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\(2x^3+10x^2=0\)
=>\(2x^2\left(x+5\right)=0\)
=>\(x^2\left(x+5\right)=0\)(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: ...
\(x^2-6x=0\Leftrightarrow x\left(x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
\(\left(\dfrac{-4}{9}\right)^2=\dfrac{16}{81}\Rightarrow x=2\)
\(\left(-\dfrac{1}{3}\right)^3=\dfrac{-1}{27}\Rightarrow x=1\)
\(\left(-\dfrac{1}{3}\right)^4=\dfrac{1}{81}\Rightarrow x=1\)
\(\left(-\dfrac{4}{9}\right)^x=\dfrac{16}{81}\\ \left(-\dfrac{4}{9}\right)^x=\left(\dfrac{4}{9}\right)^2\\ \left(-\dfrac{4}{9}\right)^x=\left(-\dfrac{4}{9}\right)^2\\ x=2\\ -----------\\ \left(-\dfrac{1}{3}\right)^{2x+1}=-\dfrac{1}{27}\\ \left(-\dfrac{1}{3}\right)^{2x+1}=\left(-\dfrac{1}{3}\right)^3\\ 2x+1=3\\ 2x=3-1=2\\ x=\dfrac{2}{2}=1\\ -----------\\ \left(-\dfrac{1}{3}\right)^{3x+1}=\dfrac{1}{81}\\\left(-\dfrac{1}{3}\right)^{3x+1}=\left(\dfrac{1}{3}\right)^4\\ \left(-\dfrac{1}{3}\right)^{3x+1}=\left(-\dfrac{1}{3}\right)^4\\ 3x+1=4\\ 3x=4-1=3\\ x=\dfrac{3}{3}=1\)
\(\left(\dfrac{7}{5}\right)^x=\dfrac{49}{25}\Leftrightarrow\left(\dfrac{7}{5}\right)^x=\left(\dfrac{7}{5}\right)^2\Leftrightarrow x=2\)
\(a,32< 2^n< 128\)
\(=>2^5< 2^n< 2^7\)
\(=>n=6\)
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) \(3^2.3^n=3^5\\ \Rightarrow3^{n+2}=3^5\\ \Rightarrow n+2=5\\ \Rightarrow n=3\left(nhận\right)\)
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
=>\(\dfrac{5}{461}< \dfrac{5}{128}\)
=>\(\dfrac{5}{461}-1< \dfrac{5}{128}-1\)
=>\(\dfrac{-456}{461}< \dfrac{-123}{128}\)