\(\sqrt{9x+9}+\sqrt{x+1}=20\)đk : x>= -1

\(\Leftrightarrow3\sqrt{x+1}+\sqrt{x+1}=20\Leftrightarrow4\sqrt{x+1}=20\)

\(\Leftrightarrow\sqrt{x+1}=5\Leftrightarrow x+1=25\Leftrightarrow x=24\)

\(\sqrt{9x+9}=3\sqrt{x+1}\)

thay vào rồi sẽ ra x=24 bạn nha

a, Với x >= 0 ; x khác 4 

\(=\frac{x-3\sqrt{x}+2-\left(x+4\sqrt{x}+3\right)-x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{-3\sqrt{x}-3-x-4\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}=\frac{-7\sqrt{x}-6-x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{-\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}=\frac{-\sqrt{x}-6}{\sqrt{x}-2}\)

b, \(Q+1>0\Leftrightarrow\frac{-\sqrt{x}-6+\sqrt{x}-2}{\sqrt{x}-2}>0\Leftrightarrow\frac{-8}{\sqrt{x}-2}>0\)

\(\Rightarrow\sqrt{x}-2< 0\Leftrightarrow x< 4\Rightarrow0\le x< 4\)

c, \(\frac{-\left(\sqrt{x}+6\right)}{\sqrt{x}-2}=\frac{-\left(\sqrt{x}-2+8\right)}{\sqrt{x}-2}=-1-\frac{8}{\sqrt{x}-2}\)

\(\Rightarrow\sqrt{x}-2\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

) Bài 1: Biến đổi tương đương thôi: \((ac+bd)^2+(ad-bc)^2=a^2c^2+b^2d^2+2abcd+a^2d^2+b^2c^2-2abcd\) \(=a^2c^2+b^2d^2+a^2d^2+b^2c^2=(a^2+b^2)(c^2+d^2)\) Ta có đpcm Bài 2: Áp dụng kết quả bài 1: \((a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2\geq (ac+bd)^2\) do \((ad-bc)^2\geq 0\) Dấu bằng xảy ra khi \(ad=bc\Leftrightarrow \frac{a}{c}=\frac{b}{d}\)

^HT^