Rút gọn :

A=1x−2+1x+2+x2+1x2−4A=1x−2+1x+2+x2+1x2−4

=x+2(x−2)(x+2)+x−2(x+2)(x−2)+x2+1(x−2)(x+2)=x+2(x−2)(x+2)+x−2(x+2)(x−2)+x2+1(x−2)(x+2)

=x+2+x−2+x2+1(x−2)(x+2)=x+2+x−2+x2+1(x−2)(x+2)

=(x+1)2(x−2)(x+2)=(x+1)2(x−2)(x+2)

=x+2x−2=x+2x−2  

Vậy : A=x+2x−2A=x+2x−2

b) Để phân thức nhận giá trị âm

⇔x+2x−2<0⇔x+2x−2<0

⇔−2<x<2⇔−2<x<2

Vậy : −2<x<2−2<x<2 

Answer:

\(\left(2x-7\right)^2-6x+21=0\)

\(\Rightarrow\left(2x-7\right)^2-\left(6x-21\right)=0\)

\(\Rightarrow\left(2x-7\right)^2-3\left(2x-7\right)=0\)

\(\Rightarrow\left(2x-7\right)\left(2x-7-3\right)=0\)

\(\Rightarrow\left(2x-7\right)\left(2x-10\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-7=0\\2x-10=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=5\end{cases}}}\)

Answer:

\((x^2+1)(6x-4)-x(2x-1)(3x-2)= -3\)

\(\Rightarrow x^2.\left(6x-4\right)+\left(6x-4\right)-\left(2x^2-x\right)\left(3x-2\right)=-3\)

\(\Rightarrow6x^3-x^2+6x-4-2x^2.\left(3x-2\right)-x\left(3x-2\right)=-3\)

\(\Rightarrow6x^3-x^2+6x-6x^3+4x^2-3x^2+2x=-3+4\)

\(\Rightarrow8x=1\)

\(\Rightarrow x=\frac{1}{8}\)

Answer:

Mình sửa lại bài nhé. Lần trước làm sai ạ.

\((x^2+1)(6x-4)-x(2x-1)(3x-2)= -3\)

\(\Rightarrow x^2.\left(6x-4\right)+\left(6x-4\right)-\left(2x^2-x\right)\left(3x-2\right)=-3\)

\(\Rightarrow6x^3-4x^2+6x-4-2x^2.\left(3x-2\right)+x\left(3x-2\right)=-3\)

\(\Rightarrow6x^3-4x^2+6x-4-2x^3+4x^2+3x^2-2x+3=0\)

\(\Rightarrow3x^2+4x-1=0\)

\(\hept{\begin{cases}a=3\\b=4\\c=-1\end{cases}}\)

\(\Delta=b^2-4ac=4^2-4.3.-1=16+12=28>0\)

Vậy phương trình có hai nghiệm phân biệt

\(\hept{\begin{cases}x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-4+\sqrt{28}}{6}=\frac{-2+\sqrt{7}}{3}\\x_1=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-4-\sqrt{28}}{6}=\frac{-2-\sqrt{7}}{3}\end{cases}}\)

Answer:

\(P=\left|x-2021\right|+\left|x-1\right|\)

\(=\left|2021-x\right|+\left|x-1\right|\ge\left|2021-x+x-1\right|\ge2020\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}2021-x\ge0\\x-1\le0\end{cases}}\Rightarrow\hept{\begin{cases}x\le2021\\x\ge1\end{cases}}\Rightarrow1\le x\le2021\)

Vậy giá trị nhỏ nhất của \(P=2020\) khi \(1\le x\le2021\)

Answer:

\((x+1)(x+2)(x+3)(x+4)-24\)

\(=[\left(x+1\right)\left(x+4\right)].[\left(x+2\right)\left(x+3\right)]-24\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)

Ta đặt \(a=x^2+5x+4\)

\(=a\left(a+2\right)-24\)

\(=a^2+2a-24\)

\(=a^2+2a+1-25\)

\(=\left(a+1\right)^2-5^2\)

\(=\left(a+1-5\right)\left(a+1+5\right)\)

\(=\left(a-4\right)\left(a+6\right)\)

\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)