ai giair giup với
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Vậy : như thế nào là số có UCLN=1, mk chưa hiểu lắm ???
Bài 1:
a) Do \(a>b\) ⇒ \(6a>6b\)
b) Do \(a>b\) ⇒ \(-5a< -5b\)
c) Do \(a>b\) ⇒ \(a+1>b+1\) Và \(3a>3b\)
\(\Rightarrow3a+1>3b+1\)
d) Do \(a>b\) ⇒ \(a-3>b-3\)
\(\Rightarrow5\left(a-3\right)>5\left(b-3\right)\)
e) Do \(a>b\) ⇒ \(-7a< -7b\)
\(\Rightarrow4-7a< 4-7b\)
g) Do \(a>b\) ⇒ \(-2a< -2b\)
\(\Rightarrow-6-2a< -6-2b\)
h) Do \(a>b\) ⇒ \(a+3>b+3\) và \(a+5>b+5\)
⇒ \(a+5>b+3\left(5>3\right)\)
Ban chia từng câu ra từng lần đăng để mọi người dễ trả lời
\(6,\\ a,P=9\left(x^2-2\cdot\dfrac{1}{9}x+\dfrac{1}{81}\right)+\dfrac{26}{9}=9\left(x-\dfrac{1}{9}\right)^2+\dfrac{26}{9}\ge\dfrac{26}{9}\\ P_{min}=\dfrac{26}{9}\Leftrightarrow x-\dfrac{1}{9}=0\Leftrightarrow x=\dfrac{1}{9}\\ b,Q=3\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}=3\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\\ Q_{min}=\dfrac{1}{4}\Leftrightarrow x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\\ c,R=\left(x^2-2xy+y^2\right)+x^2+1=\left(x-y\right)^2+x^2+1\ge1\\ R_{min}=1\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\x=0\end{matrix}\right.\Leftrightarrow x=y=0\)
1) Ta có: \(x-2\sqrt{x}=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
2) Ta có: \(6+\sqrt{x}-x=0\)
\(\Leftrightarrow x-\sqrt{x}-6=0\)
\(\Leftrightarrow\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)=0\)
\(\Leftrightarrow\sqrt{x}-3=0\)
\(\Leftrightarrow x=9\)
3) Ta có: \(x+3\sqrt{x}-4=0\)
\(\Leftrightarrow\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\sqrt{x}-1=0\)
\(\Leftrightarrow x=1\)
1. Ha went to doctor's yesterday because he was sick
2. He ate some spinach last night
3. The doctor said that the vegetable could be dirty so he must wash carefully
4. The doctor gave him some medicine
5. He felt better after take medicine
`Answer:`
Bài 1:
a. \(\frac{1}{12}+\frac{3}{4}-\frac{5}{8}\)
\(=\frac{2}{24}+\frac{18}{24}-\frac{15}{24}\)
\(=\frac{5}{24}\)
c. \(-\frac{1}{2}+\frac{3}{7}-\frac{1}{9}+-\frac{7}{18}+\frac{4}{7}\)
\(=\left(-\frac{1}{2}-\frac{1}{9}-\frac{7}{18}\right)+\left(\frac{3}{7}+\frac{4}{7}\right)\)
\(=\left(-\frac{9}{18}-\frac{2}{18}-\frac{7}{18}\right)+\frac{7}{7}\)
\(=-\frac{18}{18}+\frac{7}{7}\)
\(=0\)
Bài 2:
\(x-\frac{3}{-5}=\frac{1}{15}\)
\(\Rightarrow x=\frac{1}{15}+\frac{3}{-5}\)
\(\Rightarrow x=\frac{1}{15}-\frac{3}{5}\)
\(\Rightarrow x=-\frac{8}{15}\)