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\(\sqrt{25x}-\sqrt{16x}=9\) khi \(x\) bằng :
(A) 1 (B) 3 (C) 9 (D) 81
\(đkxđ\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne49\end{cases}}\)
\(B=\left(\frac{\sqrt{x}}{x-49}-\frac{\sqrt{x}-7}{x+7\sqrt{x}}\right):\)\(\frac{2\sqrt{x}-7}{x+7\sqrt{x}}+\frac{\sqrt{x}}{7-\sqrt{x}}\)
\(=\left(\frac{\sqrt{x}}{\left(\sqrt{x}-7\right)\left(\sqrt{x}+7\right)}-\frac{\left(\sqrt{x}-7\right)^2}{\sqrt{x}\left(\sqrt{x}+7\right)\left(\sqrt{x}-7\right)}\right)\)\(:\frac{2\sqrt{x}-7}{\sqrt{x}\left(\sqrt{x}+7\right)}-\frac{\sqrt{x}}{\sqrt{x}-7}\)
\(\frac{x-x+14\sqrt{x}-49}{\sqrt{x}\left(\sqrt{x}-7\right)\left(\sqrt{x}+7\right)}:\frac{2\sqrt{x}-7}{\sqrt{x}\left(\sqrt{x}+7\right)}\)\(-\frac{\sqrt{x}}{\sqrt{x}-7}\)
\(=\frac{7\left(2\sqrt{x}-7\right)\sqrt{x}\left(\sqrt{x}+7\right)}{\sqrt{x}\left(\sqrt{x}+7\right)\left(\sqrt{x}-7\right)\left(2\sqrt{x}-7\right)}\)\(-\frac{\sqrt{x}}{\sqrt{x}-7}\)
\(=\frac{7}{\sqrt{x}-7}-\frac{\sqrt{x}}{\sqrt{x}-7}=\frac{7-\sqrt{x}}{\sqrt{x}-7}=-1\)
Trong tam giác vuông ABC (\(\widehat{C}=90^o\)), ta có:
sinA=BC/AB=2/3⇒AB=3/2 BC
Áp dụng định lí Py-ta-go trong tam giác vuông ABC, ta có:
\(AC=\sqrt{AB^2-BC^2}=\sqrt{\left(\dfrac{3}{2}BC\right)^2-BC^2}=\dfrac{BC\sqrt{5}}{2}\)
Ta có:
\(\tan B=\dfrac{AC}{BC}=\dfrac{\dfrac{BC\sqrt{5}}{2}}{BC}=\dfrac{\sqrt{5}}{2}\)
Chọn đáp án D
a)\(\dfrac{2-\sqrt{2}}{\sqrt{2}}\)
\(\Leftrightarrow\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}}\)
\(\Leftrightarrow\sqrt{2}-1\)
b)\(\sqrt{\dfrac{x-49}{\sqrt{x}-7}}\)
\(\Leftrightarrow\sqrt{\dfrac{\left(\sqrt{x}-7\right)\left(\sqrt{x}+7\right)}{\sqrt{x}-7}}\)
\(\Leftrightarrow\sqrt{\sqrt{x}+7}\)
c)\(\sqrt{7-2\sqrt{6}}\)
\(\Leftrightarrow\sqrt{6-2\sqrt{6}+1}\)
\(\Leftrightarrow\sqrt{\left(\sqrt{6}-1\right)^2}\)
\(\Leftrightarrow\sqrt{6}-1\)
d)\(\sqrt{4+2\sqrt{3}}\)
\(\Leftrightarrow\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(\Leftrightarrow\sqrt{3}+1\)
e)\(\sqrt{13-14\sqrt{3}}\)
Câu này có dư số 1 ở chỗ 14 phải k bn ???
\(a.2\sqrt{x-2}=16\left(ĐK:x\ge2\right)\Leftrightarrow\sqrt{x-2}=8\Leftrightarrow x-2=64\Leftrightarrow x=66\)
\(b.\sqrt{x-1}>3\left(ĐK:x\ge1\right)\Leftrightarrow x-1>9\Leftrightarrow x>10\)
\(c.-5\sqrt{2x+4}\le-10\left(ĐK:x\ge2\right)\\ \Leftrightarrow\sqrt{2x+4}\ge2\\ \Leftrightarrow2x+4\ge4\\ \Leftrightarrow2x\ge0\Leftrightarrow x\ge0\)
\(a.2\sqrt{x-2}=16\left(ĐK:x>2\right)\Leftrightarrow\sqrt{x-2}=8\Leftrightarrow x-2=64\Leftrightarrow x=66\)
b.\(\sqrt{x-1}>3\left(ĐK:x>1\right)\Leftrightarrow x-1>9\Leftrightarrow x>10\)
\(c.-5\sqrt{2x+4}< -10\left(ĐK:x>-2\right)\\ \Leftrightarrow\sqrt{2x+4}>2\\ \Leftrightarrow2x+4>4\\ \Leftrightarrow2x>0\Leftrightarrow x>0\)
Hướng dẫn trả lời:
Ta có: √2+√x=32+x=3 . Vì hai vế đều dương, ta bình phương hai vế
(√2+√x)2=32⇔2+√x=9⇔√x=7⇔(√x)2=72⇔x=49(2+x)2=32⇔2+x=9⇔x=7⇔(x)2=72⇔x=49
Chọn đáp án D
(A) 1
(B) \(\sqrt{7}\)
(C) 7
(D) 49