\(26,b\\ 27,a\\ 28,d\\ 29,b\\ 30,d\)

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

Áp dụng BĐT cosi, ta có: 

\(\sqrt{x}+\dfrac{1}{\sqrt{x}}\ge2\Leftrightarrow\dfrac{1}{\sqrt{x}+\dfrac{1}{\sqrt{x}}}\le\dfrac{1}{2}\\ \Leftrightarrow\dfrac{2}{\sqrt{x}+\dfrac{1}{\sqrt{x}}}\le1\)

Vậy GTLN của \(\dfrac{2\sqrt{x}}{x+1}\) là 1. Dấu \("="\) xảy ra \(\Leftrightarrow\sqrt{x}=\dfrac{1}{\sqrt{x}}\Leftrightarrow x=1\)

\(a,2^2=4,2^3=8,2^4=16,2^5=32,2^6=64,2^7=128,2^8=256,2^9=512,2^{10}=1024\)

\(b,3^2=9,3^3=27,3^4=81,3^5=243\)

\(c,4^2=16,4^3=64,4^4=256\)

\(d,5^2=25,5^3=125,5^4=625\)

\(1,4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\\ 2,\sqrt{7-4\sqrt{3}}=\sqrt{\left(2-\sqrt{3}\right)^2}=2-\sqrt{3}\\ 3,\sqrt{5+2\sqrt{6}}=\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}=\sqrt{2}+\sqrt{3}\\ 4,\sqrt{16+6\sqrt{7}}=\sqrt{\left(3+\sqrt{7}\right)^2}=3+\sqrt{7}\\ 5,\sqrt{12+2\sqrt{35}}=\sqrt{\left(\sqrt{7}+\sqrt{5}\right)^2}=\sqrt{7}+\sqrt{5}\\ 6,\sqrt{9-4\sqrt{5}}=\sqrt{\left(\sqrt{5}-2\right)^2}=\sqrt{5}-2\)

Tick plzzz

\(f,A< 0\Leftrightarrow\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}< 0\\ \Leftrightarrow2-5\sqrt{x}< 0\left(\sqrt{x}+3>0\right)\\ \Leftrightarrow x>\dfrac{4}{25}\)

\(g,A>0\\ \Leftrightarrow\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}>0\\ \Leftrightarrow2-5\sqrt{x}>0\left(\sqrt{x}+3>0\right)\\ \Leftrightarrow x< \dfrac{4}{25}\)

\(h,A>0\Leftrightarrow x< \dfrac{4}{25}\)

Mà \(x\in N\Leftrightarrow x=0\)

\(a,\) ĐKXĐ: \(x\ge0;x\ne1\)

\(b,\) Khi \(x=0\)

\(\Leftrightarrow A=\dfrac{0-11}{0+0-3}-\dfrac{0-2}{0-1}-\dfrac{0+3}{0+3}\\ \Leftrightarrow A=\dfrac{-11}{-3}-\dfrac{-2}{-1}-\dfrac{3}{3}=\dfrac{11}{3}-2-1=-\dfrac{11}{3}-3=\dfrac{2}{3}\)

\(c,A=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\\ A=\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\\ A=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\\ A=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\\ A=\dfrac{\left(\sqrt{x}-1\right)\left(2-5\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}\)

\(d,\) Để \(A=-\dfrac{8}{5}\)

\(\Leftrightarrow\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}=-\dfrac{8}{5}\\ \Leftrightarrow10-25\sqrt{x}=-8\sqrt{x}-24\\ \Leftrightarrow17\sqrt{x}=34\\ \Leftrightarrow\sqrt{x}=2\\ \Leftrightarrow x=4\)

\(e,\) Để \(A=\sqrt{x}-\dfrac{18}{5}\)

\(\Leftrightarrow\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}=\sqrt{x}-\dfrac{18}{5}\\ \Leftrightarrow5\left(2-5\sqrt{x}\right)=5\sqrt{x}\left(\sqrt{x}+3\right)-18\left(\sqrt{x}+3\right)\\ \Leftrightarrow10-25\sqrt{x}=5x+15\sqrt{x}-18\sqrt{x}-54\\ \Leftrightarrow5x+22\sqrt{x}-64=0\\ \Leftrightarrow\left(\sqrt{x}-2\right)\left(32\sqrt{x}+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=-\dfrac{32}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x\in\varnothing\end{matrix}\right.\Leftrightarrow x=4\)

Download The Geometer's Sketchpad version Tiếng Việt nha

1. Your kitchen is bigger than mine

2. Dave is the tallest person in the class

3. Dad speaks English better than Mum

4. Hoa can't cook as well as Lan

5. This restaurant ts the best one in the city

6. That watch is better than this one

7. My sister doesn't write more carelessly than she did

8. Tom drives more carefully than Peter

Câu 7 ko chắc đâu nha

\(a,\) Gọi O là giao điểm của AH và MN

Vì AMHN là hình chữ nhật nên \(OA=OH=OM=ON\)

\(\Rightarrow\Delta AOM\) cân tại O

\(\Rightarrow\widehat{OAM}=\widehat{OMA}\)

Mà \(\widehat{OAM}=\widehat{C}\) (cùng phụ \(\widehat{CAH}\))

\(\Rightarrow\widehat{OMA}=\widehat{C}\)

Xét \(\Delta AMN\) và \(\Delta ACB\) có:

\(\widehat{OMA}=\widehat{C}\)

\(\widehat{BAC}\) chung

\(\Rightarrow\Delta AMN=\Delta ACB\left(g.g\right)\\ \Rightarrow\dfrac{AM}{AC}=\dfrac{AN}{AB}\\ \Rightarrow AM\cdot AB=AN\cdot AC\left(đfcm\right)\)

\(1,a\\ 2,d\\ 3,c\\ 4,a\\ 5,b\\ 6,b\\ 7,c\\ 8,b\\ 9,b\\ 10,c\\ 11,b\\ 12,a\\ 13,c\\ 14,d\\ 15,b\\ 16,b\\ 17,a\\ 18,c\\ 19,a\\ 20,b\)