rút gọn các biểu thức a/ A=$\sqrt{7 2\sqrt{6}} \sqrt{7-2\sqrt{6}}-2\sqrt{6}$b/ B= $\sqrt{7 2\sqrt{10}}-\sqrt{7 2\sqrt{10}}-\sqrt{8}$c/ C=$\sqrt{14 4\sqrt{6}}-\sqrt{14-4\sqrt{6}}$d/ D=\(\sqrt{3 \...

0

NU

Nguyễn Uyên Minh

14 tháng 5 2022

B 5. Rút gọn các biểu thức sau:

a)$\sqrt{7+4\sqrt{3}}$ b)$\sqrt{9-4\sqrt{5}}$

c)$\sqrt{14+6\sqrt{5}}$ d)$\sqrt{17-12\sqrt{2}}$

1

NN

Nguyễn Ngọc Huy Toàn

14 tháng 5 2022

a.$\sqrt{7+4\sqrt{3}}=\sqrt{\left(\sqrt{3}+2\right)^2}=\left|\sqrt{3}+2\right|=\sqrt{3}+2$

b.$\sqrt{9-4\sqrt{5}}=\sqrt{\left(\sqrt{5}-2\right)^2}=\left|\sqrt{5}-2\right|=\sqrt{5}-2$

c.$\sqrt{14+6\sqrt{5}}=\sqrt{\left(\sqrt{5}+3\right)^2}=\left|\sqrt{5}+3\right|=\sqrt{5}+3$

d.$\sqrt{17-12\sqrt{2}}=\sqrt{\left(2\sqrt{2}-3\right)^2}=\left|2\sqrt{2}-3\right|=3-2\sqrt{2}$

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MA

Minh Anh Vũ

1 tháng 7 2021

Rút gọn biểu thức:

a) $\sqrt{6+2\sqrt{4-2\sqrt{3}}}$

b) $\sqrt{6-2\sqrt{3+\sqrt{13+4\sqrt{3}}}}$

c) $\sqrt{\sqrt{3}+\sqrt{48-10\sqrt{7+4\sqrt{3}}}}$

d) $\sqrt{23-6\sqrt{10+4\sqrt{3-2\sqrt{2}}}}$

2

NN

Nguyễn Ngọc Lộc

1 tháng 7 2021

$a,=\sqrt{6+2\sqrt{3-2\sqrt{3}+1}}$

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

$=\sqrt{6+2\left(\sqrt{3}-1\right)}$

$=\sqrt{4+2\sqrt{3}}$

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

$b,=\sqrt{6-2\sqrt{3+\sqrt{12+2\sqrt{12}+1}}}$

$=\sqrt{6-2\sqrt{3+\sqrt{12}+1}}$

$=\sqrt{6-2\sqrt{3+2\sqrt{3}+1}}$

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

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

$c,=\sqrt{\sqrt{3}+\sqrt{48-10\sqrt{4+2.2\sqrt{3}+3}}}$

$=\sqrt{\sqrt{3}+\sqrt{48-10\left(2+\sqrt{3}\right)}}$

$=\sqrt{\sqrt{3}+\sqrt{28-10\sqrt{3}}}$

$=\sqrt{\sqrt{3}+\sqrt{25-2.5\sqrt{3}+3}}$

$=\sqrt{\sqrt{3}+5-\sqrt{3}}=\sqrt{5}$

$d,=\sqrt{23-6\sqrt{10+4\sqrt{2-2\sqrt{2}+1}}}$

$=\sqrt{23-6\sqrt{6+4\sqrt{2}}}$

$=\sqrt{23-6\sqrt{4+2.2\sqrt{2}+2}}$

$=\sqrt{23-6\sqrt{\left(2+\sqrt{2}\right)^2}}$

$=\sqrt{23-12-6\sqrt{2}}=\sqrt{11-6\sqrt{2}}$

$=\sqrt{9-2.3\sqrt{2}+2}=3-\sqrt{2}$

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NL

Nguyễn Lê Phước Thịnh

1 tháng 7 2021

a) Ta có: $\sqrt{6+2\sqrt{4-2\sqrt{3}}}$

$=\sqrt{6+2\left(\sqrt{3}-1\right)}$

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

b) Ta có: $\sqrt{6-2\sqrt{3+\sqrt{13+4\sqrt{3}}}}$

$=\sqrt{6-2\sqrt{4+2\sqrt{3}}}$

$=\sqrt{6-2\left(\sqrt{3}+1\right)}$

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

c) Ta có: $\sqrt{\sqrt{3}+\sqrt{48-10\sqrt{7+4\sqrt{3}}}}$

$=\sqrt{\sqrt{3}+\sqrt{48-10\left(2+\sqrt{3}\right)}}$

$=\sqrt{\sqrt{3}+\sqrt{28-10\sqrt{3}}}$

$=\sqrt{\sqrt{3}+5-\sqrt{3}}$

$=\sqrt{5}$

d) Ta có: $\sqrt{23-6\sqrt{10+4\sqrt{3-2\sqrt{2}}}}$

$=\sqrt{23-6\sqrt{10+4\left(\sqrt{2}-1\right)}}$

$=\sqrt{23-6\sqrt{6-4\sqrt{2}}}$

$=\sqrt{23-6\left(2-\sqrt{2}\right)}$

$=\sqrt{11+6\sqrt{2}}$

$=3+\sqrt{2}$

bài 1 rút gọn biểu thức sau:a)$\sqrt{16+6\sqrt{7}}$- $\sqrt{8-2\sqrt{7}}$ b)K=$\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{6}-\sqrt{2}}$c)$\sqrt{60-24\sqrt{6}}$+$\sqrt{40-16\sqrt{6}}$ d)B=(3+$\sqrt{3}$)$\sqrt{12-6\sqrt{13}}$e)$\sqrt{6-4\sqrt{2}}$-$\sqrt{\left(\sqrt{2}-\sqrt{6}\right)^2}$bài 2 cho biểu thức A=$\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{3}{\sqrt{x}-3}\right).\dfrac{\sqrt{x}+3}{x+9}$( với x≥0 và x≠ 9)a) rút gọn biểu thức Ab) tính giá...

DT

dương thị thanh vân

9 tháng 10 2021

1

NH

Nguyễn Hoàng Minh

9 tháng 10 2021

$1,\\ a,=\sqrt{\left(3+\sqrt{7}\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}=3+\sqrt{7}-\sqrt{7}+1=4\\ b,K=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{\sqrt{3}-1}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\\ c,=\sqrt{\left(6-2\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-4\right)^2}=6-2\sqrt{6}+2\sqrt{6}-4=2\\ e,=\sqrt{\left(2-\sqrt{2}\right)^2}-\left(\sqrt{6}-\sqrt{2}\right)=2-\sqrt{2}-\sqrt{6}+\sqrt{2}=2-\sqrt{6}$

$2,\\ a,A=\dfrac{x-3\sqrt{x}+3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}+3}{x+9}\\ A=\dfrac{x+9}{\left(\sqrt{x}-3\right)\left(x+9\right)}=\dfrac{1}{\sqrt{x}-3}\\ b,x=4+2\sqrt{3}\Leftrightarrow\sqrt{x}=\sqrt{3}+1\\ \Leftrightarrow A=\dfrac{1}{\sqrt{3}+1-3}=\dfrac{1}{\sqrt{3}+2}=2-\sqrt{3}$

DT

dương thị thanh vân

9 tháng 10 2021

cảm ơn bạn

TT

Trần Thị Quỳnh Mai

12 tháng 8 2017

Câu hỏi hay

Rút gọn biểu...

4

LT

lê thị thu huyền

23 tháng 9 2017

a) đặt $A=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}$

nhân cả hai vế với $\sqrt{2}$, ta được:

$\sqrt{2}A=\sqrt{2}\sqrt{4-\sqrt{7}}-\sqrt{2}\sqrt{4+\sqrt{7}}$

$=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}$

$=\sqrt{\left(1-\sqrt{7}\right)^2}-\sqrt{\left(1+ \sqrt{7}\right)^2}$

$=\left|1-\sqrt{7}\right|-\left|1+\sqrt{7}\right|$

$=\sqrt{7}-1-\sqrt{7}-1$

$=-2$

$\Rightarrow A=-\frac{2}{\sqrt{2}}=-\sqrt{2}$

M

Mafia

12 tháng 5 2018

a) $\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}$

$=\frac{\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}}{\sqrt{2}}$

$=\frac{\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}$

$=\frac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}$

$=\frac{-2}{\sqrt{2}}=-\sqrt{2}$

NT

Nguyễn Thị Yến

5 tháng 10 2021

rút gọn

a) $\sqrt{8+\sqrt{55}}-\sqrt{8-\sqrt{55}}-\sqrt{125}$

b) $\left(\sqrt{7-3\sqrt{5}}\right)\left(7+3\sqrt{5}\right)\left(3\sqrt{2}+\sqrt{10}\right)$

c) $\left(\sqrt{14}-\sqrt{10}\right)\left(6-\sqrt{35}\right)\left(\sqrt{6+\sqrt{35}}\right)$

2

NL

Nguyễn Lê Phước Thịnh

5 tháng 10 2021

b: Ta có: $\left(\sqrt{7-3\sqrt{5}}\right)\cdot\left(7+3\sqrt{5}\right)\cdot\left(3\sqrt{2}+\sqrt{10}\right)$

$=\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\left(7+3\sqrt{5}\right)$

$=4\left(7+3\sqrt{5}\right)$

$=28+12\sqrt{5}$

rút gọnA=$\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}$B=$\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}$C=$\left(\sqrt{7}-\sqrt{10}\right)^2+\sqrt{280}$D=$\dfrac{\sqrt{99}}{\sqrt{11}}+\sqrt{7}\cdot\sqrt{63}-\sqrt{\sqrt{81}}$E=$\sqrt{27}\left(s-\sqrt{5}\right)^2\cdot\left(3\sqrt{48}\right)$giải chi tiết ra giúp mik nha,cảm ơn...

AH

Akai Haruma

Giáo viên

5 tháng 10 2021

Lời giải:

a.

$A=\sqrt{8+\sqrt{55}}-\sqrt{8-\sqrt{55}}-\sqrt{125}$
$\sqrt{2}A=\sqrt{16+2\sqrt{55}}-\sqrt{16-2\sqrt{55}}-\sqrt{250}$

$=\sqrt{(\sqrt{11}+\sqrt{5})^2}-\sqrt{(\sqrt{11}-\sqrt{5})^2}-5\sqrt{10}$

$=|\sqrt{11}+\sqrt{5}|-|\sqrt{11}-\sqrt{5}|-5\sqrt{10}$

$=2\sqrt{5}-5\sqrt{10}$

$\Rightarrow A=\sqrt{10}-5\sqrt{5}$

b.

$B=\sqrt{7-3\sqrt{5}}.(7+3\sqrt{5})(3\sqrt{2}+\sqrt{10})$

$B\sqrt{2}=\sqrt{14-6\sqrt{5}}(7+3\sqrt{5})(3\sqrt{2}+\sqrt{10})$

$=\sqrt{(3-\sqrt{5})^2}(7+3\sqrt{5}).\sqrt{2}(3+\sqrt{5})$

$=(3-\sqrt{5})(7\sqrt{2}+3\sqrt{10})(3+\sqrt{5})$

$=(3^2-5)(7\sqrt{2}+3\sqrt{10})$

$=4(7\sqrt{2}+3\sqrt{10})=28\sqrt{2}+12\sqrt{10}$

$\Rightarrow B=28+12\sqrt{5}$

c.

$C=\sqrt{2}(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{6+\sqrt{35}}$

$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{12+2\sqrt{35}}$

$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{(\sqrt{7}+\sqrt{5})^2}

$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})(\sqrt{7}+\sqrt{5})$

$=(7-5)(6-\sqrt{35})$

$=2(6-\sqrt{35})=12-2\sqrt{35}$

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NM

Nguyễn Minh Quân

16 tháng 7 2021

1

NH

Nguyễn Huy Tú ( ✎﹏IDΣΛ亗 )

16 tháng 7 2021

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Đúng(1)

NM

Nguyễn Minh Quân

17 tháng 7 2021

đó là số 2 ko phải chữ s mik xin lỗi

....

24 tháng 6 2021

Rút gọn các biểu thức sau:

a $\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}$

b $\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}$

c $\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}$

d $\dfrac{10}{\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}}\left(\dfrac{1+\sqrt{2}}{\sqrt{4-2\sqrt{3}}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\right)$

1

LT

Lê Thị Thục Hiền

24 tháng 6 2021

a)$A=\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}$

$=\sqrt[3]{1+3\sqrt{2}+3\sqrt{2^2}+2\sqrt{2}}-\sqrt[3]{2\sqrt{2}-3\sqrt{2^2}+3\sqrt{2}-1}$

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

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

b)$B=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}$

$\Leftrightarrow B^3=5+2\sqrt{13}+3\sqrt[3]{\left(5+2\sqrt{13}\right)\left(5-2\sqrt{13}\right)}\left(\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5+2\sqrt{13}}\right)+5-2\sqrt{13}$

$\Leftrightarrow B^3=10+3.\sqrt[3]{-27}.B$

$\Leftrightarrow B^3+9B-10=0$

$\Leftrightarrow\left(B-1\right)\left(B^2+B+10\right)=0$

$\Leftrightarrow B=1$ (vì $B^2+B+10>0$)

c)$C=\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}$

$\Leftrightarrow2C=\sqrt[3]{8\sqrt{5}+16}-\sqrt[3]{8\sqrt{5}-16}=\sqrt[3]{1+3\sqrt{5}+3\sqrt{5^2}+5\sqrt{5}}-\sqrt[3]{5\sqrt{5}-3\sqrt{5^2}+3\sqrt{5}-1}$

$=\sqrt[3]{\left(1+\sqrt{5}\right)^3}-\sqrt[3]{\left(\sqrt{5}-1\right)^3}$

$=1+\sqrt{5}-\left(\sqrt{5}-1\right)$

$\Rightarrow C=1$

d) $D=\dfrac{10}{\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}}\left(\dfrac{1+\sqrt{2}}{\sqrt{4-2\sqrt{3}}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\right)$

$=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{\left(\sqrt[3]{3}+\sqrt[3]{2}\right)\left(\sqrt[3]{9^2}-\sqrt[3]{6}+\sqrt[3]{2^2}\right)}\left(\dfrac{1+\sqrt{2}}{\sqrt{\left(1-\sqrt{3}\right)^2}}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\right)$

$=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{5}.\dfrac{1+\sqrt{2}}{\left|1-\sqrt{3}\right|}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}$

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

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

$=\sqrt[3]{3}+\sqrt[3]{2}$

Vậy...

Y

Yeutoanhoc

24 tháng 6 2021

Khiếp CTV kìa sợ quá ;-;

HA

Hoài An

20 tháng 9 2021

a)$\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}$

b) $\sqrt{7+4\sqrt{3}}-\sqrt{4+2\sqrt{3}}$

c) $\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}$

d)$\sqrt{7+2\sqrt{10}}-\sqrt{3-2\sqrt{2}}$

1

LL

Lấp La Lấp Lánh

20 tháng 9 2021

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

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

c) $=\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{\left(\sqrt{7}-1\right)^2}=\sqrt{7}+1+\sqrt{7}-1=2\sqrt{7}$

d) $=\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{5}+\sqrt{2}-\sqrt{2}+1=\sqrt{5}+1$

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....

24 tháng 6 2021

rút gọn các biểu thức sau:

a $\sqrt[3]{8\sqrt{5}-16}.\sqrt[3]{8\sqrt{5}+16}$

b $\sqrt[3]{7-5\sqrt{2}}-\sqrt[6]{8}$

c $\sqrt[3]{4}.\sqrt[3]{1-\sqrt{3}}.\sqrt[6]{4+2\sqrt{3}}$

d $\dfrac{2}{\sqrt[3]{3}-1}-\dfrac{4}{\sqrt[3]{9}-\sqrt[3]{3}+1}$

2

Y

Yeutoanhoc

24 tháng 6 2021

`c)root{3}{4}.root{3}{1-sqrt3}.root{6}{(sqrt3+1)^2}`

`=root{3}{4(1-sqrt3)}.root{3}{1+sqrt3}`

`=root{3}{4(1-sqrt3)(1+sqrt3)}`

`=root{3}{4(1-3)}=-2`

`d)2/(root{3}{3}-1)-4/(root{9}-root{3}{3}+1)`

`=(2(root{3}{9}+root{3}{3}+1))/(3-1)-(4(root{3}{3}+1))/(3+1)`

`=root{3}{9}+root{3}{3}+1-root{3}{3}-1`

`=root{3}{9}`

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Y

Yeutoanhoc

24 tháng 6 2021

`a)root{3}{8sqrt5-16}.root{3}{8sqrt5+16}`

`=root{3}{(8sqrt5-16)(8sqrt5+16)}`

`=root{3}{320-256}`

`=root{3}{64}=4`

`b)root{3}{7-5sqrt2}-root{6}{8}`

`=root{3}{1-3.sqrt{2}+3.2.1-2sqrt2}-root{6}{(2)^3}`

`=root{3}{(1-sqrt2)^3}-sqrt2`

`=1-sqrt2-sqrt2=1-2sqrt2`