a) 4/7.|x| + 1/28 = 12/28

4/7.|x| = 11/28

|x| = 11/16

=> x = 11/16

b) (2)^2x-1 = 8 = 2³

=> 2x + 1 = 3

2x = 2

x = 1

c) \(\dfrac{y^4-1}{y^3+y^2+y+1}\)

\(=\dfrac{\left(y^2+1\right)\left(y^2-1\right)}{y^2\left(y+1\right)+\left(y+1\right)}\)

\(=\dfrac{\left(y^2+1\right)\left(y+1\right)\left(y-1\right)}{\left(y+1\right)\left(y^2+1\right)}\)

\(=y-1\) 

d) \(\dfrac{2x^2-9x+7}{-2x^2-x+28}\)

\(=\dfrac{2x^2-2x-7x+7}{-\left(2x^2+8x-7x-28\right)}\)

\(=\dfrac{2x\left(x-1\right)-7\left(x-1\right)}{-\left(2x-7\right)\left(x+4\right)}\)

\(=-\dfrac{\left(2x-7\right)\left(x-1\right)}{\left(2x-7\right)\left(x+4\right)}\)

\(=\dfrac{1-x}{x+4}\)

a)\(x^4-6x^2+2x+28\)

\(=\left(x^4-x^3\right)+\left(x^3-x^2\right)-\left(5x^2-5x\right)-\left(3x-3\right)+25\)

\(=\left(x-1\right)\left(x^3+x^2-5x-3\right)+25\)

=> số dư là 25

b) Cách làm tương tự câu a nhé

a: \(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\)

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

a) \(7x^2=28\Leftrightarrow x^2=7\Leftrightarrow x=\sqrt{7}\)

c) \(\left(x-1\right)\left(x+\dfrac{5}{2}\right)=0\Leftrightarrow x\in\left\{1;\dfrac{-5}{2}\right\}\)

a, ĐKXĐ : \(D=R\)

BPT \(\Leftrightarrow x^2+5x+4< 5\sqrt{x^2+5x+4+24}\)

Đặt \(x^2+5x+4=a\left(a\ge-\dfrac{9}{4}\right)\)

BPTTT : \(5\sqrt{a+24}>a\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a+24\ge0\\a< 0\end{matrix}\right.\\\left\{{}\begin{matrix}a\ge0\\25\left(a+24\right)>a^2\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-24\le a< 0\\\left\{{}\begin{matrix}a^2-25a-600< 0\\a\ge0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-24\le a< 0\\0\le a< 40\end{matrix}\right.\)

\(\Leftrightarrow-24\le a< 40\)

- Thay lại a vào ta được : \(\left\{{}\begin{matrix}x^2+5x-36< 0\\x^2+5x+28\ge0\end{matrix}\right.\)

\(\Leftrightarrow-9< x< 4\)

Vậy ....

b, ĐKXĐ : \(x>0\)

BĐT \(\Leftrightarrow2\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)< x+\dfrac{1}{4x}+1\)

- Đặt \(\sqrt{x}+\dfrac{1}{2\sqrt{x}}=a\left(a\ge\sqrt{2}\right)\)

\(\Leftrightarrow a^2=x+\dfrac{1}{4x}+1\)

BPTTT : \(2a\le a^2\)

\(\Leftrightarrow\left[{}\begin{matrix}a\le0\\a\ge2\end{matrix}\right.\)

\(\Leftrightarrow a\ge2\)

\(\Leftrightarrow a^2\ge4\)

- Thay a vào lại BPT ta được : \(x+\dfrac{1}{4x}-3\ge0\)

\(\Leftrightarrow4x^2-12x+1\ge0\)

\(\Leftrightarrow x=(0;\dfrac{3-2\sqrt{2}}{2}]\cup[\dfrac{3+2\sqrt{2}}{2};+\infty)\)

Vậy ...

a: \(x\in\left\{0;25\right\}\)

c: \(x\in\left\{0;5\right\}\)

_{Sorry mình nhầm câu a}

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

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

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

d) (x² - 1)³ - (x⁴ + x² + 1)(x² - 1) = 0

Giải:

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

\(\Leftrightarrow\) 4x² - 4x + 1 + x² + 6x + 9 - 5(x² - 49) = 0

\(\Leftrightarrow\) 4x² - 4x + 1 + x² + 6x + 9 - 5x² + 245 = 0

\(\Leftrightarrow\) 2x + 255 = 0

\(\Leftrightarrow\) 2x = - 255

\(\Leftrightarrow\) x = - 255 : 2

\(\Leftrightarrow\) x = \(-\frac{255}{2}\)

Vậy x = \(-\frac{255}{2}\)

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

\(\Leftrightarrow\) x³ + 8 - x³ - 2x = 15

\(\Leftrightarrow\) 8 - 2x = 15

\(\Leftrightarrow\) 2x = 8 - 1

\(\Leftrightarrow\) 2x = - 7

\(\Leftrightarrow\) x = - 7 : 2

\(\Leftrightarrow\) x = \(-\frac{7}{2}\)

Vậy x = \(-\frac{7}{2}\)

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

\(\Leftrightarrow\) x³ + 6x² + 27x + 27 - x(9x² + 6x + 1) + 8x³ - 1 = 28

\(\Leftrightarrow\) x³ + 6x² + 27x + 27 - 9x³ - 6x² - x + 8x³ - 1 = 28

\(\Leftrightarrow\) 26x + 26 = 28

\(\Leftrightarrow\) 26x = 28 - 26

\(\Leftrightarrow\) 26x = 2

\(\Leftrightarrow\) x = 2 : 26

\(\Leftrightarrow\) x = \(\frac{1}{13}\)

Vậy x = \(\frac{1}{13}\)

d) (x² - 1)³ - (x⁴ + x² + 1)(x² - 1) = 0

\(\Leftrightarrow\) x⁶ - 2x² + 1 - (x⁶ - 1) = 0

\(\Leftrightarrow\) x⁶ - 2x² + 1 - x⁶ + 1 = 0

\(\Leftrightarrow\) -2x² + 2 = 0

\(\Leftrightarrow\) -2x² = - 2

\(\Leftrightarrow\) x² = - 2 : (- 2)

\(\Leftrightarrow\) x² = 1

\(\Leftrightarrow\) x = 1 hoặc x = - 1

Vậy x \(\in\) {1; - 1}

cơ may sao nay gặp được Dân Chơi Đất Bắc luôn nè ::)))