a) \(x^2-2x-4y^2-4y=\left(x^2-4y^2\right)-\left(2x+4y\right)=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)

\(=\left(x+2y\right)\left(x-2y-2\right)\)

b) \(x^4+3x^3-4x-4=\left(x^4-4\right)+\left(2x^3-4x\right)=\left(x^2+2\right)\left(x^2-2\right)+2x\left(x^2-2\right)\)

\(=\left(x^2-2\right)\left(x^2+2+2x\right)\)

c) \(x^2\left(1-x\right)^2-4x-4x^2=x^2\left(x^2-2x+1\right)-4x-4x^2=x^4-2x^3+x^2-4x-4x^2\)

\(x^4-2x^3-3x^2-4x=x\left(x^3-2x^2-3x-4\right)\)

Có gì sai thì nói giùm nha vì đang vội

Lớp 7 hai cách một cách biến đổi thay một cách thay không biến đổi mình nghĩ cũng được tính
Cách 1:
\(f\left(x\right)=x^2-2ax+a^2\)
Thay \(x=a\) ta có:
\(f\left(a\right)=a^2-2a^2+a^2=0\)
Cách 2:
\(f\left(x\right)=x^2-2ax+a^2=\left(x^2-ax\right)-\left(ax-a^2\right)=x\left(x-a\right)-a\left(x-a\right)=\left(x-a\right)\left(x-a\right)=\left(x-a\right)^2\)

\(\Rightarrow f\left(a\right)=\left(a-a\right)^2=0\)

Bài này đơn giản lắm động não một tí đi

Dạ Nguyệt Có cái j đó sai sai

\(x^4+4x^3+4x^2-4=0\)
\(x^2\left(x^2+4x+4\right)=4\)
\(x^2\left(x+2\right)^2=2^2\)
\(\left[x\left(x+2\right)\right]^2=2^2\)
\(\left|x\left(x+2\right)\right|=2\)

TH1:

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

\(x_1=-1-\sqrt{3}\)

\(x_2=-1+\sqrt{3}\)

TH2:

\(x\left(x+2\right)=-2\)
\(x^2+2x+1=-1\)
\(\left(x+1\right)^2=-1\) vô nghiệm

KL: \(x_1=-1-\sqrt{3}\\ x_2=-1+\sqrt{3}\)

\(A=\dfrac{x^3+3x-4}{x^2+x+4}\)

ĐK: \(x^2+x+4\ne0\) đúng với mọi x

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

KL: \(A=x-1\)

BĐT cơ bản

\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)

\(\dfrac{ab}{c+1}=ab\dfrac{1}{c+a+b+c}=ab\dfrac{1}{\left(c+a\right)+\left(b+c\right)}\le\dfrac{ab}{4}\left[\dfrac{1}{c+a}+\dfrac{1}{b+c}\right]\)

\(\dfrac{bc}{a+1}=bc\dfrac{1}{a+a+b+c}=bc\dfrac{1}{\left(a+b\right)+\left(a+c\right)}\le\dfrac{bc}{4}\left[\dfrac{1}{a+b}+\dfrac{1}{a+c}\right]\)

\(\dfrac{ac}{b+1}=ac\dfrac{1}{b+a+b+c}=ac\dfrac{1}{\left(b+a\right)+\left(b+c\right)}\le\dfrac{ac}{4}\left[\dfrac{1}{b+a}+\dfrac{1}{b+c}\right]\)

Công lại:

\(A\le\left[\dfrac{ab+bc}{4\left(c+a\right)}+\dfrac{ab+ac}{4\left(b+c\right)}+\dfrac{bc+ac}{4\left(b+a\right)}\right]\)

\(A\le\left[\dfrac{b\left(a+c\right)}{4\left(c+a\right)}+\dfrac{a\left(b+c\right)}{4\left(b+c\right)}+\dfrac{c\left(b+a\right)}{4\left(b+a\right)}\right]\)

\(A\le\left[\dfrac{b}{4}+\dfrac{a}{4}+\dfrac{c}{4}\right]\)

\(A\le\dfrac{b+a+c}{4}=\dfrac{1}{4}\)

Đẳng thức khi \(a=b=c=\dfrac{1}{3}\)

Xong rồi đó mỏi cái lưng