Vì: \(\left(x-2\right)^2\ge0\forall x\Rightarrow\left(x-2\right)^2+9\ge9\)
\(\Rightarrow\sqrt{\left(x-2\right)^2+9}\ge\sqrt{9}=3\) (đpcm)
Dấu ''='' xảy ra khi x = 2
1) gpt \(x^2+3x\sqrt{\dfrac{x^2+1}{x}}=10x-1\)
2) ghpt \(\left\{{}\begin{matrix}x^2+y^2+2\left(x+y\right)=6\\xy\left(x+2\right)\left(y+2\right)=9\end{matrix}\right.\)
3) cho a,b,c dương thỏa abc=1
CMR \(\dfrac{2}{a^2\left(b+c\right)}+\dfrac{2}{b^2\left(c+a\right)}+\dfrac{2}{c^2\left(a+b\right)}\ge3\)
\[D=\left ( \frac{1}{3\sqrt{x}-6} +\frac{1}{x-2\sqrt{x}}\right )\left ( \frac{1}{6} +\frac{1}{2\sqrt{x}}\right )\\ D=\left ( \frac{1}{3\left ( \sqrt{x}-2 \right )} +\frac{1}{\sqrt{x}\left ( \sqrt{x}-2 \right )}\right ).\frac{\sqrt{x}+3}{6\sqrt{x}}\\ D=\frac{\sqrt{x}+3}{3\sqrt{x}\left ( \sqrt{x}-2 \right )}.\frac{\sqrt{x}+3}{6\sqrt{x}}\\ D=\frac{\left ( \sqrt{x}+3 \right )^{2}}{18x\left ( \sqrt{x}-2 \right )}\\ D=\frac{x+6\sqrt{x}+9}{18x\sqrt{x}-36x}\]
A/ Đúng
B/ Sai
rút gọn biểu thức sau
D=\(\left(\sqrt{x-\sqrt{18}}-\sqrt{x+\sqrt{18}}\right)\sqrt{x+\sqrt{x^2-18}}\) với \(x\ge18\)
A=\(\sqrt{x+6\sqrt{x-9}}+\sqrt{x-6\sqrt{x-9}}\)
C=\(\dfrac{\sqrt{2-\sqrt{2-4-x^2}}\left[\sqrt{\left(2+x\right)^3}+\sqrt{\left(2-x\right)^3}\right]}{4-\sqrt{4-x^2}}\)
\(\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right)\): \(\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
A=\(\left(\frac{\sqrt{x}-1}{x-4}-\frac{\sqrt{x+1}}{x+4\sqrt{x+4}}\right)\):\(\frac{x\sqrt{x}}{\left(4-x\right)^2}\)
B=\(\left(\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{x+9\sqrt{x}}{x-9}\right):\frac{x\sqrt{x}}{\left(4-x\right)^2}\)
rút gọn các biểu thức
Rút gọn
\(A=\left(\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}\right):\left(1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)
\(B=\left(\frac{x-3\sqrt{x}}{x-9}-1\right):\left(\frac{9-x}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
Câu 1 . Cho \(a,b\ge3.\) Chứng minh rằng
\(A=21\left(a+\dfrac{1}{b}\right)+3\left(b+\dfrac{1}{a}\right)\ge80\)
Câu 2. Giải phương trình :
\(x^2+6x-1=2\sqrt{5x^3-3x^2+3x-2}\)
Câu 3. Tìm GTNN của
\(Q=\dfrac{1}{2}\left(\dfrac{x^{10}}{y^2}+\dfrac{y^{10}}{x^2}\right)+\dfrac{1}{4}\left(x^{16}+y^{16}\right)-\left(1+x^2y^2\right)^2\)
Câu 4 . Giải phương trình
\(\dfrac{\sqrt{x-2009}-1}{x-2009}+\dfrac{\sqrt{y-2010}-1}{y-2010}+\dfrac{\sqrt{z-2011}-1}{z-2011}=\dfrac{3}{4}\)
Rút gọn P = \(\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}-\dfrac{\sqrt{x}}{3-\sqrt{x}}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right)\).\(\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
