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a)ĐK:\(-\dfrac{5}{2x+1}\ge0\) và \(2x+1\ne0\)
\(\Leftrightarrow2x+1>0\) \(\Leftrightarrow x>-\dfrac{1}{2}\)
Vậy \(x< -\dfrac{1}{2}\) thì căn thức có nghĩa
b)\(\sqrt[3]{64}+\sqrt[3]{-27}-\sqrt[3]{-4}.\sqrt[3]{2}=\sqrt[3]{4^3}+\sqrt[3]{-3^3}-\sqrt[3]{-8}\)
\(=4+\left(-3\right)-\left(-2\right)\)
\(=3\)
À không, ý a \(\Leftrightarrow2x+1< 0\Leftrightarrow x< -\dfrac{1}{2}\)
Bài 1 :
a, ĐKXĐ : \(\dfrac{2x+1}{x^2+1}\ge0\)
Mà \(x^2+1\ge1>0\)
\(\Rightarrow2x+1\ge0\)
\(\Rightarrow x\ge-\dfrac{1}{2}\)
Vậy ...
b, Ta có : \(\sqrt[3]{-27}+\sqrt[3]{64}-\sqrt[3]{-\dfrac{128}{2}}\)
\(=-3+4-\left(-4\right)=-3+4+4=5\)
Bài 2 :
\(a,=2\sqrt{5}+6\sqrt{5}+5\sqrt{5}-12\sqrt{5}\)
\(=\sqrt{5}\left(2+6+5-12\right)=\sqrt{2}\)
\(b,=\sqrt{5}+\sqrt{5}+\left|\sqrt{5}-2\right|\)
\(=2\sqrt{5}+\sqrt{5}-2=3\sqrt{5}-2\)
\(c,=\dfrac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\)
\(=\dfrac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\)
\(=3\)
1.a) Để căn thức có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x^2}{2x-1}\ge0\\2x-1\ne0\end{matrix}\right.\)
\(\Leftrightarrow2x-1>0\Leftrightarrow x>\dfrac{1}{2}\)
Vậy...
b, \(\dfrac{\sqrt[3]{625}}{\sqrt[3]{5}}-\sqrt[3]{-216}.\sqrt[3]{\dfrac{1}{27}}=\sqrt[3]{\dfrac{625}{5}}-\sqrt[3]{-\dfrac{216}{27}}=\sqrt[3]{125}-\sqrt[3]{-8}=5-\left(-2\right)=7\)
a) Để căn thức có nghĩa thì 2x-1>0
\(\Leftrightarrow2x>1\)
hay \(x>\dfrac{1}{2}\)
b) Ta có: \(\dfrac{\sqrt[3]{625}}{\sqrt[3]{5}}-\sqrt[3]{-216}\cdot\sqrt[3]{\dfrac{1}{27}}\)
\(=5-\left(-6\right)\cdot\dfrac{1}{3}\)
\(=5+6\cdot\dfrac{1}{3}=5+2=7\)
a) ĐKXĐ: \(\dfrac{2x+1}{x^2+1}\ge0\Leftrightarrow2x+1\ge0\Leftrightarrow x\ge-\dfrac{1}{2}\)
b) \(\sqrt[3]{-27}+\sqrt[3]{64}-\dfrac{\sqrt[3]{-128}}{\sqrt[3]{2}}=-3+4-\sqrt[3]{-64}=1+4=5\)
a: ĐKXĐ: \(x\ge-\dfrac{1}{2}\)
b: Ta có: \(\sqrt[3]{-27}+\sqrt[3]{64}-\dfrac{\sqrt[3]{-128}}{\sqrt[3]{2}}\)
\(=-3+4-\left(-4\right)\)
=-3+4+4
=5
a) Để căn thức bậc 2 có nghĩa \(\Rightarrow3-5x\ge0\Rightarrow x\le\dfrac{3}{5}\)
b) Để căn thức bậc 2 có nghĩa \(\Rightarrow\dfrac{5}{2x+1}\ge0\Rightarrow2x+1>0\Rightarrow x>-\dfrac{1}{2}\)
1) ĐKXĐ: \(\left[{}\begin{matrix}x\ge2\\x\le1\end{matrix}\right.\)
2) ĐKXĐ: \(\dfrac{x-6}{x-2}\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2< 0\\x-6\ge0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< 2\\x\ge6\end{matrix}\right.\)
3) ĐKXĐ: \(\dfrac{2x-4}{5-x}\ge0\)
\(\Leftrightarrow\dfrac{x-2}{x-5}\le0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2\ge0\\x-5< 0\end{matrix}\right.\Leftrightarrow2\le x< 5\)
Bài 1 :
a, ĐKXĐ : \(\dfrac{1}{2-x}\ge0\)
Mà 1 > 0
\(\Rightarrow2-x>0\)
\(\Rightarrow x< 2\)
Vậy ...
b, Ta có : \(\sqrt[3]{125}.\sqrt[3]{216}-\sqrt[3]{512}.\sqrt[3]{\dfrac{1}{8}}\)
\(=5.6-\dfrac{8.1}{2}=26\)
1a) Để căn thức bậc 2 có nghĩa thì \(\dfrac{1}{2-x}\ge0\Rightarrow2-x>0\Rightarrow x< 2\)
b) \(\sqrt[3]{125}.\sqrt[3]{-216}-\sqrt[3]{512}.\sqrt[3]{\dfrac{1}{8}}=\sqrt[3]{5^3}.\sqrt[3]{\left(-6\right)^3}-\sqrt[3]{8^3}.\sqrt[3]{\left(\dfrac{1}{2}\right)^3}\)
\(=5.\left(-6\right)-8.\dfrac{1}{2}=-34\)
\(\dfrac{\sqrt{ab}-b}{b}-\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{b}\right)^2}-\dfrac{\sqrt{a}}{\sqrt{b}}=\dfrac{\sqrt{a}-\sqrt{b}}{\sqrt{b}}-\dfrac{\sqrt{a}}{\sqrt{b}}\)
\(=-\dfrac{\sqrt{b}}{\sqrt{b}}=-1< 0\)
Để căn thức \(\sqrt{\dfrac{2x+1}{x^2+1}}\) có nghĩa thì:
\(\left\{{}\begin{matrix}\dfrac{2x+1}{x^2+1}\ge0\\x^2+1\ne0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x+1\ge0\left(vì.x^2+1>0\forall x\right)\\x^2+1\ne0\forall x\end{matrix}\right.\)
\(\Rightarrow2x\ge-1\Leftrightarrow x\ge-\dfrac{1}{2}\)
#\(Toru\)
\(\sqrt{\dfrac{2x+1}{x^2+1}}\)
Có nghĩa khi:
\(\dfrac{2x+1}{x^2+1}\ge0\)
\(\Leftrightarrow2x+1\ge0\)
\(\Leftrightarrow2x\ge-1\)
\(\Leftrightarrow x\ge-\dfrac{1}{2}\)
Vậy: ...