\(21-\left(2x-4\right)\left(x+1\right)\)

\(=21-2x^2-2x+4x+4\)

\(=-2x^2+2x+25\)

a: Ta có: \(\left|\dfrac{2}{5}-x\right|+\dfrac{1}{2}=3.5\)

\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{2}{5}=3\\x-\dfrac{2}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{5}\\x=-\dfrac{13}{5}\end{matrix}\right.\)

b: Ta có: \(\dfrac{21}{5}+3:\left|\dfrac{x}{4}-\dfrac{2}{3}\right|=6\)

\(\Leftrightarrow3:\left|\dfrac{1}{4}x-\dfrac{2}{3}\right|=6-\dfrac{21}{5}=\dfrac{9}{5}\)

\(\Leftrightarrow\left|\dfrac{1}{4}x-\dfrac{2}{3}\right|=\dfrac{5}{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{4}x-\dfrac{2}{3}=\dfrac{5}{3}\\\dfrac{1}{4}x-\dfrac{2}{3}=-\dfrac{5}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{4}x=\dfrac{7}{3}\\\dfrac{1}{4}x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{28}{3}\\x=-4\end{matrix}\right.\)

a)

ĐKXĐ: \(x\ne-4\)

Để A nguyên thì \(3x+21⋮x+4\)

\(\Leftrightarrow3x+12+9⋮x+4\)

mà \(3x+12⋮x+4\)

nên \(9⋮x+4\)

\(\Leftrightarrow x+4\inƯ\left(9\right)\)

\(\Leftrightarrow x+4\in\left\{1;-1;3;-3;9;-9\right\}\)

\(\Leftrightarrow x\in\left\{-3;-5;-1;-7;5;-13\right\}\)(nhận)

Vậy: Để A nguyên thì \(x\in\left\{-3;-5;-1;-7;5;-13\right\}\)

b) ĐKXĐ: \(x\ne\dfrac{1}{2}\)

Để B nguyên thì \(2x^3-7x^2+7x+5⋮2x-1\)

\(\Leftrightarrow2x^3-x^2-6x^2+3x+4x-2+7⋮2x-1\)

\(\Leftrightarrow x^2\left(2x-1\right)-3x\left(2x-1\right)+2\left(2x-1\right)+7⋮2x-1\)

\(\Leftrightarrow\left(2x-1\right)\left(x^2-3x+2\right)+7⋮2x-1\)

mà \(\left(2x-1\right)\left(x^2-3x+2\right)⋮2x-1\)

nên \(7⋮2x-1\)

\(\Leftrightarrow2x-1\inƯ\left(7\right)\)

\(\Leftrightarrow2x-1\in\left\{1;-1;7;-7\right\}\)

\(\Leftrightarrow2x\in\left\{2;0;8;-6\right\}\)

hay \(x\in\left\{1;0;4;-3\right\}\)(nhận)

Vậy: \(x\in\left\{1;0;4;-3\right\}\)

a) | 5/4x -7/2| - | 5/8x + 3/5| = 0

|5/4x - 7/2| = | 5/8x + 3/5|

TH1: 5/4x - 7/2 = 5/8x + 3/5

=> 5/4x - 5/8x = 3/5 +7/2

5/8x = 41/10

x = 41/10:5/8

x = 164/25

TH2: 5/4x - 7/2 = -5/8x - 3/5

=> 5/4x + 5/8x  = -3/5 +7/2

15/8x  = 29/10

x = 29/10 : 15/8

x = 116/75

KL: x = 164/25 hoặc x = 116/75

các bài cn lại b lm tương tự nha! h lm dài lắm!

a/\(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x+3\right)\left(x-3\right)=26\)

↔ \(x^3+2^3\)\(-x\left(x^2-3^2\right)\)= 26

↔\(x^3+8-x^3+9x=26\)

↔\(9x=18\leftrightarrow x=2\)

Vậy x=2

b/\(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x-4\right)\left(x+4\right)=21\)

\(\Leftrightarrow x^3-3^3-x\left(x^2-4^2\right)=21\)

\(\Leftrightarrow x^3-9-x^3+16x=21\)

\(\Leftrightarrow16x=30\)

\(\Leftrightarrow x=\frac{15}{8}\)

Vậy \(x=\frac{15}{8}\)

c/\(\left(2x-1\right)\left(4x^2+2x+1\right)-4x\left(2x^2-3\right)=23\)

↔\(\left(2x\right)^3-1^3-4x\left(2x^2-3\right)=23\)

↔\(8x^3-1-8x^3+12x=23\)

↔\(12x=24\leftrightarrow x=2\)

Vậy x=2

a, (x + 2)(x² - 2x + 4 ) - x(x + 3)(x - 3) = 26

<=> x³ + 8 - x(x² - 9) = 26

<=> x³ + 8 - x³ + 9x = 26

<=> 9x - 18 = 0

<=> 9x = 18

<=> x = 2
b, (x - 3)(x² + 3x + 9) - x(x - 4)(x + 4) = 21

<=> x³ - 27 - x(x² - 16) = 21

<=> x³ - 27 - x³ + 16x = 21

<=> 16x - 48 = 0

<=> 16x = 48

<=> x = 3
c, (2x - 1)(4x² + 2x + 1) - 4x(2x² - 3) = 23

<=> 8x³ - 1 - 8x³ + 12x = 23

<=> 12x - 24 = 0

<=> 12x = 24

<=> x = 2

\(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x+3\right)\left(x-3\right)=26\)

\(< =>x^3-2x^2+4x+2x^2-4x+8-x\left(x^2-9\right)-26=0\)

\(< =>x^3+8-x^3+9x-26=0\)

\(< =>9x-18=0< =>x=2\)