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\(\sqrt{16a}+2\sqrt{40a}-3\sqrt{90a}\)
\(=\sqrt{4^2a}+2\sqrt{4.10a}-3\sqrt{9.10a}\)
\(=\sqrt{4^2a}+2\sqrt{2^2.10a}-3\sqrt{3^2.10a}\)
\(=4\sqrt{a}+2.2\sqrt{10a}-3.3\sqrt{10a}\)
\(=4\sqrt{a}+4\sqrt{10a}-9\sqrt{10a}\)
\(=4\sqrt{a}-5\sqrt{10a}\)
.Cảm ơn!!
đk : \(x\ne4\) ; \(x\ge0\)
1) a) Q = \(\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}+\dfrac{2\sqrt{x}}{x-4}\)
Q = \(\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-\dfrac{2\sqrt{x}}{4-x}\)
Q = \(\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-\dfrac{2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
Q = \(\dfrac{2\left(2-\sqrt{x}\right)+2+\sqrt{x}-2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
Q = \(\dfrac{4-2\sqrt{x}+2+\sqrt{x}-2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
Q = \(\dfrac{6-3\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\) = \(\dfrac{3\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
Q = \(\dfrac{3}{2+\sqrt{x}}\)
b) ta có Q = \(\dfrac{6}{5}\) \(\Leftrightarrow\) \(\dfrac{3}{2+\sqrt{x}}\) = \(\dfrac{6}{5}\) \(\Leftrightarrow\) \(\dfrac{6}{4+2\sqrt{x}}\) = \(\dfrac{6}{5}\)
\(\Leftrightarrow\) \(4+2\sqrt{x}=5\) \(\Leftrightarrow\) \(2\sqrt{x}=1\) \(\Leftrightarrow\) \(\sqrt{x}=\dfrac{1}{2}\) \(\Leftrightarrow\) \(x=\dfrac{1}{4}\)
c) điều x nguyên ; x \(\ge\) 0 ; x\(\ne\) 4
ta có Q nguyên \(\Leftrightarrow\) \(\dfrac{3}{2+\sqrt{x}}\) nguyên
\(\Rightarrow\) \(2+\sqrt{x}\) là ước của 3 là 3 ; 1 ; -1 ; -3
mà \(2+\sqrt{x}\ge2\) (đk :\(x\ge0\)) vậy còn lại 3
\(\Leftrightarrow\) \(2+\sqrt{x}=3\) \(\Leftrightarrow\) x = 1 (tmđk)
vậy x = 1 nguyên thì Q nguyên
2) a) \(\sqrt{16a}+2\sqrt{40a}-3\sqrt{90a}\) = \(4\sqrt{a}+4\sqrt{10a}-9\sqrt{10a}\)
= \(4\sqrt{a}-5\sqrt{10a}\)
b) \(\left(2\sqrt{3}+5\right)\sqrt{3}-\sqrt{60}\) = \(6+5\sqrt{3}-\sqrt{60}\)
c) \(\left(\sqrt{99}-\sqrt{8}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)
= \(33-2\sqrt{22}-11+3\sqrt{22}\)
= \(22+\sqrt{22}\)
a) \(\left(2\sqrt{3}+\sqrt{5}\right)\sqrt{3}-\sqrt{60}\) = \(6+\sqrt{15}-2\sqrt{15}\)
= \(6-\sqrt{15}\)
b) \(\left(5\sqrt{2}+2\sqrt{5}\right)\sqrt{5}-\sqrt{250}\) = \(5\sqrt{10}+10-5\sqrt{10}\) = \(10\)
c) \(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\) = \(14-2\sqrt{21}-7+2\sqrt{21}\)
= \(7\)
d) \(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)
= \(33-3\sqrt{22}-11+3\sqrt{22}\) = \(22\)
a)(2√3+√5)√3-√60
=6+√15-2√15
=6-√15
b)(5√2+2√5)√5-√250
=5√10+10-5√10
=10
c)(√28-√12-√7)√7+2√21
=14-2√21-7+2√21
=7
d)(√99-√18-√11)√11+3√22
=33-3√22-11+3√22
=22
a: \(=6-\sqrt{15}+2\sqrt{15}=6+\sqrt{15}\)
b: \(=\left(\sqrt{7}-2\sqrt{3}\right)\cdot\sqrt{7}+2\sqrt{21}\)
\(=7-2\sqrt{21}+2\sqrt{21}=7\)
c: \(=10+5\sqrt{10}-5\sqrt{10}=10\)
d: \(=22-\sqrt{198}+\sqrt{198}=22\)
a) \(\sqrt{98}-\sqrt{72}+0,5\sqrt{8}=7\sqrt{2}-6\sqrt{2}+\sqrt{2}=2\sqrt{2}\)
b) \(\left(2\sqrt{3}+\sqrt{5}\right).\sqrt{3}-\sqrt{60}=6+\sqrt{15}-2\sqrt{15}=6-\sqrt{15}\)
c) \(\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)=\left(\sqrt{x}\right)^3+8\)
d) \(\left(\sqrt{x}-\sqrt{y}\right)\left(x+y+\sqrt{xy}\right)=\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3\)
a: \(=4a-4\sqrt{10a}-9\sqrt{10a}=4a-13\sqrt{10a}\)
b: \(=\sqrt{x}\left(4-\sqrt{2}\right)\cdot\sqrt{x}\left(1-\sqrt{2}\right)\)
\(=x\cdot\left(4-4\sqrt{2}-\sqrt{2}+2\right)\)
\(=\left(6-5\sqrt{2}\right)x\)
c: \(=\dfrac{2}{2x-1}\cdot x\sqrt{5}\cdot\left(2x-1\right)=2x\sqrt{5}\)
Lời giải:
a)
$\sqrt{98}-\sqrt{72}+0.5\sqrt{8}=7\sqrt{2}-6\sqrt{2}+0,5.2\sqrt{2}$
$=7\sqrt{2}-6\sqrt{2}+\sqrt{2}=2\sqrt{2}$
b)
$\sqrt{16a}+2\sqrt{40a}-3\sqrt{90a}$
$=4\sqrt{a}+4\sqrt{10}.\sqrt{a}-9\sqrt{10}.\sqrt{a}$
$=(4+4\sqrt{10}-9\sqrt{10})\sqrt{a}=(4-5\sqrt{10}).\sqrt{a}$
c)
$(2\sqrt{3}+\sqrt{5})\sqrt{3}-\sqrt{60}=2.3+\sqrt{15}-2\sqrt{15}$
$=6-\sqrt{15}$
d)
$(\sqrt{99}-\sqrt{18}-\sqrt{11})\sqrt{11}+3\sqrt{32}$
$=\sqrt{99}.\sqrt{11}-\sqrt{18}.\sqrt{11}-11+3\sqrt{32}$
$=\sqrt{9}.\sqrt{11}.\sqrt{11}-3\sqrt{2}.\sqrt{11}-11+12\sqrt{2}$
$=3.11+\sqrt{2}(12-3\sqrt{11})-11$
$=22+\sqrt{2}(12-3\sqrt{11})$