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a)đk:`2x-4>=0`
`<=>2x>=4`
`<=>x>=2.`
b)đk:`3/(-2x+1)>=0`
Mà `3>0`
`=>-2x+1>=0`
`<=>1>=2x`
`<=>x<=1/2`
c)`đk:(-3x+5)/(-4)>=0`
`<=>(3x-5)/4>=0`
`<=>3x-5>=0`
`<=>3x>=5`
`<=>x>=5/3`
d)`đk:-5(-2x+6)>=0`
`<=>-2x+6<=0`
`<=>2x-6>=0`
`<=>2x>=6`
`<=>x>=3`
e)`đk:(x^2+2)(x-3)>=0`
Mà `x^2+2>=2>0`
`<=>x-3>=0`
`<=>x>=3`
f)`đk:(x^2+5)/(-x+2)>=0`
Mà `x^2+5>=5>0`
`<=>-x+2>0`
`<=>-x>=-2`
`<=>x<=2`
a, ĐKXĐ : \(2x-4\ge0\)
\(\Leftrightarrow x\ge\dfrac{4}{2}=2\)
Vậy ..
b, ĐKXĐ : \(\left\{{}\begin{matrix}\dfrac{3}{-2x+1}\ge0\\-2x+1\ne0\end{matrix}\right.\)
\(\Leftrightarrow-2x+1>0\)
\(\Leftrightarrow x< \dfrac{1}{2}\)
Vậy ..
c, ĐKXĐ : \(\dfrac{-3x+5}{-4}\ge0\)
\(\Leftrightarrow-3x+5\le0\)
\(\Leftrightarrow x\ge\dfrac{5}{3}\)
Vậy ...
d, ĐKXĐ : \(-5\left(-2x+6\right)\ge0\)
\(\Leftrightarrow-2x+6\le0\)
\(\Leftrightarrow x\ge-\dfrac{6}{-2}=3\)
Vậy ...
e, ĐKXĐ : \(\left(x^2+2\right)\left(x-3\right)\ge0\)
\(\Leftrightarrow x-3\ge0\)
\(\Leftrightarrow x\ge3\)
Vậy ...
f, ĐKXĐ : \(\left\{{}\begin{matrix}\dfrac{x^2+5}{-x+2}\ge0\\-x+2\ne0\end{matrix}\right.\)
\(\Leftrightarrow-x+2>0\)
\(\Leftrightarrow x< 2\)
Vậy ...
1) Ta có: \(3\sqrt{12}+\dfrac{1}{2}\sqrt{48}-\sqrt{27}\)
\(=3\cdot2\sqrt{3}+\dfrac{1}{2}\cdot4\sqrt{3}-3\sqrt{3}\)
\(=6\sqrt{3}+2\sqrt{3}-3\sqrt{3}\)
\(=5\sqrt{3}\)
2) Ta có: \(\dfrac{2}{\sqrt{3}-5}\)
\(=\dfrac{2\left(\sqrt{3}+5\right)}{\left(\sqrt{3}-5\right)\left(\sqrt{3}+5\right)}\)
\(=\dfrac{2\left(\sqrt{3}+5\right)}{3-25}\)
\(=\dfrac{-2\left(\sqrt{3}+5\right)}{22}\)
\(=\dfrac{-\sqrt{3}-5}{11}\)
3) Ta có: \(\sqrt{\dfrac{2}{5}}\)
\(=\dfrac{\sqrt{2}}{\sqrt{5}}\)
\(=\dfrac{\sqrt{2}\cdot\sqrt{5}}{5}\)
\(=\dfrac{\sqrt{10}}{5}\)
Nếu em thấy các câu hỏi do lag mà bị gửi đúp (tức là rất nhiều câu hỏi giống nhau xuất hiện cùng 1 chỗ) thì xóa giúp mình nhé cho đỡ vướng. Nhưng nhớ để lại 1 câu. Cảm ơn em.
a) \(\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}\)
\(=\sqrt{\dfrac{\left(3-\sqrt{5}\right)^2}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}}\)
\(=\dfrac{\sqrt{\left(3-\sqrt{5}\right)^2}}{\sqrt{3^2-\left(\sqrt{5}\right)^2}}\)
\(=\dfrac{\left|3-\sqrt{5}\right|}{\sqrt{9-5}}\)
\(=\dfrac{3-\sqrt{5}}{2}\)
b) \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)
\(=\sqrt{\dfrac{\left(2-\sqrt{3}\right)^2}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}\)
\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)^2}}{\sqrt{2^2-\left(\sqrt{3}\right)^2}}\)
\(=\dfrac{\left|2-\sqrt{3}\right|}{\sqrt{4-3}}\)
\(=\dfrac{2-\sqrt{3}}{1}\)
\(=2-\sqrt{3}\)
a: \(=\sqrt{\dfrac{\left(3-\sqrt{5}\right)\left(3-\sqrt{5}\right)}{4}}=\dfrac{3-\sqrt{5}}{2}\)
b: \(=\sqrt{\dfrac{\left(2-\sqrt{3}\right)^2}{1}}=2-\sqrt{3}\)
d: \(=\left(-3+3\sqrt{6}+4+2\sqrt{6}-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
=(căn 6-11)(căn 6+11)
=6-121=-115
bài 1) a) \(xy\sqrt{\dfrac{x}{y}}=x\sqrt{y}\sqrt{y}\dfrac{\sqrt{x}}{\sqrt{y}}=x\sqrt{x}\sqrt{y}=\left(\sqrt{x}\right)^3\sqrt{y}\)
b) \(\sqrt{\dfrac{5a^3}{49b}}=\dfrac{\sqrt{5a^3}}{\sqrt{49b}}=\dfrac{\sqrt{5a^3}}{7\sqrt{b}}=\dfrac{\sqrt{5a^3}.\sqrt{b}}{7\sqrt{b}.\sqrt{b}}=\dfrac{\sqrt{5a^3b}}{7b}\)
bài 2) a) \(\dfrac{\sqrt{3}-3}{1-\sqrt{3}}=\dfrac{\sqrt{3}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}=\sqrt{3}\)
b) \(\dfrac{5-\sqrt{15}}{\sqrt{3}-\sqrt{5}}=\dfrac{-\sqrt{5}\left(\sqrt{3}-\sqrt{5}\right)}{\sqrt{3}-\sqrt{5}}=-\sqrt{5}\)
c) \(\dfrac{2\sqrt{2}+2}{5\sqrt{2}}=\dfrac{\sqrt{2}\left(2+\sqrt{2}\right)}{5\sqrt{2}}=\dfrac{2+\sqrt{2}}{5}\)
b: Ta có: \(B=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\cdot\left(\dfrac{x\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\left(x+\sqrt{x}+1+\sqrt{x}\right)\)
\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\sqrt{x}-1}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}-1}\)
a: Ta có: \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}-11\right)\)
\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}-11\right)\)
\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}-11\right)\)
\(=127-22\sqrt{6}\)
b: Ta có: \(\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
\(=\left(1-\sqrt{5}\right)\left(-1-\sqrt{5}\right)\)
=-1+5
=4
Bài 1:
a: \(\sqrt{27}+\dfrac{1}{2}\sqrt{48}-\sqrt{108}\)
\(=3\sqrt{3}+\dfrac{1}{2}\cdot4\sqrt{3}-6\sqrt{3}\)
\(=-3\sqrt{3}+2\sqrt{3}=-\sqrt{3}\)
b: \(\left(\sqrt{14}-\sqrt{10}\right)\cdot\sqrt{6+\sqrt{35}}\)
\(=\left(\sqrt{7}-\sqrt{5}\right)\cdot\sqrt{2}\cdot\sqrt{6+\sqrt{35}}\)
\(=\left(\sqrt{7}-\sqrt{5}\right)\cdot\sqrt{12+2\sqrt{35}}\)
\(=\left(\sqrt{7}-\sqrt{5}\right)\cdot\sqrt{\left(\sqrt{7}+\sqrt{5}\right)^2}\)
\(=\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)=7-5=2\)
c: \(\dfrac{\sqrt{15}+\sqrt{3}}{1+\sqrt{5}}-\dfrac{2}{\sqrt{3}-1}\)
\(=\dfrac{\sqrt{3}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}-\dfrac{2\left(\sqrt{3}+1\right)}{3-1}\)
\(=\sqrt{3}-\sqrt{3}-1=-1\)
Bài 2:
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >1\end{matrix}\right.\)
\(A=\dfrac{x-5}{x+2\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}+\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x-5}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}+\dfrac{1}{\sqrt{x}+3}+\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x-5+\sqrt{x}-1+2\sqrt{x}+6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
b: A=2
=>\(\sqrt{x}=2\left(\sqrt{x}-1\right)\)
=>\(2\sqrt{x}-2=\sqrt{x}\)
=>\(\sqrt{x}=2\)
=>x=4(nhận)
c: Để A là số nguyên thì \(\sqrt{x}⋮\sqrt{x}-1\)
=>\(\sqrt{x}-1+1⋮\sqrt{x}-1\)
=>\(\sqrt{x}-1\inƯ\left(1\right)\)
=>\(\sqrt{x}-1\in\left\{1;-1\right\}\)
=>\(\sqrt{x}\in\left\{2;0\right\}\)
=>\(x\in\left\{4;0\right\}\)
1)
a) \(\sqrt{2x-4}\) có nghĩa khi:
\(2x-4\ge0\)
\(\Leftrightarrow2x\ge4\)
\(\Leftrightarrow x\ge\dfrac{4}{2}\)
\(\Leftrightarrow x\ge2\)
b) \(\sqrt{\dfrac{-7}{4-x}}\) có nghĩa khi
\(\dfrac{-7}{4-x}\ge0\) mà \(-7< 0\)
\(\Rightarrow4-x\le0\)
\(\Leftrightarrow x\ge4\)
\(1,ĐKXĐ:x\ge0\\ x\sqrt{3}=-\sqrt{3x^2}\\ \Leftrightarrow3x^2=9x^2\\ \Leftrightarrow6x^2=0\\ \Leftrightarrow x=0\left(tm\right)\)
\(2,ab^2\sqrt{a}=ab^2\sqrt{a}\)
\(3,a\sqrt{\dfrac{b}{a}}=\sqrt{ab}\)
Bài 2:
\(\dfrac{2\sqrt{3}-10}{5}\cdot\sqrt{\dfrac{5+\sqrt{3}}{5-\sqrt{3}}}\)
\(=\dfrac{2\sqrt{3}-10}{5}\cdot\sqrt{\dfrac{28+10\sqrt{3}}{22}}\)
\(=\dfrac{2\sqrt{3}-10}{5}\cdot\dfrac{5+\sqrt{3}}{\sqrt{22}}\)
\(=\dfrac{2\left(\sqrt{3}-5\right)\left(\sqrt{3}+5\right)}{5\sqrt{22}}\)
\(=\dfrac{2\cdot\left(3-25\right)}{5\sqrt{22}}=\dfrac{-44}{5\sqrt{22}}=\dfrac{-2\sqrt{22}}{5}\)