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

bài này khó ak nha

Vế trái: 4/(x+2).(x+6)+7/(x+6).(x+13)

<=>1/x+2 -1/x+6 +1/x+6 -1/x+13

<=>1/x+2-1/x+13

=> 1/x+2-1/x+13=2x+1/(x+2).(x+16) -3/(x+13).(x+16)

<=>1/x+2 - 1/x+13 + 1/x+13 - 1/x+16=2x+1/(x+2).(x+16)

<=>1/x+2 - 1/x+16=2x+1/(x+2).(x+16)

<=> 14/(x+2).(x+16)= 2x+1/(x+2).(x+16)

<=> 2x+1=14

<=> 2x=14-1

<=> 2x=13

<=> x=13:2

<=> x=13/2

Vậy x=13/2

Chắc là vầy. Mk cug ko chắc nữa

Vế trái: 4/(x+2).(x+6)+7/(x+6).(x+13)

<=>1/x+2 -1/x+6 +1/x+6 -1/x+13

<=>1/x+2-1/x+13

=> 1/x+2-1/x+13=2x+1/(x+2).(x+16) -3/(x+13).(x+16)

<=>1/x+2 - 1/x+13 + 1/x+13 - 1/x+16=2x+1/(x+2).(x+16)

<=>1/x+2 - 1/x+16=2x+1/(x+2).(x+16)

<=> 14/(x+2).(x+16)= 2x+1/(x+2).(x+16)

<=> 2x+1=14

<=> 2x=14-1

<=> 2x=13

<=> x=13:2

<=> x=13/2

Vậy x=13/2

Chúc bạn học tốt

a) x²+8x+16-x²+1=16

8x+17=16

8x=-1

x=-1/8

b) 4x²-4x+1+x²+6x+9-5x²+245=0

2x+255=0

x=-255/2

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 \(\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=\dfrac{-5}{2}\end{matrix}\right.\).

\(\frac{1}{16}=\left(\frac{1}{4}\right)^2=\left(-\frac{1}{4}\right)^2\)

=> \(\left(\frac{1}{3}-2x\right)=\frac{1}{4}\)hoặc \(\left(-\frac{1}{4}\right)\)

*) \(\left(\frac{1}{3}-2x\right)=\frac{1}{4}\)

\(\Rightarrow2x=\frac{1}{3}-\frac{1}{4}=\frac{1}{12}\Rightarrow x=\frac{1}{12}:2=\frac{1}{24}\)

*) \(\left(\frac{1}{3}-2x\right)=\left(-\frac{1}{4}\right)\)

\(\Rightarrow2x=\frac{1}{3}-\left(-\frac{1}{4}\right)=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\Rightarrow x=\frac{7}{12}:2=\frac{7}{24}\)

Vậy \(x\in\left\{\frac{1}{24};\frac{7}{24}\right\}\)

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