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

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

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

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

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

a: ĐKXĐ của A là: \(\begin{cases}x+2<>0\\ x^2-4<>0\\ x^2+3x+2<>0\end{cases}\)

=>\(\begin{cases}x<>-2\\ x^2<>4\\ \left(x+1\right)\left(x+2\right)<>0\end{cases}\)

=>x∉{-2;2;-1}

ĐKXĐ cua B là \(x^3-1<>0\)

=>\(x^3<>1\)

=>x<>1

b: \(\frac{4x}{x+2}-\frac{x^3-8}{x^3+8}\cdot\frac{4x^2-8x+16}{x^2-4}\)

\(=\frac{4x}{x+2}-\frac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\cdot\frac{4\left(x^2-2x+4\right)}{\left(x-2\right)\left(x+2\right)}\)

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

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

\(A=\left(\frac{4x}{x+2}-\frac{x^3-8}{x^3+8}\cdot\frac{4x^2-8x+16}{x^2-4}\right):\frac{16}{x+2}\cdot\frac{x^2+3x+2}{x^2+x+1}\)

\(=\frac{-16}{\left.\left(x+2\right)^2\right.}\cdot\frac{x+2}{16}\cdot\frac{\left(x+1\right)\left(x+2\right)}{x^2+x+1}=\frac{-\left(x+1\right)}{x^2+x+1}\)

\(B=\frac{x^2+x-2}{x^3-1}\)

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

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

b: Đặt P=A+B

\(=\frac{x+2-x-1}{x^2+x+1}=\frac{1}{x^2+x+1}\)

\(=\frac{1}{x^2+x+\frac14+\frac34}=\frac{1}{\left(x+\frac12\right)^2+\frac34}\le1:\frac34=\frac43\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi x+1/2=0

=>x=-1/2

điều kiện xác định \(x\ne0\)

ta có : \(\dfrac{x+1}{x^2+2x+4}-\dfrac{x-2}{x^2-2x+4}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x^2-2x+4\right)-\left(x-2\right)\left(x^2+2x+4\right)}{\left(x^2+2x+4\right)\left(x^2-2x+4\right)}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\)

\(\Leftrightarrow-x^2+2x+12=\dfrac{6}{x}\Leftrightarrow x\left(-x^2+2x+12\right)=6\)

\(\Leftrightarrow-x^3+2x^2+12x=6\Leftrightarrow-x^3+2x^2+12x-6=0\)

tới đây bn bấm máy tính nha

câu b lm tương tự nha

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

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

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

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

2: Để B=0 thì -x-1=0

hay x=-1(nhận)

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)

13: \(\lim_{x\to+\infty}\left(\sqrt[3]{x^3+4x^2}-x\right)\)

\(=\lim_{x\to+\infty}\frac{x^3+4x^2-x^3}{\sqrt[3]{\left(x^3+4x^2\right)^2}+x\cdot\sqrt[3]{x^3+4x^2}+x^2}\)

\(=\lim_{x\to+\infty}\frac{4x^2}{x^2\cdot\sqrt[3]{\left(1+\frac{4}{x}\right)^2}+x^2\cdot\sqrt[3]{1+\frac{4}{x}}+x^2}\)

\(=\lim_{x\to+\infty}\frac{4}{\sqrt[3]{\left(1+\frac{4}{x}\right)^2}+\sqrt[3]{1+\frac{4}{x}}+1}=\frac{4}{\sqrt[3]{\left(1+0\right)^2}+\sqrt[3]{1+0}+1}=\frac43\)

12: \(\lim_{x\to2}\frac{4-x^2}{x^3-8}\)

\(=\lim_{x\to2}\frac{\left(2-x\right)\left(2+x\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\lim_{x\to2}\frac{-\left(x+2\right)}{x^2+2x+4}=\frac{-\left(2+2\right)}{2^2+2\cdot2+4}=-\frac{4}{4+4+4}=-\frac{4}{12}=-\frac13\)

11: \(\lim_{x\to4}\frac{4x-1}{x^2-8x+16}=\lim_{x\to4}\frac{4x-1}{\left(x-4\right)^2}=+\infty\)

vì \(\lim_{x\to4}4x-1=4\cdot4-1=15>0;\left(x-4\right)^2=\left(4-4\right)^2=0\)

\(a,ĐK:...\\ PT\Leftrightarrow x^2-6x=x^2-7x+10\\ \Leftrightarrow x=10\left(tm\right)\\ b,ĐK:...\\ PT\Leftrightarrow2x\left(4-x\right)-\left(2-2x\right)\left(8-x\right)=\left(8-x\right)\left(4-x\right)\\ \Leftrightarrow8x-2x^2+16+18x-2x^2=32-12x+x^2\\ \Leftrightarrow3x^2-38x+16=0\left(casio\right)\\ c,ĐK:...\\ PT\Leftrightarrow2x\left(x-4\right)-4x=0\\ \Leftrightarrow2x^2-12x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)

GHI RÕ DÙM MÌNH ĐK CỦA CẢ 3 CÂU LUÔN ĐC KO Á.