Bài 3: Liên hệ giữa phép nhân và phép khai phương

\(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}=2\sqrt{2}\)

\(\Leftrightarrow\sqrt{x+2\sqrt{2\left(x-2\right)}}+\sqrt{x-2\sqrt{2\left(x-2\right)}}=2\sqrt{2}\)

\(\Leftrightarrow2x+2\sqrt{\left[x+2\sqrt{2\left(x-2\right)}\text{ }\right]\left[x-2\sqrt{2\left(x-2\right)}\text{ }\right]}=8\)

\(\Leftrightarrow2\sqrt{\left[x+2\sqrt{2\left(x-2\right)}\text{ }\right]\left[x-2\sqrt{2\left(x-2\right)}\text{ }\right]}=8-2x\)

\(\Leftrightarrow4\left[x+2\sqrt{2\left(x-2\right)}\text{ }\right]\left[x-2\sqrt{2\left(x-2\right)}\text{ }\right]=64-32x+4x^2\)

\(\Leftrightarrow4x^2-32x+64=64-32x+4x^2+\)

\(\Leftrightarrow64=64\) (Đúng)

⇒ Phương trình có vô số nghiệm.

Vậy \(S=\mathbb R\).

\(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}=2\sqrt{2}\)

ĐK: \(x\ge2\), PT tương đương với:

\(x+2\sqrt{2x-4}+2\sqrt{\left(x+2\sqrt{2x-4}\right)\left(x-2\sqrt{2x-4}\right)}+x-2\sqrt{2x-4}=8\)

\(\Leftrightarrow2x+2\sqrt{x^2-4\left(2x-4\right)}=8\)

\(\Leftrightarrow2x+2\sqrt{x^2-8x+16}=8\\ \Leftrightarrow x+\left|x-4\right|=8\)

Với x < 4 => \(x+4-x=8\)

\(\Leftrightarrow4=8\) (loại)

Với \(x\ge4\) => \(x+x-4=8\)

\(\Leftrightarrow x=6\) (thỏa mãn)

`\sqrt{[27(x-1)^2]/12} +3/2 - (x - 2)\sqrt{[50x^2]/[8(x-2)^2]}`   `(1 < x < 2)`

`=\sqrt{[3(x-1)]^2 .3}/\sqrt{2^2 .3} + 3/2 - (x - 2) \sqrt{(5x)^2 . 2}/\sqrt{[2(x - 2)]^2 . 2}`

`=[3\sqrt{3}|x-1|]/[2\sqrt{3}]+3/2-(x-2)[5\sqrt{2}|x|]/[2\sqrt{2}|x-2|]`

`=[3(x-1)]/2+3/2-[5x(x-2)]/[2(2-x)]`   (Vì `1 < x < 2`)

`=3/2x - 3/2 + 3/2 + 5/2x`

`=4x`

a: =1/căn 2(căn 6-2căn 5+căn 6+2căn 5)

=1/căn2(căn 5-1+căn 5-1)

=2*căn 5/2

=căn 10

b: =1/căn 2(căn 4+2căn 3-căn 4-2căn 3)

=1/căn 2(căn 3+1-căn 3+1)

=1/căn 2*2=căn 2

c: =1/căn 2(căn 8+2 căn 7-căn 8-2căn 7)

=1/căn 2(căn 7+1-căn 7+1)

=1/căn2 *2=căn 2

d: =1/căn 2(căn 12-2căn 11+căn 12-2căn 11)

=1/căn 2(căn 11-1+căn 11-1)

=2*căn 11/2=căn 22

e: =1/căn 2(căn 8-2căn 15-căn 8+2căn 15)

=1/căn 2(căn 5-căn 3-căn 5-căn 3)

=-2*căn 3/2=-căn6

1) \(\sqrt{11-6\sqrt{2}}-\sqrt{6-4\sqrt{2}}\)

\(=\sqrt{9-6\sqrt{2}+2}-\sqrt{4-4\sqrt{2}+2}\)

\(=\sqrt{3^2-2\cdot3\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{2^2-2\cdot2\cdot\sqrt{2}\cdot\left(\sqrt{2}\right)^2}\)

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

\(=\left|3-\sqrt{2}\right|-\left|2-\sqrt{2}\right|\)

\(=3-\sqrt{2}-2+\sqrt{2}=1\)

2) \(\sqrt{\left(1-\sqrt{2}\right)^2}-\sqrt{\left(2-\sqrt{2}\right)^2}\)

\(=\left|1-\sqrt{2}\right|-\left|2-\sqrt{2}\right|\)

\(=\sqrt{2}-1+2-\sqrt{2}\)

\(=1\)

3) \(\sqrt{\left(2\sqrt{2}-1\right)^2}-\sqrt{17+12\sqrt{2}}\)

\(=\sqrt{\left(2\sqrt{2}-1\right)^2}-\sqrt{9+12\sqrt{2}+8}\)

\(=\sqrt{\left(2\sqrt{2}-1\right)^2}-\sqrt{3^2+2\cdot3\cdot2\sqrt{2}+\left(2\sqrt{2}\right)^2}\)

\(=\sqrt{\left(2\sqrt{2}-1\right)^2}-\sqrt{\left(3+2\sqrt{2}\right)^2}\)

\(=\left|2\sqrt{2}-1\right|-\left|3+2\sqrt{2}\right|\)

\(=2\sqrt{2}-1-3-2\sqrt{2}=-4\)

4) \(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\)

\(=\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{9-6\sqrt{5}+5}\)

\(=\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{3^2-2\cdot3\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}\)

\(=\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(3-\sqrt{5}\right)^2}\)

\(=\left|2-\sqrt{5}\right|+\left|3-\sqrt{5}\right|\)

\(=\sqrt{5}-2+3-\sqrt{5}=1\)

5) \(\sqrt{\left(4-3\sqrt{2}\right)^2}-\sqrt{19+6\sqrt{2}}\)

\(=\sqrt{\left(4-3\sqrt{2}\right)^2}-\sqrt{18+6\sqrt{2}+1}\)

\(=\sqrt{\left(4-3\sqrt{2}\right)^2}-\sqrt{\left(3\sqrt{2}\right)^2+2\cdot3\sqrt{2}\cdot1+1^2}\)

\(=\sqrt{\left(4-3\sqrt{2}\right)^2}-\sqrt{\left(3\sqrt{2}+1\right)^2}\)

\(=\left|4-3\sqrt{2}\right|-\left|3\sqrt{2}+1\right|\)

\(=3\sqrt{2}-4-3\sqrt{2}-1=-5\)

6) \(\sqrt{3-2\sqrt{2}}+\sqrt{\left(2-\sqrt{2}\right)^2}\)

\(=\sqrt{2-2\sqrt{2}+1}+\sqrt{\left(2-\sqrt{2}\right)^2}\)

\(=\sqrt{\left(\sqrt{2}\right)^2-2\cdot\sqrt{2}\cdot1+1^2}+\sqrt{\left(2-\sqrt{2}\right)^2}\)

\(=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}\)

\(=\sqrt{2}-1+2-\sqrt{2}=1\)

1: =3-căn 2-2+căn 2=1

2: \(=\sqrt{2}-1+2-\sqrt{2}=1\)

3: \(=2\sqrt{2}-1-3-2\sqrt{2}=-4\)

4: \(=\sqrt{5}-2+3-\sqrt{5}=1\)

5: \(=3\sqrt{2}-4-3\sqrt{2}-1=-5\)

6: \(=\sqrt{2}-1+2-\sqrt{2}=1\)

a: =>x-5=9

=>x=14

b: căn x-10=-2

=>\(x\in\varnothing\)

c: căn 2x-1=căn 5

=>2x-1=5

=>2x=6

=>x=3

d: căn 4-5x=12

=>4-5x=144

=>5x=-140

=>x=-28

e: =>7|x-1|=35

=>|x-1|=5

=>x-1=5 hoặc x-1=-5

=>x=6 hoặc x=-4

f: =>\(\sqrt{x+3}\left(\sqrt{x-3}-5\right)=0\)

=>x+3=0 hoặc x-3=25

=>x=28 hoặc x=-3

a) \(a-5\sqrt{a}\)

\(=\sqrt{a}\left(\sqrt{a}-\sqrt{5}\right)\)

b) \(a-7\)

\(=\left(\sqrt{a}-\sqrt{7}\right)\left(\sqrt{a}+\sqrt{7}\right)\)

c) \(a+4\sqrt{a}+4\)

\(=\left(\sqrt{a}+2\right)^2\)

d) \(\sqrt{xy}-4\sqrt{x}+3\sqrt{y}-12\)

\(=\sqrt{x}\left(\sqrt{y}-4\right)+3\left(\sqrt{y}-4\right)\)

\(=\left(\sqrt{x}+3\right)\left(\sqrt{y}-4\right)\)

a: \(a-5\sqrt{a}=\sqrt{a}\left(\sqrt{a}-5\right)\)

b: \(a-7=\left(\sqrt{a}-\sqrt{7}\right)\left(\sqrt{a}+\sqrt{7}\right)\)

c: \(a+4\sqrt{a}+4=\left(\sqrt{a}+2\right)^2\)

d: \(\sqrt{xy}-4\sqrt{x}+3\sqrt{y}-12\)

=căn x(căn y-4)+3(căn y-4)

=(căn y-4)(căn x+3)

\(a,a-5\sqrt{a}=\sqrt{a}\left(\sqrt{a}-5\right)\\ b,a-7=\left(\sqrt{a}\right)^2-\left(\sqrt{7}\right)^2=\left(\sqrt{a}-\sqrt{7}\right)\left(\sqrt{a}+\sqrt{7}\right)\\ c,a+4\sqrt{a}+4=\left(\sqrt{a}\right)^2+2\sqrt{a}.2+2^2=\left(\sqrt{a}+2\right)^2\\ d,Sửa.đề:\sqrt{x}.\sqrt{y}-4\sqrt{x}+3\sqrt{y}-12\\ =\sqrt{x}\left(\sqrt{y}-4\right)+3.\left(\sqrt{y}-4\right)=\left(\sqrt{x}+3\right)\left(\sqrt{y}-4\right)\)

\(\sqrt{2}\left(\sqrt{3}+1\right)\)

\(=\sqrt{6}+\sqrt{2}\)

Ta có: \(\left(\sqrt{6}+\sqrt{2}\right)^2=8+4\sqrt{3}\)

Và: \(\left(\sqrt{3}+2\right)^2=7+4\sqrt{3}\)

Ta thấy: \(8+4\sqrt{3}>7+4\sqrt{3}\)

Hay: \(\sqrt{2}\left(\sqrt{3}+1\right)>\sqrt{3}+2\) (đpcm)

cíuuu

a: =căn 3+căn 5-căn 3=căn5
b: \(=\sqrt{x-2-2\sqrt{x-2}+1}=\sqrt{\left(\sqrt{x-2}-1\right)^2}\)

\(=\left|\sqrt{x-2}-1\right|\)

a: =căn 16/5*36/5*49=7*4*9/5=252/5

b: =căn 250*2,5=căn 625=25

c: =căn 3/2*2/3=1

d: =5*căn 2*7*căn 2=70