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\(A=\left|a-3\right|-3a=3-a-3a=3-4a\)
\(B=4a+3-\left|2a-1\right|=4a+3-2a+1=2a+4\)
\(C=\dfrac{4}{a^2-4}\left|a-2\right|=\dfrac{-4\left(a-2\right)}{\left(a-2\right)\left(a+2\right)}=\dfrac{-4}{a+2}\)
\(D=\dfrac{a^2-9}{12}:\sqrt{\dfrac{\left(a+3\right)^2}{16}}=\dfrac{a^2-9}{12}:\dfrac{\left|a+3\right|}{4}=\dfrac{\left(a-3\right)\left(a+3\right).4}{-12\left(a+3\right)}=\dfrac{3-a}{3}\)
b: B=căn 49a^2+3a
=|7a|+3a
=7a+3a(a>=0)
=10a
c: C=căn16a^4+6a^2
=4a^2+6a^2
=10a^2
d: \(D=3\cdot3\cdot\sqrt{a^6}-6a^3=6\cdot\left|a^3\right|-6a^3\)
TH1: a>=0
D=6a^3-6a^3=0
TH2: a<0
D=-6a^3-6a^3=-12a^3
e: \(E=3\sqrt{9a^6}-6a^3\)
\(=3\cdot\sqrt{\left(3a^3\right)^2}-6a^3\)
=3*3a^3-6a^3(a>=0)
=3a^3
f: \(F=\sqrt{16a^{10}}+6a^5\)
\(=\sqrt{\left(4a^5\right)^2}+6a^5\)
=-4a^5+6a^5(a<=0)
=2a^5
a) \(\sqrt{9a^4}=\sqrt{\left(3a^2\right)^2}=\left|3a^2\right|=3a^2\)
b) \(2\sqrt{a^2}-5a=2\left|a\right|-5a=-2a-5a=-7a\)
c) \(\sqrt{16\left(1+4x+4x^2\right)}=\sqrt{\left[4\left(1+2x\right)\right]^2}=\left|4\left(1+2x\right)\right|=4\left(1+2x\right)\)
\(B=\sqrt{16a^4}+6a^2=4a^2+6a^2=10a^2\)\(A=\sqrt{49a^2}+3a=7a+3a=10a\)
\(C=4x-\sqrt{\left(x^2-4x+4\right)}=4x-\sqrt{\left(x-2\right)^2}=4x-x+2=3x+2\)
\(E=\sqrt{y^2+6y+9}-\sqrt{y^2-6y+9}=\sqrt{\left(y+3\right)^2}-\sqrt{\left(y-3\right)^2}=\left|y+3\right|-\left|y-3\right|=y+3-y+3=6\)
\(D=\dfrac{a-b}{\sqrt{a}-\sqrt{b}}=\dfrac{\left(a-b\right)\cdot\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\cdot\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{a\sqrt{a}+a\sqrt{b}-b\sqrt{a}-b\sqrt{b}}{a-b}=\dfrac{\sqrt{a}\cdot\left(a-b\right)+\sqrt{b}\cdot\left(a-b\right)}{a-b}=\dfrac{\left(a-b\right)\cdot\left(\sqrt{a}+\sqrt{b}\right)}{a-b}=\sqrt{a}+\sqrt{b}\)
Làm nốt ::v
\(2.3\sqrt{\left(a-2\right)^2}=3\text{ |}a-2\text{ |}=3\left(a-2\right)\left(a< 2\right)\)
\(3.\sqrt{81a^4}+3a^2=\sqrt{3^4.a^4}+3a^2=9a^2+3a^2=12a^2\)
\(4.\sqrt{64a^2}+2a=\text{ |}8a\text{ |}+2a=8a+2a=10a\left(a>=0\right)\)
\(6.\sqrt{a^2+6a+9}+\sqrt{a^2-6a+9}=\sqrt{\left(a+3\right)^2}+\sqrt{\left(a-3\right)^2}=\text{ |}a+3\text{ |}+\text{ |}a-3\text{ |}\)
\(7.\dfrac{\sqrt{1-2x+x^2}}{x-1}=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{\text{ |}x-1\text{ |}}{x-1}\)
\(8.\dfrac{\sqrt{9x^2-6x+1}}{9x^2-1}=\dfrac{\sqrt{\left(3x-1\right)^2}}{\left(3x-1\right)\left(3x+1\right)}=\dfrac{\text{ |}3x-1\text{ |}}{\left(3x-1\right)\left(3x+1\right)}\)
\(9.4-x-\sqrt{4-4x+x^2}=4-x-\sqrt{\left(x-2\right)^2}=4-x-\text{ |}x-2\text{ |}\)
Mình làm ba câu mẫu, bạn theo đó mà làm các câu còn lại.
Giải:
1) \(2\sqrt{a^2}\)
\(=2\left|a\right|\)
\(=2a\left(a\ge0\right)\)
Vậy ...
5) \(3\sqrt{9a^6}-6a^3\)
\(=3\sqrt{\left(3a^3\right)^2}-6a^3\)
\(=3.3a^3-6a^3\)
\(=9a^3-6a^3\)
\(=3a^3\)
Vậy ...
10) \(C=\sqrt{4x^2-4x+1}-\sqrt{4x^2+4x+1}\)
\(\Leftrightarrow C=\sqrt{\left(2x-1\right)^2}-\sqrt{\left(2x+1\right)^2}\)
\(\Leftrightarrow C=2x-1^2-\left(2x+1^2\right)\)
\(\Leftrightarrow C=2x-1-2x-1\)
\(\Leftrightarrow C=-2\)
Vậy ...
a) \(\sqrt{\dfrac{2a}{3}}\cdot\sqrt{\dfrac{3a}{8}}\)
\(=\sqrt{\dfrac{2a\cdot3a}{3\cdot8}}\)
\(=\sqrt{\dfrac{6a^2}{24}}\)
\(=\sqrt{\dfrac{a^2}{4}}\)
\(=\dfrac{\sqrt{a^2}}{\sqrt{4}}\)
\(=\dfrac{a}{2}\)
b) \(\sqrt{3a}\cdot\sqrt{\dfrac{52}{a}}\)
\(=\sqrt{3a\cdot\dfrac{52}{a}}\)
\(=\sqrt{3\cdot52}\)
\(=\sqrt{13\cdot3\cdot4}\)
\(=2\sqrt{39}\)
c) \(2y^2\cdot\sqrt{\dfrac{x^4}{4y^2}}\)
\(=2y^2\cdot\dfrac{\sqrt{\left(x^2\right)^2}}{\sqrt{\left(2y\right)^2}}\)
\(=2y^2\cdot\dfrac{x^2}{-2y}\)
\(=\dfrac{2y^2\cdot x^2}{-2y}\)
\(=-x^2y\)
a: \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
b: \(\sqrt{\dfrac{2a}{3}}\cdot\sqrt{\dfrac{3a}{8}}=\sqrt{\dfrac{6a^2}{24}}=\sqrt{\dfrac{a^2}{4}}=\dfrac{a}{2}\)
c: \(\sqrt{5a\cdot45a}-3a=-15a-3a=-18a\)
a) \(5\sqrt{25a^2}-25=25\left|a\right|-25==-25a-25\left(a< 0\right)\)
b) \(\sqrt{49a^2}+3a=7\left|a\right|+3a=-7a+3a\left(a< 0\right)=-4a\)
c) \(3\sqrt{9a^6}=9\left|a^3\right|-6a^3\)
Xét \(a\ge0\Rightarrow9\left|a^3\right|-6a^3=9a^3-6a^3=3a^3\)
Xét \(a< 0\Rightarrow9\left|a^3\right|-6a^3=-9a^3-6a^3=-15a^3\)
a) 5\(\sqrt{25a^2}\) - 25 với a < 0
= 5\(\sqrt{\left(5a\right)^2}\) - 25
= 5.\(\left|5a\right|\) - 25
= 5.-(5a) - 25
= -25a - 25 Vì a < 0
b) \(\sqrt{49a^2}\) + 3a với a < 0
= \(\sqrt{\left(7a\right)^2}\) + 3a
= \(\left|7a\right|\) + 3a
= -7a + 3a Vì a < 0
= -4a
c) 3\(\sqrt{9a^6}\) - 6a3 với a bất kì
= 3\(\sqrt{\left(3a^3\right)^2}\) - 6a3
= 3\(\left|3a^3\right|\) - 6a3
= 9a3 - 6a3
= 3a3
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