\(\dfrac{1}{5}\sqrt[]{25x+50}-5\sqrt[]{x+2}+\sqrt[]{9x+18}+9=0\)

\(\Leftrightarrow\dfrac{1}{5}\sqrt[]{25\left(x+2\right)}-5\sqrt[]{x+2}+\sqrt[]{9\left(x+2\right)}+9=0\)

\(\Leftrightarrow\dfrac{1}{5}.5\sqrt[]{x+2}-5\sqrt[]{x+2}+3\sqrt[]{x+2}+9=0\)

\(\Leftrightarrow\sqrt[]{x+2}-5\sqrt[]{x+2}+3\sqrt[]{x+2}+9=0\)

\(\Leftrightarrow\sqrt[]{x+2}\left(1-5+3\right)+9=0\)

\(\Leftrightarrow-\sqrt[]{x+2}+9=0\)

\(\Leftrightarrow\sqrt[]{x+2}=9\)

\(\Leftrightarrow x+2=81\)

\(\Leftrightarrow x=79\)

\(110\%x+115\%y=400\\ \Rightarrow1.1x+1.15y=400\\ x+y=360\\ \Leftrightarrow1.1\left(x+y\right)=360\cdot1.1=396\\ \Rightarrow\left(1.1x+1.15y\right)-1.1\left(x+y\right)=1.1x+1.15y-1.1x-1.1y=0.05y=4\\ \Leftrightarrow y=\dfrac{4}{0.05}=80\\ \Rightarrow x=360-80=280.\)

\(y=\left(m+4\right)x+m-1\left(1\right)\)

a) Hàm số (1) đồng biến

\(\Leftrightarrow m+4\) lớn hơn \(0\)

\(\Leftrightarrow m\) lớn hơn \(-4\)

b) Hàm số (1) nghịch biến

\(\Leftrightarrow m+4\) nhỏ hơn \(0\)

\(\Leftrightarrow m\) nhỏ hơn \(-4\)

(Điện thoại tôi không đánh dấu nhỏ lớn được)

- CuCl₂: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)

- Cu(NO₃)₂: \(CuO+2HNO_3\rightarrow Cu\left(NO_3\right)_2+H_2O\)

- Fe₂(SO₄)₃: \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

- FeCl₃: \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

PT: \(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)

Ta có: \(n_{H_2SO_4}=0,25.1=0,25\left(mol\right)\)

a, Theo PT: \(n_{NaOH}=2n_{H_2SO_4}=0,5\left(mol\right)\Rightarrow V_{NaOH}=\dfrac{0,5}{2}=0,25\left(l\right)\)

b, Theo PT: \(n_{Na_2SO_4}=n_{H_2SO_4}=0,25\left(mol\right)\)

\(\Rightarrow C_{M_{Na_2SO_4}}=\dfrac{0,25}{0,25+0,25}=0,5\left(M\right)\)

Sửa đề: \(A=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}-3}-\frac12\right)\)

ĐKXĐ: x>=0; x<>9

a: Ta có: \(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\)

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

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

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

Ta có: \(\frac{\sqrt{x}-1}{\sqrt{x}-3}-\frac12\)

\(=\frac{2\left(\sqrt{x}-1\right)-\sqrt{x}+3}{2\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+1}{2\left(\sqrt{x}-3\right)}\)

Ta có: \(A=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}-3}-\frac12\right)\)

\(=-\frac{3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{\sqrt{x}+1}{2\left(\sqrt{x}-3\right)}\)

\(=\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\frac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}=\frac{-6}{\sqrt{x}+3}\)

b: Để A nguyên thì -6⋮\(\sqrt{x}+3\)

=>\(\sqrt{x}+3\in\left\lbrace3;6\right\rbrace\)

=>\(\sqrt{x}\in\left\lbrace0;3\right\rbrace\)

=>x∈{0;9}

Kết hợp ĐKXĐ, ta được: x=0(nhận)

   (\(\sqrt{5}\) - 1)\(\sqrt{6+2\sqrt{5}}\) 

=  (\(\sqrt{5}\) - 1).\(\sqrt{\left(\sqrt{5}\right)^2+2\sqrt{5}+1}\)

= (\(\sqrt{5}\) - 1).\(\sqrt{\left(\sqrt{5}+1\right)^2}\)

= (\(\sqrt{5}\) - 1).(\(\sqrt{5}\) +1)

= 5 - 1

= 4