I . Rút gọn biểu thức
a. \(\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{3\left(x+y\right)^2}{2}}\)
b. \(\dfrac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}\)
a)\(\dfrac{2}{x^2-y^2}\sqrt{\dfrac{3\left(x+y\right)^2}{2}}\left(x,y\ge0;x\ne y\right)\)
b)\(\dfrac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}\left(a>0,5\right)\)
Rút gọn các biểu thức sau:
\(A=\dfrac{a^2-1}{3}\sqrt{\dfrac{9}{\left(1-a\right)^2}}\) với a < 1
\(B=\sqrt{\left(3a-5\right)^2}-2a+4\) với a < \(\dfrac{1}{2}\)
\(C=4a-3-\sqrt{\left(2a-1\right)^2}\) với a < 2
\(D=\dfrac{a-2}{4}\sqrt{\dfrac{16a^4}{\left(a-2\right)^2}}\) với a < 2
a) Ta có: \(A=\dfrac{a^2-1}{3}\cdot\sqrt{\dfrac{9}{\left(1-a\right)^2}}\)
\(=\dfrac{\left(a+1\right)\cdot\left(a-1\right)}{3}\cdot\dfrac{3}{\left|1-a\right|}\)
\(=\dfrac{\left(a+1\right)\left(a-1\right)}{1-a}\)
=-a-1
b) Ta có: \(B=\sqrt{\left(3a-5\right)^2}-2a+4\)
\(=\left|3a-5\right|-2a+4\)
\(=5-3a-2a+4\)
=9-5a
c) Ta có: \(C=4a-3-\sqrt{\left(2a-1\right)^2}\)
\(=4a-3-\left|2a-1\right|\)
\(=4a-3-2a+1\)
\(=2a-2\)
d) Ta có: \(D=\dfrac{a-2}{4}\cdot\sqrt{\dfrac{16a^4}{\left(a-2\right)^2}}\)
\(=\dfrac{a-2}{4}\cdot\dfrac{4a^2}{\left|a-2\right|}\)
\(=\dfrac{a^2\left(a-2\right)}{-\left(a-2\right)}\)
\(=-a^2\)
Câu 1:
\(C=\dfrac{1}{x+2}-\dfrac{x^3-4x}{x^2+4}\cdot\left(\dfrac{1}{x^2+4x+4}-\dfrac{1}{4-x^2}\right)\)
a) Rút gọn C
b) x bằng mấy để C = 1?
Câu 2:
\(B=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
a) Rút gọn B
b) x bằng mấy để \(\left|B\right|=B\)
Câu 3: Rút gọn:
\(A=\left[\dfrac{\left(1-a\right)^2}{3a+\left(a-1\right)^2}+\dfrac{2a^2-4a-1}{a^3-1}-\dfrac{1}{1-a}\right]:\dfrac{2a}{a^3+a}\)
Rút gọn biểu thức sau: C= \(\dfrac{1}{2a-1}.\sqrt{5a^4.\left(1-4a+4a^2\right)}\)
Rút gọn:
a. \(\dfrac{2}{x^2-y^2}\sqrt{\dfrac{3\left(x+y\right)^2}{2}}\) với \(x\ge0,y\ge0\) và \(x\ne y;\)
b. \(\dfrac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}\) với a > 0,5.
a) Vì nên
. Do đó:
=
b)
Vì a>0,5 nên 2a-1>0. Do đó .
Cho biểu thức:
\(B=\left(\dfrac{3}{\sqrt{x}+2}-\dfrac{1}{\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}-6}{x-2\sqrt{x}}+\dfrac{1}{\sqrt{x}}\right)\)
với x > 0 , x ≠ 4
a. Rút gọn B
b. Tìm x để B = 1/3
a) Ta có: \(B=\left(\dfrac{3}{\sqrt{x}+2}-\dfrac{1}{\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}-6}{x-2\sqrt{x}}+\dfrac{1}{\sqrt{x}}\right)\)
\(=\dfrac{3\sqrt{x}-6-\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{\sqrt{x}-6+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}-8}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+2}\)
b) Để \(B=\dfrac{1}{3}\) thì \(\dfrac{\sqrt{x}}{\sqrt{x}+2}=\dfrac{1}{3}\)
\(\Leftrightarrow3\sqrt{x}=\sqrt{x}+2\)
\(\Leftrightarrow2\sqrt{x}=2\)
\(\Leftrightarrow x=1\)(thỏa ĐK)
a) B= \(\left(\dfrac{3\left(\sqrt{x}-2\right)-1\left(\sqrt{x}+2\right)}{x-4}\right):\left(\dfrac{\sqrt{x}-6+1\left(\sqrt{x}-2\right)}{x-2\sqrt{x}}\right)\)
\(=\dfrac{2\sqrt{x}-8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\left(\sqrt{x}-2\right)\sqrt{x}}{2\sqrt{x}-8}\)=\(\dfrac{\sqrt{x}}{\sqrt{x}+2}\)
b) Để B=\(\dfrac{1}{3}\Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}+2}=\dfrac{1}{3}\Leftrightarrow\sqrt{x}+2=3\sqrt{x}\Rightarrow x=1\)
rút gọn
\(\dfrac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}\) với a > 0,5
\(=\dfrac{2\sqrt{5}\left|a\left(2a-1\right)\right|}{2a-1}=\dfrac{2a\left(2a-1\right)\sqrt{5}}{2a-1}=2a\sqrt{5}\)
\(=\dfrac{2\sqrt{5}\cdot a\left(2a-1\right)}{2a-1}=2a\sqrt{5}\)
Rút gọn
a)\(A=\dfrac{2}{x^2-y^2}.\sqrt{\dfrac{3x^2+6xy+3y^2}{4}}\)
b)\(B=\dfrac{1}{2a-1}.\sqrt{5a^4\left(1-4a+4a^2\right)}\)
\(a.A=\dfrac{2}{x^2-y^2}.\sqrt{\dfrac{3x^2+6xy+3y^2}{4}}=\dfrac{2}{\left(x-y\right)\left(x+y\right)}.\dfrac{\left(x+y\right)\sqrt{3}}{2}=\dfrac{\sqrt{3}}{x-y}\) ( x # y )
\(b.\dfrac{1}{2x-1}.\sqrt{5a^4\left(1-4x+4a^2\right)}=\dfrac{1}{2a-1}.\left(2a-1\right)a^2\sqrt{5}=a^2\sqrt{5}\) ( a # \(\dfrac{1}{2}\) )
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