Rút gọn các biểu thức:
a. \(\dfrac{\sqrt{63y^3}}{\sqrt{7y}}\) với y > 0
b. \(\dfrac{\sqrt{68a^4b^6}}{\sqrt{128a^6b^6}}\) với a < 0, b ≠ 0
Rút gọn các biểu thức
a) \(\dfrac{\sqrt{63y^3}}{\sqrt{7y}};\left(y>0\right)\)
b) \(\dfrac{\sqrt{48x^3}}{\sqrt{3x^5}};\left(x>0\right)\)
c) \(\dfrac{\sqrt{45mn2}}{\sqrt{20m}};\left(m>0;n>0\right)\)
d) \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}};\left(a< 0;b\ne0\right)\)
a. \(\sqrt{\dfrac{63y^3}{7y}}\)=\(\sqrt{9y^2}\)=3y
b.\(\sqrt{\dfrac{48x^3}{3x^5}}\)=\(\sqrt{16\cdot\dfrac{1}{X^2}}\)= \(\sqrt{16}\cdot\sqrt{\dfrac{1}{X^2}}\)=\(4\cdot\dfrac{1}{X}=\dfrac{4}{X}\)
c.\(\sqrt{\dfrac{45mn^2}{20m}}=\sqrt{\dfrac{9n^2}{4}}=\dfrac{\sqrt{9n^2}}{\sqrt{4}}=\dfrac{3n}{2}\)
d. \(\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\dfrac{1}{2\sqrt{2}a}\)
a) \(\dfrac{\sqrt{63y^3}}{\sqrt{7y}}=\sqrt{\dfrac{63y^3}{7y}}=\sqrt{9y^2}=3y\)
b) \(\dfrac{\sqrt{48x^3}}{\sqrt{3x^5}}=\sqrt{\dfrac{48x^3}{3x^5}}=\sqrt{\dfrac{16}{x^2}}=\dfrac{4}{x}\)
c) \(\dfrac{\sqrt{45mn^2}}{\sqrt{20m}}=\sqrt{\dfrac{45mn^2}{20m}}=\sqrt{\dfrac{9n^2}{4}}=\dfrac{3n}{2}\)
d) \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}=\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\dfrac{1}{2\left|a\right|\sqrt{2}}=\dfrac{-1}{2a\sqrt{2}}\)
Rút gọn các biểu thức:
a)\(\frac{\sqrt{63y^3}}{\sqrt{7y}}\) (y>0).
b)\(\frac{\sqrt{48x^3}}{\sqrt{3x^5}}\) (x>0).
c)\(\frac{\sqrt{45mn^2}}{\sqrt{20m}}\) (m>0 và n>0).
d)\(\frac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}\) (a<0 và b khác 0).
a)\(\frac{\sqrt{63y^3}}{\sqrt{7}y}=\frac{\sqrt{7\cdot3^2\cdot y^2\cdot y}}{\sqrt{7}y}=\frac{\sqrt{7}\cdot\sqrt{3^2}\cdot\sqrt{y^2}\cdot\sqrt{y}}{\sqrt{7}y}=\frac{\sqrt{7}\cdot3\cdot y\cdot\sqrt{y}}{\sqrt{7}y}=3\sqrt{y}\)
b)\(\frac{\sqrt{48x^3}}{\sqrt{3x^5}}=\frac{\sqrt{4^2\cdot3\cdot x^2\cdot x}}{\sqrt{3\cdot x^2\cdot x^3}}=\frac{\sqrt{4^2}\cdot\sqrt{3}\cdot\sqrt{x^3}}{\sqrt{3}\cdot\sqrt{x^2}\cdot\sqrt{x^3}}=\frac{4}{x}\)
c)\(\frac{\sqrt{45mn^2}}{\sqrt{20m}}=\frac{\sqrt{5\cdot3^2\cdot m\cdot n^2}}{\sqrt{5\cdot2^2\cdot m}}=\frac{\sqrt{5}\cdot\sqrt{3^2}\cdot\sqrt{m}\cdot\sqrt{n^2}}{\sqrt{5}\cdot\sqrt{2^2}\cdot\sqrt{m}}=\frac{3\left|n\right|}{2}\)
d)\(\frac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}=\frac{\sqrt{4^2\cdot a^2\cdot a^2\cdot b^2\cdot b^2\cdot b^2}}{\sqrt{4^2\cdot8\cdot a^2\cdot a^2\cdot a^2\cdot b^2\cdot b^2\cdot b^2}}=\frac{\sqrt{4^2}\cdot\sqrt{a^2}\cdot\sqrt{a^2}\cdot\sqrt{b^2}\cdot\sqrt{b^2}\cdot\sqrt{b^2}}{\sqrt{4^2}\cdot\sqrt{8}\cdot\sqrt{a^2}\cdot\sqrt{a^2}\cdot\sqrt{a^2}\cdot\sqrt{b^2}\cdot\sqrt{b^2}\cdot\sqrt{b^2}}=\frac{4\cdot a^2\cdot b^3}{4\cdot\sqrt{8}\cdot\left|a\right|^3\cdot b^3}=\frac{a^2}{\sqrt{8}\left|a\right|^3}\)
1) Rút gọn biểu thức \(\sqrt{0,81x^2}\) ta được kết quả là ....
2) Rút gọn \(\dfrac{\sqrt{63y^2}}{\sqrt{7y}}\) (với y < 0) ta được kết quả là ....
\(1,=0,9\left|x\right|\\ 2,Sửa:\dfrac{\sqrt{63y^3}}{\sqrt{7y}}=\sqrt{\dfrac{63y^3}{7y}}=\sqrt{9y^2}=3\left|y\right|=-3y\)
a)\(\sqrt{4\left(a-3\right)^2}vớia\ge3\)
b)\(\sqrt{a^2\left(a+1\right)^2}vớia>0\)
c)\(\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}vớia< 0,b\ne0\)
a) \(\sqrt{4\left(a-3\right)^2}=2\left(a-3\right)=2a-6\)
b) \(\sqrt{a^2\left(a+1\right)^2}=a\left(a+1\right)=a^2+a\)
c) \(\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\dfrac{1}{\sqrt{8}\left|a\right|}=\dfrac{1}{-\sqrt{8}a}=\dfrac{-\sqrt{8}}{8a}\)
a: \(\sqrt{4\left(a-3\right)^2}=2\cdot\left(a-3\right)=2a-6\)
b: \(\sqrt{a^2\left(a+1\right)^2}=a\left(a+1\right)=a^2+a\)
c: \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}=\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\sqrt{\dfrac{2}{16a^2}}=-\dfrac{\sqrt{2}}{4a}\)
\(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}vớia< 0,b\ne0\)
\(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}=\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\sqrt{\dfrac{2}{16a^2}}=-\dfrac{\sqrt{2}}{4a}\)
\(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}=\dfrac{4a^2b^3}{8\sqrt{2}a^3b^3}=\dfrac{1}{2\sqrt{2}a}\)
Cho các biểu thức A=\(\dfrac{6}{x-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\) và B=\(\dfrac{3}{\sqrt{x}-1}\) với x≥0, x≠1, x≠9
a) Tính giá trị của B khi x=4
b) Rút gọn biểu thức P=A-B
c) Tìm xϵN để biểu thức \(\dfrac{1}{P}\) đạt giá trị lớn nhất
a) Thay x=4 vào biểu thức \(B=\dfrac{3}{\sqrt{x}-1}\), ta được:
\(B=\dfrac{3}{\sqrt{4}-1}=\dfrac{3}{2-1}=3\)
Vậy: Khi x=4 thì B=3
b) Ta có: P=A-B
\(\Leftrightarrow P=\dfrac{6}{x-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{3}{\sqrt{x}-1}\)
\(\Leftrightarrow P=\dfrac{6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow P=\dfrac{6+x-\sqrt{x}-3\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow P=\dfrac{x-\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow P=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow P=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow P=\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
Rút gọn các biểu thức:
a) \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}\) ( a <0 ; b # 0 )
b) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\) ( x lớn hơn hoặc = 0)
c) \(\sqrt{\dfrac{\left(x-2\right)^2}{\left(3-x\right)^2}}+\dfrac{x^2-1}{x-3}\) ( x<3 tại x = 0,5)
d) \(\dfrac{x-1}{\sqrt{y}-1}.\sqrt{\dfrac{\left(y-2\sqrt{y}+1^2\right)}{\left(x-1\right)^4}}\) ( x # 1; y >= 0, y #1)
e) \(4x-\sqrt{8}+\dfrac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\) ( x > -2 tại x = -\(\sqrt{2}\))
a) \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}\)
\(=\dfrac{4a^2b^3}{8\sqrt{2}a^3b^3}\)
\(=\dfrac{1}{2\sqrt{2}a}\)
\(=\dfrac{\sqrt{2}}{4a}\)
b) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\)
chịu đấy :v
c) \(\sqrt{\dfrac{\left(x-2\right)^2}{\left(3-x\right)^2}}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{x-2}{3-x}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{x-2}{-\left(x-3\right)}+\dfrac{x^2-1}{x-3}\)
\(=-\dfrac{x-2}{x-3}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{-\left(x-2\right)+x^2-1}{x-3}\)
\(=\dfrac{-x+1+x^2}{x-3}\)
d) \(\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(y-2\sqrt{y}+1^2\right)}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{y-2\sqrt{y}+1}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y-2\sqrt{y}+1}}{\left(x-1\right)^2}\)
\(=\dfrac{1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y-2\sqrt{y}+1}}{x-1}\)
\(=\dfrac{\sqrt{y-2\sqrt{y}+1}}{\left(\sqrt{y}-1\right)\left(x-1\right)}\)
\(=\dfrac{\sqrt{y-2\sqrt{y}+1}}{x\sqrt{y}-\sqrt{y}-x+1}\)
e) \(4x-\sqrt{8}+\dfrac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\)
\(=4x-2\sqrt{2}+\dfrac{\sqrt{x^2\cdot\left(x+2\right)}}{\sqrt{x+2}}\)
\(=4x-2\sqrt{2}+\sqrt{x^2}\)
\(=4x-2\sqrt{x}+x\)
\(=5x-2\sqrt{2}\)
Rút gọn biểu thức:
\(\sqrt{\frac{2a}{3}}.\sqrt{\frac{3a}{8}}vớia\ge0\)\(\sqrt{5a}.\sqrt{45a}-3avớia\ge0\)\(4\sqrt{16a^6}-6a^3\rightarrow kq2TH\)\(\left(3-a\right)^2-\sqrt{0,2}.\sqrt{180a^4}\)\(\sqrt{\frac{27.\left(a-3\right)^2}{48}}vớia< 3\)\(\frac{\sqrt{63y^3}}{\sqrt{7y}}vớiy>0\)\(\frac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^2}}vớia< 0,b\ne0\)\(\frac{a-b}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{a^3}+\sqrt{b^3}}{a-b}\left(a\ge0;b\ge0;a\ne b\right)\)\(\frac{2a+\sqrt{ab}-3b}{2a-5\sqrt{ab}+3b}\left(a,b\ge0;4a\ne9b\right)\)*Liên hệ giữa phép chia và phép khai phương :
Rút gọn các biểu thức :
a,\(\frac{\sqrt{63y^3}}{\sqrt{7y}}\) (y>0)
b,\(\frac{\sqrt{48x^3}}{\sqrt{3x^5}}\) (x>0)
c,\(\frac{\sqrt{45mn^2}}{\sqrt{20m}}\) (m>0 và n>0)
d,\(\frac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}\) (a<0 và b khác 0)
e,\(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\) (x\(\ge\) 0)
f, \(\frac{x-1}{\sqrt{y}-1}\sqrt{\frac{\left(y-2\sqrt{y}+1\right)}{\left(x-1\right)^4}}\) (x\(\ne\)1 , y\(\ne\)1 và y\(\ge\)0)