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,\(\left|-5x\right|\)=3x-16

\(\Leftrightarrow\)\(\left[{}\begin{matrix}-5x=3x-16\\-5x=-3x+16\end{matrix}\right.\)            \(\Leftrightarrow\)\(\left[{}\begin{matrix}-8x=-16\\-2x=16\end{matrix}\right.\)                 \(\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\)

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

=>6xy-8y-9x+12=6xy-15y+2x-5 và 2y-6+16=3x+6

=>-9x-8y+12+15y-2x+5=0 và 3x+6-2y-10=0

=>-11x+7y=-17 và 3x-2y=4

=>x=6 và y=7

\(\left(x-1\right)^3+\left(2-x\right)\left(4x+2x+x^2\right)+3\left(x+2\right)=16\)

\(\Leftrightarrow-7x^2+18x+5=16\)

\(\Leftrightarrow-7x+18x+5=16-16\)

\(\Leftrightarrow-7x^2+18x-11=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{11}{7}\end{cases}}\)

Vậy:...

a, \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)

\(\Leftrightarrow x^2+8x+16-\left(x^2-x+x-1\right)=16\)

\(\Leftrightarrow8x+1=0\Leftrightarrow x=-\frac{1}{8}\)

b, \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)

\(\Leftrightarrow2x+255=0\Leftrightarrow x=-\frac{225}{2}\)

c, \(\left(x+2\right)\left(x-2\right)-x^3-2x=15\)

\(\Leftrightarrow x^2-4-x^3-2x=15\)( vô nghiệm )

d, \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)

\(\Leftrightarrow x^3+9x^2+27x+27-9x^3+6x^2-x+8x^3+1=28\)

\(\Leftrightarrow15x^2+26=0\Leftrightarrow x^2\ne-\frac{26}{15}\)( vô nghiệm )

Tính nhẩm hết á, sai bỏ quá nhá, sắp đi hc ... nên chất lượng hơi kém xíu ~~~ 

a) ( x + 4 )² - ( x + 1 )( x - 1 ) = 16

<=> x² + 8x + 16 - ( x² - 1 ) = 16

<=> x² + 8x + 16 - x² + 1 = 16

<=> 8x + 17 = 16

<=> 8x = -1

<=> x = -1/8

b) ( 2x - 1 )² + ( x + 3 )² - 5( x + 7 )( x - 7 ) = 0

<=> 4x² - 4x + 1 + x² + 6x + 9 - 5( x² - 49 ) = 0

<=> 5x² + 2x + 10 - 5x² + 245 = 0

<=> 2x + 255 = 0

<=> 2x = -255

<=> 2x = -255/2

c) ( x + 2 )( x² - 2x + 4 ) - x( x² + 2 ) = 15

<=> x³ + 2³ - x³ - 2x = 15

<=> 8 - 2x = 15

<=> 2x = -7

<=> x = -7/2

d) ( x + 3 )³ - x( 3x + 1 )² + ( 2x + 1 )( 4x² - 2x + 1 ) = 28

<=> x³ + 9x² + 27x + 27 - x( 9x² + 6x + 1 ) + [ ( 2x )³ + 1³ ] = 28

<=> x³ + 9x² + 27x + 27 - 9x³ - 6x² - x + 8x³ + 1 = 28

<=> 3x² + 26x + 28 = 28

<=> 3x² + 26x = 0

<=> x( 3x + 26 ) = 0

<=> \(\orbr{\begin{cases}x=0\\3x+26=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{26}{3}\end{cases}}\)