a) \(x^4-x^2+\dfrac{1}{4}-\dfrac{225}{4}=0\\ \left(x^2-\dfrac{1}{2}\right)^2-\dfrac{15}{2}^2=0\\ \left(x+7\right)\left(x-8\right)=0\\ \left[{}\begin{matrix}x=8\\x=-7\end{matrix}\right.\)

Vậy x = 8 hoặc x = -7

a: Ta có: \(x^4-x^2-56=0\)

\(\Leftrightarrow x^4-8x^2+7x^2-56=0\)

\(\Leftrightarrow\left(x^2-8\right)\left(x^2+7\right)=0\)

\(\Leftrightarrow x^2-8=0\)

hay \(x\in\left\{2\sqrt{2};-2\sqrt{2}\right\}\)

Lời giải:

Điều kiện: $x\neq -2; x\neq -2; x\neq -8; x\neq -14$

Đề bài 

$\Rightarrow \frac{(x+4)-(x+2)}{(x+2)(x+4)}+\frac{(x+8)-(x+4)}{(x+4)(x+8)}+\frac{(x+14)-(x+8)}{(x+8)(x+14)}=\frac{x}{(x+2)(x+14)}$

$\Rightarrow \frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{(x+2)(x+14)}$

$\Rightarrow \frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{(x+2)(x+14)}$

$\Rightarrow \frac{12}{(x+2)(x+14)}=\frac{x}{(x+2)(x+14)}$

$\Rightarrow 12=x$ (thỏa mãn)

\(S=x^6-8\)

\(S=\left(x^2\right)^3-2^3\)

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

⇒ Chọn C

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

đkxđ: x khác 0

\(\Leftrightarrow8.\left(x+\dfrac{1}{x}\right)\left(x+\dfrac{1}{x}\right)-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)+4\left(x^2+\dfrac{1}{x^2}\right)^2=x^2+8x+16\)

\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left[\left(8.x+\dfrac{1}{x}\right)-4\left(x^2+\dfrac{1}{x^2}\right)\right]+4\left(x^4+2+\dfrac{1}{x^2}\right)-x^2-8x-16=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left[\left(\dfrac{8x^2+1}{x}-4x^2-\dfrac{4}{x^2}\right)\right]+4x^4+8+\dfrac{4}{x^2}-x^2-8x-16=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left(\dfrac{x\left(8x^2+1\right)}{x^2}-\dfrac{4x^2.x^2}{x^2}-\dfrac{4}{x^2}\right)+......=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left(\dfrac{8x^3+x-4x^4-4}{x^2}\right)+...=0\)

\(\Leftrightarrow\dfrac{x^2}{x}.-\dfrac{4x^4+8x^3+x-4}{x^2}+.....=0\)

\(\Leftrightarrow-\dfrac{4x^6+8x^5+x^3-4x^2}{x^3}+\dfrac{4x^4+8+4x^2}{1}-\dfrac{x^2-8x-16}{1}=0\)

\(\Leftrightarrow......+\dfrac{x^3.\left(4x^4+8+4x^2\right)}{x^3}-\dfrac{x^3\left(x^2-8x-16\right)}{x^3}=0\)

\(\Leftrightarrow-4x^6+8x^5+x^3-4x^2+4x^7+8x^3+4x^5-x^5+8x^4+16x^3=0\)

\(\Leftrightarrow4x^7-4x^6+12x^5+8x^4+25x^3-4x^2=0\)

=> x=0 ( loại , ko tm)

Vậy pt vô nghiệm

Phương pháp:

Đặt \(x+\dfrac{1}{x}=a\Rightarrow a^2=x^2+\dfrac{1}{x^2}+2\Leftrightarrow a^2-2=x^2+\dfrac{1}{x^2}\)

Thay vào pt

\(x\ne0:đặt:x+\dfrac{1}{x}=t\)

\(pt\Leftrightarrow2t^2+4\left(t^2-2\right)^2-4\left(t^2-2\right)t^2=\left(x+4\right)^2\)

\(\Leftrightarrow2t^2+4\left(t^4-4t^2+4\right)-4\left(t^4-2t^2\right)=\left(x+4\right)^2\)

\(\Leftrightarrow2t^2+4t^4-16t^2+16-4t^4+8t^2=\left(x+4\right)^2\)

\(\Leftrightarrow-6t^2+16=\left(x+4\right)^2\)

\(\Leftrightarrow-6\left(x^2+2+\dfrac{1}{x^2}\right)+16=x^2+8x+16\)

\(\Leftrightarrow-6x^2-\dfrac{6}{x^2}-x^2-8x-12=0\Leftrightarrow-6x^4-x^4-8x^3-12x^2-6=0\Leftrightarrow-7x^4-8x^3-12x^2-6=0\left(vô-nghiệm\right)\)

(bn xem lại đề)

a: Ta có: \(\left(7x+4\right)^2-\left(7x-4\right)\left(7x+4\right)\)

\(=\left(7x+4\right)\left(7x+4-7x+4\right)\)

\(=8\left(7x+4\right)\)

=56x+32

b: Ta có: \(8\left(x-2\right)^2-3\left(x^2-4x-5\right)-5x^2\)

\(=8x^2-32x+32-3x^2+12x+15-5x^2\)

\(=-20x+47\)

c: Ta có: \(\left(x+1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3x\left(x+1\right)\)

\(=x^3+3x^2+3x+1-x^3+1-3x^2-3x\)

=2

\(A=x^2-16-6x-2x^2+x^2+6x+9=-7\\ B=\left(x^2+4\right)\left(x^2-4\right)-x^4+9\\ B=x^4-16-x^4+9=-7\)

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

\(=x^2-16-2x^2-6x+x^2+6x+9=-7\)

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

\(=\left(x^2+4\right)\left(x^2-4\right)-\left(x^4-9\right)\)

\(=x^4-16-x^4+9=-7\)

a/

\(VT=\dfrac{\left(x+4\right)-\left(x+2\right)}{\left(x+2\right)\left(x+4\right)}+\dfrac{\left(x+8\right)-\left(x+4\right)}{\left(x+4\right)\left(x+8\right)}+\dfrac{\left(x+14\right)-\left(x+8\right)}{\left(x+8\right)\left(x+14\right)}=\)

\(=\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+8}+\dfrac{1}{x+8}-\dfrac{1}{x+14}=\)

\(=\dfrac{1}{x+2}-\dfrac{1}{x+14}=\dfrac{12}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow\dfrac{12}{\left(x+2\right)\left(x+14\right)}=\dfrac{x}{\left(x+2\right)\left(x+14\right)}\left(x\ne-2;x\ne-14\right)\)

\(\Rightarrow x=12\)

\(\dfrac{x}{2023}+\dfrac{x+1}{2022}+...+\dfrac{x+2022}{1}+2023=0\)

\(\dfrac{1}{2023}x+\dfrac{1}{2022}x+\dfrac{1}{2022}\cdot1+...+\dfrac{1}{1}x+\dfrac{1}{1}\cdot2022+2023=0\)

\(x\left(\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}\right)+\left(\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023\right)=0\)

\(x\left(\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}\right)=\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023\)

\(x=\dfrac{\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023}{\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}}\)

\(x=\dfrac{\dfrac{1}{2022}+\dfrac{2022}{2022}+\dfrac{2}{2021}+\dfrac{2021}{2021}+...+\dfrac{2022}{1}+\dfrac{1}{1}}{\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}}\)

\(x=\dfrac{\dfrac{2023}{2022}+\dfrac{2023}{2021}+...+\dfrac{2023}{1}}{\dfrac{1}{2022}+\dfrac{1}{2021}+...+\dfrac{1}{1}}=2023\)

Vậy x = 2023