Bài 2:
\(\sqrt{2x-1}=5\)
=> 2x - 1 = 25
=> 2x = 26
=> x = 13
b) \(\sqrt[3]{3x+2}=-3\)
=> 3x + 2 = -27
=> 3x = -29
=> x = -29/3
P/s: Mỗi lần chỉ đc hỏi 1 bài thôi em nehs!
1. ĐK:
a, \(x\ge\dfrac{5}{2}\)
b, \(\left\{{}\begin{matrix}x\ge0\\x\ne\dfrac{9}{4}\end{matrix}\right.\)
2.
a,ĐK: \(x\ge\dfrac{1}{2}\)
PT\(\Leftrightarrow2x-1=25\Rightarrow x=13\left(tm\right)\)
b,ĐK: \(\forall x\in R\)
PT\(\Leftrightarrow3x+2=-27=>x=-\dfrac{29}{3}\) (tm)
3.
a,\(\sqrt{5}.\sqrt{1,2}.\sqrt{24}=\sqrt{120}.\sqrt{1,2}=12\)
b,\(\dfrac{\sqrt{4444}}{\sqrt{1111}}=\dfrac{\sqrt{4}.\sqrt{1111}}{\sqrt{1111}}=2\)
c,\(\sqrt{\dfrac{3}{5}}+\sqrt{\dfrac{5}{3}}-\dfrac{1}{2}\sqrt{60}=\dfrac{8}{\sqrt{15}}-\sqrt{15}=-\dfrac{7}{\sqrt{15}}\)
d,\(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}}=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\)
Bài 5:
a) Ta có: \(P=\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{x-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
b) Thay \(x=33-8\sqrt{2}\) vào P, ta được:
\(P=\dfrac{4\sqrt{2}-1}{33-8\sqrt{2}+4\sqrt{2}-1+1}\)
\(=\dfrac{\left(4\sqrt{2}-1\right)\left(33+4\sqrt{2}\right)}{1057}=\dfrac{132\sqrt{2}+32-33-4\sqrt{2}}{1057}\)
\(=\dfrac{128\sqrt{2}-1}{1057}\)
Bài 4
\(x-4\sqrt{x}-1\) \(=\left(x-4\sqrt{x}+4\right)-5=\left(\sqrt{x}-2\right)^2-5\)
Biểu thức đặt min khi \(\left(\sqrt{x}-2\right)^2\text{ m}\text{i}\text{n}\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2=0\Leftrightarrow x=0\)
Vậy y đặt giá trị nhỏ nhất là -5 khi x=0
