Toán Giải giúp tôi bài tập lớp 9 môn toán đại bài 33 24/09/2021 By Allison Giải giúp tôi bài tập lớp 9 môn toán đại bài 33
a) \(\sqrt{2}.x – \sqrt{50} = 0\) \(\Leftrightarrow \sqrt{2}x=\sqrt{50}\) \(\Leftrightarrow x=\dfrac{\sqrt{50}}{\sqrt{2}}\) \(\Leftrightarrow x =\sqrt{\dfrac{50}{2}}\) \(\Leftrightarrow x= \sqrt{25}\) \(\Leftrightarrow x= \sqrt{5^2}\) \(\Leftrightarrow x=5\). Vậy \(x=5\). b) \(\sqrt{3}.x + \sqrt{3} = \sqrt{12} + \sqrt{27}\) \( \Leftrightarrow \sqrt{3}.x = \sqrt{12} + \sqrt{27} – \sqrt{3}\) \(\Leftrightarrow \sqrt{3}.x=\sqrt{4.3}+\sqrt{9.3}- \sqrt{3}\) \(\Leftrightarrow \sqrt{3}.x=\sqrt{4}. \sqrt{3}+\sqrt{9}. \sqrt{3}- \sqrt{3}\) \(\Leftrightarrow \sqrt{3}.x=\sqrt{2^2}. \sqrt{3}+\sqrt{3^2}. \sqrt{3}- \sqrt{3}\) \(\Leftrightarrow \sqrt{3}.x=2 \sqrt{3}+3\sqrt{3}- \sqrt{3}\) \(\Leftrightarrow \sqrt{3}.x=(2+3-1).\sqrt{3}\) \(\Leftrightarrow \sqrt{3}.x=4\sqrt{3}\) \(\Leftrightarrow x=4\). Vậy \(x=4\). Hai câu còn lại em làm tương tự nhé. Trả lời
a) 2√.x−50−−√=0 ⇔2√.x=50−−√⇔x=50−−√2√=502−−−√=25−−√=5 Vậy S={5} b) 3√.x+3√=12−−√+27−−√ ⇔3√.x+3√=4√.3√+9√.3√⇔3√.x=23√+33√−3√⇔3√.x=43√⇔x=4 Vậy S={4} c) 3√.x2−12−−√=0 ⇔x2=12−−√3√=123−−−√=4√=2⇔x=±2√ Vậy S={±2√} d) x25√−20−−√=0 ⇔x2=20−−√.5√=20.5−−−√=100−−−√=10⇔x=±10−−√ Vậy S={±10} Trả lời
a)
\(\sqrt{2}.x – \sqrt{50} = 0\)
\(\Leftrightarrow \sqrt{2}x=\sqrt{50}\)
\(\Leftrightarrow x=\dfrac{\sqrt{50}}{\sqrt{2}}\)
\(\Leftrightarrow x =\sqrt{\dfrac{50}{2}}\)
\(\Leftrightarrow x= \sqrt{25}\)
\(\Leftrightarrow x= \sqrt{5^2}\)
\(\Leftrightarrow x=5\).
Vậy \(x=5\).
b)
\(\sqrt{3}.x + \sqrt{3} = \sqrt{12} + \sqrt{27}\)
\( \Leftrightarrow \sqrt{3}.x = \sqrt{12} + \sqrt{27} – \sqrt{3}\)
\(\Leftrightarrow \sqrt{3}.x=\sqrt{4.3}+\sqrt{9.3}- \sqrt{3}\)
\(\Leftrightarrow \sqrt{3}.x=\sqrt{4}. \sqrt{3}+\sqrt{9}. \sqrt{3}- \sqrt{3}\)
\(\Leftrightarrow \sqrt{3}.x=\sqrt{2^2}. \sqrt{3}+\sqrt{3^2}. \sqrt{3}- \sqrt{3}\)
\(\Leftrightarrow \sqrt{3}.x=2 \sqrt{3}+3\sqrt{3}- \sqrt{3}\)
\(\Leftrightarrow \sqrt{3}.x=(2+3-1).\sqrt{3}\)
\(\Leftrightarrow \sqrt{3}.x=4\sqrt{3}\)
\(\Leftrightarrow x=4\).
Vậy \(x=4\).
Hai câu còn lại em làm tương tự nhé.
a) 2√.x−50−−√=0
⇔2√.x=50−−√⇔x=50−−√2√=502−−−√=25−−√=5
Vậy S={5}
b) 3√.x+3√=12−−√+27−−√
⇔3√.x+3√=4√.3√+9√.3√⇔3√.x=23√+33√−3√⇔3√.x=43√⇔x=4
Vậy S={4}
c) 3√.x2−12−−√=0
⇔x2=12−−√3√=123−−−√=4√=2⇔x=±2√
Vậy S={±2√}
d) x25√−20−−√=0
⇔x2=20−−√.5√=20.5−−−√=100−−−√=10⇔x=±10−−√
Vậy S={±10}