a: \(A=\sqrt{x}+\dfrac{\sqrt{x}\left(1+2\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=\sqrt{x}+\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}\)

Khi x=4 thì \(A=2+\dfrac{2\cdot2+1}{2+1}=2+\dfrac{5}{3}=\dfrac{11}{3}\)

b: Khi x=(2-căn 3)^2 thì \(A=2-\sqrt{3}+\dfrac{2\left(2-\sqrt{3}\right)+1}{2-\sqrt{3}+1}\)

\(=2-\sqrt{3}+\dfrac{4-2\sqrt{3}+1}{3-\sqrt{3}}\)

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

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

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

\(=\dfrac{14-7\sqrt{3}}{3-\sqrt{3}}\)

d: A=2

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

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

=>\(x+\sqrt{x}-1=0\)

=>\(\left[{}\begin{matrix}\sqrt{x}=\dfrac{-1+\sqrt{5}}{2}\left(nhận\right)\\\sqrt{x}=\dfrac{-1-\sqrt{5}}{2}\left(loại\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{6-2\sqrt{5}}{4}=\dfrac{3-\sqrt{5}}{2}\)

Bài 4 :

a) Ta có :

 \(BC^2=AB^2+AC^2\left(Pitago\right)\)

\(\Leftrightarrow AC^2=BC^2-AB^2=100-36=64\)

\(\Leftrightarrow AC=8\left(cm\right)\)

\(\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}\)

\(\Leftrightarrow\dfrac{1}{AH^2}=\dfrac{AB^2+AC^2}{AB^2.AC^2}\)

\(\Leftrightarrow AH^2=\dfrac{AB^2.AC^2}{AB^2+AC^2}=\dfrac{6^2.8^2}{36+64}=\dfrac{6^2.8^2}{100}\)

\(\Leftrightarrow AH=\dfrac{6.8}{10}=\dfrac{24}{5}\left(cm\right)\)

b: Xét ΔABC vuông tại A có

sin C=AB/BC=3/5

nên góc C=37 độ

=>góc B=53 độ

c: ΔHAB vuông tại H có HE là đường cao

nên AE*BE=HE^2

ΔAHC vuông tại H có HF là đường cao

nên AF*FC=HF^2

Xét tứ giác AEHF có

góc AEH=góc AFH=góc FAE=90 độ

=>AEHF là hình chữ nhật

=>AH=EF

AE*BE+AF*FC

=HE^2+HF^2

=EF^2

=AH^2

=HB*HC

d: \(\dfrac{EB}{FC}=\dfrac{BH^2}{AB}:\dfrac{CH^2}{AC}=\dfrac{BH^2}{AB}\cdot\dfrac{AC}{CH^2}\)

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

a) \(x-4\sqrt{x-2}+2\left(x\ge2\right)\) 

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

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

\(=\left(\sqrt{x-2}\right)^2-2\cdot2\cdot\sqrt{x-2}+2^2\)

\(=\left(\sqrt{x-2}-2\right)^2\)

b) \(x+4\sqrt{x-2}+2\left(x\ge2\right)\)

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

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

\(=\left(\sqrt{x-2}\right)^2+2\cdot2\cdot\sqrt{x-2}+2^2\)

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

a) \(\sqrt[]{x-9}+2\sqrt[]{y-2}+3\sqrt[]{z-3}=\dfrac{x+y+z}{2}\left(1\right)\)

\(Đkxđ:\left\{{}\begin{matrix}x\ge9\\y\ge2\\z\ge3\end{matrix}\right.\)

Áp dụng Bất đẳng thức Bunhiacopxki :

\(\left(1\sqrt[]{x-9}+2\sqrt[]{y-2}+3\sqrt[]{z-3}\right)^2\le\left(1^2+2^2+3^2\right)\left(x-9+y-2+z-3\right)=14\left(x+y+z-14\right)\)

Dấu "=" xảy ra khi và chỉ khi :

\(\dfrac{x-9}{1}=\dfrac{y-2}{2}=\dfrac{z-3}{3}\left(a\right)\)

\(\left(1\right)\Leftrightarrow\)\(14\left(x+y+z-14\right)=\dfrac{\left(x+y+z\right)^2}{4}\left(2\right)\)

Đặt \(t=x+y+z\)

\(\Leftrightarrow14t-196=\dfrac{t^2}{4}\)

\(\Leftrightarrow t^2+56t-784=0\)

\(\Leftrightarrow\left(t-28\right)^2=0\)

\(\Leftrightarrow t=28\)

\(\Leftrightarrow x+y+z=28\)

\(\left(a\right)\Leftrightarrow\dfrac{x-9}{1}=\dfrac{y-2}{2}=\dfrac{z-3}{3}=\dfrac{x+y+z-14}{6}=\dfrac{28-14}{6}=\dfrac{7}{3}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-9=1.\dfrac{7}{3}=\dfrac{7}{3}\\y-2=2.\dfrac{7}{3}=\dfrac{14}{3}\\z-3=3.\dfrac{7}{3}=7\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{34}{3}\\y=\dfrac{20}{3}\\z=10\end{matrix}\right.\)

a) \(\sqrt{\dfrac{1}{8}}\cdot\sqrt{2}\cdot\sqrt{125}\cdot\sqrt{\dfrac{1}{5}}\)

\(=\dfrac{1}{\sqrt{8}}\cdot\sqrt{2}\cdot5\sqrt{5}\cdot\dfrac{1}{\sqrt{5}}\)

\(=\dfrac{\sqrt{2}}{2\sqrt{2}}\cdot\dfrac{5\sqrt{5}}{\sqrt{5}}\)

\(=\dfrac{1}{2}\cdot5\)

\(=\dfrac{5}{2}\)

b) \(4\sqrt{50}+2\sqrt{8}-4\sqrt{72}-\sqrt{32}\)

\(=4\cdot5\sqrt{2}+2\cdot2\sqrt{2}-4\cdot6\sqrt{2}-4\sqrt{2}\)

\(=20\sqrt{2}+4\sqrt{2}-24\sqrt{2}-4\sqrt{2}\)

\(=\left(20+4-24-4\right)\sqrt{2}\)

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

c) \(2\sqrt{20}-3\sqrt{45}+5\sqrt{80}-5\sqrt{5}\)

\(=2\cdot2\sqrt{5}-3\cdot3\sqrt{5}+5\cdot4\sqrt{5}-5\sqrt{5}\)

\(=4\sqrt{5}-9\sqrt{5}+20\sqrt{5}-5\sqrt{5}\)

\(=\left(20-9-5+4\right)\sqrt{5}\)

\(=10\sqrt{5}\)

d) \(2ab\sqrt{a^2b}-5a^2\sqrt{b^3}\) (\(a,b\ge0\)) 

\(=2ab\cdot\left|a\right|\sqrt{b}-5a^2\left|b\right|\sqrt{b}\)

\(=2a^2b\sqrt{b}-5a^2b\sqrt{b}\)

\(=\left(2a^2b-5a^2b\right)\sqrt{b}\)

\(=-3a^2b\sqrt{b}\)

e) \(\sqrt{40}+\sqrt{\dfrac{2}{5}}-\sqrt{\dfrac{5}{2}}\)

\(=2\sqrt{10}+\dfrac{\sqrt{10}}{5}-\dfrac{\sqrt{10}}{2}\)

\(=\dfrac{20\sqrt{10}}{10}+\dfrac{2\sqrt{10}}{10}-\dfrac{5\sqrt{10}}{10}\)

\(=\dfrac{\left(20+2-5\right)\sqrt{10}}{10}\)

\(=\dfrac{17\sqrt{10}}{10}\)

Câu d đúng đề chưa bạn 

\(sin15^o+sin75^o-cos15^o-cos75^o+sin30^o\)

\(=\left(sin15+sin75^o\right)-\left(cos15^o+cos75^o\right)+sin30^o\)

\(=\dfrac{\sqrt{6}}{2}-\dfrac{\sqrt{6}}{2}+\dfrac{1}{2}\)

\(=0+\dfrac{1}{2}\)

\(=\dfrac{1}{2}\)

\(sin15^o+sin75^o-cos15^0-cos75^o+sin30^o\)

\(=cos75^o+cos15^0-cos15^0-cos75^o+sin30^o\)

\(=sin30^o=\dfrac{1}{2}\)

\(2+\dfrac{3\left(x+1\right)}{3}\le3-\dfrac{x-1}{4}\)

\(\Leftrightarrow2+x+1\le\dfrac{12}{4}-\dfrac{x-1}{4}\)

\(\Leftrightarrow x+3\le\dfrac{13-x}{4}\)

\(\Leftrightarrow\dfrac{4x+12}{4}\le\dfrac{13-x}{4}\)

\(\Leftrightarrow4x+12\le13-x\)

\(\Leftrightarrow4x+x\le13-12\)

\(\Leftrightarrow5x\le1\)

\(\Leftrightarrow x\le\dfrac{1}{5}\)

Vậy: \(x\le\dfrac{1}{5}\) 

\(2+\dfrac{3\left(x+1\right)}{3}\le3-\dfrac{x-1}{4}\)

\(\Leftrightarrow\dfrac{12x+36}{12}\le\dfrac{33-3x}{12}\)

\(\Leftrightarrow12x+36\le33-3x\)

\(\Leftrightarrow12x+3x\le-36+33\)

\(\Leftrightarrow15x\le-3\)

\(\Leftrightarrow x\le\dfrac{-1}{5}\)

\(a,\dfrac{2x-1}{3}< \dfrac{x+6}{2}\)

\(\Leftrightarrow\dfrac{4x-2}{6}< \dfrac{3x+18}{6}\)

\(\Leftrightarrow4x-2< 3x+18\)

\(\Leftrightarrow4x-3x< 2+18\)

\(\Leftrightarrow x< 20\)

\(b,\dfrac{5\left(x-1\right)}{6}-1>\dfrac{2\left(x+1\right)}{3}\)

\(\Leftrightarrow\dfrac{5x-11}{6}>\dfrac{4x+4}{6}\)

\(\Leftrightarrow5x-11>4x+4\)

\(\Leftrightarrow5x-4x>11+4\)

\(\Leftrightarrow x>15\)