\(b,=\left(x+8-x+2\right)^2=100\\ c,=x^2\left(x^2-16\right)-x^4+1=x^4-16x^2-x^4+1=1-16x^2\\ d,=x^3+1-x^3+1=2\)

b) \(=\left(x+8-x+2\right)^2=10^2=100\)

c) \(=x^2\left(x^2-16\right)-\left(x^4-1\right)=x^4-16x^2-x^4+1=1-16x^2\)

d) \(=x^3+1-x^3+1=2\)

a: \(\frac{4\left(x+3\right)}{3x^2-x}:\frac{x^2+3x}{1-3x}\)

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

b: \(\frac{x+1}{x^2-2x-8}\cdot\frac{4-x}{x^2+x}\)

\(=\frac{x+1}{\left(x-4\right)\left(x+2\right)}\cdot\frac{-\left(x-4\right)}{x\left(x+1\right)}=\frac{-1}{x\left(x+2\right)}\)

c: \(\frac{9x+5}{2\left(x-1\right)\left(x+3\right)^2}-\frac{5x-7}{2\left(x-1\right)\left(x+3\right)^2}\)

\(=\frac{9x+5-5x+7}{2\left(x-1\right)\left(x+3\right)^2}=\frac{4x+12}{2\left(x-1\right)\left(x+3\right)^2}\)

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

d: \(\frac{18}{\left(x-3\right)\left(x^2-9\right)}-\frac{3}{x^2-6x+9}-\frac{x}{x^2-9}\)

\(=\frac{18}{\left(x+3\right)\left(x-3\right)^2}-\frac{3}{\left(x-3\right)^2}-\frac{x}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{18-3\left(x+3\right)-x\left(x-3\right)}{\left(x-3\right)^2\cdot\left(x+3\right)}=\frac{18-3x-9-x^2+3x}{\left(x-3\right)^2\cdot\left(x+3\right)}=\frac{-x^2+9}{\left(x-3\right)^2\left(x+3\right)}=\frac{-1}{x-3}\)

e: \(\frac{1}{x^2-x+1}+\frac{1}{1-x^2}+\frac{2}{x^3+1}\)

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

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

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

\(A=x^2+4x-21-x^2-4x+5=-16\\ B=-2\left(4x^2+20x+25\right)-\left(1-16x^2\right)\\ B=-8x^2-40x-50-1+16x^2=8x^2-40x-51\\ C=x^2\left(x^2-16\right)-\left(x^4-1\right)=x^4-16x^2-x^4+1=1-16x^2\\ D=x^3+1-\left(x^3-1\right)=2\\ E=x^3-3x^2+3x-1-x^3+1-9x^2+1=-12x^2+3x+1\)

\(7,D\\ 8,D\\ 9,B\\ 10,B\\ 11,B\\ 12,C\\ 13,B\\ 14,A\)

a) \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=\left(x^2+3x+1\right)^2+x\)

\(\Leftrightarrow\left(x^2+3x\right)\left(x^2+3x+2\right)=\left(x^2+3x+1\right)^2+x\)

\(\Leftrightarrow\left(t-1\right)\left(t+1\right)=t^2+x\) (với \(t=x^2+3x+1\))

\(\Leftrightarrow t^2-1=t^2+x\)

\(\Leftrightarrow x=-1\).

b) \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)=\left(x^2+8x+11\right)^2+2x\)

\(\Leftrightarrow\left(x^2+8x+7\right)\left(x^2+8x+15\right)=\left(x^2+8x+11\right)^2+2x\)

\(\Leftrightarrow\left(t-4\right)\left(t+4\right)=t^2+2x\) (với \(t=x^2+8x+11\))

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

\(\Leftrightarrow x=-8\)

c) \(\left(x^2-x+1\right)\left(x^2+x+1\right)\left(x-1\right)\left(x+1\right)=63\)

\(\Leftrightarrow\left(x^3-1\right)\left(x^3+1\right)=63\)

\(\Leftrightarrow x^6-1=63\)

\(\Leftrightarrow x^6=64\)

\(\Leftrightarrow x=\pm2\)

https://onlinemath.vn/cau-hoi/viet-1-doan-van-tong-phan-hop-khoang-12-cau-phan-tich-kho-tho-thu-2-bai-que-huong-trong-do-su-dung-1-cau-cam-than-vs-cau-ghep-chi-ro.8109170456376 help

\(a,x^2+4x-21-x^2-4x+5=-16\\ b,=\left(x+8-x+2\right)^2=10^2=100\\ c,=x^2\left(x^2-16\right)-\left(x^4-1\right)\\ =x^4-16x^2-x^4+1=1-16x^2\\ d,=x^3+1-x^3+1=2\)

\(a,=x^2+4x-21-x^2-4x+5=-16\\ b,=\left(x+8-x+2\right)^2=10^2=100\\ c,=x^2\left(x^2-16\right)-\left(x^4-1\right)\\ =x^4-16x^2-x^4+1=1-16x^2\\ d,=x^3+1-x^3+1=2\)

\(a,=\left(x+8-x+2\right)^2=10^2=100\\ b,=x^2\left(x^2-16\right)-\left(x^4-1\right)=x^4-16x^2-x^4+1=1-16x^2\\ c,=x^3+1-x^3+1=2\)

a) Ta có: \(x^2\left(x+1\right)+x+1=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow x+1=0\)

hay x=-1

b) Ta có: \(x^2-x=-2x^2+2x\)

\(\Leftrightarrow3x^2-3x=0\)

\(\Leftrightarrow3x\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

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

\(\Leftrightarrow2x^2\left(x-1\right)+x^2-x=0\)

\(\Leftrightarrow2x^2\left(x-1\right)+x\left(x-1\right)=0\)

\(\Leftrightarrow x\left(x-1\right)\cdot\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{-1}{2}\end{matrix}\right.\)

d) Ta có: \(\left(x-2\right)\left(x^2+4\right)=x^2-2x\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+4\right)-x\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2-x+4\right)=0\)

\(\Leftrightarrow x-2=0\)

hay x=2

1: ĐKXĐ: x<>0

\(\sqrt{x^2+x+2} + \frac{1}{x} = \frac{13-7x}{2}\)

\(\Leftrightarrow \sqrt{x^2+x+2} = \frac{13}{2} - \frac{7x}{2} - \frac{1}{x} = \frac{13x - 7x^2 - 2}{2x}\)

=>\(\left(\sqrt{x^2+x+2} - 2\right) + \left(\frac{1}{x} - 1\right) + \frac{7x - 7}{2} = 0\)

=>\(\frac{x^2+x-2}{\sqrt{x^2+x+2} + 2}+\frac{1-x}{x}+\frac{7(x-1)}{2}=0\)

=>\((x-1)\left[\frac{x+2}{\sqrt{x^2+x+2} + 2}-\frac{1}{x}+\frac{7}{2}\right]=0\)

=>x-1=0

=>x=1(nhận)

7: ĐKXĐ: \(16-x^2>0\)

=>\(x^2<16\)

=>-4<x<4

\(\frac{x^3}{\sqrt{16-x^2}} + x^2 - 16 = 0\)

\(\Leftrightarrow \frac{x^3}{\sqrt{16-x^2}} - \left(\sqrt{16-x^2}\right)^2 = 0\)

\(\Leftrightarrow x^3 - \left(\sqrt{16-x^2}\right)^3 = 0\)

\(\Leftrightarrow x^3 = \left(\sqrt{16-x^2}\right)^3\)

\(\Leftrightarrow x = \sqrt{16-x^2} \quad (x > 0)\)

\(\Leftrightarrow x^2 = 16 - x^2\) và x>0

=>\(2x^2=16\) và x>0

=>\(x^2=8\) và 0<x<4

=>\(x=2\sqrt2\)