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NX: 2x+3; 5(2x+3) và 2(2x+3) cùng dấu
+TH1: 2x+3 \(\ge\)0 => x \(\ge\frac{-3}{2}\)
=> 5(2x+3), 2(2x+3) \(\ge\)0
=> |5(2x+3)| = 5(2x+3); |2(2x+3)| = 2(2x+3); |2x+3| = 2x+3
=> (2x+3)(5+2+1) = 16
=> 2x+3 = 2
=> 2x = -1
=> x = -1/2 (t/m)
+ TH2: 2x+3 < 0 => x < -3/2
cmtt => -5(2x+3) - 2(2x+3) - (2x+3) = 16
=> (2x+3)(-5-2-1) = 16
=> 2x+3 = -2
=> 2x = -5
=> x = -5/2 (t/m)
/8(2x+3/ = 16
/2x+3/=2
2x+3=2 hoặc 2x+3=-2
2x=-1 hoặc 2x=-5
x=-1/2 hoặc x=-5/2
bạn trả lời nhé
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
\(\left(x+3\right)\left(x^2-3x+9\right)=x^3+27\)
\(\left(2x+y^2\right)\left(2x-y^2\right)=4x^2-y^4\)
a: \(=-2x^2\cdot3x+2x^2\cdot4X^3-2x^2\cdot7+2x^2\cdot x^2\)
\(=8x^5+2x^4-6x^3-14x^2\)
b: \(=2x^3-3x^2-5x+6x^2-9x-15\)
\(=2x^3+3x^2-14x-15\)
c: \(=\dfrac{-6x^5}{3x^3}+\dfrac{7x^4}{3x^3}-\dfrac{6x^3}{3x^3}=-2x^2+\dfrac{7}{3}x-2\)
d: \(=\dfrac{\left(3x-2\right)\left(3x+2\right)}{3x+2}=3x-2\)
e: \(=\dfrac{2x^4-8x^3-6x^2-5x^3+20x^2+15x+x^2-4x-3}{x^2-4x-3}\)
=2x^2-5x+1
Ta có: \(\left|5\left(2x+3\right)\right|+\left|2\left(2x+3\right)\right|+\left|2x+ 3\right|=16\)
\(\Rightarrow5\left|2x+3\right|+2\left|2x+3\right|+\left|2x+3\right|=16\)
\(\Rightarrow\left|2x+3\right|\left(5+2+1\right)=16\)
\(\Rightarrow\left|2x+3\right|.8=16\)
\(\Rightarrow\left|2x+3\right|=2\)
\(\Rightarrow\left[{}\begin{matrix}2x+3=2\\2x+3=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=-1\\2x=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=\dfrac{-5}{2}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=\dfrac{-5}{2}\end{matrix}\right.\).