\(x\left(x-5\right)+3\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

Vậy tập nghiệm pt là: \(S=\left\{5;-3\right\}\)

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

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

=>\(\left[{}\begin{matrix}x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

a: \(\dfrac{1}{4003}>0;0>-\dfrac{75}{106}\)

Do đó: \(\dfrac{1}{4003}>-\dfrac{75}{106}\)

b: \(-19< -17\)

=>\(-\dfrac{19}{31}< -\dfrac{17}{31}\)

c: \(\dfrac{-33}{37}>\dfrac{-34}{37}\)

mà \(-\dfrac{34}{37}>-\dfrac{34}{35}\)

nên \(\dfrac{-33}{37}>-\dfrac{34}{35}\)

d: \(\dfrac{-13}{77}=\dfrac{-13\cdot205}{77\cdot205}=\dfrac{-2665}{77\cdot205}\)

\(\dfrac{-34}{205}=\dfrac{-34\cdot77}{205\cdot77}=\dfrac{-2618}{205\cdot77}\)

mà -2665<-2618

nên \(\dfrac{-13}{77}< \dfrac{-34}{205}\)

e: \(\dfrac{-456}{461}=-1+\dfrac{5}{461};\dfrac{-123}{128}=-1+\dfrac{5}{128}\)

461>128

=>\(\dfrac{5}{461}< \dfrac{5}{128}\)

=>\(\dfrac{5}{461}-1< \dfrac{5}{128}-1\)

=>\(\dfrac{-456}{461}< \dfrac{-123}{128}\)

\(2x^3+10x^2=0\)

=>\(2x^2\left(x+5\right)=0\)

=>\(x^2\left(x+5\right)=0\)(Vì 2>0)

=>\(\left[{}\begin{matrix}x^2=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

\(2x^3+10x^2=0\Leftrightarrow x^2\left(2x+10\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

`x^2-6x=0`

`<=>x(x-6)=0`

TH1: `x =0 `

TH2: `x - 6=0<=>x=6`

Vậy: ... 

\(x^2-6x=0\Leftrightarrow x\left(x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

a: A={x∈N|x=3k+1; k∈N; 0<=k<=6}

b: B={x∈N|x=k³; 1<=k<=5}

a)Mỗi phần tử đều cách nhau 3 đơn vị

b)ko biết làm

??

2 bạn nhé.

\(\left(\dfrac{-4}{9}\right)^2=\dfrac{16}{81}\Rightarrow x=2\) 

\(\left(-\dfrac{1}{3}\right)^3=\dfrac{-1}{27}\Rightarrow x=1\)

\(\left(-\dfrac{1}{3}\right)^4=\dfrac{1}{81}\Rightarrow x=1\)

\(\left(-\dfrac{4}{9}\right)^x=\dfrac{16}{81}\\ \left(-\dfrac{4}{9}\right)^x=\left(\dfrac{4}{9}\right)^2\\ \left(-\dfrac{4}{9}\right)^x=\left(-\dfrac{4}{9}\right)^2\\ x=2\\ -----------\\ \left(-\dfrac{1}{3}\right)^{2x+1}=-\dfrac{1}{27}\\ \left(-\dfrac{1}{3}\right)^{2x+1}=\left(-\dfrac{1}{3}\right)^3\\ 2x+1=3\\ 2x=3-1=2\\ x=\dfrac{2}{2}=1\\ -----------\\ \left(-\dfrac{1}{3}\right)^{3x+1}=\dfrac{1}{81}\\\left(-\dfrac{1}{3}\right)^{3x+1}=\left(\dfrac{1}{3}\right)^4\\ \left(-\dfrac{1}{3}\right)^{3x+1}=\left(-\dfrac{1}{3}\right)^4\\ 3x+1=4\\ 3x=4-1=3\\ x=\dfrac{3}{3}=1\)

\(\left(\dfrac{7}{5}\right)^x=\dfrac{49}{25}\Leftrightarrow\left(\dfrac{7}{5}\right)^x=\left(\dfrac{7}{5}\right)^2\Leftrightarrow x=2\)

Không sử dụng dấu tương đương nhé!

c:

\(a,32< 2^n< 128\)

\(=>2^5< 2^n< 2^7\)

\(=>n=6\)

Vậy...

\(b,2.16\ge2^n>4\)

\(=>2^5\ge2^n>2^2\)

\(=>n\in\left\{3;4;5\right\}\)

Vậy...

\(c,3^2.3^n=3^5\)

        \(3^n=3^5:3^2\)

        \(3^n=3^3\)

\(=>n=3\)

Vậy...

\(d,\left(2^2:4\right).2^n=4\)

     \(\left(2^2:2^2\right).2^n=4\)

                 \(1.2^n=4\)

                    \(2^n=4:1\)

                    \(2^n=4\)

              \(=>2^n=2^2\)

             \(=>n=2\)

Vậy ...

\(e,\dfrac{1}{9}.3^4.3^n=3^7\)

   \(\dfrac{1}{9}.81.3^n=3^7\)

       \(3^2.3^n=3^7\)

           \(3^n=3^7:3^2\)

           \(3^n=3^5\)

\(=>n=5\)

Vậy...

\(g,\dfrac{1}{2}.2^n+4.2^n=9.2^5\)

 \(\left(\dfrac{1}{2}+4\right).2^n=9.2^5\)

             \(\dfrac{9}{2}.2^n=9.32\)

              \(\dfrac{9}{2}.2^n=288\)

                  \(2^n=288:\dfrac{9}{2}\)

                  \(2^n=2^6\)

\(=>n=6\)

Vậy...

a) \(32< 2^n< 128\\ \Rightarrow2^5< 2^n< 2^7\\ \Rightarrow5< n< 7\)

Mà: \(n\inℕ^∗\)

\(\Rightarrow n=6\)

b) \(2.16\ge2^n>4\\ \Rightarrow2^1.2^4\ge2^n>2^2\\ \Rightarrow2^5\ge2^n>2^2\\ \Rightarrow5\ge n>2\)

Mà: \(n\inℕ^∗\)

\(\Rightarrow n\in\left\{5;4;3\right\}\)

c) \(3^2.3^n=3^5\\ \Rightarrow3^{n+2}=3^5\\ \Rightarrow n+2=5\\ \Rightarrow n=3\left(nhận\right)\)