a: Xét ΔABC vuông tại A có AH là đường cao

nên \(AH^2=HB\cdot HC\)

Xét tứ giác AMHN có \(\hat{AMH}=\hat{ANH}=\hat{MAN}=90^0\)

nên AMHN là hình chữ nhật

=>\(HA^2=HM^2+HN^2\)

\(BM^2+3\cdot AH^2+CN^2\)

\(=BM^2+HM^2+CN^2+HN^2+2\cdot HA^2\)

\(=BH^2+CH^2+2\cdot AH^2=BH^2+CH^2+2\cdot BH\cdot CH\)

\(=\left(BH+CH\right)^2=BC^2\)

b: Xét ΔBHA vuông tại H có HM là đường cao

nên \(BM\cdot BA=BH^2\)

=>\(BM=\frac{BH^2}{BA}\)

Xét ΔCHA vuông tại H có HN là đường cao

nên \(CN\cdot CA=CH^2\)

=>\(CN=\frac{CH^2}{CA}\)

Xét ΔABC vuông tại A có AH là đường cao

nên \(AB^2=BH\cdot BC;AC^2=CH\cdot CB;AH\cdot BC=AB\cdot AC\)

=>\(\frac{AB^2}{AC^2}=\frac{BH\cdot BC}{CH\cdot BC}=\frac{BH}{CH}\)

=>\(\frac{BH^2}{CH^2}=\left(\frac{AB^2}{AC^2}\right)^2=\frac{AB^4}{AC^4}\)

\(BM\cdot CN\cdot BC\)

\(=\frac{BH^2}{BA}\cdot\frac{CH^2}{AC}\cdot BC=\frac{\left(BH\cdot CH\right)^2}{AB\cdot AC}\cdot BC\)

\(=\frac{AH^4}{AH\cdot BC}\cdot BC=AH^3\)

c: \(\frac{BM}{CN}=\frac{BH^2}{BA}:\frac{CH^2}{CA}\)

\(=\frac{BH^2}{AB}\cdot\frac{AC}{CH^2}=\frac{BH^2}{CH^2}\cdot\frac{AC}{AB}\)

\(=\frac{AB^4}{AC^4}\cdot\frac{AC}{AB}=\frac{AB^3}{AC^3}\)

a: Ta có: \(\hat{BAI}+\hat{DAI}=\hat{BAD}=90^0\)

\(\hat{BAI}+\hat{KBA}=90^0\) (ΔKBA vuông tại K)

Do đó: \(\hat{DAI}=\hat{KBA}\)

Xét ΔHAD vuông tại H và ΔKBA vuông tại K có

AD=BA

\(\hat{HAD}=\hat{KBA}\)

Do đó; ΔHAD=ΔKBA

=>AH=BK

b: ΔHAD=ΔKBA

=>DH=AK

Xét ΔABI vuông tại B có BK là đường cao

nên \(AK\cdot AI=AB^2\)

=>\(DH\cdot AI=AB^2\) không đổi

A = \(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}\) - \(\sqrt{5}\)

A =  -\(\sqrt{2}\)(\(\dfrac{1-\sqrt{3}}{1-\sqrt{3}}\)) - \(\sqrt{5}\)

A = -\(\sqrt{2}\) - \(\sqrt{5}\)

Mỗi bạn có 10 viên bi hồi tring em nhé chúc em học tốt 

Đặt BC=a

Xét ΔABC vuông tại A có cos B=\(\frac{BA}{BC}\)

=>\(\frac{BA}{a}=cos60=\frac12\)

=>BA=0,5a

ΔABC vuông tại A

=>\(BC^2=BA^2+AC^2\)

=>\(AC^2=a^2-\left(0,5a\right)^2=0,75a^2=\frac34a^2\)

=>\(AC=\frac{a\sqrt3}{2}\)

Xét ΔABC vuông tại A có AH là đường cao

nên \(AB\cdot AC=AH\cdot BC\)

=>\(AH\cdot a=\frac12a\cdot\frac{a\sqrt3}{2}=\frac{a^2\sqrt3}{4}\)

=>\(AH=\frac{a\sqrt3}{4}\)

\(\frac{AH}{BC}=\frac{a\sqrt3}{4}:a=\frac{\sqrt3}{4}\)

54. In Nepal, literacy increased by approximately 40 percent from the early 1950s to the end of the century. A. 40 percent of the people of Nepal became literate during the latter half of the 20th century. B. At the end of the 20th century, about 40 percent of the Nepalese were literate, much higher than in the carly 1950s. C. About 40 percent more of the Nepalese population became literate during the second half of the 20th century D. From the middle to the end of 20th century, only about 40 percent of Nepalese people could read and write.   55. Everyone in my family keeps on telling me to eat my food more slowly. A. To get me to eat my food more slowly, my whole family has to tell me firmly to do so. B. Everyone in my family takes turn telling me to eat my food more slowly. C. My whole family is constantly saying that I should eat my food more slowly. D. All of my family members cat slowly, and they're always telling me that I ought to do so as well.   56. This is the second time I've warned you, and I don't intend to do so again. A. If you don't heed this warning, you could be in big trouble. B. Two warnings are all I'm going to give you, and this is the second one. C. I hope you don't expect another warning after this one. D. You've already had two warnings, and the next one will be your last

đk x >= 0 ; x khác 4 ; 9 

\(M=\dfrac{x-9-\left(x-4\right)+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\dfrac{1}{\sqrt{x}+1}=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\dfrac{1}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)

b, \(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}=\dfrac{\sqrt{x}-2+3}{\sqrt{x}-2}=1+\dfrac{3}{\sqrt{x}-2}\Rightarrow\sqrt{x}-2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

c, \(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-2\ge0\Leftrightarrow\dfrac{\sqrt{x}+1-2\sqrt{x}+4}{\sqrt{x}-2}\ge0\)

\(\Leftrightarrow\dfrac{5-\sqrt{x}}{\sqrt{x}-2}\ge0\Leftrightarrow\left\{{}\begin{matrix}x\le25\\x>4\end{matrix}\right.\Leftrightarrow4< x\le25\)

Kết hợp đk vậy 4 < x =< 25  ; x khác 9 

a, \(M=\sqrt{2}-\sqrt{6-2\sqrt{5}}=\sqrt{2}-\left(\sqrt{5}-1\right)=\sqrt{2}-\sqrt{5}+1\)

b, \(N=-\sqrt{2}-\sqrt{5}\left(\sqrt{5}-\sqrt{2}\right)=-\sqrt{2}-5+\sqrt{10}\)

c, \(P=\sqrt{\left(2\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{3}\right)^2}\)

\(=2\sqrt{2}+\sqrt{3}+2\sqrt{2}-\sqrt{3}=4\sqrt{2}\)

- Đặt \(\left\{{}\begin{matrix}\sqrt{4x^2-20x+28}=a>0\\3x^2-15x+20=b>0\end{matrix}\right.\)

- Khi đó, ta có hệ:

\(\left\{{}\begin{matrix}a=b\\a^2-\dfrac{4}{3}b=\dfrac{4}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=b\\a^2-\dfrac{4}{3}a=\dfrac{4}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=b\\a^2-\dfrac{4}{3}a-\dfrac{4}{3}=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=b\\\left(a^2-\dfrac{4}{3}a+\dfrac{4}{9}\right)-\dfrac{16}{9}=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=b\\\left(a-\dfrac{2}{3}\right)^2-\dfrac{16}{9}=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=b\\\left(a-\dfrac{2}{3}-\dfrac{4}{3}\right)\left(a-\dfrac{2}{3}+\dfrac{4}{3}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=b\\\left[{}\begin{matrix}a=2\\a=-\dfrac{2}{3}\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}a=b=2\left(nhận\right)\\a=b=-\dfrac{2}{3}\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{4x^2-20x+28}=2\\3x^2-15x+20=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x^2-20x+24=0\\3x^2-15x+18=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4\left(x-2\right)\left(x-3\right)=0\\3\left(x-2\right)\left(x-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

- Vậy \(S=\left\{2;3\right\}\)

\(2\sqrt{x^2-5x+7}=3x^2-15x+20\)

đk x^2 - 5x + 7 > 0 

Đặt \(\sqrt{x^2-5x+7}\) = t 

\(2t=3t^2-1\Leftrightarrow3t^2-2t-1=0\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-\dfrac{1}{3}\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow x^2-5x+6=0\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)(tm)