a) Tìm Min

\(A=\frac{4x+3}{x^2+1}=A+1=\frac{x^2+4x+1}{x^2+1}\)\(=\frac{\left(x+2\right)^2}{x^2+1}\)\(=0\)

\(=MinA=-1\)

Dấu '' = '' xảy ra khi \(x+2=0=x=-2\)

Tìm Max

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

\(=MaxA=4\)

Dấu '' = '' xảy ra khi \(2x-1=0=0=x=\frac{1}{2}\)

Ko biết -Ko hiểu ( oke )

Answer:

Bài 1:

a) \(3x\left(2x^2+5x-4\right)\)

\(=3x.2x^2+3x.5x-3x.4\)

\(=6x^3+15x^2-12x\)

b) Áp dụng hằng đẳng thức số hai: \(\left(A-B\right)^2=A^2-2AB+B^2\)

Bài 2:

\(x^2+6x=x.\left(x+6\right)\)

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

\(x^2-2xy+3x+6y=\left(x^2-2xy\right)+\left(3x-6y\right)=\left(x-2y\right)\left(x+3\right)\)

Bài 3:

\(\left(x-2\right)^2-x^2=0\)

\(\Rightarrow x^2-4x+4-x^2=0\)

\(\Rightarrow-4x+4=0\)

\(\Rightarrow x=1\)

\(3x\left(2x-1\right)-24x+12=0\)

\(\Rightarrow x\left(2x-1\right)-8x+4=0\)

\(\Rightarrow\left(2x-1\right)\left(x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-1=0\\x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=4\end{cases}}\)

Bài 4:

Có: \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^o\)

\(\Rightarrow\widehat{A}+60^o+90^o+80^o=360^o\)

\(\Rightarrow x=\widehat{A}=230^o\)

Answer:

Bài 4:

Bạn sửa lại giúp mình đáp án:

\(\Rightarrow x=\widehat{A}=130^o\)