Toán Liên hệ giữa phép nhân và phép khai phương

duclk said:

a) Điều kiện x>−34x>−34x > \frac{-3}{4}
√4x+3−12=0⇔√4x+3=12⇔(√4x+3)2=14⇔(4x+3)=144x+3−12=0⇔4x+3=12⇔(4x+3)2=14⇔(4x+3)=14\sqrt{4x + 3} -\frac{1}{2} = 0 \Leftrightarrow \sqrt{4x + 3} = \frac{1}{2} \Leftrightarrow (\sqrt{4x + 3})^{2} = \frac{1}{4} \Leftrightarrow (4x + 3) = \frac{1}{4}
b) √x2−x+1−x=2⇔√(x−1)2=x+2⇔|x−1|=x+2x2−x+1−x=2⇔(x−1)2=x+2⇔|x−1|=x+2\sqrt{x^{2} - x +1} - x = 2\Leftrightarrow \sqrt{(x-1)^{2}} = x +2 \Leftrightarrow \left | x -1 \right | = x+2
c) √x2−x+14−2x=1⇔√(x−12)2=2x+1⇔∣∣x−12∣∣=2x+1

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quynhphamdq said:

Câu 3:
a.,(√x−1)(√x−2)−(√x+1)2(x−1)(x−2)−(x+1)2\left ( \sqrt{x}-1 \right )\left ( \sqrt{x}-2 \right )-\left ( \sqrt{x+1} \right )^2
(√x−1)(√x−2)−(√x+1)2=x−√x−2√x+2−x−1=1−3√x(x−1)(x−2)−(x+1)2=x−x−2x+2−x−1=1−3x\left ( \sqrt{x}-1 \right )\left ( \sqrt{x}-2 \right )-\left ( \sqrt{x+1} \right )^2 = x-\sqrt{x}-2\sqrt{x}+2-x-1 =1-3\sqrt{x}
b.ĐKXĐ :x≠4x≠4x\neq 4
Ta có:
2√x+4x−42x+4x−4\frac{2\sqrt{x}+4}{x-4}=2√x+4x−4=2(√x+2)(√x+2)(√x−2)=2√x−2

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