Bài 6: Biến đối đơn giản biểu thức chứa căn bậc hai

Để A là số nguyên thì \(3\sqrt{x}+6+5⋮\sqrt{x}+2\)

\(\Leftrightarrow\sqrt{x}+2\in\left\{1;-1;5;-5\right\}\)

\(\Leftrightarrow\sqrt{x}=3\)

hay x=9

Đề có sai ko v???

Biểu thức nguyên đề thế này:

\(\dfrac{2x}{x+3\sqrt{x}+2}+\dfrac{5\sqrt{x}}{x+4\sqrt{x}+3}+\dfrac{\sqrt{x}+10}{x+5\sqrt{x}+6}\)

các đại ca xem... thế nào ạ??.....

Mỹ Duyên, nguyen van tuan

\(A=\dfrac{1}{5+2\sqrt{6}}-\dfrac{1}{5-2\sqrt{6}}=\dfrac{5-2\sqrt{6}}{25-24}-\dfrac{5+2\sqrt{6}}{25-24}=5-2\sqrt{6}-5-2\sqrt{6}=-4\sqrt{6}\)

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

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\(C=\dfrac{3}{\sqrt{3}}+\dfrac{2\sqrt{3}}{\sqrt{3}+1}=\sqrt{3}+\dfrac{2\sqrt{3}\left(\sqrt{3}-1\right)}{3-1}=\sqrt{3}+\sqrt{3}\left(\sqrt{3}-1\right)=\sqrt{3}+3-\sqrt{3}=3\)

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\(D=\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}-\dfrac{1}{2-\sqrt{3}}==\dfrac{\left(\sqrt{15}-\sqrt{12}\right)\left(\sqrt{5}+2\right)}{1}-\dfrac{2+\sqrt{3}}{1}=5\sqrt{3}+2\sqrt{15}-2\sqrt{15}-4\sqrt{3}-2+\sqrt{3}=-2\)

a) \(\dfrac{14}{2\sqrt{3}-\sqrt{5}}\)

\(=\dfrac{14\left(2\sqrt{3}+\sqrt{5}\right)}{\left(2\sqrt{3}-\sqrt{5}\right)\left(2\sqrt{3}+\sqrt{5}\right)}\)

\(=\dfrac{14\left(2\sqrt{3}+\sqrt{5}\right)}{\left(2\sqrt{3}\right)^2-\sqrt{5^2}}=\dfrac{14\left(2\sqrt{3}+\sqrt{5}\right)}{12-5}\)

\(=\dfrac{14\left(2\sqrt{3}+\sqrt{5}\right)}{7}=2\left(2\sqrt{3}+\sqrt{5}\right)\)

\(=4\sqrt{3}+2\sqrt{5}\)

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

x lớn hơn 2

⇒(x - 6)⁶ ≥ 0

Do (x - 6)⁶ = ((x-6)³)² ≥ 0 với mọi x

⇒\(\sqrt{\left(x-6\right)^6}\) luôn có nghĩa với mọi x

√(x-6)⁶=√[(x-6)³]²=(x-6)³ có nghĩa ∀

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

\(\Leftrightarrow\left(\sqrt{3x+1}+\sqrt{2-x}\right)^2=9\)

\(\Leftrightarrow3x+1+2.\sqrt{\left(3x+1\right)\left(2-x\right)}+2-x=0\)

\(\Leftrightarrow2x+3+2\sqrt{\left(3x+1\right)\left(2-x\right)}=9\)

\(\Leftrightarrow2x+2\sqrt{\left(3x+1\right)\left(2-x\right)}=6\)

\(\Leftrightarrow2\left(x+\sqrt{6x-3x^2+2-x}\right)=6\)

\(\Leftrightarrow x+\sqrt{5x-3x^2+2}\) = 3

\(\Leftrightarrow\sqrt{5x-3x^2+2}=3-x\)

\(\Leftrightarrow5x-3x^2+2=\left(3-x\right)^2\)

\(\Leftrightarrow5x-3x^2+2=9-6x+x^2\)

\(\Leftrightarrow5x-3x^2+2-9+6x-x^2=0\)

\(\Leftrightarrow11x-4x^2-7=0\)

\(\Leftrightarrow11x-11x^2+7x^2-7=0\)

\(\Leftrightarrow11x\left(1-x\right)-7\left(1-x^2\right)=0\)

\(\Leftrightarrow11x\left(1-x\right)-7\left(1+x\right)\left(1-x\right)=0\)

\(\Leftrightarrow\left(1-x\right)\left(11x-7-7x\right)=0\)

\(\Leftrightarrow\left(1-x\right)\left(4x-7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}1-x=0\\4x-7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{7}{4}\end{matrix}\right.\)

Sai thì thông cảm nha!!!hate

Ta chứng minh công thức:

\(1+\dfrac{1}{n^2}+\dfrac{1}{\left(n+1\right)^2}=\left(1+\dfrac{1}{n}+\dfrac{1}{n+1}\right)^2\) bằng cách quy đồng biểu thức ở vế phải rồi áp dụng vào bài tập

b: \(\Leftrightarrow6\sqrt{x-2}-15\cdot\dfrac{\sqrt{x-2}}{5}=20+4\sqrt{x-2}\)

\(\Leftrightarrow\sqrt{x-2}\cdot6-3\sqrt{x-2}-4\sqrt{x-2}=20\)

\(\Leftrightarrow-\sqrt{x-2}=20\)(vô lý)