\(\dfrac{a-4}{\sqrt{a+2}}vớia\ge0\)
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HC
2
J
28 tháng 8 2021
\(\dfrac{a-4\sqrt{a}+4}{\sqrt{a}-2}=\dfrac{\left(\sqrt{a}-2\right)^2}{\sqrt{a}-2}=\sqrt{a}-2\)
17 tháng 7 2017
\(\dfrac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\dfrac{\sqrt{a^3}-\sqrt{b^3}}{\sqrt{a}-\sqrt{b}}\)
\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\sqrt{a}-\sqrt{b}}\)
\(=a+\sqrt{ab}+b\)
AT
21 tháng 10 2018
a) \(\sqrt{16x-8}+\sqrt{36x-18}-\sqrt{64x-32}=\sqrt{10}\)
\(\Leftrightarrow\sqrt{8\left(2x-1\right)}+\sqrt{18\left(2x-1\right)}-\sqrt{32\left(2x-1\right)}=\sqrt{10}\)
\(\Leftrightarrow\sqrt{8}.\sqrt{2x-1}+\sqrt{18}.\sqrt{2x-1}-\sqrt{32}.\sqrt{2x-1}=\sqrt{10}\)
\(\Leftrightarrow\sqrt{2x-1}.\left(\sqrt{8}+\sqrt{18}-\sqrt{32}\right)=\sqrt{10}\)
\(\Leftrightarrow\sqrt{2x-1}.\sqrt{2}=\sqrt{10}\)
\(\Leftrightarrow\sqrt{2x-1}=\sqrt{5}\)
\(\Leftrightarrow2x-1=5\)
\(\Leftrightarrow x=3\)
Vậy ...
b) \(\sqrt{x^2-6x+9}=x+3\)
\(\Leftrightarrow\sqrt{x^2-2.x.3+3^2}=x+3\)
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=x+3\)
\(\Leftrightarrow\left|x-3\right|=x+3\)
\(\Leftrightarrow x-3=x+3\) hoặc \(x-3=-x-3\)
\(\Leftrightarrow x=0\)
Vậy ...
21 tháng 10 2018
bài 2 :
A = \(\left(\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}-\dfrac{4\sqrt{ab}}{a-b}\right)\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}-\left(a+b\right)}\right)\)
\(=\left(\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}-\dfrac{4\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a+\sqrt{b}}\right)}\right)\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}-\left(a+b\right)}\right)\)
\(=\left(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\right)\left(\dfrac{\sqrt{a^3}+\sqrt{b^3}}{\sqrt{ab}-a-b}\right)\)
\(=\left(\dfrac{a+2\sqrt{ab}+b-4\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\right)\left(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{-a+\sqrt{ab}-b}\right)\)
\(=\dfrac{a-2\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}.\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{-\left(a-\sqrt{ab}+b\right)}\)
\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}.\left(-\left(\sqrt{a}+\sqrt{b}\right)\right)\)
\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right).\left(-1\right).\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}\)
\(=-\left(\sqrt{a}-\sqrt{b}\right)=\sqrt{b}-\sqrt{a}\)
cuối cùng cũng xong, mong bn phù hộ độ trì cho mk
S
27 tháng 8 2021
a) \(\sqrt{4\left(a-3\right)^2}=2\left(a-3\right)=2a-6\)
b) \(\sqrt{a^2\left(a+1\right)^2}=a\left(a+1\right)=a^2+a\)
c) \(\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\dfrac{1}{\sqrt{8}\left|a\right|}=\dfrac{1}{-\sqrt{8}a}=\dfrac{-\sqrt{8}}{8a}\)
27 tháng 8 2021
a: \(\sqrt{4\left(a-3\right)^2}=2\cdot\left(a-3\right)=2a-6\)
b: \(\sqrt{a^2\left(a+1\right)^2}=a\left(a+1\right)=a^2+a\)
c: \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}=\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\sqrt{\dfrac{2}{16a^2}}=-\dfrac{\sqrt{2}}{4a}\)
5 tháng 9 2022
a: \(=6\sqrt{a}+\dfrac{1}{3}\sqrt{a}-3\sqrt{a}+\sqrt{7}=\dfrac{10}{3}\sqrt{a}+\sqrt{7}\)
b: \(=5a\cdot5b\sqrt{ab}+\sqrt{3}\cdot2\sqrt{3}\cdot ab\sqrt{ab}+9ab\cdot3\sqrt{ab}-5b\cdot9a\sqrt{ab}\)
\(=25ab\sqrt{ab}+12ab\sqrt{ab}+27ab\sqrt{ab}-45ab\sqrt{ab}\)
\(=19ab\sqrt{ab}\)
c: \(=\dfrac{\sqrt{ab}}{b}+\sqrt{ab}-\dfrac{a}{b}\cdot\dfrac{\sqrt{b}}{\sqrt{a}}\)
\(=\sqrt{ab}\left(\dfrac{1}{b}+1\right)-\dfrac{\sqrt{a}}{\sqrt{b}}\)
\(=\sqrt{ab}\)
d: \(=11\sqrt{5a}-5\sqrt{5a}+2\sqrt{5a}-12\sqrt{5a}+9\sqrt{a}\)
\(=-4\sqrt{5a}+9\sqrt{a}\)
12 tháng 6 2017
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 !!!
HH
1
15 tháng 2 2022
đề bài sai nha bạn vì nếu a=0 thì ko thực hiện được phép tính nha
DQ
1
15 tháng 7 2021
a) \(\sqrt{27\cdot48\cdot\left(1-a\right)^2}\)
\(=3\sqrt{3}\cdot4\sqrt{3}\cdot\left|1-a\right|\)
\(=36\cdot\left(a-1\right)=36a-36\)
b) \(\dfrac{1}{a-b}\cdot\sqrt{a^4\left(a-b\right)^2}\)
\(=\dfrac{1}{a-b}\cdot\left(a-b\right)\cdot a^2\)
\(=a^2\)
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