Question

Bài 10 (trang 11 SGK Toán 9 Tập 1)

Chứng minh:

a) $(\sqrt{3}-1)^2=4-2\sqrt{3}$  ;                b) $\sqrt{4-2\sqrt{3}}-\sqrt{3}=-1$.

Answer 1

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

b) \(\sqrt{4-2\sqrt{3}}\)=\(\sqrt{3}\)-1 >0

Bình phương 2 vế, ta có:

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

Answer 2

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

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

 

Answer 3

a. ( √3 -1)^2 = 3- 2√3 +1 = 4-2√3 (=VP)

b. √(√3-1)^2 - √3 = |√3-1| - √3 = √3-1-√3 = -1 (=VP)

 

Answer 4

a)  \((\sqrt{3}-1)^2=3-2\times\sqrt{3}+1=4-2\sqrt{3}\)

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

Answer 5

Answer 6

Answer 7

Answer 8

Answer 9

a. ta có 
VT = (\(\sqrt{3}\) - 1)² = (\(\sqrt{3}\) )² - 2\(\sqrt{3}\) + 1
      = 3 - 2\(\sqrt{3}\) + 1 
     = 4 - 2\(\sqrt{3}\)
     = VP(đpcm)
b. theo câu a ta có
VT= \(\sqrt{4-2\sqrt{3}}\) - \(\sqrt{3}\) = \(\sqrt{\left(\sqrt{3}-1\right)^2}\) - \(\sqrt{3}\)
    = |\(\sqrt{3}\) - 1| - \(\sqrt{3}\) -1 - \(\sqrt{3}\)
    = -1 = VP ( vì \(\sqrt{3}\) - 1 > 0) (đpcm)
 

Answer 10

Answer 11

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

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

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

b) theo yêu cầu a ta có:

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

= \(|\sqrt{3}\)-1\(|\) -\(\sqrt{3}\) =\(\sqrt{3}\)-1-\(\sqrt{3}\)

= -1 =VT(vì \(\sqrt{3}\)-1>0  ( đpcm)

 

Answer 12

Answer 13

a) VT = \(\sqrt{3}^2\)- 2\(\sqrt{3}\) + 1 = 4 - 2\(\sqrt{3}\) = VP ( đpcm )

b) VT = (\(\sqrt{3}-1\)⁾² - \(\sqrt{3}\) = \(\left|\sqrt{3}-1\right|\) - \(\sqrt{3}\) = \(\sqrt{3}\) - 1 - \(\sqrt{3}\) = -1 = VP (đpcm)

 

Answer 14

Answer 15

a) Ta có:

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

        $=3-2\sqrt{3}+1=4-2\sqrt{3}=VP$

Vậy $(\sqrt{3}-1{)}^2=4-2\sqrt{3}$ (đpcm)

b) Theo câu a) ta có:

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

        $=|\sqrt{3}-1|-\sqrt{3}=\sqrt{3}-1-\sqrt{3}$

        $=-1=VP$ 

Answer 16

a) 

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

        $=3-2\sqrt{3}+1=4-2\sqrt{3}=VP$

Vậy $(\sqrt{3}-1{)}^2=4-2\sqrt{3}$ 

b) Theo câu a) ta có

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

        $=|\sqrt{3}-1|-\sqrt{3}=\sqrt{3}-1-\sqrt{3}$

        $=-1=VP$ (vì $\sqrt{3}-1>0$) 

Answer 17

Answer 18

Answer 19

 

Answer 20

a) Ta có: (\(\sqrt{3}-1\))\(^2\) = 3 - 2\(\sqrt{3}\) + 1 = 4 - 2\(\sqrt{3}\)

b) Ta có: \(\sqrt{4-2\sqrt{3}}\) - \(\sqrt{3}\) = \(\sqrt{\left(\sqrt{3}-1\right)^2}\) - \(\sqrt{3}\) = \(|\sqrt{3}-1|\)- \(\sqrt{3}\) = \(\sqrt{3}-1\) - \(\sqrt{3}\) = -1

Answer 21

a)Ta có: 

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

$=3-2\sqrt{3}+1$

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

$=4-2\sqrt{3}$ = VP

Vậy  $(\sqrt{3}-1{)}^2=4-2\sqrt{3}$  (đpcm)

b)Ta có: 

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

 $=\sqrt{3-2\sqrt{3}+1}-\sqrt{3}$

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

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

$=\left|\sqrt{3}-1\right|-\sqrt{3}$

$=\sqrt{3}-1-\sqrt{3}$ 

$=(\sqrt{3}-\sqrt{3})-1=-1$ = VP. 

(do $3>1\iff \sqrt{3}>\sqrt{1}\iff \sqrt{3}>1$$\iff \sqrt{3}-1>0$

$\Rightarrow \left|\sqrt{3}-1\right|=\sqrt{3}-1$)  

 

 

Answer 22

Answer 23

a) Ta có: Vế trái =${\left(\sqrt{3}-1\right)}^2={\left(\sqrt{3}\right)}^2-2.\sqrt{3}.1+1^2$

$=3-2\sqrt{3}+1$

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

$=4-2\sqrt{3}$ = Vế phải

Vậy  $(\sqrt{3}-1{)}^2=4-2\sqrt{3}$  (đpcm)

b) Ta có: 

$\mathrm{Vế\; trái}=\sqrt{4-2\sqrt{3}}-\sqrt{3}$$=\sqrt{\left(3+1\right)-2\sqrt{3}}-\sqrt{3}$

 $=\sqrt{3-2\sqrt{3}+1}-\sqrt{3}$

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

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

$=\left|\sqrt{3}-1\right|-\sqrt{3}$

$=\sqrt{3}-1-\sqrt{3}$ 

$=(\sqrt{3}-\sqrt{3})-1=-1$ = Vế phải 

(do $3>1\iff \sqrt{3}>\sqrt{1}\iff \sqrt{3}>1$$\iff \sqrt{3}-1>\mathrm{0)}$

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


 

Answer 24

Answer 25

Answer 26

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

Ta có:

VT = (\(\sqrt{3}-1\))² = (\(\sqrt{3}\))² - 2.\(\sqrt{3}\).1 + 1² = 3 - 2\(\sqrt{3}\) + 1 = 4 - 2\(\sqrt{3}\)

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

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

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

Vì phần a VP = 4 - 2\(\sqrt{3}\) mà phần b có bểu thức \(\sqrt{4-2\sqrt{3}}\) nên ta thay \(\sqrt{4-2\sqrt{3}}\) = ( \(\sqrt{3}\) - 1 )² ta đc:

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

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

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

Answer 27

a) Ta có:

VT = (\sqrt{3}-1)^2 = (\sqrt{3})^2- 2\sqrt{3} + 1VT=(3​−1)2=(3​)2−23​+1

        = 3 - 2\sqrt{3} + 1 = 4 - 2\sqrt{3} = VP=3−23​+1=4−23​=VP

Vậy (\sqrt{3} - 1)^2 = 4 - 2\sqrt{3}(3​−1)2 =4−23​ (đpcm)

b) Theo câu a) ta có:

VT=\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{(\sqrt{3}-1)^2}-\sqrt{3}VT=4−23​​−3​=(3​−1)2​−3​

        = |\sqrt{3} - 1| - \sqrt{3} = \sqrt{3} - 1 - \sqrt{3}=∣3​−1∣−3​=3​−1−3​

        $=-1=VP$ (vì \sqrt{3} - 1 > 03​−1>0) (đpcm)

Answer 28

a) Ta có :

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

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

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

b) Theo câu a) ta có :

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

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

\(=-1=VP\) ( vì \(\sqrt{3}-1>0\) ) 

Answer 29

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

⇔\(3-2\sqrt{3}+1=4+2\sqrt{3}\)

⇔\(4+2\sqrt{3}=VP\)

b, \(\sqrt{4-2\sqrt{3}}-\sqrt{3}=-1\)

⇔\(\sqrt{1-2\sqrt{3}+3}-\sqrt{3}=-1\)

\(\Leftrightarrow(1-\sqrt{3})^2-\sqrt{3}=-1\)

\(\Leftrightarrow\sqrt{3}-1-\sqrt{3}=-1\)

\(\Leftrightarrow-1=VP\)

 

 

Answer 30

a) vế trái = (căn 3)^2 -2*căn3*1 +1^2