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\(a.\sqrt{1-4a+4a^2}-2a=\sqrt{\left(1-2a\right)^2}-2a=\left|1-2a\right|-2a\)
*\(a>\dfrac{1}{2}\Rightarrow\left|1-2a\right|-2a=2a-1-2a=4a-1\)
* \(a\le\dfrac{1}{2}\Rightarrow\left|1-2a\right|-2a=1-2a-2a=1-4a\)
\(b.x-2y-\sqrt{x^2-4xy+4y^2}=x-2y-\sqrt{\left(x-2y\right)^2}=x-2y-\left|x-2y\right|\)
* \(x\ge2y\Rightarrow x-2y-\left|x-2y\right|=x-2y-x+2y=2x\)
* \(x< 2y\Rightarrow x-2y-\left|x-2y\right|=x-2y-2y+x=2x-4y\)
\(c.x^2+\sqrt{x^4-8x^2+16}=x^2+\sqrt{\left(x^2-4\right)^2}=x^2+\left|x^2-4\right|\)
* \(x^2-4\ge0\Rightarrow x^2+\left|x^2-4\right|=x^2+x^2-4=2x^2-4\)
* \(x^2-4< 0\Rightarrow x^2+\left|x^2-4\right|=x^2+4-x^2=4\)
\(d.2x-1-\dfrac{\sqrt{x^2-10x+25}}{x-5}=2x-1-\dfrac{\sqrt{\left(x-5\right)^2}}{x-5}=2x-1-\dfrac{\left|x-5\right|}{x-5}\)
* \(x\ge5\Rightarrow2x-1-\dfrac{\left|x-5\right|}{x-5}=2x-1-1=2x-2\)
* \(x< 5\Rightarrow2x-1-\dfrac{\left|x-5\right|}{x-5}=2x-1+1=2x\)
\(e.\dfrac{\sqrt{x^4-4x^2+4}}{x^2-2}=\dfrac{\sqrt{\left(x^2-2\right)^2}}{x^2-2}=\dfrac{\left|x^2-2\right|}{x^2-2}\)
* \(x^2\ge2\Rightarrow\dfrac{\left|x^2-2\right|}{x^2-2}=1\)
* \(x^2< 2\Rightarrow\dfrac{\left|x^2-2\right|}{x^2-2}=-1\)
\(f.\sqrt{\left(x-4\right)^2}+\dfrac{x-4}{\sqrt{x^2-8x+16}}=\left|x-4\right|+\dfrac{x-4}{\sqrt{\left(x-4\right)^2}}=\left|x-4\right|+\dfrac{x-4}{\left|x-4\right|}\)
* \(x\ge4\Rightarrow\left|x-4\right|+\dfrac{x-4}{\left|x-4\right|}=x-4+\dfrac{x-4}{x-4}=x-5\)
* \(x< 4\Rightarrow\left|x-4\right|+\dfrac{x-4}{\left|x-4\right|}=4-x-1=5-x\)
\(a,\sqrt{1-4a+4a^2}-2a\)
\(=\sqrt{\left(1-2a\right)^2}-2a\)
\(=1-2a-2a\)
\(=1-4a\)
\(b,x-2y-\sqrt{x^2-4xy+4y^2}\)
\(=x-2y-\sqrt{\left(x-2y\right)^2}\)
\(=x-2y-\left(x-2y\right)\)
\(=x-2y-x+2y\)
\(=0\)
\(c,x^2+\sqrt{x^4-8x^2+16}\)
\(=x^2+\sqrt{\left(x^2-4\right)^2}\)
\(=x^2+x^2-4\)
\(=2x^2-4\)
Các câu còn lại tương tự nha
\(a,\sqrt{1-4a+4a^2}-2a\)
\(=\sqrt{\left(1-2a\right)^2}-2a\)
\(=\left(1-2a\right)-2a\)
\(=1-4a\)
\(b,x-2y-\sqrt{x^2-4xy+4y^2}\)
\(=x-2y-\sqrt{\left(x-2y\right)^2}\)
\(=x-2y-\left(x-2y\right)\)
\(=x-2y-x+2y\)
\(=0\)
\(c,x^2+\sqrt{x^4-8x^2+16}\)
\(=x^2+\sqrt{\left(x^2-2^2\right)^2}\)
\(=x^2+\left(x^2-4\right)\)
\(=x^2+x^2-4\)
\(=2x^2-4\)
\(d,2x-1-\frac{\sqrt{x^2-10x+25}}{x-5}\)
\(=2x-1-\frac{\sqrt{\left(x-5\right)^2}}{x-5}\)
\(=2x-1-\frac{x-5}{x-5}\)
\(=2x-1-1\)
\(=2x-2\)
\(=2\left(x-1\right)\)
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 $|x^2-4|=4-x^2$)
Bài 1:
a: \(A=\left|2a-1\right|-2a\)
TH1: a>=1/2
A=2a-1-2a=-1
TH2: a<1/2
A=1-2a-2a=1-4a
b: \(B=x-2y-\left|x-2y\right|\)
TH1: x>=2y
A=x-2y-x+2y=0
TH2: x<2y
A=x-2y+x-2y=2x-4y
c: \(=x^2+\left|x^2-4\right|\)
TH1: x>=2 hoặc x<=-2
\(A=x^2+x^2-4=2x^2-4\)
TH2: -2<x<2
\(A=x^2+4-x^2=4\)
d: \(D=2x-1-\dfrac{\left|x-5\right|}{x-5}\)
TH1: x>5
\(D=2x-1-1=2x-2\)
TH2: x<5
D=2x-1+1=2x
\(a,5\sqrt{4a^6}-3a^3=5\left|2a^3\right|-3a^2=-10a^3-3a^3=-13a^3\)(vì a<0)
b)\(\sqrt{9a^4}+3a^2=\left|3a^2\right|+3a^2=3a^2+3a^2=6a^2\)
c)\(\frac{\sqrt{x^2-10x+25}}{x-5}=\frac{\left|x-5\right|}{x-5}\)
Với x-5>0 => x>5 => \(\frac{\sqrt{x^2-10x+25}}{x-5}=1\)
Với x-5<0=>x<5 =>\(\frac{\sqrt{x^2-10x+25}}{x-5}=-1\)
1.
a) \(A=\sqrt{1}-4a+4a^2-2a\)
\(A=4a^2-6a+1\)
b) \(B=\frac{5-x}{x^2-10x+25}=\frac{-\left(x-5\right)}{\left(x-5\right)^2}=\frac{-1}{x-5}\)
c) \(C=\sqrt{\left(x-1\right)^2}+\frac{x-1}{\sqrt{x^2-2x+1}}\)
\(C=\left|x-1\right|+\frac{x-1}{\sqrt{\left(x-1\right)^2}}=\left|x-1\right|+\frac{x-1}{\left|x-1\right|}\)
+) Xét \(x-1>0\Leftrightarrow x>1\)ta có \(C=x-1+\frac{x-1}{x-1}=x-1+1=x\)
+) Xét \(x-1< 0\Leftrightarrow x< 1\)ta có \(C=1-x+\frac{x-1}{1-x}=1-x-1=-x\)
2.
a) \(\sqrt{2-\sqrt{3}}\cdot\sqrt{2+\sqrt{3}}\)
\(=\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=\sqrt{4-3}=1\)
b) \(\sqrt{3\sqrt{2}-2\sqrt{3}}\cdot\sqrt{3\sqrt{2}+2\sqrt{3}}\)
\(=\sqrt{\left(3\sqrt{2}-2\sqrt{3}\right)\left(3\sqrt{2}+2\sqrt{3}\right)}\)
\(=\sqrt{\left(3\sqrt{2}\right)^2-\left(2\sqrt{3}\right)^2}\)
\(=\sqrt{18-12}=\sqrt{6}\)
c) Sửa luôn đề \(\sqrt{13-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}\)
\(=\sqrt{\left(2\sqrt{3}\right)^2-2\cdot2\sqrt{3}\cdot1+1}+\sqrt{2^2+2\cdot2\cdot\sqrt{3}+3}\)
\(=\sqrt{\left(2\sqrt{3}-1\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=\left|2\sqrt{3}-1\right|+\left|2+\sqrt{3}\right|\)
\(=2\sqrt{3}-1+2+\sqrt{3}\)
\(=3\sqrt{3}+1\)
a)=1-4a
b) = 2x - 4y
c) = 2x - 2 (nếu x>5)
=2x(nếu x<5)
-1. 2x. 2x
a) A= 1 - 4a
b) B=2x-4y
c) C= 2x-2 (nếu x>5)
= 2x (nếu x< 5)
a) A= 1-4A
b ) B =2x-4y
c) C = 2x-2( nếu x>5 )
=2x(nếu x< 5)
\(a,A=\sqrt{1-4a+4a^2}-2a\\ =\sqrt{\left(1-2a\right)^2}-2a\\ =\left|1-2a\right|-2a\\nếu1-2a\ge0\Leftrightarrow a\le\dfrac{1}{2}thì\left|1-2a\right|=1-2a\\ khiđóA=1-2a-2a=1-4a\\ nếu1-2a< 0\Leftrightarrow a>\dfrac{1}{2}thì\left|1-2a\right|=2a-1\\ khiđóA=2a-1-2a=-1\)
\(b,B=x-2y-\sqrt{x^2-4xy+4y^2}=x-2y-\sqrt{\left(x-2y\right)^2=x-2y-\left|x-2y\right|}\\ nếux-2y\ge0\Leftrightarrow x\ge2ythì\left|x-2y\right|=x-2y\\ khiđóB=x-2y-\left(x-2y\right)=0\\ nếux-2y< 0thì\left|x-2y\right|=2y-x\\ khiđóB=x-2y-\left(2y-x\right)=x-2y-2y+x=2x\)
a\()\)A=2a - 1 - 2a = -1
b\()\)B=x - 2y \((2y-x)\)= x - 2y -2y + x= 2x
c\()\)C=2x - 1 =\(\dfrac{-(x-5)}{x-5}\)= 2x - 1 + 1 = 2x
A=\(\sqrt{1-4a+4a^2}-2a\)
=\(\sqrt{(1-2a)^2}-2a=\left|1-2a\right|-2a\)
\(TH_1:1-2a\ge0\Leftrightarrow a\le\dfrac{1}{2}\Rightarrow\left|1-2a\right|-2a=1-2a-2a=1-4a\)
\(TH_2:1-2a< 0\Leftrightarrow a>\dfrac{1}{2}\Rightarrow\left|1-2a\right|-2a=2a-1-2a=-1\)
B=\(x-2y-\sqrt{x^2-4xy+4y^2}\)
=\(x-2y-\sqrt{(x-2y)^2}=x-2y-\sqrt{(x-2y)^2}\)=\(x-2y-\left|x-2y\right|\)
\(TH_1:x-2y\ge0\Leftrightarrow x\ge2y\Rightarrow x-2y-\left|x-2y\right|=x-2y-x+2y=0\)
\(TH_2:x-2y< 0\Leftrightarrow x< 2y\Rightarrow x-2y-\left|x-2y\right|=x-2y+2y+x=2x\)
C=\(2x-1-\dfrac{\sqrt{x^2-10x+25}}{x-5}\)ĐKXĐ:\(x\ne5\)
=\(2x-1-\dfrac{\sqrt{(x-5)^2}}{x-5}=2x-1-\dfrac{\left|x-5\right|}{x-5}\)
\(TH_1:x-5>0\Leftrightarrow x>5\Rightarrow2x-1-\dfrac{\left|x-5\right|}{x-5}=2x-1-\dfrac{x-5}{x-5}=2x-1-1=2x-2\)
\(TH_2:x-5< 0\Leftrightarrow x< 5\Rightarrow2x-1-\dfrac{\left|x-5\right|}{x-5}=2x-1-\dfrac{-(x-5)}{x-5}=2x-1+1=2x\)
a) A=1-4a b) B=2x-4y c) C=2x-2( nếu x>5) =2x( nếu x<5)
a) A =1-4a
b)B =2x-4y
c) C= 2x-2(nếu x>5)
x= 2x(nếu x<5)
a,A= \(-1\)
b,B= 2x
c,C=2x
a) A = -1
b) B = 2x
c) với x - 5 > 0 ⇔ C = 2x - 2
với x-5 < 0 ⇔ C = 2x
a) A = -1
b) B = 2x
c) C = 2x
A=-1
B=2x
C=2x
a) A = -1
b) B = 2x
c) C = 2x
a = 6
b = 1
c = 2x + 2
a) -1
b) 2x
c) 2x/ 2x-2
a) A=-1
b) B=2
c) C=2x
a) trường hợp 1 : A = 1 - 4a
trường hợp 2 : A = -1
b) trường hợp 1 : B = 0
trường hợp 2 : B = 2x
c) trường hợp 1 : C = 2x - 2
trường hợp 2 : C= 2x
a) A = -1
b) B = 2x
c) C = 2x
a)A=1-4a nếu a≤\(\dfrac{1}{2}\) hoặc = -1 nếu a lớn hơn \(\dfrac{1}{2}\)
b)B=0 nếu y ≤\(\dfrac{x}{2}\)hoặc = 2x - 4y nếu y lớn hơn \(\dfrac{x}{2}\)
c)C=2x - 2 nếu x≥5 hoặc = 2x nếu x bé hơn 5
a) A = 1 - 4a khi 1 - 2a \(\ge\)0
A = -1 khi 1 - 2a < 0
b) B = 0 khi x - 2y \(\ge\)0
B = 2x khi x - 2y < 0
c) C = 2x - 2 khi x - 5 > 0
C = 2x khi x - 5 > 0
a) A=\(\sqrt{1-4a+4a^2}-2a\)=\(\sqrt{\left(1-2a\right)^2}-2a\)= \(|1-2a|-2a\)
TH1: với \(1-2a\)≥ 0 ⇔ \(a\)≤\(\dfrac{1}{2}\)thì \(|1-2a|=1-2a\)
⇒A=\(|1-2a|-2a\)=\(1-2a-2a\)=\(1-4a\)
TH2 : với \(1-2a< 0\)⇔ \(a\)> \(\dfrac{1}{2}\)thì \(|1-2a|=2a-1\)
⇒A=\(|1-2a|-2a\)= \(2a-1-2a\)=-1
b) B= \(x-2y-\sqrt{x^2-4xy+4y^2}\)= \(x-2y-\sqrt{\left(x-2y\right)^2}\)= \(x-2y-|x-2y|\)
TH1: với \(x-2y\)≥0 ⇒\(|x-2y|=x-2y\)
⇒B= \(x-2y-\left(x-2y\right)\)=\(x-2y-x+2y\)=\(0\)
TH2 : với \(x-2y< 0\)⇒\(|x-2y|=2y-x\)
⇒B= \(x-2y-|x-2y|\)=\(x-2y-\left(2y-x\right)\)=\(x-2y-2y+x\)=\(2x-4y\)
c) C=\(2x-1-\dfrac{\sqrt{x^2-10x+25}}{x-5}\)= \(2x-1-\dfrac{\sqrt{\left(x-5\right)^2}}{x-5}\)=\(2x-1-\dfrac{|x-5|}{x-5}\)
TH1:\(x-5>0\Rightarrow|x-5|=x-5\)
⇒C=\(2x-1-\dfrac{|x-5|}{x-5}\)=\(2x-1-\dfrac{x-5}{x-5}\)=\(2x-1-1\)=\(2x-2\)
TH2: \(x-5< 0\Rightarrow|x-5|=5-x\)
⇒C= \(2x-1-\dfrac{5-x}{x-5}\)=\(2x-1+\dfrac{x-5}{x-5}\)=\(2x-1+1\)=\(2x\)
a) \(A=\sqrt{1-4a+4a^2}-2a=\sqrt{4a^2-4a+1}-2a=\sqrt{\left(2a-1\right)^2}-2a=|2a-1|-2a\)
TH1: \(2a-1\ge0\Rightarrow|2a-1|=2a-1\Rightarrow A=2a-1-2a=-1\)
TH2:\(2a-1< 0\Rightarrow|2a-1|=-\left(2a-1\right)=-2a+1=1-2a\Rightarrow A=1-2a-2a=1-4a\)
b)\(B=x-2y-\sqrt{x^2-4xy+4y^2}=x-2y-\sqrt{\left(x-2y\right)^2}=x-2y-|x-2y|\)
TH1:\(x-2y\ge0\Rightarrow|x-2y|=x-2y\Rightarrow B=x-2y-\left(x-2y\right)=x-2y-x+2y=0\)
TH2:\(x-2y< 0\Rightarrow|x-2y|=-\left(x-2y\right)=2y-x\Rightarrow B=x-2y-\left(2y-x\right)=x-2y-2y+x\)
\(=2x-4y\)
c) \(C=2x-2-\dfrac{\sqrt{x^2-10x+25}}{x-5}=2x-2-\dfrac{\sqrt{\left(x-5\right)^2}}{x-5}=\dfrac{|x-5|}{x-5}\)
TH1: \(x-5\ge0\Rightarrow|x-5|=x-5\Rightarrow C=\dfrac{x-5}{x-5}=1\)
TH2:\(x-5< 0\Rightarrow|x-5|=-\left(x-5\right)\Rightarrow C=\dfrac{-\left(x-5\right)}{x-5}=-1\)
a) -1
b) bằng x -2y với x ≥ 2y hoặc bằng 2x nếu x < 2y
c) bằng 2x -2 với x > 5 hoặc bằng 2x nếu x <5
a,A=\(\sqrt{1-4a+4a^2}-2a\)
=\(\sqrt{\left(1-2a\right)^2}-2a\)=\(\left|1-2a\right|-2a\)
nếu \(1-2a\ge0\Leftrightarrow a\le\dfrac{1}{2}\) thì:\(\left|1-2a\right|=1-2a\)
Ta có:\(1-2a-2a\)=\(1-4a\)
nếu \(1-2a\le0\Leftrightarrow a\ge\dfrac{1}{2}\) thì:\(\left|1-2a\right|=2a-1\)
Ta có:\(2a-1-2a\)=\(-1\)
b,B=\(x-2y-\sqrt{x^2-4xy+4y^2}\)
=\(x-2y-\sqrt{\left(x-2y\right)^2}\)=\(x-2y-\left|x-2y\right|\)
nếu\(x-2y\ge0\Leftrightarrow x\ge2y\) thì:\(\left|x-2y\right|=x-2y\)
Ta có:\(x-2y-x+2y=0\)
nếu\(x-2y\le0\Leftrightarrow x\le2y\) thì:\(\left|x-2y\right|=2y-x\)
Ta có:\(x-2y-2y+x=2x-4y\)
c,C=\(2x-1-\dfrac{\sqrt{x^2-10x+25}}{x-5}\)
=\(2x-1-\dfrac{\sqrt{\left(x-5\right)^2}}{x-5}\)=\(2x-1-\dfrac{\left|x-5\right|}{x-5}\)
nếu\(x-5>0\Leftrightarrow x>5\) thì\(\left|x-5\right|=x-5\)
Ta có:\(2x-1-\dfrac{x-5}{x-5}=2x-1-1=2x-2\)
nếu\(x-5< 0\Leftrightarrow x< 5\) thì\(\left|x-5\right|=5-x\)
Ta có:\(2x-1-\dfrac{5-x}{x-5}=2x-1+1=2x\)
\(A=1-4\text{a};B=0;C=2\text{x}-2\)
A.=\(\sqrt{\left(1-2a\right)^2}-2a=\left|1-2a\right|-2a\)
+,Nếu 1-2a\(\ge0\Leftrightarrow a\le\dfrac{1}{2}thì\left|1-2a\right|=1-2a\Leftrightarrow1-2a-2a=1-4a\)
+,Nếu 1-2a<0\(\Leftrightarrow a>\dfrac{1}{2}thì\left|1-2a\right|=2a-1\Leftrightarrow2a-1-2a=-1\)
B.=\(x-2y-\sqrt{\left(x-2y\right)^2}=x-2y-\left|x-2y\right|\)
+,Nếu x-2y\(\ge0\Leftrightarrow x\ge2y=x-2y-\left(x-2y\right)=x-2y-x+2y=0\)
+,Nếu x-2y < 0\(\Leftrightarrow x-2ythì\left|x-2y\right|=2y-x=x-2y-\left(2y-x\right)=x-2y-2y+x=2x\)
C.=2x -1 -\(\dfrac{\sqrt{\left(x-5\right)^2}}{x-5}=2x-1-\dfrac{\left|x-5\right|}{x-5}\)
+,Nếu x-5\(\ge0\Leftrightarrow x>5=2x-1-\dfrac{x-5}{x-5}=2x-1-1=2x-2\)
+,Nếu x-5 <0\(\Leftrightarrow x< 5=2x-1-\dfrac{-\left(x-5\right)}{x-5}=2x-1+1=2x\)
a. -1
b. 2x
c. 2x
a)\(A=\sqrt{1-4a+4a^2}-2a=\sqrt{\left(1-2a\right)^2}-2a=\left|1-2a\right|-2a\)
+ Nếu \(1-2a\ge0\Leftrightarrow a\le\dfrac{1}{2}\) thì \(\left|1-2a\right|=1-2a.\) Khi đó \(A=1-2a-2a=1-4a.\)
+ Nếu \(1-2a< 0\Leftrightarrow a>\dfrac{1}{2}\) thì \(\left|1-2a\right|=2a-1\). Khi đó \(A=2a-1-2a=-1\).
b)\(B=x-2y-\sqrt{x^2-4xy+4y^2}=x-2y-\sqrt{\left(x-2y\right)^2}=x-2y-\left|x-2y\right|.\)
+ Nếu \(x-2y\ge0\Leftrightarrow x\ge2y\) thì \(\left|x-2y\right|=x-2y\). Khi đó \(B=x-2y-\left(x-2y\right)=0\)
+ Nếu \(x-2y< 0\Leftrightarrow x< 2y\) thì \(\left|x-2y\right|=2y-x\). Khi đó \(B=x-2y-\left(2y-x\right)=x-2y-2y+x=2x.\)