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a: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{15}{3x+y}+\dfrac{12}{2x-3y}=-6\\\dfrac{15}{3x+y}-\dfrac{25}{2x-3y}=105\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{37}{2x-3y}=-111\\\dfrac{4}{2x-3y}+\dfrac{5}{3x+y}=-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=-\dfrac{1}{3}\\3x+y=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-3y=-\dfrac{1}{3}\\9x+3y=\dfrac{3}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}11x=\dfrac{7}{6}\\3x+y=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7}{66}\\y=\dfrac{2}{11}\end{matrix}\right.\)

 

8 tháng 1 2022

hình đây nha mnundefined

Em tách nhỏ ra rồi hỏi nhe!! VD như 1 bài hỏi 1 lần á

3) Ta có: \(\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{8}+\sqrt{12}}-\dfrac{3\sqrt{5}}{4}\)

\(=\dfrac{\sqrt{5}\left(\sqrt{2}+\sqrt{3}\right)}{2\left(\sqrt{2}+\sqrt{3}\right)}-\dfrac{3\sqrt{5}}{4}\)

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

\(=\dfrac{-\sqrt{5}}{4}\) 

1) Ta có: \(\dfrac{1}{\sqrt{3}-1}+\dfrac{1}{4+2\sqrt{3}}-\dfrac{2}{\sqrt{3}}-\dfrac{3}{2}\)

\(=\dfrac{\sqrt{3}+1}{2}+\dfrac{2-\sqrt{3}}{2}-\dfrac{2\sqrt{3}}{3}-\dfrac{3}{2}\)

\(=\dfrac{\sqrt{3}+1+2-\sqrt{3}-3}{2}-\dfrac{2\sqrt{3}}{3}\)

\(=-\dfrac{2\sqrt{3}}{3}\)

3) Ta có: \(\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{8}+\sqrt{12}}-\dfrac{3\sqrt{5}}{4}\)

\(=\dfrac{\sqrt{5}\left(\sqrt{2}+\sqrt{3}\right)}{2\left(\sqrt{2}+\sqrt{3}\right)}-\dfrac{3\sqrt{5}}{4}\)

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

\(=\dfrac{-\sqrt{5}}{4}\) 

16 tháng 12 2021

\(ĐK:x\ge0;x\ne1\\ 1,P=\dfrac{x-2\sqrt{x}+1-x-\sqrt{x}+5\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{2}{\sqrt{x}-1}\\ 2,P< 0\Leftrightarrow\sqrt{x}-1< 0\left(2>0\right)\\ \Leftrightarrow\sqrt{x}< 1\Leftrightarrow0\le x< 1\)

7 tháng 8 2021

hjbjkrjeanjk ikbhnbalihbs 

16 tháng 7 2021
ext-9bosssssssssssssssss
NV
10 tháng 7 2021

\(M=\left(\dfrac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right).\dfrac{\sqrt{x}-2}{2}\)

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

2. Ta có: 

\(\sqrt{x}>0\Rightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}+2}>0\) hay \(M>0\)

Lại có: \(M=\dfrac{\sqrt{x}+2-1}{\sqrt{x}+2}=1-\dfrac{1}{\sqrt{x}+2}< 1\)

\(\Rightarrow0< M< 1\Rightarrow M>M^2\)

1) Ta có: \(M=\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{1}{\sqrt{x}-2}\right)\cdot\dfrac{\sqrt{x}-2}{2}\)

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

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

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

3 tháng 7 2021

\(P=\left[\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a+\sqrt{a}}{a-1}\right]:\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{1}{\sqrt{a}-1}\right)\) (đk:\(a\ge0;a\ne1\))

\(=\left[\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right]:\dfrac{\sqrt{a}-1+\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)

\(=\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{\sqrt{a}-1}\right).\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{2\sqrt{a}}\)

\(=\dfrac{1}{\sqrt{a}-1}.\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{2\sqrt{a}}=\dfrac{\sqrt{a}+1}{2\sqrt{a}}\)

2) \(\dfrac{1}{P}\ge\dfrac{\sqrt{a}+9}{8}\)

\(\Leftrightarrow\dfrac{2\sqrt{a}}{\sqrt{a}+1}\ge\dfrac{\sqrt{a}+9}{8}\)

\(\Leftrightarrow16\sqrt{a}\ge\left(\sqrt{a}+9\right)\left(\sqrt{a}+1\right)\)

\(\Leftrightarrow a-6\sqrt{a}+9\le0\)

\(\Leftrightarrow\left(\sqrt{a}-3\right)^2\le0\)

Dấu "=" xảy ra khi \(\sqrt{a}-3=0\Leftrightarrow a=9\) (tm)

Vậy...

1) ĐKXĐ: \(a\ge0;a\ne1\)

\(P=\left[\dfrac{a+\sqrt{a}+2\sqrt{a}+2}{\left(\sqrt{a}+2\right).\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}.\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right).\left(\sqrt{a}+1\right)}\right]\)\(:\left[\dfrac{\sqrt{a}-1+\sqrt{a}+1}{\left(\sqrt{a}-1\right).\left(\sqrt{a}+1\right)}\right]\)

\(\Leftrightarrow P=\left[\dfrac{\sqrt{a}.\left(\sqrt{a}+1\right)+2.\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+2\right).\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}}{\sqrt{a}-1}\right]\)\(:\dfrac{2\sqrt{a}}{\left(\sqrt{a}-1\right).\left(\sqrt{a}+1\right)}\)

\(\Leftrightarrow P=\left[\dfrac{\left(\sqrt{a}+2\right).\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+2\right).\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}}{\sqrt{a}-1}\right].\dfrac{\left(\sqrt{a}-1\right).\left(\sqrt{a}+1\right)}{2\sqrt{a}}\)

\(\Leftrightarrow P=\dfrac{\sqrt{a}+1-\sqrt{a}}{\sqrt{a}-1}.\dfrac{\left(\sqrt{a}-1\right).\left(\sqrt{a}+1\right)}{2\sqrt{a}}\)

\(\Leftrightarrow P=\dfrac{\sqrt{a}+1}{2\sqrt{a}}\)

2) Có : \(\dfrac{1}{P}\ge\dfrac{\sqrt{a}+9}{8}\)

\(\Leftrightarrow\dfrac{2\sqrt{a}}{\sqrt{a}+1}\ge\dfrac{\sqrt{a}+9}{8}\)

\(\Leftrightarrow\dfrac{2\sqrt{a}}{\sqrt{a}+1}-\dfrac{\sqrt{a}+9}{8}\ge0\)

\(\Leftrightarrow\dfrac{16\sqrt{a}-\left(\sqrt{a}+9\right).\left(\sqrt{a}+1\right)}{8.\left(\sqrt{a}+1\right)}\ge0\)

\(\Leftrightarrow\dfrac{16\sqrt{a}-a-10\sqrt{a}-9}{8.\left(\sqrt{a}+1\right)}\ge0\)

\(\Leftrightarrow\dfrac{-\left(a-6\sqrt{a}+9\right)}{8.\left(\sqrt{a}+1\right)}\ge0\)

\(\Leftrightarrow\dfrac{\left(\sqrt{a}-3\right)^2}{8.\left(\sqrt{a}+1\right)}\le0\)

Vì \(\sqrt{a}\ge0\Rightarrow8.\left(\sqrt{a}+1\right)>0\)  mà \(\left(\sqrt{a}-3\right)^2\) \(\ge0\) 

\(\Rightarrow\) \(\dfrac{\left(\sqrt{a}-3\right)^2}{8.\left(\sqrt{a}+1\right)}=0\) \(\Rightarrow\left(\sqrt{a}-3\right)^2=0\) \(\Leftrightarrow\sqrt{a}-3=0\Leftrightarrow\sqrt{a}=3\Leftrightarrow a=9\)

Vậy để\(\dfrac{1}{P}\ge\dfrac{\sqrt{a}+9}{8}\) thì \(a=9\)