ủa sao ra \(-\sqrt{6}\) ddc v :V

\(Đặt.A=\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\\ \Rightarrow A^2=\left(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\right)^2\\ \Rightarrow A^2=12-3\sqrt{7}+12+3\sqrt{7}-2\sqrt{\left(12-3\sqrt{7}\right)\left(12+3\sqrt{7}\right)}\\ \Rightarrow A^2=24-2\sqrt{12^2-\left(3\sqrt{7}\right)^2}\\ \Rightarrow A^2=24-2\sqrt{144-63}\\ \Rightarrow A^2=24-2\sqrt{81}\\ \Rightarrow A^2=24-2.9\\ \Rightarrow A^2=24-18\\ \Rightarrow A^2=6\\ \Rightarrow A=-\sqrt{6}\)

`x` tỉ lệ thuận với `y` theo hệ số tỉ lệ là `3/2` hay `x/3 = y/2`

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

`x/3 = y/2 = (x+y)/(3+2) = 10,5/5=2,1`

`x/3 = 2,1 => x= 2,1 .3=6,3`

`y/2 = 2,1 => y= 2,1.2=4,2`

\(\dfrac{x}{\sqrt{x}-1}=\dfrac{9}{2}\left(ĐKXĐ:x\ge0;x\ne1\right)\\ \Leftrightarrow2x=9\left(\sqrt{x}-1\right)\\ \Leftrightarrow2x=9\sqrt{x}-9\\ \Leftrightarrow2x-9\sqrt{x}+9=0\\ \Leftrightarrow2x-6\sqrt{x}-3\sqrt{x}+9=0\\ \Leftrightarrow2\sqrt{x}\left(\sqrt{x}-3\right)-3\left(\sqrt{x}-3\right)=0\\ \Leftrightarrow\left(\sqrt{x}-3\right)\left(2\sqrt{x}-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-3=0\\2\sqrt{x}-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=\dfrac{3}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=9\left(tm\right)\\x=\dfrac{9}{4}\left(tm\right)\end{matrix}\right.\)

\(\dfrac{4}{9}:\left(-\dfrac{1}{7}\right)+6\dfrac{5}{9}:\left(-\dfrac{1}{7}\right)\\ =\dfrac{4}{9}.\left(-7\right)+\dfrac{59}{9}.\left(-7\right)\\ =\left(\dfrac{4}{9}+\dfrac{59}{9}\right)\left(-7\right)\\ =\dfrac{63}{9}.\left(-7\right)\\ =7.\left(-7\right)\\ =-49\)

\(\dfrac{-1}{4}.13\dfrac{9}{11}-0,25.6\dfrac{2}{11}\\ =\dfrac{-1}{4}.\dfrac{152}{11}+\left(-\dfrac{1}{4}\right).\dfrac{68}{11}\\ =\dfrac{-1}{4}\left(\dfrac{152}{11}+\dfrac{68}{11}\right)\\ =\dfrac{-1}{4}.\dfrac{220}{11}\\ =\dfrac{-1}{4}.20\\ =-5\)

\(A=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)+7\\ =\left[\left(x-1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]+7\\ =\left(x^2+5x-6\right)\left(x^2+5x+6\right)+7\\ =\left(x^2+5x\right)^2-6^2+7\\ =\left(x^2+5x\right)^2-36+7\\ =\left(x^2+5x\right)^2-29\ge-29\)

Dấu "=" xảy ra `<=> x^2 +5x= 0`

`<=>x(x+5)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

Vậy GTNN của biểu thức là `-29` \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

\(\dfrac{-4}{7}.\dfrac{2}{9}+\dfrac{-4}{7}.\dfrac{7}{9}+2\dfrac{4}{7}\\ =\dfrac{-4}{7}.\left(\dfrac{2}{9}+\dfrac{7}{9}\right)+2+\dfrac{4}{7}\\ =\dfrac{-4}{7}.1+2+\dfrac{4}{7}\\ =\left(\dfrac{-4}{7}+\dfrac{4}{7}\right)+2\\ =0+2\\ =2\)

\(\dfrac{-1}{6}+\dfrac{2}{3}.\left(-\dfrac{3}{4}\right)+\dfrac{\left(-2\right)^2}{5}\\ =\dfrac{-1}{6}-\dfrac{1}{2}+\dfrac{4}{5}\\ =\dfrac{-5}{30}-\dfrac{15}{30}+\dfrac{24}{30}\\ =\dfrac{4}{30}\\ =\dfrac{2}{15}\)