Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1: \(=3\left(x+\dfrac{2}{3}\sqrt{x}+\dfrac{1}{3}\right)\)
\(=3\left(x+2\cdot\sqrt{x}\cdot\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{2}{9}\right)\)
\(=3\left(\sqrt{x}+\dfrac{1}{3}\right)^2+\dfrac{2}{3}>=3\cdot\dfrac{1}{9}+\dfrac{2}{3}=1\)
Dấu '=' xảy ra khi x=0
2: \(=x+3\sqrt{x}+\dfrac{9}{4}-\dfrac{21}{4}=\left(\sqrt{x}+\dfrac{3}{2}\right)^2-\dfrac{21}{4}>=-3\)
Dấu '=' xảy ra khi x=0
3: \(A=-2x-3\sqrt{x}+2< =2\)
Dấu '=' xảy ra khi x=0
5: \(=x-2\sqrt{x}+1+1=\left(\sqrt{x}-1\right)^2+1>=1\)
Dấu '=' xảy ra khi x=1
Bài 3: \(3\left(\sqrt{2x^2+1}-1\right)=x\left(1+3x+8\sqrt{2x^2+1}\right)\)
\(\Leftrightarrow\left(3-8x\right)\sqrt{2x^2+1}=3x^2+x+3\)
\(\Rightarrow\left(3-8x\right)^2\left(2x^2+1\right)=\left(3x^2+x+3\right)^2\)
\(\Leftrightarrow119x^4-102x^3+63x^2-54x=0\)
\(\Leftrightarrow x\left(7x-6\right)\left(17x^2+9\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{6}{7}\end{cases}}\)
Thử lại, ta nhận được \(x=0\)là nghiệm duy nhất của phương trình
bài 1) a) \(xy\sqrt{\dfrac{x}{y}}=x\sqrt{y}\sqrt{y}\dfrac{\sqrt{x}}{\sqrt{y}}=x\sqrt{x}\sqrt{y}=\left(\sqrt{x}\right)^3\sqrt{y}\)
b) \(\sqrt{\dfrac{5a^3}{49b}}=\dfrac{\sqrt{5a^3}}{\sqrt{49b}}=\dfrac{\sqrt{5a^3}}{7\sqrt{b}}=\dfrac{\sqrt{5a^3}.\sqrt{b}}{7\sqrt{b}.\sqrt{b}}=\dfrac{\sqrt{5a^3b}}{7b}\)
bài 2) a) \(\dfrac{\sqrt{3}-3}{1-\sqrt{3}}=\dfrac{\sqrt{3}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}=\sqrt{3}\)
b) \(\dfrac{5-\sqrt{15}}{\sqrt{3}-\sqrt{5}}=\dfrac{-\sqrt{5}\left(\sqrt{3}-\sqrt{5}\right)}{\sqrt{3}-\sqrt{5}}=-\sqrt{5}\)
c) \(\dfrac{2\sqrt{2}+2}{5\sqrt{2}}=\dfrac{\sqrt{2}\left(2+\sqrt{2}\right)}{5\sqrt{2}}=\dfrac{2+\sqrt{2}}{5}\)
bài 2:
a: \(\dfrac{25}{5-2\sqrt{3}}=\dfrac{125+10\sqrt{3}}{13}\)
b: \(\dfrac{8}{\sqrt{5}+2}=8\sqrt{5}-32\)
c: \(\dfrac{6}{2\sqrt{3}-\sqrt{7}}=\dfrac{12\sqrt{3}+6\sqrt{7}}{5}\)
d: \(=\dfrac{\sqrt{3}\left(3\sqrt{3}-2\right)}{\sqrt{2}\left(3\sqrt{3}-2\right)}=\dfrac{\sqrt{6}}{2}\)
bài 1 :
Hình : 2,43 2,43 0,9 A B C H
ta có : \(sin\widehat{BAH}=\dfrac{0,9}{2,43}=\dfrac{10}{27}\Rightarrow\widehat{BAH}\simeq21^o44'\)
\(\Rightarrow\widehat{ABC}=180^o-2\left(21^o44'\right)=136^o32'\)
vậy .....................................................................................................................
bài 2 : \(\dfrac{4}{3+\sqrt{5}+\sqrt{2+2\sqrt{5}}}=\dfrac{4\left(1-\sqrt{\sqrt{5}-2}\right)}{\left(3+\sqrt{5}+\sqrt{2+2\sqrt{5}}\right)\left(1-\sqrt{\sqrt{5}-2}\right)}\)
\(=\dfrac{4\left(1-\sqrt{\sqrt{5}-2}\right)}{3+\sqrt{5}+\sqrt{2+2\sqrt{5}}-3\sqrt{\sqrt{5}-2}-\sqrt{5}\sqrt{\sqrt{5}-2}-\sqrt{6-2\sqrt{5}}}\)
\(=\dfrac{4\left(1-\sqrt{\sqrt{5}-2}\right)}{4+\sqrt{2+2\sqrt{5}}-\left(3+\sqrt{5}\right)\sqrt{\sqrt{5}-2}}\) \(=\dfrac{4\left(1-\sqrt{\sqrt{5}-2}\right)}{4+\sqrt{2+2\sqrt{5}}-\sqrt{\left(\sqrt{5}-2\right)\left(14+6\sqrt{5}\right)}}\)\(=\dfrac{4\left(1-\sqrt{\sqrt{5}-2}\right)}{5}=1-\sqrt{\sqrt{5}-2}\)
bài 3 : 1) ta có : \(A=x+3\sqrt{x}-3=x+3\sqrt{x}+\dfrac{9}{4}-\dfrac{21}{4}\)
\(=\left(\sqrt{x}+\dfrac{3}{2}\right)^2-\dfrac{21}{4}\ge\left(\dfrac{3}{2}\right)^2-\dfrac{21}{4}\ge-3\)
dâu "=" xảy ra khi \(x=0\)
2) ta có : \(A=-2x-3\sqrt{x}+2=-2\left(x+\dfrac{3}{2}\sqrt{x}\right)+2\le2\)
dâu "=" xảy ra khi \(x=0\)
3) ta có : \(A=-4x-5\sqrt{x}-3=-4\left(x+\dfrac{5}{4}\sqrt{x}\right)-3\le-3\)
dâu "=" xảy ra khi \(x=0\)
Vy Lan Lê xin cái địa chỉ điii :))