ĐKXĐ : \(\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)

Ta có :

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

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

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

\(=\dfrac{\left(\sqrt{a}-1\right)^2\left(\sqrt{a}+1\right)^2}{-8\sqrt{a}}\)

Vậy....

a/ \(A=\dfrac{6}{5.8}+\dfrac{22}{8.19}+\dfrac{24}{19.31}+\dfrac{198}{101.200}\)

\(=2\left(\dfrac{3}{5.8}+\dfrac{11}{8.19}+\dfrac{12}{19.31}+...+\dfrac{99}{101.200}\right)\)

\(=2\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{19}+....+\dfrac{1}{101}-\dfrac{1}{200}\right)\)

\(=2\left(\dfrac{1}{5}-\dfrac{1}{200}\right)\)

\(=\dfrac{39}{100}\)

b/ \(A=\dfrac{1}{2^2}+\dfrac{1}{4^2}+.....+\dfrac{1}{100^2}\)

Ta có :

\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)

...........

\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)

\(\Leftrightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{99.100}\)

\(\Leftrightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{99}-\dfrac{1}{100}\)

\(\Leftrightarrow A< 1-\dfrac{1}{100}< 1\left(đpcm\right)\)

Phương trình hoành độ giao điểm là :

\(-x^2=mx+2\)

\(\Leftrightarrow x^2+mx+2=0\)

Lại có : \(\Delta=m^2-8>0\)

Theo định lí Vi - et ta có :

\(\left\{{}\begin{matrix}x1+x2=-m\\x1x2=2\end{matrix}\right.\)

\(\left(x1+1\right)\left(x2+1\right)=0\)

\(\Leftrightarrow x1x2+x1+x1+1=0\)

\(\Leftrightarrow2-m+1=0\Leftrightarrow m=3\)

a/ Xét \(\Delta ABC\) và \(\Delta HAC\) có :

\(\left\{{}\begin{matrix}\widehat{C}chung\\\widehat{BAC}=\widehat{AHC}=90^0\end{matrix}\right.\)

\(\Leftrightarrow\Delta ABC\sim HAC\left(g-g\right)\)

b/ \(BC=\sqrt{AB^2+AC^2}=10cm\)

\(AH.BC=AB.AC\Leftrightarrow AH=\dfrac{AB.AC}{BC}=4,8cm\)

c/ \(\Delta HEA\sim\Delta CEH\left(g-g\right)\)

\(\Leftrightarrow\dfrac{HE}{CE}=\dfrac{EA}{HE}\Leftrightarrow HE^2=EA.EC\left(đpcm\right)\)

ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)

\(A=\left(\dfrac{1}{\sqrt{x}-2}-\dfrac{2\sqrt{x}}{4-x}+\dfrac{1}{2+\sqrt{x}}\right)\left(\dfrac{2}{\sqrt{x}}-1\right)\)

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

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

\(=\dfrac{4\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\dfrac{2-\sqrt{x}}{\sqrt{x}}\)

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

b/ \(A< -1\)

\(\Leftrightarrow\dfrac{-2}{\sqrt{x}+2}+\dfrac{\sqrt{x}+2}{\sqrt{x}+2}< 0\)

\(\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}< 0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}>0\\\sqrt{x}-2< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>0\\x< 4\end{matrix}\right.\)

Vậy..

Để hàm số đc xác định :

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge m-1\\x\le2m\end{matrix}\right.\)

\(\Leftrightarrow m-1\le x\le2m\)

Mà \(x\in\left(-1;3\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}m-1\le-1\\2m\ge3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m\le0\\x\ge\dfrac{3}{2}\end{matrix}\right.\)

Gọi quãng đường AB là \(x\left(x>0\right)\)

Lúc đi : \(x=50t\left(km\right)\)

Lúc về : \(x=60\left(t-0,5\right)\)

Từ đó ta có pt :

\(\Leftrightarrow50t=60t-30\Leftrightarrow t=\left(3h\right)\)

Khi đó : \(x=150\left(km\right)\)

Ta có :

\(\dfrac{U_1}{U_2}=\dfrac{N_1}{N_2}\)

\(\Leftrightarrow\dfrac{150}{12}=\dfrac{6000}{N_2}\)

\(\Leftrightarrow N_2=480\) (vòng)

đại gia cho người nghèo xin ít tiền nào @@

\(\left(3x+3\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+3=0\\x-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=6\end{matrix}\right.\)

Vậy..