\(\dfrac{1}{x-1}-\dfrac{1}{x+1}-\dfrac{2}{x^2+1}-\dfrac{4}{x^4+1}-\dfrac{8}{x^8+1}-\dfrac{16}{x^{16}+1}\)

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

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

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

\(=\dfrac{4}{x^4-1}-\dfrac{4}{x^4+1}-\dfrac{8}{x^8+1}-\dfrac{16}{x^{16}+1}\)

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

\(=\dfrac{8}{x^8-1}-\dfrac{8}{x^8+1}-\dfrac{16}{x^{16}+1}\)

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

\(=\dfrac{16}{x^{16}-1}-\dfrac{16}{x^{16}+1}\)

\(=\dfrac{16\left(x^{16}+1-x^{16}+1\right)}{x^{32}-1}\)

\(=\dfrac{32}{x^{32}-1}\)

b) ĐKXĐ : \(x\ne\pm1\)

\(P=\dfrac{x}{x-1}+\dfrac{3}{x+1}-\dfrac{6x-4}{x^2-1}\)

\(=\dfrac{x\left(x+1\right)+3\left(x-1\right)-\left(6x-4\right)}{\left(x-1\right)\left(x+1\right)}\)

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

c) ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

\(A=\dfrac{1}{x+\sqrt{x}}+\dfrac{2\sqrt{x}}{x-1}-\dfrac{1}{x-\sqrt{x}}\)

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

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

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

a) ĐKXĐ : \(x\ge0;x\ne16\)

\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}+4}+\dfrac{4}{\sqrt{x-4}}\right):\dfrac{x+16}{\sqrt{x}+2}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-4\right)+4\left(\sqrt{x}+4\right)}{x-16}:\dfrac{x+16}{\sqrt{x}+2}\)

\(=\dfrac{x+16}{x-16}:\dfrac{x+16}{\sqrt{x}+2}=\dfrac{\sqrt{x}+2}{x-16}\)

\(=\left(\dfrac{\sqrt{x}.\left(\sqrt{x}-4\right)}{x-4}+\dfrac{4.\left(\sqrt{x}+4\right)}{x-4}\right):\dfrac{x+16}{\sqrt{x}+2}\)

\(=\left(\dfrac{x-4\sqrt{x}+4\sqrt{x}+16}{x-4}\right).\dfrac{\sqrt{x}+2}{x+16}\)

\(=\dfrac{x+16}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+2}{x+16}\)

\(=\dfrac{1}{\sqrt{x}-2}\)

\(A=\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(A=\left(\dfrac{1+x}{\left(1-x\right)\left(1+x\right)}+\dfrac{1-x}{\left(1+x\right)\left(1-x\right)}\right)+...+\dfrac{16}{1+x^{16}}\)

\(A=\dfrac{1+x+1-x}{1-x^2}+\dfrac{2}{1+x^2}+...+\dfrac{16}{1+x^{16}}\)

\(A=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+...+\dfrac{16}{1+x^{16}}\)

Tiếp tục các bước như ở dòng 2 và 3 ta có :

\(A=\dfrac{16}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)

\(A=\dfrac{16\left(1+x^{16}\right)}{\left(1-x^{16}\right)\left(1+x^{16}\right)}+\dfrac{16\left(1-x^{16}\right)}{\left(1+x^{16}\right)\left(1-x^{16}\right)}\)

\(A=\dfrac{16+16x^{16}+16-16x^{16}}{1-x^{32}}\)

\(A=\dfrac{32}{1-x^{32}}\)

\(A=\dfrac{2}{x^2+2x}+\dfrac{2}{x^2+6x+8}+\dfrac{2}{x^2+10x+24}+\dfrac{2}{x^2+14x+48}\)

\(A=\dfrac{2}{x\left(x+2\right)}+\dfrac{2}{\left(x+2\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}+\dfrac{2}{\left(x+6\right)\left(x+8\right)}\)

\(A=\dfrac{1}{x}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+8}\)

\(A=\dfrac{1}{x}-\dfrac{1}{x+8}=\dfrac{x+8}{x\left(x+8\right)}-\dfrac{x}{\left(x+8\right)}=\dfrac{8}{x\left(x+8\right)}\)

\(B=\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(B=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(B=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(B=\dfrac{8}{1-x^8}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(B=\dfrac{16}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)

\(B=\dfrac{32}{1-x^{32}}\)

a: \(=\dfrac{4x^3+8x^2-11x+3-\left(x^2-5\right)\left(2x-1\right)-2x^3-5x^2+x+1}{\left(2x-1\right)^3}\)

\(=\dfrac{2x^3+3x^2-10x+4-2x^3+x^2+10x-5}{\left(2x-1\right)^3}\)

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

b: \(=\dfrac{1+x+1-x}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

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

\(=\dfrac{4+4x^4+4-4x^4}{1-x^8}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{8+8x^8+8-8x^8}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{32}{1+x^{32}}\)

TK

https://lazi.vn/edu/exercise/giai-phuong-trinh-4x-5-x-1-2-x-x-1-7-x-2-3-x-5

a: \(\Leftrightarrow4x-5=2x-2+x\)

=>4x-5=3x-2

=>x=3(nhận)

b: =>7x-35=3x+6

=>4x=41

hay x=41/4(nhận)

c: \(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)

\(\Leftrightarrow\dfrac{28}{6\left(x-4\right)}-\dfrac{6\left(x+2\right)}{6\left(x-4\right)}=\dfrac{-9}{6\left(x-4\right)}-\dfrac{5\left(x-4\right)}{6\left(x-4\right)}\)

\(\Leftrightarrow28-6x-12=-9-5x+20\)

=>-6x+16=-5x+11

=>-x=-5

hay x=5(nhận)

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

\(\Leftrightarrow4x=16\)

hay x=4(nhận)

Đây là bài giải pt chứ có phải biểu thức đâu mà thu gọn hả bạn?

Lời giải:

a. ĐKXĐ: $x\neq 1$

PT $\Leftrightarrow \frac{x^2+x+1}{(x-1)(x^2+x+1)}+\frac{2x(x-1)}{(x-1)(x^2+x+1)}=\frac{3x^2}{(x-1)(x^2+x+1)}$

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

$\Leftrightarrow 3x^2-x+1=3x^2$

$\Leftrightarrow x=1$ (không thỏa đkxđ)

Vậy pt vô nghiệm.

b. ĐKXĐ: $x\neq \pm 3$

PT $\Leftrightarrow \frac{(x+2)(x+3)}{(x-3)(x+3)}=\frac{x^2+3x}{(x-3)(x+3)}$

$\Leftrightarrow (x+2)(x+3)=x^2+3x$

$\Leftrightarrow x^2+5x+6=x^2+3x$

$\Leftrightarrow 2x+6=0$

$\Leftrightarrow x=-3$ (không thỏa mãn đkxđ)

Do đó pt vô nghiệm.

c. ĐKXĐ: $x\neq \pm 2$

PT $\Leftrightarrow \frac{(x-2)^2-(x+2)^2}{(x+2)(x-2)}=\frac{-16}{(x-2)(x+2)}$

$\Leftrightarrow (x-2)^2-(x+2)^2=-16$

$\Leftrightarrow -8x=-16$

$\Leftrightarrow x=2$ (vi phạm đkxđ)

Do đó pt vô nghiệm.