Bài 1: Rút gọn biểu thức
a) \(\left|x-2\right|+\dfrac{\sqrt{x^2-4x+4}}{x-2}\)
b) \(\sqrt{1-4a+4a^2}-2a\)
c) \(x-2y-\sqrt{x^2-4xy+4y^2}\)
d) \(x^2+\sqrt{x^4-8x^2+16}\)
Rút gọn các biểu thức:
a. \(\sqrt{1-4a+4a^2}-2a\)
b. \(x-2y-\sqrt{x^2-4xy+4y^2}\)
c. \(x^2+\sqrt{x^4-8x^2+16}\)
d. \(2x-1-\dfrac{\sqrt{x^2-10x+25}}{x-5}\)
e. \(\dfrac{\sqrt{x^4-4x^2+4}}{x^2-2}\)
f. \(\sqrt{\left(x-4\right)^2}+\dfrac{x-4}{\sqrt{x^2-8x+16}}\)
\(a.\sqrt{1-4a+4a^2}-2a=\sqrt{\left(1-2a\right)^2}-2a=\left|1-2a\right|-2a\)
*\(a>\dfrac{1}{2}\Rightarrow\left|1-2a\right|-2a=2a-1-2a=4a-1\)
* \(a\le\dfrac{1}{2}\Rightarrow\left|1-2a\right|-2a=1-2a-2a=1-4a\)
\(b.x-2y-\sqrt{x^2-4xy+4y^2}=x-2y-\sqrt{\left(x-2y\right)^2}=x-2y-\left|x-2y\right|\)
* \(x\ge2y\Rightarrow x-2y-\left|x-2y\right|=x-2y-x+2y=2x\)
* \(x< 2y\Rightarrow x-2y-\left|x-2y\right|=x-2y-2y+x=2x-4y\)
\(c.x^2+\sqrt{x^4-8x^2+16}=x^2+\sqrt{\left(x^2-4\right)^2}=x^2+\left|x^2-4\right|\)
* \(x^2-4\ge0\Rightarrow x^2+\left|x^2-4\right|=x^2+x^2-4=2x^2-4\)
* \(x^2-4< 0\Rightarrow x^2+\left|x^2-4\right|=x^2+4-x^2=4\)
\(d.2x-1-\dfrac{\sqrt{x^2-10x+25}}{x-5}=2x-1-\dfrac{\sqrt{\left(x-5\right)^2}}{x-5}=2x-1-\dfrac{\left|x-5\right|}{x-5}\)
* \(x\ge5\Rightarrow2x-1-\dfrac{\left|x-5\right|}{x-5}=2x-1-1=2x-2\)
* \(x< 5\Rightarrow2x-1-\dfrac{\left|x-5\right|}{x-5}=2x-1+1=2x\)
\(e.\dfrac{\sqrt{x^4-4x^2+4}}{x^2-2}=\dfrac{\sqrt{\left(x^2-2\right)^2}}{x^2-2}=\dfrac{\left|x^2-2\right|}{x^2-2}\)
* \(x^2\ge2\Rightarrow\dfrac{\left|x^2-2\right|}{x^2-2}=1\)
* \(x^2< 2\Rightarrow\dfrac{\left|x^2-2\right|}{x^2-2}=-1\)
\(f.\sqrt{\left(x-4\right)^2}+\dfrac{x-4}{\sqrt{x^2-8x+16}}=\left|x-4\right|+\dfrac{x-4}{\sqrt{\left(x-4\right)^2}}=\left|x-4\right|+\dfrac{x-4}{\left|x-4\right|}\)
* \(x\ge4\Rightarrow\left|x-4\right|+\dfrac{x-4}{\left|x-4\right|}=x-4+\dfrac{x-4}{x-4}=x-5\)
* \(x< 4\Rightarrow\left|x-4\right|+\dfrac{x-4}{\left|x-4\right|}=4-x-1=5-x\)
rút gọn các biểu thức sau
a)x-2y-\(\sqrt{x^2-4xy+4y^2}\) d)\(\sqrt{\dfrac{x^4-4x^2+4}{x^2-2}}\)
B)\(x^2+\sqrt{x^4-8x^2+16}\) e)\(\sqrt{\left(x^2-4\right)^2}+\dfrac{x-4}{\sqrt{x^2-8x+16}}\)
C)\(2x-1-\sqrt{\dfrac{x^2-10x+25}{x-5}}\)
a) \(x-2y-\sqrt{x^2-4xy+4y^2}\)
\(=x-2y-\sqrt{\left(x-2y\right)^2}\)
\(=x-2y-\left|x-2y\right|\)
TH1: \(x-2y--\left(x-2y\right)\)
\(=x-2y+x-2y\)
\(=2x-4y\)
TH2: \(x-2y-\left(x-2y\right)\)
\(=x-2y-x+2y\)
\(=0\)
b) \(x^2+\sqrt{x^4-8x^2+16}\)
\(=x^2+\sqrt{\left(x^2-4\right)^2}\)
\(=x^2+\left|x^2-4\right|\)
TH1:
\(x^2+-\left(x^2-4\right)\)
\(=x^2-x^2+4\)
\(=4\)
TH2:
\(x^2+\left(x^2-4\right)\)
\(=x^2+x^2-4\)
\(=2x^2-4\)
c) \(2x-1-\sqrt{\dfrac{x^2-10x+25}{x-5}}\) (x>5)
\(=2x-1-\sqrt{\dfrac{\left(x-5\right)^2}{x-5}}\)
\(=2x-1-\sqrt{x-5}\)
d) \(\sqrt{\dfrac{x^4-4x^2+4}{x^2-2}}\) (\(x>\sqrt{2}\))
\(=\sqrt{\dfrac{\left(x^2-2\right)^2}{x^2-2}}\)
\(=\sqrt{x^2-2}\)
e) \(\sqrt{\left(x^2-4\right)^2}+\dfrac{x-4}{\sqrt{x^2-8x+16}}\)
\(=\left|x^2-4\right|+\dfrac{x-4}{\sqrt{\left(x-4\right)^2}}\)
\(=\left|x^2-4\right|+\sqrt{\dfrac{\left(x-4\right)^2}{\left(x-4\right)^2}}\)
\(=\left|x^2-4\right|+1\)
TH1:
\(x^2-4+1\)
\(=x^2-3\)
TH2:
\(-\left(x^2-4\right)+1\)
\(=-x^2+4+1\)
\(=-x^2+5\)
a: \(A=x-2y-\sqrt{x^2-4xy+4y^2}\)
=x-2y-|x-2y|
Khi x>=2y thì A=x-2y-x+2y=0
Khi x<2y thì A=x-2y+x-2y=2x-4y
b: \(B=x^2+\sqrt{x^4-8x^2+16}\)
\(=x^2+\left|x^2-4\right|\)
TH1: x>=2 hoặc x<=-2
B=x^2+x^2-4=2x^2-4
TH2: -2<=x<=2
B=x^2+4-x^2=4
c: \(C=2x-1-\sqrt{\dfrac{x^2-10x+25}{x-5}}\)
\(=2x-1-\sqrt{\dfrac{\left(x-5\right)^2}{x-5}}=2x-1-\sqrt{x-5}\)
d: \(D=\sqrt{\dfrac{x^4-4x^2+4}{x^2-2}}=\sqrt{\dfrac{\left(x^2-2\right)^2}{x^2-2}}=\sqrt{x^2-2}\)
Rút gọn các biểu thức sau:
a) \(\sqrt{1-4a+4a^2}-2a\)
b) \(x-2y-\sqrt{x^2-4xy+4y^2}\)
c) \(x^2+\sqrt{x^4-8x^2+16}\)
d) \(2x-1-\frac{\sqrt{x^2-10x+25}}{x-5}\)
e) \(\frac{\sqrt{x^4-4x^2+4}}{x^2-2}\)
f) \(\sqrt{\left(x-4\right)^2}+\frac{x-4}{\sqrt{x^2}-8x+16}\)
Giúp em với mọi người ơi!!! Pls!
\(a,\sqrt{1-4a+4a^2}-2a\)
\(=\sqrt{\left(1-2a\right)^2}-2a\)
\(=1-2a-2a\)
\(=1-4a\)
\(b,x-2y-\sqrt{x^2-4xy+4y^2}\)
\(=x-2y-\sqrt{\left(x-2y\right)^2}\)
\(=x-2y-\left(x-2y\right)\)
\(=x-2y-x+2y\)
\(=0\)
\(c,x^2+\sqrt{x^4-8x^2+16}\)
\(=x^2+\sqrt{\left(x^2-4\right)^2}\)
\(=x^2+x^2-4\)
\(=2x^2-4\)
Các câu còn lại tương tự nha
\(a,\sqrt{1-4a+4a^2}-2a\)
\(=\sqrt{\left(1-2a\right)^2}-2a\)
\(=\left(1-2a\right)-2a\)
\(=1-4a\)
\(b,x-2y-\sqrt{x^2-4xy+4y^2}\)
\(=x-2y-\sqrt{\left(x-2y\right)^2}\)
\(=x-2y-\left(x-2y\right)\)
\(=x-2y-x+2y\)
\(=0\)
\(c,x^2+\sqrt{x^4-8x^2+16}\)
\(=x^2+\sqrt{\left(x^2-2^2\right)^2}\)
\(=x^2+\left(x^2-4\right)\)
\(=x^2+x^2-4\)
\(=2x^2-4\)
\(d,2x-1-\frac{\sqrt{x^2-10x+25}}{x-5}\)
\(=2x-1-\frac{\sqrt{\left(x-5\right)^2}}{x-5}\)
\(=2x-1-\frac{x-5}{x-5}\)
\(=2x-1-1\)
\(=2x-2\)
\(=2\left(x-1\right)\)
Bài 1 rút gọn biểu thức
a, A =\(\sqrt{1-4a+4a^2}-2a\)
b, B=\(x-2y-\sqrt{x_{ }^2-4xy+4y^2}\)
c, C=\(x^2+\sqrt{x^4-8x^2+16}\)
d,D=\(2x-1-\dfrac{\sqrt{x^2-10x+25}}{x-5}\)
e, E=\(\dfrac{\sqrt{x^4-4x^2+4}}{x^2-2}\)
f, F=\(\sqrt{\left(x-4\right)^2}+\dfrac{x-4}{\sqrt{x^2-8x+16}}\)
Bài 2 cho biểu thức A=\(\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}\)
a, Với giá trị nào của x thì A có nghĩa
b,Tính A nếu x lớn hơn hoặc bằng \(\sqrt{2}\)
Bài 3 cho 3 số dương x, y, z thỏa mãn điều kiện : xy+yz+zx=1 tính
A= \(x\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}+y\sqrt{\dfrac{\left(1+z^2\right)\left(1+x^2\right)}{1+y^2}}+z\sqrt{\dfrac{\left(1+x^2\right)\left(1+y^2\right)}{1+z^2}}\)
BTVN nhiều nhỉ?
a,A=-1
b,B=2x-4y
c,C=2x^2-4
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
Rút gọn các biểu thức sau:
a) \(A=\sqrt{1-4a+4a^2}-2a\)
b) \(B=x-2y-\sqrt{x^2-4xy+4y^2}\)
c) \(C=2x-1-\dfrac{\sqrt{x^2-10x+25}}{x-5}\)
a)=1-4a
b) = 2x - 4y
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)
Rút gọn các biểu thức sau:
a, \(\sqrt{1-4a+4a^2}\) -2a với a ≥ \(\frac{1}{2}\)
b, x- 2y- \(\sqrt{x^2-4xy+4y^2}\) với x<2y
c, x2 + \(\sqrt{x^4-8x^2+16}\) với x2<4
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 $|x^2-4|=4-x^2$)
Rút gọn biểu thức:
a) \(\dfrac{\sqrt{x^2+4x+4}}{x-1}\)
b) \(x-2y-\sqrt{x^2-4xy+4y^2}\) ( x>= 0; y>=0)
c) \(\dfrac{\sqrt{x^2+4x+4}}{x^2-4}\)
d) \(\dfrac{\sqrt{x^2+4x+4}}{x^2-2}\)
a: \(=\dfrac{\left|x+2\right|}{x-1}\)
b: \(=x-2y-\left|x-2y\right|\)\(=\left[{}\begin{matrix}x-2y-x+2y=0\\x-2y+x-2y=2x-4y\end{matrix}\right.\)
c: \(=\dfrac{\left|x+2\right|}{\left(x+2\right)\left(x-2\right)}=\pm\dfrac{1}{x-2}\)
Cho biểu thức: \(A=\dfrac{\sqrt{x}-1}{2\sqrt{x}+1}-\dfrac{3}{1-2\sqrt{x}}-\dfrac{4\sqrt{x}+4}{4x-1}\) và \(B=\dfrac{\sqrt{x}-4}{\sqrt{x}}\)với x > 0 , x = 1/4
a. TÍnh giá trị của biểu thức B biết \(x=\sqrt{28-16\sqrt{3}}+2\sqrt{3}\)
b. Rút gọn biểu thức A
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:
\(B=\dfrac{2-4}{2}=-1\)
Cho biểu thức:
\(A=\left(\dfrac{\sqrt{x}-2}{\sqrt{x}+2}-\dfrac{8\sqrt{x}}{4-x}\right):\dfrac{\sqrt{x}+2}{1-2\sqrt{x}}\)
với x ≥ 0 , x ≠ 4 . x ≠ 1/4
a. Rút gọn A
b. Tìm x để A = -1/3