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a: \(VT=\dfrac{3\sqrt{6}}{2}+\dfrac{2\sqrt{6}}{3}-\dfrac{4\sqrt{6}}{2}\)
\(=\dfrac{-\sqrt{6}}{2}+\dfrac{2\sqrt{6}}{3}=\dfrac{-3\sqrt{6}+4\sqrt{6}}{6}=\dfrac{\sqrt{6}}{6}\)
b: \(VT=\dfrac{\left(\sqrt{6x}+\dfrac{\sqrt{6x}}{3}+\sqrt{6x}\right)}{\sqrt{6x}}\)
\(=1+\dfrac{1}{3}+1=2\dfrac{1}{3}\)
+ Ta có:
2√6−√5=2(√6+√5)(√6−√5)(√6+√5)26−5=2(6+5)(6−5)(6+5)
=2(√6+√5)(√6)2−(√5)2=2(√6+√5)6−5=2(6+5)(6)2−(5)2=2(6+5)6−5
=2(√6+√5)1=2(√6+√5)=2(6+5)1=2(6+5).
+ Ta có:
3√10+√7=3(√10−√7)(√10+√7)(√10−√7)310+7=3(10−7)(10+7)(10−7)
=3(√10−√7)(√10)2−(√7)2=3(10−7)(10)2−(7)2=3(√10−√7)10−7=3(10−7)10−7
=3(√10−√7)3=√10−√7=3(10−7)3=10−7.
+ Ta có:
1√x−√y=1.(√x+√y)(√x−√y)(√x+√y)1x−y=1.(x+y)(x−y)(x+y)
=√x+√y(√x)2−(√y)2=√x+√yx−y=x+y(x)2−(y)2=x+yx−y
+ Ta có:
2ab√a−√b=2ab(√a+√b)(√a−√b)(√a+√b)2aba−b=2ab(a+b)(a−b)(a+b)
=2ab(√a+√b)(√a)2−(√b)2=2ab(√a+√b)a−b=2ab(a+b)(a)2−(b)2=2ab(a+b)a−b.
\(\frac{2}{\sqrt{6}-\sqrt{5}}=\frac{2\left(\sqrt{6}+\sqrt{5}\right)}{\left(\sqrt{6}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{5}\right)}=\frac{2\left(\sqrt{6}+\sqrt{5}\right)}{6-5}=2\left(\sqrt{6}+\sqrt{5}\right)\)
\(\frac{3}{\sqrt{10}+\sqrt{7}}=\frac{3\left(\sqrt{10}-\sqrt{7}\right)}{\left(\sqrt{10}-\sqrt{7}\right)\left(\sqrt{10}+\sqrt{7}\right)}=\frac{3\left(\sqrt{10}-\sqrt{7}\right)}{10-7}=\sqrt{10}-\sqrt{7}\)
\(\frac{1}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{x}+\sqrt{y}}{x-y}\)
\(\frac{2ab}{\sqrt{a}-\sqrt{b}}=\frac{2ab\left(\sqrt{a}+\sqrt{b}\right)}{a-b}\)
a: \(A=\dfrac{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}-\dfrac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}-3}-6\)
\(=-\sqrt{x}+3-\sqrt{x}+3-6=-2\sqrt{x}\)
b: \(\left(\dfrac{2\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right):\left(\dfrac{2\sqrt{x}}{\sqrt{x}+1}-1\right)\)
\(=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x+1\right)}-\dfrac{1}{\sqrt{x}+1}\right):\dfrac{2\sqrt{x}-\sqrt{x}-1}{\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}-x-1}{\left(\sqrt{x}+1\right)\left(x+1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{1}{x+1}\)
g: \(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\left(\dfrac{x-1}{\sqrt{x}+1}-2\right)\)
\(=\dfrac{\sqrt{x}+1+\sqrt{x}-1}{x-1}\cdot\left(\sqrt{x}-1-2\right)\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{x-1}\)
a) Ta có: \(\left(2-\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\right)\left(2+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)=\left[2-\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}\right]\left[2+\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right]\)\(=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=2^2-\left(\sqrt{3}\right)^2=4-3=1\) (đpcm)
b) Ta có \(A=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+1}{x-4\sqrt{x}+4}\)\(=\left[\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}\right].\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}\)\(=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}.\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
a) VT=\left(\dfrac{2 \sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right) \cdot \dfrac{1}{\sqrt{6}}VT=(8−223−6−3216)⋅61
=\left(\dfrac{\sqrt{2} \cdot \sqrt{2} \cdot \sqrt{3}-\sqrt{6}}{\sqrt{2^{2} \cdot 2}-2}-\dfrac{\sqrt{6^{2} .6}}{3}\right) \cdot \dfrac{1}{\sqrt{6}}=(22⋅2−22⋅2⋅3−6−362.6)⋅
d) Ta có: \(D=\left(\dfrac{5\sqrt{x}-6}{x-9}-\dfrac{2}{\sqrt{x}+3}\right):\left(1+\dfrac{6}{x-9}\right)\)
\(=\dfrac{5\sqrt{x}-6-2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{x-9+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{5\sqrt{x}-6-2\sqrt{x}+6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{x-3}\)
\(=\dfrac{3\sqrt{x}}{x-3}\)
f) Ta có: \(\left(\dfrac{3}{\sqrt{1+x}}+\sqrt{1-x}\right):\left(\dfrac{3}{\sqrt{1-x^2}}+1\right)\)
\(=\dfrac{3+\sqrt{1-x^2}}{\sqrt{1+x}}:\dfrac{3+\sqrt{1-x^2}}{\sqrt{1-x^2}}\)
\(=\dfrac{\sqrt{1-x^2}}{\sqrt{1+x}}=\sqrt{1-x}\)
a)\(=4\sqrt{6}-3\sqrt{6}+1-\sqrt{6}\)
\(=1\)
b)ĐK: \(x>0,x\ne9\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)-x-9}{x-9}\right):\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3\sqrt{x}-9}{\sqrt{x}+3}.\dfrac{\sqrt{x}}{2\sqrt{x}+4}\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
a) -17√3/3 b) 11√6
c) 21 d) 11
a) a) Biến đổi vế trái thành 32√6+23√6−42√6326+236−426 và làm tiếp.
b) Biến đổi vế trái thành (√6x+13√6x+√6x):√6x(6x+136x+6x):6x và làm tiếp
a) \(\frac{3}{2}\sqrt{6}+2\sqrt{\frac{2}{3}}-4\sqrt{\frac{3}{2}}\)
\(=\frac{3}{2}\sqrt{6}+\frac{2}{3}\sqrt{6}-\frac{4}{2}\sqrt{6}\)
\(=\left(\frac{3}{2}+\frac{2}{3}-\frac{4}{2}\right)\sqrt{6}\)
\(=\frac{1}{6}\cdot\sqrt{6}=\frac{\sqrt{6}}{6}\left(đpcm\right)\)
b) \(\left(x\sqrt{\frac{6}{x}}+\sqrt{\frac{2x}{3}}+\sqrt{6x}\right):\sqrt{6x}\)
\(=\left(\sqrt{6x}+\frac{1}{3}\sqrt{6x}+\sqrt{6x}\right):\sqrt{6x}\)
\(=\left[\left(1+\frac{1}{3}+1\right)\sqrt{6x}\right]:\sqrt{6x}\)
\(=\frac{7}{3}\sqrt{6x}:\sqrt{6x}=\frac{7}{3}=2\frac{1}{3}\left(đpcm\right)\)
a) \(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}=\dfrac{3}{2}\sqrt{6}+\sqrt{6}.\dfrac{2}{3}-\dfrac{4}{2}\sqrt{6}=\sqrt{6}.\left(\dfrac{3}{2}+\dfrac{2}{3}-2\right)=\dfrac{\sqrt{6}}{6}\)
b) \(\left(x\sqrt{\dfrac{6}{x}}+\sqrt{\dfrac{2x}{3}}+\sqrt{6x}\right):\sqrt{6x}=\left(\sqrt{6x}+\sqrt{6x}.\dfrac{1}{3}+\sqrt{6x}\right):\sqrt{6x}=2\dfrac{1}{3}\)
a) =\(\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}+\dfrac{4}{2}\sqrt{6}\)
=\(\left(\dfrac{3}{2}+\dfrac{2}{3}+\dfrac{4}{2}\right)\).\(\sqrt{6}\)
=\(\dfrac{1}{6}.\sqrt{6}\) = \(\dfrac{\sqrt{6}}{6}\) ( đpcm)
b) =\(\left(x\sqrt{\dfrac{6}{x}}+\sqrt{\dfrac{2x}{3}}+\sqrt{6x}\right)\):\(\sqrt{6x}\)
=\(\left(\sqrt{6x}+\dfrac{1}{3}\sqrt{6x}+\sqrt{6x}\right)\):\(\sqrt{6x}\)
= \(\dfrac{7}{3}\)\(\sqrt{6}\) : \(\sqrt{6}\)
=\(\dfrac{7}{3}\) = \(2\dfrac{1}{3}\) (đpcm)
a. VT = \(\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{6}\)
= \(\sqrt{6}\left(\dfrac{3}{2}+\dfrac{2}{3}-2\right)\)= \(\sqrt{6}.\dfrac{1}{6}\)= \(\dfrac{\sqrt{6}}{6}\)
b. VT = \(\left(\sqrt{6x}+\dfrac{1}{3}\sqrt{6x}+\sqrt{6x}\right):\sqrt{6x}\)= \(2\dfrac{1}{3}\)
a)
a) \(\dfrac{3}{2}\sqrt{6}\) + \(\dfrac{2}{3}\sqrt{6}\) - \(\dfrac{4}{2}\sqrt{6}\) = \(\left(\dfrac{3}{2}+\dfrac{2}{3}-\dfrac{4}{2}\right)\)\(\sqrt{6}\) = \(\dfrac{1}{6}\sqrt{6}=\dfrac{\sqrt{6}}{6}\)=VT
b) \(\left(\sqrt{6x}+\dfrac{1}{3}\sqrt{6x}+\sqrt{6x}\right):\sqrt{6x}=\dfrac{7}{3}\sqrt{6x}\) x \(\dfrac{1}{\sqrt{6x}}\)= \(\dfrac{7\sqrt{6}}{3\sqrt{6}}=\dfrac{7}{3}\) = \(2\dfrac{1}{3}\)=VT
a)\(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}=\dfrac{\sqrt{6}}{6}\)
Xét VT=\(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}\)
⇔VT=\(\dfrac{3}{2}\sqrt{6}+2\dfrac{\sqrt{6}}{3}-4\dfrac{\sqrt{6}}{2}\)
⇔VT=\(\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{6}\)
⇔VT=\(\left(\dfrac{3}{2}+\dfrac{2}{3}-2\right)\cdot\sqrt{6}\)
⇔VT=\(\dfrac{1}{6}\cdot\sqrt{6}\)
⇔VT=\(\dfrac{\sqrt{6}}{6}\)=VP
→\(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}=\dfrac{\sqrt{6}}{6}\left(ĐPCM\right)\)
b)\(\left(x\sqrt{\dfrac{6}{x}}+\sqrt{\dfrac{2x}{3}}+\sqrt{6x}\right)\div\sqrt{6x}=2\dfrac{1}{3}\) với x>0
Xét VT=\(\left(x\sqrt{\dfrac{6}{x}}+\sqrt{\dfrac{2x}{3}}+\sqrt{6x}\right)\div\sqrt{6x}\)
⇔VT=\(\left(x\cdot\dfrac{\sqrt{6x}}{x}+\dfrac{\sqrt{6x}}{3}+\sqrt{6x}\right)\div\sqrt{6x}\)
⇔VT=\(\left(\sqrt{6x}+\dfrac{1}{3}\sqrt{6x}+\sqrt{6x}\right)\div\sqrt{6x}\)
⇔VT=\(\left(1+\dfrac{1}{3}+1\right)\cdot\sqrt{6x}\cdot\dfrac{1}{\sqrt{6x}}\)
⇔VT=\(\dfrac{7}{3}\)=\(2\dfrac{1}{3}\)=VP
→\(\left(x\sqrt{\dfrac{6}{x}}+\sqrt{\dfrac{2x}{3}}+\sqrt{6x}\right)\div\sqrt{6x}=2\dfrac{1}{3}\left(ĐPCM\right)\)
a/ \(\dfrac{3}{2}\)\(\sqrt{6}\) + 2\(\sqrt{\dfrac{2}{3}}\) - 4\(\sqrt{\dfrac{3}{2}}\)
= \(\dfrac{3}{2}\)\(\sqrt{6}\) + \(\dfrac{2}{3}\)\(\sqrt{6}\) - 2\(\sqrt{6}\)
= (\(\dfrac{3}{2}\) + \(\dfrac{2}{3}\) -2)\(\sqrt{6}\)
=\(\dfrac{1}{6}\)\(\sqrt{6}\)
= \(\dfrac{\sqrt{6}}{6}\)
b/ (x\(\sqrt{\dfrac{6}{x}}\) + \(\sqrt{\dfrac{2x}{3}}\) + \(\sqrt{6x}\)) : \(\sqrt{6x}\)
= (\(\sqrt{\dfrac{6x^2}{x}}\) + \(\sqrt{\dfrac{6x}{3^2}}\) + \(\sqrt{6x}\)) : \(\sqrt{6x}\)
=(1 + \(\dfrac{1}{3}\) + 1)\(\sqrt{6x}\) : \(\sqrt{6x}\)
=2\(\dfrac{1}{3}\)
a) \(\dfrac{3}{2}\)\(\sqrt{6}\) +2\(\sqrt{\dfrac{2}{3}}\)- 4\(\sqrt{\dfrac{3}{2}}\)=\(\dfrac{\sqrt{6}}{6}\)
Biến đổi VT ta có:
\(\dfrac{3}{2}\)\(\sqrt{6}\)+2\(\sqrt{\dfrac{2}{3}}\)-4\(\sqrt{\dfrac{3}{2}}\)
=\(\dfrac{3}{2}\)\(\sqrt{6}\)+\(\dfrac{2}{3}\)\(\sqrt{6}\)-2\(\sqrt{6}\)
=\(\dfrac{\sqrt{6}}{6}\)=VP;Vậy VT=VP
b)(x\(\sqrt{\dfrac{6}{x}}\)+\(\sqrt{\dfrac{2x}{3}}\)+\(\sqrt{6x}\)):\(\sqrt{6x}\)=2\(\dfrac{1}{3}\) x>o
Biến đổi VT ta có:
=(x\(\dfrac{1}{x}\)\(\sqrt{6x}\)+\(\dfrac{1}{3}\)\(\sqrt{6x}\)+\(\sqrt{6x}\)):\(\sqrt{6x}\)
=2\(\dfrac{1}{3}\)\(\sqrt{6x}\):\(\sqrt{6x}\)
=2\(\dfrac{1}{3}\)(TMDK) =VP
Vậy VT=VP
a. biến đổi vế trái ta có :
VT = \(\dfrac{3}{2}\sqrt{6}\) + 2\(\sqrt{\dfrac{2}{3}}\) - 4\(\sqrt{\dfrac{3}{2}}\)
VT = \(\dfrac{3}{2}\sqrt{6}\) + \(\dfrac{2}{3}\sqrt{6}\) - \(\dfrac{4}{2}\sqrt{6}\)
VT = \(\dfrac{1}{6}\sqrt{6}\)
VT = \(\dfrac{\sqrt{6}}{6}\) = VP (đpcm)
b. biến đổi vế trái ta có :
VT = (\(\sqrt{6x}\) + \(\dfrac{1}{3}\sqrt{6x}\) + \(\sqrt{6x}\) ) \(\div\) \(\sqrt{6x}\)
VT = \(\dfrac{7}{3}\sqrt{6x}\) \(\div\sqrt{6x}\)
VT = \(\dfrac{7}{3}\)
VT = \(2\dfrac{1}{3}\) = VP (vs x>0) (đpcm)
a) Biến đổi vế trái thành \dfrac{3}{2} \sqrt{6}+\dfrac{2}{3} \sqrt{6}-\dfrac{4}{2} \sqrt{6}236+326−246 và làm tiếp.
b) Biến đổi vế trái thành \left(\sqrt{6 x}+\dfrac{1}{3} \sqrt{6 x}+\sqrt{6 x}\right): \sqrt{6x}(6x+316x+6x):6x và làm tiếp.
bài 4
a) BĐVT
b) BĐVT
b,
a)Biến đổi vế trái ta có :
\(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}\)=\(\dfrac{3\sqrt{6}}{2}+\sqrt{6}-2\sqrt{6}\)
=\(\dfrac{\sqrt{6}}{2}\)( bằng vế phải)
b) Biến đổi vế trái ta có
\(\left(x\sqrt{\dfrac{6}{x}}+\sqrt{\dfrac{2x}{3}}+\sqrt{6x}\right):\sqrt{6x}\)=\(\left(\sqrt{6x}+\dfrac{\sqrt{6x}}{3}+\sqrt{6x}\right):\sqrt{6x}\)
=\(\dfrac{7}{3}\sqrt{6x}:\sqrt{6x}\)
=\(\dfrac{7}{3}\)( bằng vế phải)
\(VT=\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{3^2.\dfrac{2}{3}}-2\sqrt{2^2.\dfrac{3}{2}}=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{6}=\left(\dfrac{3}{2}+\dfrac{2}{3}-2\right)\sqrt{6}=\left(\dfrac{9}{6}+\dfrac{4}{6}-\dfrac{12}{6}\right)\sqrt{6}=\dfrac{1}{6}.\sqrt{6}=\dfrac{\sqrt{6}}{6}=VP\)
Vậy\(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}=\dfrac{\sqrt{6}}{6}\)
a)VT=\(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}\)=\(3\sqrt{\dfrac{3}{2}}+\dfrac{4}{3}\sqrt{\dfrac{3}{2}}-4\sqrt{\dfrac{3}{2}}\)=\(\dfrac{1}{3}\sqrt{\dfrac{3}{2}}\)=\(\sqrt{\dfrac{3}{18}}\)=\(\sqrt{\dfrac{1}{6}}\)=\(\dfrac{\sqrt{6}}{6}\)=VT
a,\(\dfrac{\sqrt{6}}{6}\)
b,\(\dfrac{7}{3\sqrt{2x}}\)
a)
b)