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ĐK x khác 4 và x không âm
\(=\frac{4\sqrt{x}\left(2-\sqrt{x}\right)+8x}{4-x}\\ =\frac{8\sqrt{x}+4x}{4-x}\\ =\frac{4\sqrt{x}\left(2+\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\\ =\frac{4\sqrt{x}}{2-\sqrt{x}}\)
câu b đk x>= -1/4
\(x+\sqrt{x+\dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}}}=2\)
\(x+\sqrt{\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2}=2\)
\(\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2=2\)
\(x+\dfrac{1}{4}=\left(\sqrt{2}-\dfrac{1}{2}\right)^2\)
\(x=\left(\sqrt{2}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)
\(x=\left(\sqrt{2}-\dfrac{1}{2}-\dfrac{1}{2}\right)\left(\sqrt{2}-\dfrac{1}{2}+\dfrac{1}{2}\right)\)
\(x=\sqrt{2}\left(\sqrt{2}-1\right)=2-\sqrt{2}\)
\(a,\sqrt{\frac{5.\left(38^2-17^2\right)}{8.\left(47^2-19^2\right)}}\)
\(=\sqrt{\frac{5.\left(38-17\right)\left(38+17\right)}{8.\left(47-19\right)\left(47+19\right)}}\)
\(=\sqrt{\frac{5.21.55}{8.28.66}}\)
\(=\sqrt{\frac{5775}{14784}}=\frac{5\sqrt{231}}{2\sqrt{4370}}\)
\(A=\left(\frac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right).\frac{\left(1-x\right)^2}{2}\)
\(A=\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\frac{\left(1-x\right)^2}{2}\)
\(A=\left(\frac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\frac{\left(1-x\right)^2}{2}\)
\(A=\frac{2}{\left(\sqrt{x}+1\right)^2.\left(\sqrt{x}-1\right)}.\frac{\left(1+\sqrt{x}\right)^2\left(\sqrt{x}-1\right)^2}{2}\)
\(A=\sqrt{x}-1\)
ý b,c dễ rồi nha
\(A=\sqrt{9x^2-6x+1}+\sqrt{25-30x+9x^2}\)
\(=\sqrt{9x^2-6x+1}+\sqrt{9x^2-30x+25}\)
\(=\sqrt{\left(3x-1\right)^2}+\sqrt{\left(3x-5\right)^2}\)
\(=\left|3x-1\right|+\left|3x-5\right|\)
\(=\left|3x-1\right|+\left|5-3x\right|\)
\(\ge\left|3x-1+5-3x\right|=4\)
Xảy ra khi \(\dfrac{1}{3}\le x\le\dfrac{5}{3}\)
\(A=\sqrt{9x^2-6x+1}+\sqrt{25-30x+9x^2}\)
\(=\sqrt{\left(3x-1\right)^2}+\sqrt{\left(3x-5\right)^2}\)
\(=\left|3x-1\right|+\left|3x-5\right|=\left|3x-1\right|+\left|5-3x\right|\)
Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) có:
\(A\ge\left|3x-1+5-3x\right|=\left|4\right|=4\)
Dấu " = " khi \(\left\{{}\begin{matrix}3x-1\ge0\\5-3x\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{3}\\x\le\dfrac{5}{3}\end{matrix}\right.\)
Vậy \(MIN_A=4\) khi \(\dfrac{1}{3}\le x\le\dfrac{5}{3}\)
ĐKXĐ : \(x^2-6x+9\ge0\)
=> \(\left(x-3\right)^2\ge0\) ( Luôn đúng )
- Ta có : \(A=2x+\sqrt{x^2-6x+9}\)
=> \(A=2x+\sqrt{\left(x-3\right)^2}\)
=> \(A=2x+\left|x-3\right|\)
- Thay x = -5 vào \(A=2x+\left|x-3\right|\) ta được :
\(A=2\left(-5\right)+\left|\left(-5\right)-3\right|\)
=> \(A=-10+8=-2\)