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

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

\(P=\left(3xy\right)^2-6xy\)

\(P=\left(3xy\right)^2-2.3xy.1+1-1\)

\(P=\left(3xy-1\right)^2-1\ge-1\)

dấu \("="\) xảy ra \(\Leftrightarrow3xy-1=0\Leftrightarrow xy=\dfrac{1}{3}\)

vậy MIN \(P=-1\Leftrightarrow xy=\dfrac{1}{3}\)

câu a) thì đơn giản rồi

chỉ cần xét 2 \(\Delta ACB\) và \(\Delta AHC\), 2 tam giác này đồng dạng với nhau rồi \(\Rightarrow\) các cạnh tương ứng tỉ lệ là xong

b) cách làm tương tự câu a)

c) c/m \(\Delta AHC\infty\Delta AEH\)

\(\Rightarrow\dfrac{AH}{AE}=\dfrac{AC}{AH}\)

\(\Rightarrow AE.AC=AH^2\)

theo bài ra \(AD.AB+AE.AC\le\dfrac{BC^2}{2}\)

\(\Leftrightarrow\)\(AH^2+AH^2\le\dfrac{BC^2}{2}\) vì theo b) \(AH^2=AD.AB\)

\(\Leftrightarrow2AH^2-\dfrac{BC^2}{2}\le0\)

\(\Leftrightarrow\dfrac{4AH^2}{2}-\dfrac{OB.OC}{2}\le0\)

\(\Leftrightarrow\dfrac{\left(2AH\right)^2-BC^2}{2}\le0\)

đến đây chị bí rồi

xét pt \(x^2-2\left(m-2\right)x+m^2+2m-3=0\) (1)

có \(\Delta'=\left[-\left(m-2\right)\right]^2-m^2-2m+3\)

\(\Delta'=m^2-4m+4-m^2-2m+3\)

\(\Delta'=-6m+7\)

để pt (1) có 2 nghiệm pb thì \(\Delta'>0\Leftrightarrow-6m+7>0\)

\(\Leftrightarrow-6m>-7\Leftrightarrow m< \dfrac{7}{6}\)

có vi -ét \(\left\{{}\begin{matrix}x_1+x_2=2\left(m-2\right)\\x_1x_2=m^2+2m-3\end{matrix}\right.\)

theo bài ra ta có \(\dfrac{1}{x_1.x_2}=\dfrac{1}{5}\)

\(\Rightarrow x_1.x_2=5\) \(\left(x_1x_2\ne0\right)\)

\(m^2+2m-3=5\)

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

\(\Delta'=1^2-\left(-8\right)=1+8=9>0\Rightarrow\sqrt{\Delta'}=3\)

vì \(\Delta'>0\) nên phương trình (2) có 2 nghiệm phân biệt

\(m_1=-1+3=-2\) ( TM

\(m_2=-1-3=-4\) \(m< \dfrac{7}{6}\) )

vậy .....

xét pt \(x^2-2\left(m-1\right)x+2m-3=0\) (1)

từ (1) có \(\Delta'=\left[-\left(m-1\right)\right]^2-\left(2m-3\right)\)

\(\Delta'=m^2-2m+1-2m+3\)

\(\Delta'=m^2-4m+4\)

\(\Delta'=\left(m-2\right)^2>0\forall m\ne2\)

\(\Rightarrow pt\left(1\right)\) luôn có 2 nghiệm phân biệt \(\forall m\ne2\)

có vi - ét \(\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)\\x_1.x_2=2m-3\end{matrix}\right.\)

theo bài ra ta có \(\left|x_1-x_2\right|=5\)

\(\Leftrightarrow\left(\left|x_1-x_2\right|\right)^2=25\)

\(\Leftrightarrow\left(x_1-x_2\right)^2=25\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2-25=0\)

\(\Leftrightarrow\left[2\left(m-1\right)\right]^2-4\left(2m-3\right)-25=0\)

\(\Leftrightarrow4\left(m^2-2m+1\right)-8m+12-25=0\)

\(\Leftrightarrow4m^2-8m+4-8m-13=0\)

\(\Leftrightarrow4m^2-16m-9=0\) \(\left(2\right)\)

từ (2) có \(\Delta'=\left(-8\right)^2-4.\left(-9\right)=64+36=100>0\Rightarrow\sqrt{\Delta'}=10\)

vì \(\Delta'>0\) nên pt (2) có 2 nghiệm phân biệt

\(m_1=\dfrac{8+10}{4}=\dfrac{9}{2};m_2=\dfrac{8-10}{4}=\dfrac{-1}{2}\) ( TM \(\forall m\ne2\))

vậy \(m_1=\dfrac{9}{2};m_2=\dfrac{-1}{2}\) là các giá trị cần tìm

ta có \(2m^4+2m+1\)

\(=2m^4-2m^2+\dfrac{1}{2}+2m^2+2m+\dfrac{1}{2}\)

\(=\left(2m^4-2m^2+\dfrac{1}{2}\right)+\left(2m^2+2m+\dfrac{1}{2}\right)\)

\(=2\left(m^4-m^2+\dfrac{1}{4}\right)+2\left(m^2+m+\dfrac{1}{4}\right)\)

\(=2\left(m^4+2.\dfrac{1}{2}m^2+\dfrac{1}{4}\right)+2\left(m^2+2.\dfrac{1}{2}m+\dfrac{1}{4}\right)\)

\(=2\left(m^2-\dfrac{1}{2}\right)^2+2\left(m+\dfrac{1}{2}\right)^2\ge\forall m\) ( đpcm)

\(B=\dfrac{\sqrt{x^3}-\sqrt{x}+2x-2}{\sqrt{x}+2}\)

\(B=\dfrac{\sqrt{x}\left(x-1\right)+2\left(x-1\right)}{\sqrt{x}+2}\)

\(B=\dfrac{\left(\sqrt{x}+2\right)\left(x-1\right)}{\sqrt{x}+2}\)

\(B=x-1\)

\(B=A+1\Leftrightarrow\sqrt{x}-1+1=x-1\)

\(\Leftrightarrow x-\sqrt{x}-1=0\)

\(\Leftrightarrow x-2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}-\dfrac{1}{4}-1=0\)

\(\Leftrightarrow\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{5}{4}=0\)

\(\Leftrightarrow\left(\sqrt{x}-\dfrac{1}{2}-\dfrac{\sqrt{5}}{2}\right)\left(\sqrt{x}-\dfrac{1}{2}+\dfrac{\sqrt{5}}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\left(\sqrt{5}+1\right)^2}{4}\\x=\dfrac{\left(1-\sqrt{5}\right)^2}{4}\end{matrix}\right.\)

câu A sửa lại đề 1 chút

\(A=\dfrac{x-3\sqrt{x}+2}{\sqrt{x}-2}\)

\(A=\dfrac{x-2\sqrt{x}-\sqrt{x}+2}{\sqrt{x}-2}\)

\(A=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)-\left(\sqrt{x}-2\right)}{\sqrt{x}-2}\)

\(A=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\sqrt{x}-2}\)

\(A=\sqrt{x}-1\)

có \(x=4-2\sqrt{3}\)

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

\(\Leftrightarrow\sqrt{x}=\sqrt{3}-1\)

khi đó \(A=\sqrt{x}-1\Leftrightarrow A=\sqrt{3}-1-1=\sqrt{3}-2\)

\(A=\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\left(\sqrt{a}+\dfrac{1}{\sqrt{a}}\right)\)

a) rút gọn A

b) tính A với \(a=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\left(\sqrt{4-\sqrt{15}}\right)\)