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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 :
\(a,=2\sqrt{5}+6\sqrt{5}+5\sqrt{5}-12\sqrt{5}\)
\(=\sqrt{5}\left(2+6+5-12\right)=\sqrt{2}\)
\(b,=\sqrt{5}+\sqrt{5}+\left|\sqrt{5}-2\right|\)
\(=2\sqrt{5}+\sqrt{5}-2=3\sqrt{5}-2\)
\(c,=\dfrac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\)
\(=\dfrac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\)
\(=3\)
a, để ý a có nghĩa thì 2x+1 \(\ge\)0 vì (\(x^2\) + 1\(\ge\)1, \(\forall\) x)\(\Rightarrow\)
\(\Rightarrow\) \(x\text{}\text{}\ge\)\(\frac{-1}{2}\)
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 \(\Leftrightarrow2x+1< 0\Leftrightarrow x< -\dfrac{1}{2}\)
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: \(\dfrac{\sqrt[3]{625}}{\sqrt[3]{5}}-\sqrt[3]{-216}\cdot\sqrt[3]{\dfrac{1}{27}}\)
\(=5-\left(-6\right)\cdot\dfrac{1}{3}\)
\(=5+6\cdot\dfrac{1}{3}=5+2=7\)
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) \(\sqrt[3]{125}.\sqrt[3]{-216}-\sqrt[3]{512}.\sqrt[3]{\dfrac{1}{8}}=\sqrt[3]{5^3}.\sqrt[3]{\left(-6\right)^3}-\sqrt[3]{8^3}.\sqrt[3]{\left(\dfrac{1}{2}\right)^3}\)
\(=5.\left(-6\right)-8.\dfrac{1}{2}=-34\)
\(\dfrac{\sqrt{ab}-b}{b}-\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{b}\right)^2}-\dfrac{\sqrt{a}}{\sqrt{b}}=\dfrac{\sqrt{a}-\sqrt{b}}{\sqrt{b}}-\dfrac{\sqrt{a}}{\sqrt{b}}\)
\(=-\dfrac{\sqrt{b}}{\sqrt{b}}=-1< 0\)
a, \(\left\{{}\begin{matrix}2x-1\ne0\\\frac{x^2}{2x-1}\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{1}{2}\\2x-1>0\end{matrix}\right.\Leftrightarrow x>\frac{1}{2}\)
b, \(\frac{\sqrt[3]{625}}{\sqrt[3]{5}}-\sqrt[3]{-216}.\sqrt[3]{\frac{1}{27}}=\frac{\sqrt[3]{5^3.5}}{\sqrt[3]{5}}-\sqrt[3]{\left(-6\right)^3}.\sqrt[3]{\left(\frac{1}{3}\right)^3}\)
\(=\frac{5\sqrt[3]{5}}{\sqrt[3]{5}}+6.\frac{1}{3}=5+2=7\)
a/
\(\left(\dfrac{2}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\\ =\dfrac{2\left(\sqrt{x}+3\right)+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\dfrac{2\sqrt{x}+6+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\\ =\dfrac{3\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\\ =\dfrac{\left(3\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)^2\left(\sqrt{x}+3\right)}\\ =\dfrac{3\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-3\right)^2\left(\sqrt{x}+3\right)}\\ =\dfrac{3\left(x+2\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)^2\left(\sqrt{x}+3\right)}\\ =\dfrac{3x+6\sqrt{x}+3}{\left(\sqrt{x}-3\right)^2\left(\sqrt{x}+3\right)}\)
2, \(\dfrac{1}{2}\sqrt{x+1}-\dfrac{3}{2}\sqrt{9x+9}+24\sqrt{\dfrac{x+1}{64}}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x+1}-\dfrac{9}{2}\sqrt{x+1}+3\sqrt{x+1}=-17\)
\(\Leftrightarrow-\sqrt{x+1}=-17\)
\(\Leftrightarrow x+1=289\left(x>0\right)\)
\(\Leftrightarrow x=288\)
Vậy x = 288
3, \(-5x+7\sqrt{x}+12=0\)
\(\Leftrightarrow-5x+12\sqrt{x}-5\sqrt{x}+12=0\)
\(\Leftrightarrow\sqrt{x}\left(12-5\sqrt{x}\right)+\left(12-5\sqrt{x}\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(12-5\sqrt{x}\right)=0\)
Do \(\sqrt{x}+1>0\)
\(\Rightarrow12-5\sqrt{x}=0\Leftrightarrow x=\dfrac{144}{25}\)
Vậy...
1. (Đề có chút sai sai nên mình sửa lại nhé) \(\sqrt{36x-36}-\sqrt{9x-9}-\sqrt{4x-4}=16-\sqrt{x-1}\)
(ĐK: \(x\ge1\))
\(\Leftrightarrow\sqrt{36\left(x-1\right)}-\sqrt{9\left(x-1\right)}-\sqrt{4\left(x-1\right)}=16-\sqrt{x-1}\)
\(\Leftrightarrow6\sqrt{x-1}-3\sqrt{x-1}-2\sqrt{x-1}+\sqrt{x-1}=16\)
\(\Leftrightarrow2\sqrt{x-1}=16\)
\(\Leftrightarrow\sqrt{x-1}=8\)
\(\Leftrightarrow x-1=64\)
\(\Leftrightarrow x=65\left(tm\right)\)
Vậy pt đã cho có nghiệm x=65.
2. \(\dfrac{1}{2}\sqrt{x+1}-\dfrac{3}{2}\sqrt{9x+9}+24\sqrt{\dfrac{x+1}{64}}=-17\)
(ĐK: \(x\ge-1\))
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x+1}-\dfrac{3}{2}\sqrt{9\left(x+1\right)}+3\sqrt{x+1}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x+1}-\dfrac{9}{2}\sqrt{x+1}+3\sqrt{x+1}=-17\)
\(\Leftrightarrow-\sqrt{x+1}=-17\)
\(\Leftrightarrow\sqrt{x+1}=17\)
\(\Leftrightarrow x+1=289\)
\(\Leftrightarrow x=288\left(tm\right)\)
Vậy \(S=\left\{288\right\}\)
3. \(-5x+7\sqrt{x}+12=0\) (ĐK: \(x\ge0\))
\(\Leftrightarrow5x-7\sqrt{x}-12=0\)
\(\Leftrightarrow5x+5\sqrt{x}-12\sqrt{x}-12=0\)
\(\Leftrightarrow5\sqrt{x}\left(\sqrt{x}+1\right)-12\left(\sqrt{x}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(5\sqrt{x}-12\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+1=0\\5\sqrt{x}-12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=-1\left(vô.lý\right)\\5\sqrt{x}=12\end{matrix}\right.\Leftrightarrow\sqrt{x}=\dfrac{12}{5}\Leftrightarrow x=\dfrac{144}{25}\left(tm\right)\)
Vậy pt có nghiệm \(x=\dfrac{144}{25}\)
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