\(\left(\sqrt{x}+\sqrt{y}\right)\left(\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}\right)\)

\(=\left(\sqrt{x}+\sqrt{y}\right).\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}\)

\(=\frac{-y+\sqrt{x}.\sqrt{y}}{\sqrt{y}}.\left(\sqrt{x}+\sqrt{y}\right)\)

\(=\frac{\left(\sqrt{x}.\sqrt{y}-y\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{y}}\)

\(=\frac{xy-y^2}{y}\)

\(=\frac{y\left(x-y\right)}{y}\)

= x - y (đpcm)

\(A=\frac{\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

\(A=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(A=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

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

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

\(A=\frac{4}{x-1}\)

b) \(\frac{4}{x-1}=7\)

\(\Leftrightarrow4=7.\left(x-1\right)\)

\(\Leftrightarrow\frac{4}{7}=x-1\)

\(\Leftrightarrow\frac{4}{7}+1=x\)

\(\Leftrightarrow\frac{11}{7}=x\)

\(\Rightarrow x=\frac{11}{7}\)

      ĐK :\(\hept{\begin{cases}x>=0\\x\ne1\end{cases}}\)

Ta có: \(A=\left[\frac{1}{\sqrt{x}+1}-\frac{2\left(x-1\right)}{\sqrt{x}\left(x-1\right)+x-1}\right]:\left[\frac{\sqrt{x}+1}{x-1}-\frac{2}{x-1}\right]\)

ĐKXĐ: \(\dfrac{3}{2}\le x\le3\)

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

\(A\ge\sqrt{2x-3+6-2x}+\left(2-\sqrt{2}\right)\sqrt{3-x}\ge\sqrt{3}\)

\(A_{min}=\sqrt{3}\) khi \(3-x=0\Rightarrow x=3\)

\(A=1.\sqrt{2x-3}+\sqrt{2}.\sqrt{6-2x}\le\sqrt{\left(1+2\right)\left(2x-3+6-2x\right)}=3\)

\(A_{max}=3\) khi \(2x-3=\dfrac{6-2x}{2}\Rightarrow x=2\)

câu 1:

a)3Fe+2O₂-->Fe₃O₄            HH

b)4P+5O₂-->2P₂O₅        HH

c)2Al+6HCl-->2AlCl₃+3H₂        Thế

d)2K+2H₂O-->2KOH+H₂         Thế

f) Cu+2AgNO₃-->Cu(NO₃)₂ +2Ag            Thế

\(\sqrt{x-1}+x^2-1=0\)DK: \(x\ge1\)\(\Leftrightarrow\sqrt{x-1}\left[1+\left(x+1\right)\sqrt{x-1}\right]=0\Leftrightarrow\)

*\(\sqrt{x-1}=0=>x=1\)

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

KL: x=1

b)

\(\sqrt{x^2+3}=!x^2+1!\)  đặt x^2+1=t=> t>=1

\(\sqrt{t+2}=t\Leftrightarrow t^2-t-2=0=>t=-1\left(hoacloai\right)\&t=2\)

=>\(x=+-1\)

c)

\(x^3+4=4x\sqrt{x}\)  dk x>=0 

\(x^3+4=4\sqrt{x^3}\) \(Dat..\sqrt{x^3}=t=>t\ge0\)

t^2+4=4t<=>t^2-4t+4=0=> t=2=> x=\(\sqrt[3]{4}\)

 nếu bạn  muốn minh trả lời tiếp hay gui link truc tiep den minh.

xem bài và kiểm tra lại số liệu rất có thể sai lỗi số học.

 sao không thấy ai giải/

thấy có loi roi vào copy pass linh tinh

Cảm ơn bạn nhiều lắm! Đừng giận nha, mình gấp quá nên quên nhấn chọn cho bài giải cho bạn!

ĐKXĐ : \(x\ge2\)

Ta có : \(A=\dfrac{x+3\sqrt{x-2}}{x+4\sqrt{x-2}+1}\) . Đặt t = \(\sqrt{x-2}\ge0\) \(\Rightarrow x=t^2+2\)

Khi đó : \(A=\dfrac{t^2+2+3t}{t^2+4t+3}=\dfrac{\left(t+2\right)\left(t+1\right)}{\left(t+3\right)\left(t+1\right)}=\dfrac{t+2}{t+3}=1-\dfrac{1}{t+3}\ge1-\dfrac{1}{3}=\dfrac{2}{3}\)

" = " \(\Leftrightarrow t=0\Leftrightarrow x=2\)

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