\(x-\sqrt{x}=m\Leftrightarrow x-\sqrt{x}-m=0\)

\(\Delta=1-4\left(-m\right)=1+4m\)

Để phương trình trên có nghiệm khi \(\Delta\ge0\)

\(1+4m\ge0\Leftrightarrow m\ge-\frac{1}{4}\)

sửa đề : 

\(N=\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{x+\sqrt{x}}\right)\)đk : x >= 0 

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

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

n = (\(\sqrt{yyx}\)-r554-56

​\(\frac{1-a\sqrt{a}}{1-\sqrt{a}}=\frac{\left(1-a\sqrt{a}\right)\left(1+\sqrt{a}\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\)​​​

                         \(=\frac{1-a^2+\sqrt{a}-a\sqrt{a}}{1-a}\)

                           \(=\frac{\left(1-a\right)\left(1+a\right)+\sqrt{a}\left(1-a\right)}{1-a}\)

                            \(=\frac{\left(1-a\right)\left(1+a+\sqrt{a}\right)}{1-a}\)

                               \(=1+a+\sqrt{a}\)

1+a=\(\sqrt{a}\)

Ta có: \(P=\left(a^2+\frac{1}{16a^2}\right)+\left(b^2+\frac{1}{16b^2}\right)+\frac{15}{16}\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\)

sử dụng bđt cô-si có: \(a^2+\frac{1}{16a^2}\ge\frac{1}{2};b^2+\frac{1}{16b^2}\ge\frac{1}{2};\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab}=\frac{4}{2ab}\)

Lại có: \(\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{4}{a^2+b^2}\)

\(\Rightarrow2\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\ge4\left(\frac{1}{a^2+b^2}+\frac{1}{2ab}\right)\ge4\frac{4}{a^2+b^2+2ab}=\frac{16}{\left(a+b\right)^2}=16\)

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}\ge8\)

\(\Rightarrow P\ge\frac{1}{2}+\frac{1}{2}+\frac{15}{2}=\frac{17}{2}\)

Dấu '=' xảy ra <=> \(\hept{\begin{cases}a=b\\a+b=1\end{cases}\Leftrightarrow a=b=\frac{1}{2}}\)

địt ko em

ko biết bạn nhắn làm j ?

\(\sqrt{a^2b}=\left|a\right|\sqrt{b}=a\sqrt{b}\)( vì a >= 0 )

hok

tốt 

nha

a) Ta có ${\left(ac+bd\right)}^2+{\left(ad-bc\right)}^2$$=a^2{c}^2+2acbd+b^2{d}^2+a^2{d}^2-2adbc+b^2{c}^2$

$=\left(a^2{c}^2+b^2{c}^2\right)+\left(a^2{d}^2+b^2{d}^2\right)$$=c^2\left(a^2+b^2\right)+d^2\left(a^2+b^2\right)$$=\left(a^2+b^2\right)\left(c^2+d^2\right)$

b) Ta có $0\le {\left(ad-bc\right)}^2$$\iff {\left(ac+bd\right)}^2\le {\left(ac+bd\right)}^2+{\left(ad-bc\right)}^2$

Mà theo câu a, ta có ${\left(ac+bd\right)}^2+{\left(ad-bc\right)}^2=\left(a^2+b^2\right)\left(c^2+d^2\right)$

Nên ${\left(ac+bd\right)}^2\le \left(a^2+b^2\right)\left(c^2+d^2\right)$