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a. \(\sqrt{\dfrac{2}{3}}=\sqrt{\dfrac{2.3}{3^2}}=\dfrac{1}{3}.\sqrt{6}\)
b. \(\sqrt{\dfrac{x^2}{5}}=\sqrt{\dfrac{5x^2}{5^2}}=\dfrac{x}{5}.\sqrt{5}\) (vì x \(\ge\) 0)
c. \(\sqrt{\dfrac{3}{x}}=\sqrt{\dfrac{3.x}{x^2}}=\dfrac{1}{x}.\sqrt{3x}\) (vì x > 0)
d. \(\sqrt{x^2-\dfrac{x^2}{7}}=\sqrt{\dfrac{6x^2}{7}}=\sqrt{\dfrac{6x^2.7}{7.7}}=\sqrt{\dfrac{42.x^2}{7^2}}=-\dfrac{x}{7}.\sqrt{42}\) (vì x < 0)
Câu đầu tiên: \(\sqrt{18-\sqrt{128}}=\sqrt{16-2\sqrt[]{16}\sqrt{2}+2}=\sqrt{\left(\sqrt{16}-\sqrt{2}\right)^2}=\sqrt{16}-\sqrt{2}=4-\sqrt{2}\)
CM\(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=2\)
Biến đổi vế trái ta có:
\(VT^2=\left(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\right)^2=4+\sqrt{7}-2\sqrt{\left(4+\sqrt{7}\right)\left(\sqrt{4-\sqrt{7}}\right)}+4-\sqrt{7}=8-2\sqrt{16-7}=8-2\sqrt{9}=8-2.3=2\Rightarrow VT=\sqrt{2}\)
a: \(=\sqrt{\left(2-a\right)^2\cdot\dfrac{2a}{a-2}}=\sqrt{2a\left(a-2\right)}\)
b: \(=\sqrt{\left(x-5\right)^2\cdot\dfrac{x}{\left(5-x\right)\left(5+x\right)}}\)
\(=\sqrt{\left(x-5\right)\cdot\dfrac{x}{x+5}}\)
c: \(=\sqrt{\left(a-b\right)^2\cdot\dfrac{3a}{\left(b-a\right)\left(b+a\right)}}=\sqrt{\dfrac{3a\left(b-a\right)}{b+a}}\)
B=\(\dfrac{\sqrt{a.6}}{\sqrt{6.6}}+\dfrac{\sqrt{2a.3}}{\sqrt{3.3}}+\dfrac{\sqrt{3a.2}}{\sqrt{2.2}}\)
=\(\dfrac{\sqrt{6a}}{6}+\dfrac{\sqrt{6a}}{3}+\dfrac{\sqrt{6a}}{2}\)
=\(\dfrac{\sqrt{6a}+2\sqrt{6a}+3\sqrt{6a}}{6}\)
=\(\dfrac{6\sqrt{6a}}{6}=\sqrt{6a}\)
b: \(B=\dfrac{\sqrt{6}}{6}\cdot\sqrt{a}+\dfrac{\sqrt{6}}{3}\cdot\sqrt{a}+\dfrac{\sqrt{6}}{2}\cdot\sqrt{a}\)
\(=\sqrt{a}\cdot\sqrt{6}=\sqrt{6a}\)
e: \(=2-x-x=2-2x\)
i: \(=\left|x-\left(1-x\right)\right|-2x=\left|x-1+x\right|-2x\)
\(=\left|2x-1\right|-2x\)
=1-2x-2x=1-4x
a,\(P=\dfrac{\sqrt{x}-1-2}{\sqrt{x}-1}=1-\dfrac{2}{\sqrt{x-1}}\)
P<\(\dfrac{1}{2}\)\(\Leftrightarrow1-\dfrac{2}{\sqrt{x}-1}< \dfrac{1}{2}
\)
\(\Leftrightarrow\dfrac{1}{2}< \dfrac{2}{\sqrt{x}-1}\)\(\Leftrightarrow\dfrac{2}{4}< \dfrac{2}{\sqrt{x}-1}\)
\(\Rightarrow4>\sqrt{x}-1
\Leftrightarrow5>\sqrt{x}\)
\(\Leftrightarrow25>x\)
b, x=\(\sqrt{4+2.2.\sqrt{3}+3}+\sqrt{4-2.2.\sqrt{3}+3}\)
= \(\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
= \(|2+\sqrt{3}|+|2-\sqrt{3}|\)
= \(2+\sqrt{3}+2-\sqrt{3}=4\)
suy ra P=\(\dfrac{\sqrt{4}-3}{\sqrt{4}-1}=\dfrac{-1}{1}=-1\)
Bài 1:
a: \(A=2x-\left|2x-1\right|=\left[{}\begin{matrix}2x-2x+1=1\left(x>=\dfrac{1}{2}\right)\\2x-1+2x=4x-1\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\)
b: Vì x=-2<1/2 nên \(A=4x-1=-8-1=-9\)
c: A=7 nên |2x-1|=2x-7
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{7}{2}\\\left(2x-7-2x+1\right)\left(2x-7+2x-1\right)=0\end{matrix}\right.\Leftrightarrow x=2\left(loại\right)\)
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