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\(2.3\sqrt{\left(a-2\right)^2}=3\text{ |}a-2\text{ |}=3\left(a-2\right)\left(a< 2\right)\)
\(3.\sqrt{81a^4}+3a^2=\sqrt{3^4.a^4}+3a^2=9a^2+3a^2=12a^2\)
\(4.\sqrt{64a^2}+2a=\text{ |}8a\text{ |}+2a=8a+2a=10a\left(a>=0\right)\)
\(6.\sqrt{a^2+6a+9}+\sqrt{a^2-6a+9}=\sqrt{\left(a+3\right)^2}+\sqrt{\left(a-3\right)^2}=\text{ |}a+3\text{ |}+\text{ |}a-3\text{ |}\)
\(7.\dfrac{\sqrt{1-2x+x^2}}{x-1}=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{\text{ |}x-1\text{ |}}{x-1}\)
\(8.\dfrac{\sqrt{9x^2-6x+1}}{9x^2-1}=\dfrac{\sqrt{\left(3x-1\right)^2}}{\left(3x-1\right)\left(3x+1\right)}=\dfrac{\text{ |}3x-1\text{ |}}{\left(3x-1\right)\left(3x+1\right)}\)
\(9.4-x-\sqrt{4-4x+x^2}=4-x-\sqrt{\left(x-2\right)^2}=4-x-\text{ |}x-2\text{ |}\)
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 ...
Lời giải:
a)
\(\sqrt{1-4a+4a^2}-2a=\sqrt{1-2.2a+(2a)^2}-2a\)
\(=\sqrt{(2a-1)^2}-2a=|2a-1|-2a=(2a-1)-2a=-1\)
(do $a\geq \frac{1}{2}$ nên $|2a-1|=2a-1$)
b)
\(x-2y-\sqrt{x^2-4xy+4y^2}=x-2y-\sqrt{(x-2y)^2}=x-2y-|x-2y|\)
\(=x-2y-(2y-x)=2(x-2y)\)
(do $x< 2y$ nên $|x-2y|=-(x-2y)=2y-x$)
c)
\(x^2+\sqrt{x^4-8x^2+16}=x^2+\sqrt{(x^2)^2-2.4.x^2+4^2}\)
\(=x^2+\sqrt{(x^2-4)^2}=x^2+|x^2-4|=x^2+(4-x^2)=4\)
(do $x^2< 4$ nên $|x^2-4|=4-x^2$)
\(a,\sqrt{4x^4}+6x^2=2x^2+6x^2=8x^2\)
\(b,\sqrt{25a^4}-2a^2=5a^2-2a^2=3a^2\)
\(c,\sqrt{36a^4}+8a=6a^2+8a\)
\(d,\sqrt{\left(x-3\right)^4}-x^2+3x-1=\left(x-3\right)^2-x^2+3x-1=x^2-6x+9-x^2+3x-1=-3x+8\)
a: \(M=\dfrac{x+6\sqrt{x}-3\sqrt{x}+18-x}{x-36}\)
\(=\dfrac{3\left(\sqrt{x}+6\right)}{x-36}=\dfrac{3}{\sqrt{x}-6}\)
b: \(N=\dfrac{x^2}{y}\cdot\sqrt{xy\cdot\dfrac{y}{x}}-x^2\)
\(=\dfrac{x^2}{y}\cdot y-x^2=0\)
a. Ta có:\(\frac{x}{y}\sqrt{\frac{y^2}{x^4}=}\) \(\frac{x}{y}.\frac{\left|y\right|}{x^2}=\frac{x.y}{x^2y}\)\(=\frac{1}{x}\)(Vì \(x\ne0;y>0\))
b \(3x^2\sqrt{\frac{8}{x^2}}=3x^2\frac{2\sqrt{2}}{\left|x\right|}=\frac{6x^2\sqrt{2}}{-x}=-6x\sqrt{2}\)( Vì \(x< 0\))
2.
a/ Áp dụgn hệ quả bđt cô si,ta có :
\(A=xy+yz+zx\le\dfrac{\left(x+y+z\right)}{3}=\dfrac{a^2}{3}\)
Vậy GTLN A =a^2/3 khi x= y =z =a/3
b/Áp dụng BĐT Cô-Si dạng Engel,ta có :
\(B=\dfrac{x^2}{1}+\dfrac{y^2}{1}+\dfrac{z^2}{z}\ge\dfrac{\left(x+y+z\right)^2}{3}=\dfrac{a^2}{3}\)
Vậy GTNN của B = a^2/2 khi x=y=z =a/3
\(B=\dfrac{3x}{1-x}+\dfrac{4\left(1-x\right)}{x}+7\ge2\sqrt{\dfrac{3x}{1-x}.\dfrac{4\left(1-x\right)}{x}}+7=7+4\sqrt{3}=\left(2+\sqrt{3}\right)^2\)
Vậy min B = \(\left(2+\sqrt{3}\right)^2\) khi \(\dfrac{3x}{1-x}=\dfrac{4\left(1-x\right)}{x}\Leftrightarrow x=\left(\sqrt{3}-1\right)^2\)
2) a) \(x^2-3=\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\)
b) \(x^2-6=\left(x-\sqrt{6}\right).\left(x+\sqrt{6}\right)\)
c) = \(x^2+2x.\sqrt{3}+\left(\sqrt{3}\right)^2=\left(x+\sqrt{3}\right)^2\)
d) = \(x^2-2x\sqrt{5}+\left(\sqrt{5}\right)^2=\left(x-\sqrt{5}\right)^2\)