Tam giác ABC, mình nhầm

\(\frac{2}{3}\sqrt{3}\)- \(\frac{1}{4}\sqrt{18}\)+ \(\frac{2}{5}\sqrt{2}-\frac{1}{4}\sqrt{12}\)

= \(\frac{2}{3}\sqrt{3}-\frac{3}{4}\sqrt{2}+\frac{2}{5}\sqrt{2}-\frac{2}{4}\sqrt{3}\)

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

= \(\frac{\sqrt{3}}{6}\)- \(\frac{7}{20}\sqrt{2}\)

kq ra hơi kì

#mã mã#

chăm vl

\(\sqrt{25x^2-10x+1}=4x+9\)

\(\Leftrightarrow\sqrt{\left(5x-1\right)^2}=4x+9\)

\(\Leftrightarrow\left|5x-1\right|=4x+9\)

\(\Leftrightarrow\orbr{\begin{cases}5x-1=4x+9\\5x-1=-4x-9\end{cases}\Leftrightarrow\orbr{\begin{cases}x=10\\x=-\frac{8}{9}\end{cases}}}\)

Vậy ... 

\(\sqrt{x^2+2x+1}=\sqrt{x+1}\)

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

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

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

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

Vậy ...

P = \(\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right)\). \(\frac{\left(x-1\right)^2}{2}\)( x\(\ge0\); x\(\ne\)1)

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

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

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

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

= -x² + \(\sqrt{x}\)+ 2

b. tự tính nha

c, P = -x² + \(\sqrt{x}+2\) 

           =  - (x² - 2.x.1/2 + 1/4) +2 +1/4

          = - (x-1/2)²+ 9/4

          ta có  (x - 1/2)² \(\ge0\forall x\)\(\Rightarrow-\left(x-\frac{1}{2}\right)^2\le0\forall x\)

\(\Rightarrow-\left(x-\frac{1}{2}\right)^2+\frac{9}{4}\le\frac{9}{4}\forall x\)

dấu "=" xảy ra khi và chỉ khi x-1/2 = 0

                                               x=1/2

vậy GTLN của P= 9/4 khi và chỉ khi x=1/2

#mã mã#