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\(a.A=\sqrt{75}+\sqrt{48}-\sqrt{300}=\sqrt{25.3}+\sqrt{16.3}-\sqrt{100.3}=5\sqrt{3}+4\sqrt{3}-10\sqrt{3}=-\sqrt{3}\) \(b.B=\sqrt{98}-\sqrt{72}+0,5\sqrt{8}=\sqrt{49.2}-\sqrt{36.2}+0,5\sqrt{4.2}=7\sqrt{2}-6\sqrt{2}+\sqrt{2}=2\sqrt{2}\) \(c.\dfrac{5\sqrt{5}-\sqrt{15.5}+5\sqrt{5}}{\sqrt{5}}=\dfrac{\sqrt{5}\left(10-\sqrt{15}\right)}{\sqrt{5}}=10-\sqrt{15}\)
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)
$(\sqrt{99}-\sqrt{18}-\sqrt{11})\sqrt{11}+3\sqrt{32}$
$=\sqrt{99}.\sqrt{11}-\sqrt{18}.\sqrt{11}-11+3\sqrt{32}$
$=\sqrt{9}.\sqrt{11}.\sqrt{11}-3\sqrt{2}.\sqrt{11}-11+12\sqrt{2}$
$=3.11+\sqrt{2}(12-3\sqrt{11})-11$
$=22+\sqrt{2}(12-3\sqrt{11})$
a: \(=2\sqrt{2}+30\sqrt{2}-3\sqrt{2}+6\sqrt{2}=26\sqrt{2}\)
b: \(=\dfrac{1}{2}\cdot4\sqrt{3}-2\cdot5\sqrt{3}+\sqrt{3}+\dfrac{5}{2}\sqrt{3}=-\dfrac{9}{2}\sqrt{3}\)
a) \(2x-\sqrt{4x^2+4x+1}=2x-\sqrt{\left(2x+1\right)^2}=2x-\left|2x+1\right|\)
Vì \(x< -\frac{1}{2}\)nên \(\left|2x+1\right|=-\left(2x+1\right)\)
\(\Rightarrow2x+2x+1=4x+1\)
b) \(3x+2-\sqrt{9x^2-12x+4}=3x+2-\sqrt{\left(3x-2\right)^2}=3x+2-\left|3x-2\right|\)
Khi \(x\ge\frac{2}{3}\)thì \(\left|3x-2\right|=3x-2\)
\(\Leftrightarrow3x+2-\left|3x-2\right|=3x+2-3x+2=4\)
Khi \(x< \frac{2}{3}\) thì \(\left|3x-2\right|=2-3x\)
\(\Leftrightarrow3x+2-\left|3x-2\right|=3x+2-\left(2-3x\right)=6x\)
c) \(\sqrt{9a}-\sqrt{16a}+\sqrt{49a}=3\sqrt{a}-4\sqrt{a}+7\sqrt{a}\)
Đặt \(\sqrt{a}=x\) ta được : \(3x-4x+7x=6x\)\(=6\sqrt{a}\)( Do \(a\ge0\))
d) \(\sqrt{160a}+2\sqrt{40a}-3\sqrt{90a}=4\sqrt{10a}+4\sqrt{10a}-9\sqrt{10a}\)\(=-\sqrt{10}\)
TK NKA !!!
Bài 6:
a: \(\Leftrightarrow\sqrt{x^2+4}=\sqrt{12}\)
=>x^2+4=12
=>x^2=8
=>\(x=\pm2\sqrt{2}\)
b: \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>x+1=1
=>x=0
c: \(\Leftrightarrow3\sqrt{2x}+10\sqrt{2x}-3\sqrt{2x}-20=0\)
=>\(\sqrt{2x}=2\)
=>2x=4
=>x=2
d: \(\Leftrightarrow2\left|x+2\right|=8\)
=>x+2=4 hoặcx+2=-4
=>x=-6 hoặc x=2
a, \(\sqrt{75}+\sqrt{48}-\sqrt{300}\)
\(=5\sqrt{3}+4\sqrt{3}-10\sqrt{3}\)
\(=-\sqrt{3}\)
b, \(\sqrt{81a}-\sqrt{36a}+\sqrt{144a}\)
\(=9\sqrt{a}-6\sqrt{a}+12\sqrt{a}\)
\(=15\sqrt{a}\)
c, \(\dfrac{4}{\sqrt{5}-2}-\dfrac{4}{\sqrt{5}+2}\)
\(=\dfrac{4\sqrt{5}+8-4\sqrt{5}+8}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\)
\(=\dfrac{16}{5-4}=16\)
d, \(\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{a}-\sqrt{b}}=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}=\sqrt{ab}\)
Nguyễn Huy Tú anh sinh năm 2004 là lên lớp 8 mà sao lại tl được bài lớp 9
ĐS: a) 3√5;35;
b) 9√22;922;
c) 15√2−√5;152−5;
d) 17√25.
a) \(\sqrt{75}+\sqrt{48}-\sqrt{300}\) = \(5\sqrt{3}+4\sqrt{3}-10\sqrt{3}\) = \(-\sqrt{3}\)
b) \(\sqrt{98}-\sqrt{72}+0,5\sqrt{8}\) = \(7\sqrt{2}-6\sqrt{2}+\sqrt{2}\) = \(2\sqrt{2}\)
c) \(\sqrt{9a}-\sqrt{16a}+\sqrt{49a}\) = \(3\sqrt{a}-4\sqrt{a}+7\sqrt{a}\) = \(6\sqrt{a}\)
d) \(\sqrt{16b}+2\sqrt{40b}-3\sqrt{90b}\) = \(4\sqrt{b}+4\sqrt{10b}-9\sqrt{10b}\)
= \(4\sqrt{b}-5\sqrt{10b}\)