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) 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:

a: =x^2-7x+49/4-5/4

=(x-7/2)^2-5/4>=-5/4

Dấu = xảy ra khi x=7/2

b: =x^2+x+1/4-13/4

=(x+1/2)^2-13/4>=-13/4

Dấu = xảy ra khi x=-1/2

e: =x^2-x+1/4+3/4=(x-1/2)^2+3/4>=3/4

Dấu = xảy ra khi x=1/2

f: x^2-4x+7

=x^2-4x+4+3

=(x-2)^2+3>=3

Dấu = xảy ra khi x=2

2:

a: A=2x^2+4x+9

=2x^2+4x+2+7

=2(x^2+2x+1)+7

=2(x+1)^2+7>=7

Dấu = xảy ra khi x=-1

b: x^2+2x+4

=x^2+2x+1+3

=(x+1)^2+3>=3

Dấu = xảy ra khi x=-1

Giúp mk vs

a: \(=x^3-2x^5\)

e: \(=x^4+2x^3-x^2-2x\)

Bạn chú ý đăng lẻ câu hỏi! 1/

a/ \(=x^3-2x^5\)

b/\(=5x^2+5-x^3-x\)

c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)

d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)

e/ \(=x^4-x^2+2x^3-2x\)

f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)

cmm

c: \(=\dfrac{8}{\left(x^2+3\right)\left(x-1\right)\left(x+1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x-1\right)\left(x+1\right)}+\dfrac{\left(x^2+3\right)\left(x-1\right)}{\left(x^2+3\right)\left(x-1\right)\left(x+1\right)}\)

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

Bài 1: 

a: \(\Leftrightarrow x^2-5x+6< =0\)

=>(x-2)(x-3)<=0

=>2<=x<=3

b: \(\Leftrightarrow\left(x-6\right)^2< =0\)

=>x=6

c: \(\Leftrightarrow x^2-2x+1>=0\)

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

hay \(x\in R\)

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

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