\(\sqrt{5-3x}=\sqrt{2x+8}\)

\(\Leftrightarrow5-3x=2x+8\)

\(\Leftrightarrow-3x-2x=8-5\)

\(\Leftrightarrow-5x=3\)

\(\Leftrightarrow x=\frac{-3}{5}\)

P/S" ko chắc

Mk sửa đề lại 1 chút ( chả bt mk nhìn thế nào mak vt lộn hết cả đề )......

BÀI 1: Rút gọn

\(C=a\sqrt{\frac{4a^2-4ab+b^2}{a^2}}-2a-b\)

\(ĐKXĐ:x^2-12\ge0\Rightarrow x^2\ge12\Rightarrow x\ge-2\sqrt{3}\)

\(\sqrt{x^2-12}=2\)

\(\Leftrightarrow x^2-12=4\)

\(\Leftrightarrow x^2=16\)

\(\Leftrightarrow x=\pm4\)

cảm ơn bạn

\(\frac{\sqrt{2}+\sqrt{3}}{2+\sqrt{6}}=\frac{\left(\sqrt{2}+\sqrt{3}\right)\left(2-\sqrt{6}\right)}{\left(2-\sqrt{6}\right)\left(2+\sqrt{6}\right)}\)

\(=\frac{2\sqrt{2}-2\sqrt{3}+2\sqrt{3}-3\sqrt{2}}{4-6}\)

\(=\frac{-\sqrt{2}}{-2}\)

\(=\frac{\sqrt{2}}{2}\)

cảm ơn bạn

cộng hai vế ta được: 2tan\(\alpha\)=\(\frac{31}{12}\)\(\Rightarrow\)tan\(\alpha\)=\(\frac{31}{24}\)

=> cot\(\alpha\)=\(\frac{17}{24}\)

mik nham r . hai cau nay rieng biet nha , ko lien quan j toi nhau 

Đặt \(\hept{\begin{cases}a=\frac{x}{y}\\b=\frac{y}{z}\\c=\frac{z}{x}\end{cases}}\) Ta có: \(A=\frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}=\frac{1}{\frac{x}{y}+2}+\frac{1}{\frac{y}{z}+2}+\frac{1}{\frac{z}{x}+2}\)

\(=\frac{y}{x+2y}+\frac{z}{y+2z}+\frac{x}{z+2x}\)

Cần cm \(A\le1\Leftrightarrow2A\le2\)

\(\Leftrightarrow\frac{2y}{x+2y}+\frac{2z}{y+2z}+\frac{2x}{z+2x}\le2\)

\(\Leftrightarrow\left(1-\frac{2y}{x+2y}\right)+\left(1-\frac{2z}{y+2z}\right)+\left(1-\frac{2x}{z+2x}\right)\ge1\)

\(\Leftrightarrow\frac{x}{x+2y}+\frac{y}{y+2z}+\frac{z}{z+2x}\ge1\)

\(\Leftrightarrow\frac{x^2}{x^2+2xy}+\frac{y^2}{y^2+2yz}+\frac{z^2}{z^2+2xz}\ge1\)

bđt này đúng theo cauchy-schwarz. dấu bằng xảy ra khi a=b=c=1

Thanks bạn nha Girl:>>

\(\hept{\begin{cases}\frac{3}{5}x-\frac{2}{5}y+\frac{5}{3}x-y-x=1\\\frac{2}{3}x-y+2x-\frac{3}{2}y-y=1\end{cases}}\)<=>\(\hept{\begin{cases}\frac{19}{15}x-\frac{7}{5}y=1\\\frac{8}{3}x-\frac{7}{2}y=1\end{cases}}\)<=>x=3;y=2