a) \(A=\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}+\dfrac{x^2-1}{9-x^2}\right):\left(2-\dfrac{x+5}{x+3}\right)\) (ĐK: \(x\ne\pm3\))

\(A=\left[\dfrac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x^2-1}{\left(x+3\right)\left(x-3\right)}\right]:\left(2+\dfrac{x+5}{x+3}\right)\)

\(A=\dfrac{x^2-3x-2x-6-x^2+1}{\left(x+3\right)\left(x-3\right)}:\dfrac{2\left(x+3\right)-\left(x+5\right)}{x+3}\)

\(A=\dfrac{-5x-5}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{x+3}{x+1}\)

\(A=\dfrac{-5\left(x+1\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)\left(x+1\right)}\)

\(A=\dfrac{-5}{x-3}\)

b) Ta có: \(\left|x\right|=1\)

TH1: \(\left|x\right|=-x\) với \(x< 0\)

Pt trở thành:

\(-x=1\) (ĐK: \(x< 0\)) 

\(\Leftrightarrow x=-1\left(tm\right)\)

Thay \(x=-1\) vào A ta có:

\(A=\dfrac{-5}{x-3}=\dfrac{-5}{-1-3}=\dfrac{5}{4}\)

TH2: \(\left|x\right|=x\) với \(x\ge0\)

Pt trở thành:

\(x=1\left(tm\right)\) (ĐK: \(x\ge0\)) 

Thay \(x=1\) vào A ta có:

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

c) \(A=\dfrac{1}{2}\) khi:

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

\(\Leftrightarrow-10=x-3\)

\(\Leftrightarrow x=-10+3\)

\(\Leftrightarrow x=-7\left(tm\right)\)

d) \(A\) nguyên khi:

\(\dfrac{-5}{x-3}\) nguyên

\(\Rightarrow x-3\inƯ\left(-5\right)\)

\(\Rightarrow x\in\left\{8;-2;2;4\right\}\)

a: \(A=\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}+\dfrac{x^2-1}{9-x^2}\right):\left(2-\dfrac{x+5}{x+3}\right)\)

\(=\dfrac{x\left(x-3\right)-2\left(x+3\right)-x^2+1}{\left(x-3\right)\left(x+3\right)}:\dfrac{2x+6-x-5}{x+3}\)

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

\(=\dfrac{-5x-5}{\left(x-3\right)}\cdot\dfrac{1}{x+1}=\dfrac{-5}{x-3}\)

b: |x|=1

=>x=-1(loại) hoặc x=1(nhận)

Khi x=1 thì \(A=\dfrac{-5}{1-3}=-\dfrac{5}{-2}=\dfrac{5}{2}\)

c: A=1/2

=>x-3=-10

=>x=-7

d: A nguyên

=>-5 chia hết cho x-3

=>x-3 thuộc {1;-1;5;-5}

=>x thuộc {4;2;8;-2}

\(A=2^3\times2^4\times2^5\times...\times2^9\)

\(=2^{3+4+5+...+9}\)

\(=2^{42}\)

mk giúp sau kết bạn với mình nhé:

A=2^{3+4+5+6+7+8+9}

A=2⁴²

có cần tính chi tiết ko bạn

a) Rút gọn VT = 45x + 8. Từ đó tìm được x = 2 15 .  

b) Rút gọn VT = -25x – 8. Từ đó tìm được x = − 11 25 .

a) x × 7/9 = 2/3 + 2 1/2

x × 7/9 = 2/3 + 5/2

x × 7/9 = 19/6

x = 19/6 : 7/9

x = 57/14

b) 5/7 + x : 9/4 = 4/3

x : 9/4 = 4/3 - 5/7

x : 9/4 = 13/21

x = 13/21 × 9/4

x = 39/28

Một tốp thợ có 120 người dự địnhnlamf trong 50 ngày . Khi bắt đầu làm có một số người đến làm thêm nên làm xong công việc đó trong 30 ngày . Hỏi có bao nhiêu người đến thêm?

a: A<1

=>A-1<0

=>\(\dfrac{\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}-3}< 0\)

=>\(\dfrac{4}{\sqrt{x}-3}< 0\)

=>\(\sqrt{x}-3< 0\)

=>\(\sqrt{x}< 3\)

=>0<=x<9

b: Để A<=2 thì A-2<=0

=>\(\dfrac{\sqrt{x}+1-2\sqrt{x}+6}{\sqrt{x}-3}< =0\)

=>\(\dfrac{-\sqrt{x}+7}{\sqrt{x}-3}< =0\)

=>\(\dfrac{\sqrt{x}-7}{\sqrt{x}-3}>=0\)

TH1: \(\left\{{}\begin{matrix}\sqrt{x}-7>=0\\\sqrt{x}-3>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\sqrt{x}>=7\\\sqrt{x}>3\end{matrix}\right.\)

=>\(\sqrt{x}>=7\)

=>x>=49

TH2: \(\left\{{}\begin{matrix}\sqrt{x}-7< =0\\\sqrt{x}-3< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\sqrt{x}< =7\\\sqrt{x}< 3\end{matrix}\right.\)

=>\(\sqrt{x}< 3\)

=>0<=x<9

a) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)^2=27.\)

\(\Leftrightarrow x^3+27-x\left(x^2-4x+4\right)-27=0.\)

\(\Leftrightarrow x^3-x^3+4x^2-4x=0.\)

\(\Leftrightarrow4x\left(x-1\right)=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=0.\\x-1=0.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0.\\x=1.\end{matrix}\right.\)

Vậy \(S=\left\{0;1\right\}.\)

Ta có: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\)

\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+9\left(x^2+2x+1\right)=15\)

\(\Leftrightarrow-9x^2+27x+9x^2+18x+9=15\)

\(\Leftrightarrow45x=6\)

hay \(x=\dfrac{2}{15}\)

Bài 1:

a) \(x.\dfrac{3}{4}=\dfrac{9}{14}\)

\(\Rightarrow x=\dfrac{9}{14}:\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{6}{7}\)

b) \(x:\dfrac{5}{9}=\dfrac{3}{10}\)

\(\Rightarrow x=\dfrac{3}{10}.\dfrac{5}{9}\)

\(\Rightarrow x=\dfrac{1}{6}\)

help me

Bài 2:

a) \(\dfrac{3}{10}+\dfrac{2}{5}:\dfrac{4}{15}=\dfrac{3}{10}+\dfrac{3}{2}=\dfrac{9}{5}\)

b) \(\dfrac{4}{7}.3+\dfrac{2}{3}.2=\dfrac{12}{7}+\dfrac{4}{3}=\dfrac{64}{21}\)

1.a.

\(\left(x+3\right)\left(x-2\right)< 0\)

\(TH1:\hept{\begin{cases}x+3< 0\\x-2>0\end{cases}}\Rightarrow\hept{\begin{cases}x< -3\\x>2\end{cases}}\)

\(TH2:\hept{\begin{cases}x+3>0\\x-2< 0\end{cases}\Rightarrow\hept{\begin{cases}x>-3\\x< 2\end{cases}}}\)

không biết có đúng không nữa!

\(a,A=\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}+\dfrac{x^2-1}{9-x^2}\right):\left(2-\dfrac{x+5}{3+x}\right)\\ =\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}-\dfrac{x^2-1}{x^2-9}\right):\left(\dfrac{2\left(3+x\right)}{3+x}-\dfrac{x+5}{3+x}\right)\\ =\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}-\dfrac{x^2-1}{\left(x-3\right)\left(x+3\right)}\right):\dfrac{2\left(3+x\right)-\left(x+5\right)}{3+x}\\ =\left(\dfrac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{x^2-1}{\left(x-3\right)\left(x+3\right)}\right):\dfrac{6+2x-x-5}{3+x}\)

\(=\dfrac{x^2-3x-\left(2x+6\right)-\left(x^2-1\right)}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+1}{3+x}\\ =\dfrac{x^2-3x-2x-6-x^2+1}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{3+x}{x+1}\\ =\dfrac{-5x-5}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{3+x}{x+1}\\ =\dfrac{-5\left(x+1\right).\left(3+x\right)}{\left(x-3\right)\left(x+3\right).\left(x+1\right)}\\ =\dfrac{-5}{x-3}\)

\(b,A=x^2-x-2=0\\ \Leftrightarrow x^2+x-2x-2=0\\ \Leftrightarrow x\left(x+1\right)-2\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)

\(c,\dfrac{-5}{x-3}=\dfrac{1}{2}\\ \Leftrightarrow-10=x-3\\ \Leftrightarrow-x+3=10\\ \Leftrightarrow-x=7\\ \Leftrightarrow x=7\)

Để `A=1/2` thì `x=7`