\(\left|x+5\right|=x+5\)

\(\Rightarrow x+5\ge0\Leftrightarrow x\ge-5\)

Vậy...

Có \(\widehat{B}+\widehat{C}=180^0\) (hai góc trong cùng phía do AB//CD)

mà \(\widehat{B}-\widehat{C}=30^0\)

\(\Rightarrow\widehat{B}=105^0;\widehat{C}=75^0\)

Có \(\widehat{A}+\widehat{D}=180^0\) (hai góc trong cùng phía do AB//CD)

mà \(\widehat{A}=3\widehat{D}\)\(\Rightarrow4\widehat{D}=180^0\Leftrightarrow\widehat{D}=45^0\)

\(\Rightarrow\widehat{A}=135^0\)

Vậy\(\widehat{B}=105^0;\widehat{C}=75^0\);\(\widehat{A}=135^0\);\(\widehat{D}=45^0\)

\(\left(x+3\right)^2=x^2+6x+9\)

\(\left(2+x\right)^2=4+4x+x^2\)

b)\(P=a-\sqrt{a}=a-2.\dfrac{1}{2}\sqrt{a}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

Dấu "=" xảy ra khi \(\sqrt{a}=\dfrac{1}{2}\Leftrightarrow a=\dfrac{1}{4}\)

Vậy \(P_{min}=-\dfrac{1}{4}\)

Này, mình nói bạn Thịnh là bài này mk nghĩ ý b bạn vẫn làm được mà bạn chỉ làm mỗi ý a là sao? Làm ý a bỏ ý b hả, zì kì thế

ĐK:\(\left\{{}\begin{matrix}x+3\ge0\\1-x\ge0\end{matrix}\right.\)\(\Leftrightarrow-3\le x\le1\)

a) \(P\left(x\right)=5x^3-3x+7-x=5x^3-4x+7\)

\(Q\left(x\right)=-5x^3+2x-3+2x-x^2-2=-5x^3-x^2+4x-5\)

b) \(M\left(x\right)=5x^3-4x+7-5x^3-x^2+4x-5=-x^2+2\)

\(N\left(x\right)=5x^3-4x+7-\left(-5x^3-x^2+4x-5\right)=10x^3+x^2-8x+12\)

a) ĐK:\(x\ge0;x\ne9\)

\(P=\left[\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]:\dfrac{2\sqrt{x}-2-\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\)

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\)

b)\(P=-\dfrac{3}{\sqrt{x}+3}\) 

Có \(\sqrt{x}+3\ge3;\forall x\ge0\)

\(\Leftrightarrow-\dfrac{3}{\sqrt{x}+3}\ge-\dfrac{1}{3}\)

\(P_{min}=-\dfrac{1}{3}\Leftrightarrow x=0\)

\(\left(9x^2-1\right)^2\left|x-\dfrac{1}{3}\right|=0\)

\(\Leftrightarrow\left[{}\begin{matrix}9x^2-1=0\\x-\dfrac{1}{3}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x^2=\dfrac{1}{9}\\x=\dfrac{1}{3}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy...

\(F=2x\left(x^2+3x+2\right)+x\left(5-2x^2-7x\right)+x^2-3\)

\(=2x^3+6x^2+4x+5x-2x^3-7x^2+x^2-3\)

\(=9x-3\)

\(G=\left(3x+7\right)\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\)

\(=6x^2+23x+21-\left(6x^2+23x-55\right)=76\)

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

\(=x^3-8-\left(x^3+8\right)=-16\)