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1.b) \(\left(\left|x\right|-3\right)\left(x^2+4\right)< 0\)
\(\Rightarrow\hept{\begin{cases}\left|x\right|-3\\x^2+4\end{cases}}\) trái dấu
\(TH1:\hept{\begin{cases}\left|x\right|-3< 0\\x^2+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|< 3\\x^2>-4\end{cases}}\Leftrightarrow x\in\left\{0;\pm1;\pm2\right\}\)
\(TH1:\hept{\begin{cases}\left|x\right|-3>0\\x^2+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|>3\\x^2< -4\end{cases}}\Leftrightarrow x\in\left\{\varnothing\right\}\)
Vậy \(x\in\left\{0;\pm1;\pm2\right\}\)
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a) \(\left|3-2x\right|+\frac{3}{4}=\left|-2\frac{3}{4}\right|\)
⇔ | 3 - 2x | + 3/4 = 11/4
⇔ | 3 - 2x | = 8/4 = 2
⇔ \(\orbr{\begin{cases}3-2x=2\\3-2x=-2\end{cases}}\text{⇔}\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{5}{2}\end{cases}}\)
b) 2x+2 - 2x = 96
⇔ 2x( 22 - 1 ) = 96
⇔ 2x.3 = 96
⇔ 2x = 32
⇔ 2x = 25
⇔ x = 5
c) ( 2x + 5 )3 = -27
⇔ ( 2x + 5 )3 = (-3)3
⇔ 2x + 5 = -3
⇔ 2x = -8
⇔ x = -4
a. \(\left|3-2x\right|+\frac{3}{4}=\left|-2\frac{3}{4}\right|\)
\(\Rightarrow\left|3-2x\right|+\frac{3}{4}=\left|-\frac{11}{4}\right|\)
\(\Rightarrow\left|3-2x\right|+\frac{3}{4}=\frac{11}{4}\)
\(\Rightarrow\left|3-2x\right|=2\)
\(\Leftrightarrow\orbr{\begin{cases}3-2x=2\\3-2x=-2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{5}{2}\end{cases}}\)
b. 2x+2 - 2x = 96
<=> 2x . 22 - 2x = 96
<=> 2x ( 22 - 1 ) = 96
<=> 2x . 3 = 96
<=> 2x = 32 = 25
<=> x = 5
c. ( 2x + 5 )3 = - 27
<=> ( 2x + 5 )3 = ( - 3 )3
<=> 2x + 5 = - 3
<=> 2x = - 8
<=> x = - 4
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B1:
a) \(\frac{x+4}{x+3}=\frac{x+9}{x+4}\)
-->(x+4)(x+4)=(x+3)(x+9)
\(x^2\)+4x+4x+16=\(x^2\)+9x+3x+27
\(x^2-x^2\)+4x+4x-9x-3x= - 16+27
- 4x=11
x=\(\frac{-4}{11}\)
b) \(\frac{x-5}{x+3}=\frac{x-4}{x+6}\)
-->(x-5)(x+6)=(x+3)(x-4)
\(x^2\)+6x-5x-30=\(x^2\)-4x+3x-12
\(x^2-x^2\)+6x-5x+4x-3x=30-12
2x=18
x=9
c)\(\frac{3x-1}{3x}=\frac{2x-1}{2x+1}\)
--> (3x-1)(2x+1)=3x.(2x-1)
\(6x^2\)+3x-2x-1=\(6x^2\)-3x
\(6x^2-6x^2\)+3x-2x+3x=1
4x=1
x=\(\frac{1}{4}\)
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a) Ta có \(|5\left(2x+3\right)\ge0\)
\(|2\left(2x+3\right)|\ge0\)
\(|2x+3|\ge0\)
\(\Rightarrow|5\left(2x+3\right)|+|\left(2x+3\right)|+|2x+3|\ge0\)
\(\Rightarrow5\left(2x+3\right)+2\left(2x+3\right)+2x+3=16\)
\(\Rightarrow10x+15+4x+6+2x+3=16\)
\(\Rightarrow\left(10x+4x+2x\right)+\left(15+6+3\right)=16\)
\(\Rightarrow16x+24=16\)
\(\Rightarrow24=16x-16\)
\(\Rightarrow24=x\)
Vậy x=24
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