LG a

(1−a√a1−√a+√a).(1−√a1−a)2=1(1−aa1−a+a).(1−a1−a)2=1 với a≥0a≥0 và a≠1a≠1

Phương pháp giải:

+ Biến đối vế trái thành vế phải ta sẽ có điều cần chứng minh.

+ √A2=|A|A2=|A|. 

+ |A|=A|A|=A    nếu    A≥0A≥0,

    |A|=−A|A|=−A     nếu    A<0A<0.

+ Sử dụng các hằng đẳng thức:

         a2+2ab+b2=(a+b)2a2+2ab+b2=(a+b)2

         a2−b2=(a+b).(a−b)a2−b2=(a+b).(a−b).

         a3−b3=(a−b)(a2+ab+b2)a3−b3=(a−b)(a2+ab+b2).

Lời giải chi tiết:

Biến đổi vế trái để được vế phải.

Ta có: 

VT=(1−a√a1−√a+√a).(1−√a1−a)2VT=(1−aa1−a+a).(1−a1−a)2

       =(1−(√a)31−√a+√a).(1−√a(1−√a)(1+√a))2=(1−(a)31−a+a).(1−a(1−a)(1+a))2

       =((1−√a)(1+√a+(√a)2)1−√a+√a).(11+√a)2=((1−a)(1+a+(a)2)1−a+a).(11+a)2

       =[(1+√a+(√a)2)+√a].1(1+√a)2=[(1+a+(a)2)+a].1(1+a)2

       =[(1+2√a+(√a)2)].1(1+√a)2=[(1+2a+(a)2)].1(1+a)2

       =(1+√a)2.1(1+√a)2=1=VP=(1+a)2.1(1+a)2=1=VP.

LG b

a+bb2√a2b4a2+2ab+b2=|a|a+bb2a2b4a2+2ab+b2=|a| với a+b>0a+b>0 và b≠0b≠0

Phương pháp giải:

+ Biến đối vế trái thành vế phải ta sẽ có điều cần chứng minh.

+ √A2=|A|A2=|A|. 

+ |A|=A|A|=A    nếu    A≥0A≥0,

    |A|=−A|A|=−A     nếu    A<0A<0.

+ Sử dụng các hằng đẳng thức:

         a2+2ab+b2=(a+b)2a2+2ab+b2=(a+b)2

         a2−b2=(a+b).(a−b)a2−b2=(a+b).(a−b).

         a3−b3=(a−b)(a2+ab+b2)a3−b3=(a−b)(a2+ab+b2).

Lời giải chi tiết:

Ta có:

VT=a+bb2√a2b4a2+2ab+b2VT=a+bb2a2b4a2+2ab+b2

      =a+bb2√(ab2)2(a+b)2=a+bb2(ab2)2(a+b)2

     =a+bb2√(ab2)2√(a+b)2=a+bb2(ab2)2(a+b)2

     =a+bb2|ab2||a+b|=a+bb2|ab2||a+b|

     =a+bb2.|a|b2a+b=|a|=VP=a+bb2.|a|b2a+b=|a|=VP

Vì a+b>0⇒|a+b|=a+ba+b>0⇒|a+b|=a+b.

11+a)2 và làm tiếp.

a+bb2⋅|a|b2|a+b|; với $a+b>0$ và $b\ne 0$, sẽ rút gọn tiếp được kết quả.

a) Ta có : Vì \(x\ge0\)và \(y\ge0\)nên \(x+y\ge0\)\(\Leftrightarrow\left|x+y\right|=x+y\)

\(\frac{2}{x^2-y^2}\sqrt{\frac{3\left(x+y\right)^2}{2}}\)

\(=\frac{2}{x^2-y^2}\sqrt{\frac{3}{2}.\left(x+y\right)^2}\)

\(=\frac{2}{x^2-y^2}.\sqrt{\frac{3}{2}}.\left|x+y\right|\)

\(=\frac{2}{\left(x-y\right)\left(x+y\right)}.\sqrt{\frac{3}{2}}.\left(x+y\right)\)

\(=\frac{2}{x-y}.\sqrt{\frac{3}{2}}\)

\(=\frac{1}{x-y}.2.\sqrt{\frac{3}{2}}\)

\(=\frac{1}{x-y}.\sqrt{\frac{2^2.3}{2}}\)

\(=\frac{1}{x-y}.\sqrt{6}=\frac{\sqrt{6}}{x-y}\)

a, \(\frac{2}{x^2-y^2}\sqrt{\frac{3\left(x+y\right)^2}{2}}=\frac{2}{x^2-y^2}\frac{\sqrt{3}\left|x+y\right|}{\sqrt{2}}=\frac{2\sqrt{3}\left(x+y\right)}{\left(x-y\right)\left(x+y\right)\sqrt{2}}\)

do \(x\ge0;y\ge0\)

\(=\frac{2\sqrt{3}}{\sqrt{2}\left(x-y\right)}=\frac{2\sqrt{6}}{2\left(x-y\right)}=\frac{\sqrt{6}}{x-y}\)

b) Ta có :

\(\frac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}\)

\(=\frac{2}{2a-1}\sqrt{5a^2\left(1-2.2a+2^2.a^2\right)}\)

\(=\frac{2}{2a-1}\sqrt{5a^2\left[1^2-2.1.2a+\left(2a\right)^2\right]}\)

\(=\frac{2}{2a-1}\sqrt{5a^2\left(1-2a\right)^2}\)

\(=\frac{2}{2a-1}.\sqrt{5}.\sqrt{a^2}.\sqrt{\left(1-2a\right)^2}\)

\(=\frac{2}{2a-1}\sqrt{5}.\left|a\right|.\left|1-2a\right|\)

Vì a > 0,5 <=> a > 0 <=> | a | = a

Vì a > 0,5 <=> 2a > 2 . 0,5 <=> 2a > 1 hay 1 < 2a

<=> 1 - 2a < 0 <=> | 1 - 2a | = - ( 1 - 2a )

= -1 + 2a = 2a - 1

Thay vào trên ta được : 

\(=\frac{2}{2a-1}\sqrt{5}.\left|a\right|.\left|1-2a\right|=\frac{2}{2a-1}\sqrt{5}.a.\left(2a-1\right)=2a\sqrt{5}\)

Vậy : \(\frac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}=2a\sqrt{5}\)

a)

$\begin{array}{c}Q=\frac{a}{\sqrt{a^2-b^2}}-\left(1+\frac{a}{\sqrt{a^2-b^2}}\right):\frac{b}{a-\sqrt{a^2-b^2}}\\ =\frac{a}{\sqrt{a^2-b^2}}-\frac{a^2-\left(a^2-b^2\right)}{b\sqrt{a^2-b^2}}\\ =\frac{a}{\sqrt{a^2-b^2}}-\frac{a^2-a^2+b^2}{b\sqrt{a^2-b^2}}\\ =\frac{a}{}\end{array}$

  

2:

\(VT=\dfrac{a^2b}{a-b}\cdot\dfrac{2\sqrt{2}\left(a-b\right)}{5\sqrt{3}\cdot a^2\sqrt{b}}=\dfrac{2}{15}\cdot\sqrt{6b}=VP\)
1: \(=9\sqrt{ab}+\dfrac{7\sqrt{ab}}{b}-\dfrac{5\sqrt{ab}}{a}-3\sqrt{ab}=\)6căn ab+căn ab(7/b-5/a)

=căn ab(6+7/b-5/a)

) V T = ( 2 √ 3 − √ 6 √ 8 − 2 − √ 216 3 ) ⋅ 1 √ 6 = ( √ 2 ⋅ √ 2 ⋅ √ 3 − √ 6 √ 2 2 ⋅ 2 − 2 − √ 6 2 .6 3 ) ⋅ 1 √ 6 = ( √ 2 ⋅ √ 6 − √ 6 2 √ 2 − 2 − 6 . √ 6 3 ) ⋅ 1 √ 6 = [ √ 6 ( √ 2 − 1 ) 2 ( √ 2 − 1 ) − 6 √ 6 3 ] ⋅ 1 √ 6 = ( √ 6 2 − 2 √ 6 ) ⋅ 1 √ 6 = ( √ 6 2 − 4 √ 6 2 ) ⋅ 1 √ 6 = ( − 3 2 √ 6 ) ⋅ 1 √ 6 = − 3 2 = − 1 , 5 = V P . b) V T = ( √ 14 − √ 7 1 − √ 2 + √ 15 − √ 5 1 − √ 3 ) : 1 √ 7 − √ 5 = ( √ 7 ⋅ √ 2 − √ 7 1 − √ 2 + √ 5 ⋅ √ 3 − √ 5 1 − √ 3 ) : 1 √ 7 − √ 5 = [ √ 7 ( √ 2 − 1 ) 1 − √ 2 + √ 5 ( √ 3 − 1 ) 1 − √ 3 ] : 1 √ 7 − √ 5 = ( − √ 7 − √ 5 ) ( √ 7 − √ 5 ) = − ( √ 7 + √ 5 ) ( √ 7 − √ 5 ) = − ( 7 − 5 ) = − 2 = V P . c) V T = a √ b + b √ a √ a b : 1 √ a − √ b = √ a ⋅ √ a ⋅ √ b + √ b ⋅ √ b ⋅ √ a √ a b : 1 √ a − √ b = √ a ⋅ √ a b + √ b ⋅ √ a b √ a b : 1 √ a − √ b = √ a b ( √ a + √ b ) √ a b ⋅ ( √ a − √ b ) = ( √ a + √ b ) ⋅ ( √ a − √ b ) = a − b = V P . d) V T = ( 1 + a + √ a √ a + 1 ) ( 1 − a − √ a √ a − 1 ) = ( 1 + √ a ⋅ √ a + √ a √ a + 1 ) ( 1 − √ a ⋅ √ a − √ a √ a − 1 ) = [ 1 + √ a ( √ a + 1 ) √ a + 1 ] [ 1 − √ a ( √ a − 1 ) √ a − 1 ] = ( 1 + √ a ) ( 1 − √ a ) = 1 − ( √ a ) 2 = 1 − a = V P

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​​)⋅6​1​

=\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​​)⋅6​1​

=\left(\dfrac{\sqrt{2} \cdot \sqrt{6}-\sqrt{6}}{2 \sqrt{2}-2}-\dfrac{6 . \sqrt{6}}{3}\right) \cdot \dfrac{1}{\sqrt{6}}=(22​−22​⋅6​−6​​−36.6​​)⋅6​1​

=\left[\dfrac{\sqrt{6}(\sqrt{2}-1)}{2(\sqrt{2}-1)}-\dfrac{6 \sqrt{6}}{3}\right] \cdot \dfrac{1}{\sqrt{6}}=[2(2​−1)6​(2​−1)​−366​​]⋅6​1​

=\left(\dfrac{\sqrt{6}}{2}-2 \sqrt{6}\right) \cdot \dfrac{1}{\sqrt{6}}=(26​​−26​)⋅6

a) \(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}=\dfrac{1}{2}-2=-\dfrac{3}{2}\)

 

a: Ta có: \(B=\left(\dfrac{6}{a-1}+\dfrac{10-2\sqrt{a}}{a\sqrt{a}-a-\sqrt{a}+1}\right)\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}\)

\(=\dfrac{6\sqrt{a}-6+10-2\sqrt{a}}{\left(\sqrt{a}-1\right)^2\cdot\left(\sqrt{a}+1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}\)

\(=\dfrac{4\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\cdot\dfrac{1}{4\sqrt{a}}\)

\(=\dfrac{1}{\sqrt{a}}\)

a) \(B=\left(\dfrac{6}{a-1}+\dfrac{10-2\sqrt{a}}{a\sqrt{a}-a-\sqrt{a}+1}\right).\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}=\left(\dfrac{6}{a-1}+\dfrac{10-2\sqrt{a}}{\left(a-1\right)\left(\sqrt{a}-1\right)}\right).\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}=\dfrac{6\left(\sqrt{a}-1\right)+10-2\sqrt{a}}{\left(a-1\right)\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}=\dfrac{4\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)^2\left(\sqrt{a}+1\right)}.\dfrac{\left(\sqrt{a}-1\right)^2}{4\sqrt{a}}=\dfrac{1}{\sqrt{a}}\)

b) \(C=B.\left(a-\sqrt{a}+1\right)=\dfrac{a-\sqrt{a}+1}{\sqrt{a}}=\sqrt{a}-1+\dfrac{1}{\sqrt{a}}\ge2\sqrt{\sqrt{a}.\dfrac{1}{\sqrt{a}}}-1=1\)(bất đẳng thức Cauchy cho 2 số dương)

\(1,\\ a,ĐK:\left\{{}\begin{matrix}x\ge0\\x+5\ge0\end{matrix}\right.\Leftrightarrow x\ge0\\ b,Sửa:B=\left(\sqrt{3}-1\right)^2+\dfrac{24-2\sqrt{3}}{\sqrt{2}-1}\\ B=4-2\sqrt{3}+\dfrac{2\sqrt{3}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}\\ B=4-2\sqrt{3}+2\sqrt{3}=4\\ 3,\\ =\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{1+\sqrt{x}}\right]\cdot\dfrac{\sqrt{x}-3+2-2\sqrt{x}}{\left(1-\sqrt{x}\right)\left(\sqrt{x}-3\right)}-2\\ =\left(1-\sqrt{x}\right)\cdot\dfrac{-\sqrt{x}-1}{\left(1-\sqrt{x}\right)\left(\sqrt{x}-3\right)}-2\\ =\dfrac{-\sqrt{x}-1}{\sqrt{x}-3}-2=\dfrac{-\sqrt{x}-1-2\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{-3\sqrt{x}+5}{\sqrt{x}-3}\)

LG a

√18(√2−√3)2;18(2−3)2;

Phương pháp giải:

+ √ab=√a.√bab=a.b,  với a, b≥0a, b≥0.

+ |a|=a|a|=a,  nếu a≥0a≥0 

     |a|=−a|a|=−a  nếu a<0a<0.

+ Sử dụng định lí so sánh hai căn bậc hai số học:  Với hai số a, ba, b không âm, ta có:

a<b⇔√a<√ba<b⇔a<b

Lời giải chi tiết:

Ta có:

√18(√2−√3)2=√18.√(√2−√3)218(2−3)2=18.(2−3)2

                               =√9.2.|√2−√3|=√32.2.|√2−√3|=9.2.|2−3|=32.2.|2−3|

                               =3√2.|√2−√3|=3√2(√3−√2)=32.|2−3|=32(3−2)

                               =3√2.3−3(√2)2=32.3−3(2)2

                               =3√6−3.2=3√6−6=36−3.2=36−6.

(Vì  2<3⇔√2<√3⇔√2−√3<02<3⇔2<3⇔2−3<0

Do đó: |√2−√3|=−(√2−√3)=−√2+√3|2−3|=−(2−3)=−2+3=√3−√2=3−2).

LG b

ab√1+1a2b2ab1+1a2b2

Phương pháp giải:

+ √ab=√a.√bab=a.b,  với a, b≥0a, b≥0.

+ √ab=√a√bab=ab,  với a≥0, b>0a≥0, b>0.

+ |a|=a|a|=a,  nếu a≥0a≥0 

     |a|=−a|a|=−a  nếu a<0a<0.

Lời giải chi tiết:

Ta có: 

ab√1+1a2b2=ab√a2b2a2b2+1a2b2=ab√a2b2+1a2b2ab1+1a2b2=aba2b2a2b2+1a2b2=aba2b2+1a2b2

                         =ab√a2b2+1√a2b2=ab√a2b2+1√(ab)2=aba2b2+1a2b2=aba2b2+1(ab)2

                         =ab√a2b2+1|ab|=aba2b2+1|ab|

Nếu ab>0ab>0 thì |ab|=ab|ab|=ab

          ⇒ab√a2b2+1|ab|=ab√a2b2+1ab=√a2b2+1⇒aba2b2+1|ab|=aba2b2+1ab=a2b2+1.

Nếu ab<0ab<0 thì |ab|=−ab|ab|=−ab

           ⇒ab√a2b2+1|ab|=ab√a2b2+1−ab=−√a2b2+1⇒aba2b2+1|ab|=aba2b2+1−ab=−a2b2+1.

LG c

√ab3+ab4ab3+ab4

Phương pháp giải:

+ √ab=√a.√bab=a.b,  với a, b≥0a, b≥0.

+ √ab=√a√bab=ab,  với a≥0, b>0a≥0, b>0.

+ |a|=a|a|=a,  nếu a≥0a≥0 

     |a|=−a|a|=−a  nếu a<0a<0.

Lời giải chi tiết:

Ta có: 

√ab3+ab4=√a.bb3.b+ab4=√abb4+ab4ab3+ab4=a.bb3.b+ab4=abb4+ab4

=√ab+ab4=√ab+a√(b2)2=√ab+a|b2|=√ab+ab2=ab+ab4=ab+a(b2)2=ab+a|b2|=ab+ab2.

(Vì b2>0b2>0 với mọi b≠0b≠0 nên |b2|=b2|b2|=b2).

LG d

a+√ab√a+√ba+aba+b

Phương pháp giải:

+ √ab=√a.√bab=a.b,  với a, b≥0a, b≥0.

+ √ab=√a√bab=ab,  với a≥0, b>0a≥0, b>0.

+ |a|=a|a|=a,  nếu a≥0a≥0 

     |a|=−a|a|=−a  nếu a<0a<0.

Lời giải chi tiết:

Ta có:

a+√ab√a+√b=(√a)2+√a.√b√a+√b=√a(√a+√b)√a+√ba+aba+b=(a)2+a.ba+b=a(a+b)a+b

=√a=a.

Cách khác:

a+√ab√a+√b=(a+√ab)(√a−√b)(√a+√b)(√a−√b)=a√a−a√b+√ab.√a−√ab.√b(√a)2−(√b)2=a√a−a√b+a√b−b√aa−b=a√a−b√aa−b=√a(a−b)a−b=√a

a) $2\sqrt{3}.(\sqrt{3}-\sqrt{2})=6-2\sqrt{6}.$

ab|ab|1+a2 b2. Rút gọn hơn, ta có kết quả

1a2b2=1+a2 b2.

1a2b2=−1+a2 b2.
c) 

a) \(\sqrt{18\left(\sqrt{2}-\sqrt{3}\right)^2}=3\left(\sqrt{2}-\sqrt{3}\right)\sqrt{2}=6-3\sqrt{6}\)

b) \(ab\sqrt{1+\dfrac{1}{a^2b^2}}=ab\sqrt{\dfrac{a^2b^2+1}{a^2b^2}}=ab.\dfrac{\sqrt{a^2b^2+1}}{ab}=\sqrt{a^2b^2+1}\)

c) \(\sqrt{\dfrac{a}{b^3}+\dfrac{a}{b^4}}=\sqrt{\dfrac{ab+a}{b^4}}=\dfrac{\sqrt{a\left(b+1\right)}}{b^2}\)

d) \(\dfrac{a+\sqrt{ab}}{\sqrt{a}+\sqrt{b}}=\dfrac{\left(a+\sqrt{ab}\right)\left(\sqrt{a}-\sqrt{b}\right)}{a-b}=\dfrac{a\sqrt{a}-a\sqrt{b}+a\sqrt{b}-b\sqrt{a}}{a-b}=\dfrac{\sqrt{a}\left(a-b\right)}{a-b}=\sqrt{a}\)