\(x^2+2y^2-2xy+2x-4y+2=0\)

\(\Rightarrow x^2-2xy+y^2+2\left(x-y\right)+1+y^2-2y+1=0\)

\(\Rightarrow\left(x-y\right)^2+2\left(x-y\right)+1+\left(y-1\right)^2=0\)

\(\Rightarrow\left(x-y+1\right)^2+\left(y-1\right)^2=0\)

=>................

Ta có : \(m< 1\Rightarrow m-1< 0\Leftrightarrow\left(\sqrt{m}-1\right)\left(\sqrt{m}+1\right)< 0\)

Vì \(\sqrt{m}+1>0\Rightarrow\sqrt{m}-1< 0\Rightarrow\sqrt{m}< 1\)\(\left(1\right)\)

Ta lại có : \(\sqrt{m}-1< 0\left(cmt\right)\)

Mà \(\sqrt{m}>0\left(m\ne0\right)\Rightarrow\sqrt{m}\left(\sqrt{m}-1\right)< 0\)

\(\Rightarrow m-\sqrt{m}< 0\Leftrightarrow m< \sqrt{m}\left(2\right)\)

Từ ( 1 ) và ( 2 ) \(\Rightarrow m< \sqrt{m}< 1\)khi \(0< m< 1\)\(\left(đpcm\right)\)

cậu biếtt cách áp dụng vào k?

ĐK: \(-\frac{7}{2}\le x\le5.\)

pt <=> \(x+25-4\sqrt{5-x}-6\sqrt{2x-7}=0\)

<=> \(5-x-4\sqrt{5-x}+4+2x+7-6\sqrt{2x+7}+9=0\)

<=> \(\left(\sqrt{5-x}-2\right)^2+\left(\sqrt{2x+7}-3\right)^2=0\)

<=> \(\hept{\begin{cases}\sqrt{5-x}-2=0\\\sqrt{2x+7}-3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}5-x=4\\2x+7=9\end{cases}\Leftrightarrow}x=1\left(tmdk\right)\)

A C B H

\(\tan\widehat{B}=\frac{AH}{BH}=\frac{AH}{5}\)

\(\tan\widehat{C}=\frac{AH}{HC}=\frac{AH}{20}\)

=> \(\frac{\tan\widehat{B}}{\tan\widehat{C}}=\frac{AH}{5}:\frac{AH}{20}=4\Rightarrow\tan\widehat{B}=4.\tan\widehat{C}\)

<=>x^2-y^2+x^2-xy=8

<=>(x-y)(2x+y)=8

2x+y>x-y

tự xét tiếp

lớp 9 kém thế

\(4-2xy+\left(2-x-y\right)^2=0\)

\(\Leftrightarrow y^2+x^2-4x-4y+8=0\)

\(\Leftrightarrow\left(y-2\right)^2+\left(x-2\right)^2=0\)