`#3107.101107`

`a)`

- Tổng của 2 số hữu tỉ khác dấu: \(-\dfrac{4}{15}=-\dfrac{13}{15}+\dfrac{9}{15}\)

`b)`

- Tích cảu 2 số hữu tỉ: \(-\dfrac{4}{15}=-\dfrac{8}{15}\cdot\dfrac{1}{2}\)

`c)`

Thương của 2 số hữu tỉ: \(-\dfrac{4}{15}=-\dfrac{16}{15}\div2\)

a) \(\dfrac{2x+1}{9}=\dfrac{5}{x+1}\left(x\ne-1\right)\)

\(\Rightarrow\left(2x+1\right)\left(x+1\right)=9\cdot5=45\)

\(\Rightarrow2x^2+2x+x+1=45\)

\(\Rightarrow2x^2+3x-44=0\)

\(\Rightarrow2x^2+11x-8x-44=0\)

\(\Rightarrow x\left(2x+11\right)-4\left(2x+11\right)=0\)

\(\Rightarrow\left(x-4\right)\left(2x+11\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\2x=-11\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{11}{2}\end{matrix}\right.\)

b) \(\dfrac{2x-1}{21}=\dfrac{3}{2x+1}\left(x\ne-\dfrac{1}{2}\right)\)

\(\Rightarrow\left(2x-1\right)\left(2x+1\right)=21\cdot3=63\)

\(\Rightarrow4x^2-1=63\)

\(\Rightarrow4x^2=64\)

\(\Rightarrow\left(2x\right)^2=8^2\)

\(\Rightarrow\left[{}\begin{matrix}2x=8\\2x=-8\end{matrix}\right.\)

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

c) \(\dfrac{2x-1}{2}=\dfrac{5}{x}\left(x\ne0\right)\)

\(\Rightarrow x\left(2x-1\right)=5\cdot2=10\)

\(\Rightarrow2x^2-x=10\)

\(\Rightarrow2x^2-x-10=0\)

\(\Rightarrow2x^2+4x-5x-10=0\)

\(\Rightarrow2x\left(x+2\right)-5\left(x+2\right)=0\)

\(\Rightarrow\left(2x-5\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x=5\\x=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-2\end{matrix}\right.\)

d) \(\dfrac{x-3}{3}=\dfrac{2x+1}{5}\)

\(\Rightarrow15\cdot\dfrac{x-3}{3}=15\cdot\dfrac{2x+1}{5}\)

\(\Rightarrow5\left(x-3\right)=3\left(2x+1\right)\)

\(\Rightarrow5x-15=6x+3\)

\(\Rightarrow6x-5x=-18\)

\(\Rightarrow x=-18\)

Bài 1:

\(\dfrac{a}{b}-\dfrac{a+2009}{b+2009}=\dfrac{a\left(b+2009\right)-b\left(a+2009\right)}{b\left(b+2009\right)}\)

\(=\dfrac{2009a-2009b}{b\left(b+2009\right)}=\dfrac{2009\left(a-b\right)}{b\left(b+2009\right)}\)

Vì a>b>0 nên a-b>0; b>0; b+2009>0

=>\(\dfrac{2009\left(a-b\right)}{b\left(b+2009\right)}>0\)

=>\(\dfrac{a}{b}>\dfrac{a+2009}{b+2009}\)

Đặt 222=a

=>\(\dfrac{222}{222^2+1}=\dfrac{a}{a^2+1};\dfrac{223}{223^2+1}=\dfrac{\left(a+1\right)^2}{\left(a+1\right)^2+1}\)

\(\dfrac{a^2}{a^2+1}-\dfrac{\left(a+1\right)^2}{\left(a+1\right)^2+1}\)

\(=\dfrac{a^2\left[\left(a+1\right)^2+1\right]-\left(a+1\right)^2\left(a^2+1\right)}{\left(a^2+1\right)\left[\left(a+1\right)^2+1\right]}\)

\(=\dfrac{a^2\left(a^2+2a+2\right)-\left(a^2+2a+1\right)\left(a^2+1\right)}{\left(a^2+1\right)\left[\left(a+1\right)^2+1\right]}\)

\(=\dfrac{a^4+2a^3+2a^2-a^4-a^2-2a^3-2a-a^2-1}{\left(a^2+1\right)\left[\left(a+1\right)^2+1\right]}\)

\(=\dfrac{-2a-1}{\left(a^2+1\right)\left[\left(a+1\right)^2+1\right]}< 0\)

=>\(\dfrac{222}{222^2+1}< \dfrac{223}{223^2+1}\)

   Bài 3

a; m - 2021 = 0 ⇒ m = 2021

Lập bảng ta có:

Theo bảng trên ta có \(x\) là số hữu tỉ dương khi và chỉ khi  m > 2021

Vậy m > 2021 

Bài 3b;

  Bài 3

a; m - 2021 = 0 ⇒ m = 2021

Lập bảng ta có:

Theo bảng trên ta có \(x\) là số hữu tỉ âm khi và chỉ khi  m < 2021

Vậy m <  2021 

Còn thiếu dữ liệu em nhé.

Bài 2a:

(2\(x\) + 5).3⁵ = 3⁸

2\(x\) + 5 = 3⁸ : 3⁵

2\(x\) + 5 = 3³

2\(x\) + 5 = 27

2\(x\)       = 27 - 5

2\(x\)      = 22

  \(x\)     = 22: 2

    \(x\)    = 11

Vậy \(x\) = 11

b; 5^\(x+2\) - 5^\(x+1\) = 2500

    5^\(x+1\).(5 - 1) = 2500

    5^\(x+1\). 4 = 2500

   5^\(x+1\)      = 2500 : 4

   5^\(x+1\)    = 625

   5^\(x+1\)   = 5⁴

   \(x+1\)   = 4

   \(x\)          = 4 - 1

   \(x\)         = 3

   Vậy \(x=3\)

f; (\(x\) + 4).(\(x-2\)) = 0

   \(\left[{}\begin{matrix}x+4=0\\x-2=0\end{matrix}\right.\)

    \(\left[{}\begin{matrix}x=-4\\x=2\end{matrix}\right.\)

    Vậy \(x\) \(\in\) {-4; 2}

g; (\(x\) - 2).(\(x\) + 3) < 0

    \(x\) - 2 = 0 ⇒ \(x\) = 2; \(x\) + 3  = 0 ⇒ \(x\) = -3

Lập bảng ta có:

Theo bảng trên ta có   -3 < \(x\) <  2

Vậy -3 < \(x\) < 2

e) \(x^2-3x=0\)

\(\Leftrightarrow x\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

Vậy: ... 

f) \(\left(x+4\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=2\end{matrix}\right.\)

Vậy: ...

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

TH1: \(\left\{{}\begin{matrix}x-2>0\\x+3< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>2\\x< -3\end{matrix}\right.\) 

\(\Rightarrow x\in\varnothing\)

TH2: \(\left\{{}\begin{matrix}x-2< 0\\x+3>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 2\\x>-3\end{matrix}\right.\)

\(\Rightarrow-3< x< 2\)

Vậy: ... 

h) \(\left(x-1\right)\left(x+2\right)>0\)

TH1: \(\left\{{}\begin{matrix}x-1>0\\x+2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>1\\x>-2\end{matrix}\right.\)

\(\Rightarrow x>1\)

TH2: \(\left\{{}\begin{matrix}x-1< 0\\x+2< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 1\\x< -2\end{matrix}\right.\)

\(\Rightarrow x< -2\)

Vậy: ... 

a) \(3\left(2x+7\right)-2x=9\Leftrightarrow6x+21-2x=9\)

\(\Leftrightarrow4x+21=9\)

\(\Leftrightarrow4x=9-21=-12\)

\(\Leftrightarrow x=\dfrac{-12}{4}=-3\)

Vậy: ... 

b) \(\left[\left(7x-4\right):2-2\right]\cdot13=221\)

\(\Leftrightarrow\left(7x-4\right):2-2=\dfrac{221}{13}=17\)

\(\Leftrightarrow\left(7x-4\right):2=17+2\)

\(\Leftrightarrow\left(7x-4\right):2=19\)

\(\Leftrightarrow7x-4=19\cdot2=38\)

\(\Leftrightarrow7x=42\)

\(\Leftrightarrow x=\dfrac{42}{7}=6\)

Vậy: ... 

c) \(x^2-9=0\)

\(\Leftrightarrow x^2=9\)

\(\Leftrightarrow x=\pm\sqrt{9}\)

\(\Leftrightarrow x=\pm3\)

Vậy: ....

d) \(5< x^2< 16\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2>5\\x^2< 16\end{matrix}\right.\)

Với \(x^2>5\Rightarrow\left[{}\begin{matrix}x< -\sqrt{5}\\x>\sqrt{5}\end{matrix}\right.\) (1)  

Với \(x^2< 16\Rightarrow-4< x< 4\) (2)

Từ (1) và (2) \(\left[{}\begin{matrix}-4< x< -\sqrt{5}\\\sqrt{5}< x< 4\end{matrix}\right.\)