=\frac{1}{40}\)