\(x\left(x-2\right)\left(x+2\right)\left(x+4\right)=m\)

\(\Leftrightarrow\left(x^2+2x\right)\left(x^2+2x-8\right)=m\)

Đặt \(x^2+2x=a\)

Để PT \(x^2+2x-a=0\)có 2 nghiệm phân biệt thì:

\(\Delta'=1+4a>0\)

\(\Leftrightarrow a>-0,25\)

Ta có:

\(a\left(a-8\right)=m\)

\(\Leftrightarrow a^2-8a-m=0\)

Chỉ cần phương trình này có 2 nghiệm dương phân biệt là xong.

Tự làm nhé.

hình vẽ có đường cao AH = 32 à?

ko BH=32

1/ \(x-6\sqrt{x}-8=\left(\sqrt{x}-3+\sqrt{17}\right)\left(\sqrt{x}-3-\sqrt{17}\right)\)

2/ Bài này làm gì còn phân tích được nữa.

2/ Giả sử:

\(\sqrt{n+2}-\sqrt{n+1}>\sqrt{n+1}-\sqrt{n}\)

\(\Leftrightarrow\sqrt{n+2}+\sqrt{n}>2\sqrt{n+1}\)

\(\Leftrightarrow2n+2+2\sqrt{n^2+2n}>4n+4\)

\(\Leftrightarrow\sqrt{n^2+2n}>n+1\)

\(\Leftrightarrow n^2+2n>n^2+2n+1\)

\(\Leftrightarrow0>1\) (sai)

Vậy \(\sqrt{n+2}-\sqrt{n+1}< \sqrt{n+1}-\sqrt{n}\)

\(=2,35693368\)

cách làm kìa ai chẳng biết kết quả

a)\(\sqrt{\left(x+3\right)\left(x+2\right)}+\sqrt{\left(x+3\right)\left(x-1\right)}=2\sqrt{\left(x+3\right)^2}\)

\(\Leftrightarrow\sqrt{\left(x+3\right)\left(x+2\right)}+\sqrt{\left(x+3\right)\left(x-1\right)}-2\sqrt{\left(x+3\right)^2}=0\)

\(\Leftrightarrow\sqrt{x+3}\left(\sqrt{x+2}+\sqrt{x-1}-2\sqrt{x+3}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\sqrt{x+3}=0\\\sqrt{x+2}+\sqrt{x-1}=2\sqrt{x+3}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x+3=0\\2x+1+2\sqrt{\left(x-1\right)\left(x+2\right)}=4\left(x+3\right)\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-3\\2\sqrt{\left(x-1\right)\left(x+2\right)}=2x+11\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-3\\4\left(x-1\right)\left(x+2\right)=4x^2+44x+121\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-3\\-40x=129\end{cases}}\Rightarrow x=-3\) (thỏa)

b)\(\frac{3x}{\sqrt{3x+10}}=\sqrt{3x+1}-1\)

Đk:\(x\ge-\frac{1}{3}\)

\(pt\Leftrightarrow\frac{3x}{\sqrt{3x+10}}+1=\sqrt{3x+1}\)

\(\Leftrightarrow\frac{3x}{\sqrt{3x+10}}+1-\left(\frac{3}{5}x+1\right)=\sqrt{3x+1}-\left(\frac{3}{5}x+1\right)\)

\(\Leftrightarrow\frac{3x}{\sqrt{3x+10}}-\frac{3}{5}x=\frac{3x+1-\left(\frac{3}{5}x+1\right)^2}{\sqrt{3x+1}+\frac{3}{5}x+1}\)

\(\Leftrightarrow\frac{3x\left(5-\sqrt{3x+10}\right)}{5\sqrt{3x+10}}=\frac{-\frac{9}{25}x\left(x-5\right)}{\sqrt{3x+1}+\frac{3}{5}x+1}\)

\(\Leftrightarrow\frac{3x\cdot\frac{25-3x-10}{5+\sqrt{3x+10}}}{5\sqrt{3x+10}}-\frac{-\frac{9}{25}x\left(x-5\right)}{\sqrt{3x+1}+\frac{3}{5}x+1}=0\)

\(\Leftrightarrow\frac{3x\cdot\frac{-3\left(x-5\right)}{5+\sqrt{3x+10}}}{5\sqrt{3x+10}}-\frac{-\frac{9}{25}x\left(x-5\right)}{\sqrt{3x+1}+\frac{3}{5}x+1}=0\)

\(\Leftrightarrow x\left(x-5\right)\left(\frac{\frac{-9}{5+\sqrt{3x+10}}}{5\sqrt{3x+10}}-\frac{-\frac{9}{25}}{\sqrt{3x+1}+\frac{3}{5}x+1}\right)=0\)

Dễ thấy: \(\frac{\frac{-9}{5+\sqrt{3x+10}}}{5\sqrt{3x+10}}-\frac{-\frac{9}{25}}{\sqrt{3x+1}+\frac{3}{5}x+1}< 0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x-5=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)