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2) \(\sqrt{98}-\sqrt{72}+0,5\sqrt{8}\)
\(=7\sqrt{2}-6\sqrt{2}+\sqrt{2}\)
\(=\left(7-6+1\right)\sqrt{2}\)
\(=2\sqrt{2}\)
3) \(\sqrt{9a}-\sqrt{16a}+\sqrt{49a}\)
\(=3\sqrt{a}-4\sqrt{a}+7\sqrt{a}\)
\(=\left(3-4+7\right)\sqrt{a}\)
\(=6\sqrt{a}\)
4) \(\sqrt{16b}+2\sqrt{40b}-3\sqrt{90b}\)
\(=4\sqrt{b}+4\sqrt{10b}-9\sqrt{10b}\)
\(=4\sqrt{b}-5\sqrt{10b}\)
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) \(2x-\sqrt{4x^2+4x+1}=2x-\sqrt{\left(2x+1\right)^2}=2x-\left|2x+1\right|\)
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 !!!
\(A=\sqrt{4^2b}+2\sqrt{2^2\cdot10b}-3\sqrt{3^2\cdot10b}=4\sqrt{b}+4\sqrt{10}\cdot\sqrt{b}-9\sqrt{10}\cdot\sqrt{b}\)
\(=4\sqrt{b}-5\sqrt{10}\sqrt{b}=\left(4-5\sqrt{10}\right)\sqrt{b}\)
Rut gon A = √16b+2√40b−3√90bva`b≥0
A=√42b+2√22·10b−3√32·10b=4√b+4√10·√b−9√10·√b
=4√b−5√10√b=(4−5√10)√b
\(a.A=\sqrt{75}+\sqrt{48}-\sqrt{300}=\sqrt{25.3}+\sqrt{16.3}-\sqrt{100.3}=5\sqrt{3}+4\sqrt{3}-10\sqrt{3}=-\sqrt{3}\) \(b.B=\sqrt{98}-\sqrt{72}+0,5\sqrt{8}=\sqrt{49.2}-\sqrt{36.2}+0,5\sqrt{4.2}=7\sqrt{2}-6\sqrt{2}+\sqrt{2}=2\sqrt{2}\) \(c.\dfrac{5\sqrt{5}-\sqrt{15.5}+5\sqrt{5}}{\sqrt{5}}=\dfrac{\sqrt{5}\left(10-\sqrt{15}\right)}{\sqrt{5}}=10-\sqrt{15}\)
\(=7\sqrt{b}+4\sqrt{10b}-9\sqrt{10b}=7\sqrt{b}-5\sqrt{10b}\)
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
đk : \(a\ge0;b\ge0;a\ne b\)
a) \(\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}+\dfrac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}\) = \(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2+\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
= \(\dfrac{a+2\sqrt{ab}+b+a-2\sqrt{ab}+b}{a-b}\) = \(\dfrac{2\left(a+b\right)}{a-b}\)
b) đk : \(a\ge0;b\ge0;a\ne b\)
\(\dfrac{a-b}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{a^3}-\sqrt{b^3}}{a-b}\)
= \(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
= \(\dfrac{\sqrt{a}+\sqrt{b}}{1}-\dfrac{a+\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}\) = \(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(a+\sqrt{ab}+b\right)}{\sqrt{a}+\sqrt{b}}\)
= \(\dfrac{a+2\sqrt{ab}+b-a-\sqrt{ab}-b}{\sqrt{a}+\sqrt{b}}\) = \(\dfrac{\sqrt{ab}}{\sqrt{a}+\sqrt{b}}=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{a+b}\)
a: \(=2\sqrt{2}+30\sqrt{2}-3\sqrt{2}+6\sqrt{2}=26\sqrt{2}\)
b: \(=\dfrac{1}{2}\cdot4\sqrt{3}-2\cdot5\sqrt{3}+\sqrt{3}+\dfrac{5}{2}\sqrt{3}=-\dfrac{9}{2}\sqrt{3}\)
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) \(\sqrt{16b}+2\sqrt{40b}-3\sqrt{90b}\) = \(4\sqrt{b}+4\sqrt{10b}-9\sqrt{10b}\)
= \(4\sqrt{b}-5\sqrt{10b}\)