Rút gọn các biểu thức sau
a, \(\sqrt{1-4a+4a^2-2a}\)
b,\(x^2+\sqrt{x^4-8x^2+16}\)
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HH
14 tháng 7 2017
\(a,\sqrt{1-4a+4a^2}-2a\)
\(=\sqrt{\left(1-2a\right)^2}-2a\)
\(=1-2a-2a\)
\(=1-4a\)
\(b,x-2y-\sqrt{x^2-4xy+4y^2}\)
\(=x-2y-\sqrt{\left(x-2y\right)^2}\)
\(=x-2y-\left(x-2y\right)\)
\(=x-2y-x+2y\)
\(=0\)
\(c,x^2+\sqrt{x^4-8x^2+16}\)
\(=x^2+\sqrt{\left(x^2-4\right)^2}\)
\(=x^2+x^2-4\)
\(=2x^2-4\)
Các câu còn lại tương tự nha
14 tháng 7 2017
\(a,\sqrt{1-4a+4a^2}-2a\)
\(=\sqrt{\left(1-2a\right)^2}-2a\)
\(=\left(1-2a\right)-2a\)
\(=1-4a\)
\(b,x-2y-\sqrt{x^2-4xy+4y^2}\)
\(=x-2y-\sqrt{\left(x-2y\right)^2}\)
\(=x-2y-\left(x-2y\right)\)
\(=x-2y-x+2y\)
\(=0\)
\(c,x^2+\sqrt{x^4-8x^2+16}\)
\(=x^2+\sqrt{\left(x^2-2^2\right)^2}\)
\(=x^2+\left(x^2-4\right)\)
\(=x^2+x^2-4\)
\(=2x^2-4\)
\(d,2x-1-\frac{\sqrt{x^2-10x+25}}{x-5}\)
\(=2x-1-\frac{\sqrt{\left(x-5\right)^2}}{x-5}\)
\(=2x-1-\frac{x-5}{x-5}\)
\(=2x-1-1\)
\(=2x-2\)
\(=2\left(x-1\right)\)
AD
23 tháng 7 2023
\(a,=\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2}\\ =\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\\ =\left|\sqrt{3}+\sqrt{2}\right|-\left|\sqrt{3}-\sqrt{2}\right|\\ =\sqrt{3}+\sqrt{2}-\left(\sqrt{3}-\sqrt{2}\right)\\ =\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}\\=2\sqrt{2} \)
\(b,=\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1}+\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1}\\ =\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =\left|\sqrt{3}+1\right|+\left|\sqrt{3}-1\right|\\ =\sqrt{3}+1+\sqrt{3}-1\\ =2\sqrt{3}\)
\(c,=x-4+\sqrt{\left(4^2-2.4.x+x^2\right)}\\ =x-4+\sqrt{\left(4-x\right)^2}\\ =x-4+\left|4-x\right|\\ =x-4+x-4=2x-8\) (vì \(x>4\) )
@seven
H9
HT.Phong (9A5)
CTVHS
2 tháng 11 2023
a) \(2\sqrt{32}+3\sqrt{72}-7\sqrt{50}+\sqrt{2}\)
\(=2\cdot4\sqrt{2}+3\cdot6\sqrt{2}-7\cdot5\sqrt{2}+\sqrt{2}\)
\(=8\sqrt{2}+18\sqrt{2}-35\sqrt{2}+\sqrt{2}\)
\(=-8\sqrt{2}\)
b) \(\sqrt{\left(3-\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\left|3-\sqrt{3}\right|+\left|2-\sqrt{3}\right|\)
\(=3-\sqrt{3}+\sqrt{3}-2\)
\(=1\)
c) \(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}\)
\(=\sqrt{3^2+2\cdot3\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}-3+\sqrt{2}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}-3+\sqrt{2}\)
\(=3+\sqrt{2}-3+\sqrt{2}\)
\(=2\sqrt{2}\)
d) \(x-4+\sqrt{16-8x+x^2}\left(x>4\right)\)
\(=x-4+\sqrt{x^2-8x+16}\)
\(=x-4+\sqrt{\left(x-4\right)^2}\)
\(=x-4+\left|x-4\right|\)
\(=x-4+x-4\)
\(=2x-8\)
e) \(\dfrac{1}{a-b}\sqrt{a^4\left(a-b\right)^2}\left(a< b\right)\)
\(=\dfrac{1}{a-b}\sqrt{\left[a^2\left(a-b\right)\right]^2}\)
\(=\dfrac{1}{a-b}\left|a^2\left(a-b\right)\right|\)
\(=\dfrac{-a^2\left(a-b\right)}{a-b}\)
\(=-a^2\)
8 tháng 11 2021
a) \(=5\left|a\right|+3a=5a+3a=8a\)
b) \(=3\left|a^2\right|+3a^2=3a^2+3a^2=6a^2\)
c) \(=5.2\left|a^3\right|-3a^3=-10a^3-3a^3=-13a^3\)
ND
11 tháng 8 2018
\(a.\sqrt{1-4a+4a^2}-2a=\sqrt{\left(1-2a\right)^2}-2a=\left|1-2a\right|-2a\)
*\(a>\dfrac{1}{2}\Rightarrow\left|1-2a\right|-2a=2a-1-2a=4a-1\)
* \(a\le\dfrac{1}{2}\Rightarrow\left|1-2a\right|-2a=1-2a-2a=1-4a\)
\(b.x-2y-\sqrt{x^2-4xy+4y^2}=x-2y-\sqrt{\left(x-2y\right)^2}=x-2y-\left|x-2y\right|\)
* \(x\ge2y\Rightarrow x-2y-\left|x-2y\right|=x-2y-x+2y=2x\)
* \(x< 2y\Rightarrow x-2y-\left|x-2y\right|=x-2y-2y+x=2x-4y\)
\(c.x^2+\sqrt{x^4-8x^2+16}=x^2+\sqrt{\left(x^2-4\right)^2}=x^2+\left|x^2-4\right|\)
* \(x^2-4\ge0\Rightarrow x^2+\left|x^2-4\right|=x^2+x^2-4=2x^2-4\)
* \(x^2-4< 0\Rightarrow x^2+\left|x^2-4\right|=x^2+4-x^2=4\)
\(d.2x-1-\dfrac{\sqrt{x^2-10x+25}}{x-5}=2x-1-\dfrac{\sqrt{\left(x-5\right)^2}}{x-5}=2x-1-\dfrac{\left|x-5\right|}{x-5}\)
* \(x\ge5\Rightarrow2x-1-\dfrac{\left|x-5\right|}{x-5}=2x-1-1=2x-2\)
* \(x< 5\Rightarrow2x-1-\dfrac{\left|x-5\right|}{x-5}=2x-1+1=2x\)
\(e.\dfrac{\sqrt{x^4-4x^2+4}}{x^2-2}=\dfrac{\sqrt{\left(x^2-2\right)^2}}{x^2-2}=\dfrac{\left|x^2-2\right|}{x^2-2}\)
* \(x^2\ge2\Rightarrow\dfrac{\left|x^2-2\right|}{x^2-2}=1\)
* \(x^2< 2\Rightarrow\dfrac{\left|x^2-2\right|}{x^2-2}=-1\)
\(f.\sqrt{\left(x-4\right)^2}+\dfrac{x-4}{\sqrt{x^2-8x+16}}=\left|x-4\right|+\dfrac{x-4}{\sqrt{\left(x-4\right)^2}}=\left|x-4\right|+\dfrac{x-4}{\left|x-4\right|}\)
* \(x\ge4\Rightarrow\left|x-4\right|+\dfrac{x-4}{\left|x-4\right|}=x-4+\dfrac{x-4}{x-4}=x-5\)
* \(x< 4\Rightarrow\left|x-4\right|+\dfrac{x-4}{\left|x-4\right|}=4-x-1=5-x\)
16 tháng 10 2021
a: Ta có: \(x=\sqrt{28-16\sqrt{3}}+2\sqrt{3}\)
\(=4-2\sqrt{3}+2\sqrt{3}\)
=4
Thay x=4 vào B, ta được:
\(B=\dfrac{2-4}{2}=-1\)
AH
Akai Haruma
Giáo viên
1 tháng 9 2019
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 $|x^2-4|=4-x^2$)
17 tháng 6 2017
\(\frac{\sqrt{3x^2+6xy+3y^2}}{x^2-y^2}\)
<=>\(\frac{\sqrt{3.\left(x+y\right)^2}}{\left(x-y\right).\left(x+y\right)}\)
<=>\(\frac{\sqrt{3}\left|x+y\right|}{\left(x-y\right).\left(x+y\right)}.\)
<=>\(\frac{\sqrt{3}}{x-y}\)