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Lời giải:
ĐK để tồn tại các biểu thức là $x\geq 0$
a) Ta thấy: $\sqrt{x}\geq 0\Rightarrow \sqrt{x}+5\geq 5$
$\Rightarrow A=\frac{2}{\sqrt{x}+5}\leq \frac{2}{5}$
Vậy $A_{\max}=\frac{2}{5}$ khi $x=0$
b) $\sqrt{x}+7\geq 7$
$\Rightarrow \frac{1}{\sqrt{x}+7}\leq \frac{1}{7}$
$\Rightarrow B=\frac{-3}{\sqrt{x}+7}\geq \frac{-3}{7}$
Vậy $B_{\min}=\frac{-3}{7}$ khi $x=0$
c)
$2\sqrt{x}+1\geq 1\Rightarrow C=\frac{5}{2\sqrt{x}+1}\leq 5$
Vậy $C_{\max}=5$ khi $x=0$
d)
$3\sqrt{x}+2\geq 2\Rightarrow \frac{1}{3\sqrt{x}+2}\leq \frac{1}{2}$
$\Rightarrow D=\frac{-7}{3\sqrt{x}+2}\geq \frac{-7}{2}$
Vậy $B_{\min}=\frac{-7}{2}$ khi $x=0$
1) \(A=\sqrt{17-12\sqrt{2}}=\sqrt{\left(2\sqrt{2}-3\right)^2}=3-2\sqrt{2}\)
\(B=\sqrt{4-2\sqrt{3}}+\sqrt{7-4\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\sqrt{3}-1+2-\sqrt{3}=1\)
\(C=\sqrt{63}-\sqrt{28}-\sqrt{7}=3\sqrt{7}-2\sqrt{7}-\sqrt{7}=0\)
\(D=\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}=\frac{2\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)}{3-1}=\frac{4}{2}=2\)
\(M=\left(\frac{1}{3-\sqrt{5}}-\frac{1}{3+\sqrt{5}}\right):\frac{5-\sqrt{5}}{\sqrt{5}-1}=\frac{3+\sqrt{5}-3+\sqrt{5}}{9-5}.\frac{\sqrt{5}-1}{\sqrt{5}\left(\sqrt{5}-1\right)}=\frac{2}{4}=\frac{1}{2}\)
a, ĐKXĐ: \(x\ge0;x\ne9\)
\(A=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}-3-\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x-\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+2}{\sqrt{x}+3}\)
b, \(x=5+2\sqrt{6}=2+3+2\sqrt{3}.\sqrt{2}=\left(\sqrt{3}+\sqrt{2}\right)^2\)
\(\Rightarrow\sqrt{x}=\sqrt{3}+\sqrt{2}\)
\(\Rightarrow A=\frac{\sqrt{x}+2}{\sqrt{x}+3}=\frac{\sqrt{3}+\sqrt{2}+2}{\sqrt{3}+\sqrt{2}+3}\)
c, \(A=\frac{\sqrt{x}+2}{\sqrt{x}+3}=\frac{3}{5}\Leftrightarrow5\sqrt{x}+10=3\sqrt{x}+9\)
\(\Leftrightarrow2\sqrt{x}=-1\Rightarrow\) không tồn tại giá trị \(x\) thỏa mãn
d, \(A=\frac{\sqrt{x}+2}{\sqrt{x}+3}\Leftrightarrow\sqrt{x}.A+3A=\sqrt{x}+2\)
\(\Leftrightarrow\sqrt{x}\left(A-1\right)=2-3A\)
\(\Leftrightarrow\frac{2-3A}{A-1}=\sqrt{x}\ge0\Rightarrow\frac{2-3A}{A-1}\ge0\)
Do \(A=\frac{\sqrt{x}+2}{\sqrt{x}+3}< 1\Rightarrow A-1< 0\) nên \(2-3A\le0\Leftrightarrow A\ge\frac{2}{3}\)
\(\Rightarrow MinA=\frac{2}{3}\Leftrightarrow\frac{\sqrt{x}+2}{\sqrt{x}+3}=\frac{2}{3}\Leftrightarrow x=0\)
a/ ĐKXĐ:...
\(E=\left(\frac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2+4\sqrt{x}\left(x-1\right)}{x-1}\right):\left(\frac{x-1}{\sqrt{x}}\right)\)
\(E=\left(\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1+4x\sqrt{x}-4\sqrt{x}}{x-1}\right).\frac{\sqrt{x}}{x-1}\)
\(E=\frac{4x^2}{\left(x-1\right)^2}\)
Bn ơi! Kia là chia \(\sqrt{x}-\frac{1}{\sqrt{x}}\) hay nhân z? Bn xem lại đề bài nhé! Theo mk là nhân thì nó sẽ ra kết quả ngắn gọn hơn nhìu :D
Bài 1:
a/ ĐKXĐ: \(x\ge2;x\ne11\)
b/ \(P=\frac{\left(x-5\right)\left(\sqrt{x-2}+\sqrt{3}\right)}{x-2-3}=\sqrt{x-2}+\sqrt{3}\)
c/ \(\sqrt{x-2}\ge0\forall x\in R\Rightarrow P=\sqrt{x-2}+\sqrt{3}\ge\sqrt{3}\forall x\in R\)
"="\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
a) \(A=\sqrt{81}.\sqrt{\frac{9}{4}}+2\sqrt{16}-3=\sqrt{9^2}.\sqrt{\left(\frac{3}{2}\right)^2}+2\sqrt{4^2}-3=9.\frac{3}{2}+2.4-3=\frac{37}{2}\)
b) \(B=\sqrt{9-2\sqrt{14}}=\sqrt{\left(\sqrt{7}-\sqrt{2}\right)^2}=\sqrt{7}-\sqrt{2}\)
c) Không rút gọn được.
Bài 2 : Mình hướng dẫn thôi nhé ^^
a) \(M=x^2-10x+30=\left(x^2-10x+25\right)+5=\left(x-5\right)^2+5\ge5\)
b) \(N=4x^2-12x+1=\left[\left(2x\right)^2-12x+9\right]-8=\left(2x-3\right)^2-8\ge-8\)
c) \(P=x^2-x-1=\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}\right)-\frac{1}{4}-1=\left(x-\frac{1}{2}\right)^2-\frac{5}{4}\ge-\frac{5}{4}\)
d) \(Q=16x^2-8x+3=\left[\left(4x\right)^2-8x+1\right]+2=\left(4x-1\right)^2+2\ge2\)
e) \(H=\frac{1}{9}x^2+3x-1=\left[\left(\frac{x}{3}\right)^2+2.\frac{x}{3}.\frac{9}{2}+\frac{81}{4}\right]-\frac{81}{4}-1=\left(\frac{x}{3}+\frac{9}{2}\right)^2-\frac{85}{4}\ge-\frac{85}{4}\)
Giải :
a) Điều kiện : x\(\ne\)5 ; x\(\ge\)2.
A = \(\frac{x-5}{\sqrt{x-2}-\sqrt{3}}\) = \(\frac{\left(x-5\right)\left(\sqrt{x-2}+\sqrt{3}\right)}{x-5}\) = \(\sqrt{x-2}+\sqrt{3}\) \(\ge\) 0 + \(\sqrt{3}\) = \(\sqrt{3}\) (vì \(\sqrt{x-2}\)\(\ge\)0).
Vậy GTNN của A là \(\sqrt{3}\) khi x = 2.
Tích rồi mình làm câu b) cho