a) \(\dfrac{x 1}{35}\) \(\dfrac{x 3}{33}\)=\(\dfrac{x 5}{31}\) \(\dfrac{x 7}{29}\)Hd: cộng thêm 1 vào các hạng tửb) \(\dfrac{x-10}{1994}\) \(\dfrac{x-8}{1996}\) \(\dfrac{x-6}{1998}\) \(\dfrac{x-4}{200...

2

TA

Trần An

8 tháng 2 2021

Ai cíu dới

BH

Bé Heo

8 tháng 2 2021

bạn có hướng dẫn rùi thây

Đúng(1)

DN

Đặng Ngọc Hà

23 tháng 1 2019

Giải các phương trình sau:

a) \(\dfrac{x+1}{35}+\dfrac{x+3}{33}=\dfrac{x+5}{31}+\dfrac{x+7}{29}\)

b) \(\dfrac{x-10}{1994}+\dfrac{x-8}{1996}+\dfrac{x-6}{1998}+\dfrac{x-4}{2000}+\dfrac{x-2}{2002}=\dfrac{x-2002}{2}+\dfrac{x-2000}{4}+\dfrac{x-1998}{6}+\dfrac{x-1996}{8}+\dfrac{x-1994}{10}\)

1

NL

Nguyễn Lê Phước Thịnh

4 tháng 1 2023

a: \(\Rightarrow\left(\dfrac{x+1}{35}+1\right)+\left(\dfrac{x+3}{33}+1\right)=\left(\dfrac{x+5}{31}+1\right)+\left(\dfrac{x+7}{29}+1\right)\)

=>x+36=0

=>x=-36

b: \(\Leftrightarrow\left(\dfrac{x-10}{1994}-1\right)+\left(\dfrac{x-8}{1996}-1\right)+\left(\dfrac{x-6}{1998}-1\right)+\left(\dfrac{x-4}{2000}-1\right)+\left(\dfrac{x-2}{2002}-1\right)=\left(\dfrac{x-2002}{2}-1\right)+\left(\dfrac{x-2000}{4}-1\right)+\left(\dfrac{x-1998}{6}-1\right)+\left(\dfrac{x-1996}{8}-1\right)+\left(\dfrac{x-1994}{10}-1\right)\)

=>x-2004=0

=>x=2004

H

hello

11 tháng 3 2021

GIẢI PHƯƠNG...

1

TM

Trần Minh Hoàng

11 tháng 3 2021

1) PT \(\Leftrightarrow\left(\dfrac{x+1}{35}+1\right)+\left(\dfrac{x+3}{33}+1\right)=\left(\dfrac{x+5}{31}+1\right)+\left(\dfrac{x+7}{29}+1\right)\)

\(\Leftrightarrow\dfrac{x+36}{35}+\dfrac{x+36}{33}=\dfrac{x+36}{31}+\dfrac{x+36}{29}\)

\(\Leftrightarrow\left(x+36\right)\left(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}\right)=0\)

\(\Leftrightarrow x+36=0\) (Do \(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}>0\))

\(\Leftrightarrow x=-36\).

Vậy nghiệm của pt là x = -36.

Đúng(4)

NN

Nguyễn Ngô Minh Trí

12 tháng 2 2019

giải phương trình : \(\dfrac{x+1}{35}+\dfrac{x+3}{33}=\dfrac{x+5}{31}+\dfrac{x+7}{29}\)

2

QD

Quốc Đạt

12 tháng 2 2019

\(\dfrac{x+1}{35}+\dfrac{x+3}{33}=\dfrac{x+5}{31}+\dfrac{x+7}{29}\)

\(\Leftrightarrow\dfrac{\left(x+1\right).33}{35.33}+\dfrac{\left(x+3\right).35}{33.35}=\dfrac{\left(x+5\right).29}{31.29}+\dfrac{\left(x+7\right).31}{29.31}\)

\(\Leftrightarrow\dfrac{33\left(x+1\right)+35\left(x+3\right)}{1155}=\dfrac{29\left(x+5\right)+31\left(x+7\right)}{899}\)

\(\Leftrightarrow\dfrac{33x+33+35x+35.3}{1155}=\dfrac{29x+29.5+31x+31.7}{899}\)

\(\Leftrightarrow\dfrac{68x+138}{1155}=\dfrac{60x+362}{899}\)

\(\Leftrightarrow\dfrac{\left(68x+138\right).899}{1155.899}=\dfrac{\left(60x+362\right).1155}{899.115}\)

\(\Rightarrow\left(68x+138\right).899=\left(60x+362\right).1155\)

<=> 61132x+124062=69300x+418110

<=> 61132x-69300x=418110-124062

<=> -8168x=294048

<=> x=-36

Vậy phương trình có nghiệm x=-36

TV

Truong Viet Truong

12 tháng 2 2019

\(PT\Leftrightarrow\dfrac{x+1}{35}+\dfrac{x+3}{33}=\dfrac{x+5}{31}+\dfrac{x+7}{29}\)

\(\Leftrightarrow\dfrac{x+1}{35}+1+\dfrac{x+3}{33}+1=\dfrac{x+5}{31}+1+\dfrac{x+7}{29}+1\)

\(\Leftrightarrow\dfrac{x+36}{35}+\dfrac{x+36}{33}=\dfrac{x+36}{31}+\dfrac{x+36}{29}\)

\(\Leftrightarrow\left(x+36\right)\left(\dfrac{1}{35}+\dfrac{1}{33}-\dfrac{1}{31}-\dfrac{1}{29}\right)=0\)

\(\Leftrightarrow x=-36\)

Bài 1 : Giải PT \(\dfrac{x-10}{1994}\)+\(\dfrac{x-8}{1996}\)+\(\dfrac{x-6}{1998}\)= \(\dfrac{x-2002}{2}\)+\(\dfrac{x-2000}{4}\)+\(\dfrac{x-1998}{6}\) Bài 2: Giải...

BH

Bich Hong

13 tháng 1 2018

1

HA

Hoàng Anh Thư

13 tháng 1 2018

bài 1:

\(\dfrac{x-10}{1994}+\dfrac{x-8}{1996}+\dfrac{x-6}{1998}=\dfrac{x-2002}{2}+\dfrac{x-2000}{4}+\dfrac{x-1998}{6}\)

<=>\(\left(\dfrac{x-10}{1994}-1\right)+\left(\dfrac{x-8}{1996}+-1\right)+\left(\dfrac{x-6}{1998}-1\right)=\left(\dfrac{x-2002}{2}-1\right)+\left(\dfrac{x-2000}{4}-1\right)+\left(\dfrac{x-1998}{6}-1\right)\)

<=>\(\dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}=\dfrac{x-2004}{2}+\dfrac{x-2004}{4}+\dfrac{x-2004}{6}\)

<=>\(\dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}-\dfrac{x-2004}{2}-\dfrac{x-2004}{4}-\dfrac{x-2004}{6}=0\)

<=>(x-2004)\(\left(\dfrac{1}{1994}+\dfrac{1}{1996}+\dfrac{1}{1998}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{6}\right)\)

vì 1/1994+1/1996+1/1998-1/2-1/4-1/6 khác 0

nên x-2004=0=>x=2004

vyaj.......

bài 2:

\(\dfrac{x-85}{15}+\dfrac{x-74}{13}+\dfrac{x-67}{11}+\dfrac{x-64}{9}=10\)

<=>\(\left(\dfrac{x-85}{15}-1\right)+\left(\dfrac{x-74}{13}-2\right)+\left(\dfrac{x-67}{11}-3\right)+\left(\dfrac{x-64}{9}-4\right)=0\)

<=>\(\dfrac{x-100}{15}+\dfrac{x-100}{13}+\dfrac{x-100}{11}+\dfrac{x-100}{9}=0\)

<=>\(\left(x-100\right)\left(\dfrac{1}{15}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{9}\right)=0\)

vì 1/15+1/13+1/11+1/9 khác 0

=>x-100=0<=>x=100

NN

Nguyễn Ngọc Hân

3 tháng 5 2023

a)(\(\dfrac{31}{35}\)-\(\dfrac{4}{7}\)) x \(\dfrac{8}{7}\) : 2

b)(1 - \(\dfrac{1}{2}\)) x ( 1- \(\dfrac{1}{3}\)) x (1-\(\dfrac{1}{4}\)) x ( 1-\(\dfrac{1}{5}\))

nhanh giúp mik với!!

lời giải chi tiết ạ!!!

#Toán lớp 4

2

H

Hquynh

3 tháng 5 2023

\(a,\left(\dfrac{31}{35}-\dfrac{4}{7}\right)\times\dfrac{8}{7}:2\\ =\left(\dfrac{31}{35}-\dfrac{4\times5}{35}\right)\times\dfrac{8}{7}:2\\ =\dfrac{11}{35}\times\dfrac{8}{7}:2\\ =\dfrac{88}{245}:2\\ =\dfrac{44}{245}\\ b,\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times\left(1-\dfrac{1}{5}\right)\\ =\left(\dfrac{2-1}{2}\right)\times\left(\dfrac{3-1}{3}\right)\times\left(\dfrac{4-1}{4}\right)\times\left(\dfrac{5-1}{5}\right)\\ =\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\\ =\dfrac{1}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\\ =\dfrac{1}{4}\times\dfrac{4}{5}=\dfrac{1}{5}\)

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NT

Nguyễn Thị Thương Hoài

Giáo viên VIP

3 tháng 5 2023

a, ( \(\dfrac{31}{35}\) - \(\dfrac{4}{7}\)) \(\times\) \(\dfrac{8}{7}\): 2

= \(\left(\dfrac{31}{35}-\dfrac{20}{35}\right)\) \(\times\) \(\dfrac{8}{7}\) : 2

= \(\dfrac{11}{35}\) \(\times\) \(\dfrac{8}{7}\) \(\times\) \(\dfrac{1}{2}\)

= \(\dfrac{44}{35}\) \(\times\) \(\dfrac{4}{7}\)

= \(\dfrac{44}{245}\)

b, ( 1 - \(\dfrac{1}{2}\)) \(\times\) ( 1 - \(\dfrac{1}{3}\)) \(\times\) ( 1 - \(\dfrac{1}{4}\)) \(\times\) ( 1 - \(\dfrac{1}{5}\))

= \(\dfrac{1}{2}\) \(\times\) \(\dfrac{2}{3}\) \(\times\) \(\dfrac{3}{4}\) \(\times\) \(\dfrac{4}{5}\)

= \(\dfrac{1}{5}\) \(\times\) \(\dfrac{2\times3\times4}{2\times3\times4}\)

= \(\dfrac{1}{5}\)

NT

Ngọc Thư

28 tháng 4 2017

a. \(\dfrac{x-1}{x^{2^{ }}-4}=\dfrac{3}{2-x}\)

b. \(\dfrac{1}{x-1}-\dfrac{7}{x-2}=\dfrac{1}{\left(x-1\right)\left(2-x\right)}\)

c. \(\dfrac{2x+3}{2x-3}-\dfrac{3}{4x-6}=\dfrac{2}{5}\)

d. \(\dfrac{x+29}{31}-\dfrac{x+27}{33}=\dfrac{x+17}{43}-\dfrac{x+15}{45}\)

1

B

BW_P&A

28 tháng 4 2017

a) \(\dfrac{x-1}{x^2-4}=\dfrac{3}{2-x}\)

\(\Leftrightarrow\dfrac{x-1}{\left(x-2\right)\left(x+2\right)}=-\dfrac{3}{\left(x-2\right)\left(x+2\right)}\)

\(ĐKXĐ:\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)

\(\Rightarrow x-1=-3\)

\(\Leftrightarrow x=1-3=-2\)

Vậy: \(x=-2\)

b) \(\dfrac{1}{x-1}-\dfrac{7}{x-2}=\dfrac{1}{\left(x-1\right)\left(2-x\right)}\)

\(\Leftrightarrow\dfrac{1}{x-1}-\left(-\dfrac{7}{2-x}\right)=\dfrac{1}{\left(x-1\right)\left(2-x\right)}\)

\(ĐKXĐ:\left\{{}\begin{matrix}x\ne1\\x\ne2\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{2-x}{\left(x-1\right)\left(2-x\right)}+\dfrac{7\left(x-1\right)}{\left(x-1\right)\left(2-x\right)}=\dfrac{1}{\left(x-1\right)\left(2-x\right)}\)

\(\Rightarrow2-x+7x-7=1\)

\(\Leftrightarrow-x+7x=1-2+7=6\)

\(\Leftrightarrow6x=6\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

c) \(\dfrac{2x+3}{2x-3}-\dfrac{3}{4x-6}=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{2x+3}{2x-3}-\dfrac{3}{2\left(2x-3\right)}=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{10\left(2x+3\right)}{10\left(2x-3\right)}-\dfrac{3.5}{10\left(2x-3\right)}=\dfrac{4\left(2x-3\right)}{10\left(2x-3\right)}\)

\(ĐKXĐ:x\ne\dfrac{3}{2}\)

\(\Leftrightarrow10\left(2x+3\right)-15=4\left(2x-3\right)\)

\(\Leftrightarrow20x+30-15=8x-12\)

\(\Leftrightarrow20x-8x=15-12-30\)

\(\Leftrightarrow12x=-27\)

\(\Leftrightarrow x=-\dfrac{27}{12}=-\dfrac{9}{4}\)

Vậy: \(x=-\dfrac{9}{4}\)

d) \(\dfrac{x+29}{31}-\dfrac{x+27}{33}=\dfrac{x+17}{43}-\dfrac{x+15}{45}\)

\(\Leftrightarrow\left(\dfrac{x+29}{31}+1\right)-\left(\dfrac{x+27}{33}+1\right)=\left(\dfrac{x+17}{43}+1\right)-\left(\dfrac{x+15}{45}+1\right)\)

\(\Leftrightarrow\dfrac{x+60}{31}-\dfrac{x+60}{33}=\dfrac{x+60}{43}-\dfrac{x+60}{45}\)

\(\Leftrightarrow\dfrac{x+60}{31}-\dfrac{x+60}{33}-\dfrac{x+60}{43}+\dfrac{x+60}{45}\)

\(\Leftrightarrow\left(x+60\right)\left(\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{43}+\dfrac{1}{45}\right)=0\)

\(\Leftrightarrow x+60=0\) vì \(\left(\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{43}+\dfrac{1}{45}\ne0\right)\)

\(\Leftrightarrow x=-60\)

Vậy: \(x=-60\)

_Good luck to you_

HH

hoan hang

6 tháng 5 2018

Tìm \(x\)

1. \(\dfrac{1}{7}=\dfrac{8}{-x}\)

2. \(\dfrac{3x}{9}=\dfrac{2}{6}\)

3. \(4\dfrac{2}{5}:\left(\dfrac{-33}{10}\right)+x=-1\dfrac{5}{6}\)

4. \(45\%.x-2\dfrac{3}{8}=-1\dfrac{31}{40}\)

5. \(\left(x-2\dfrac{1}{4}\right):\left(-\dfrac{5}{6}\right)=3\)

#Toán lớp 6

1

CT

chú tuổi gì

6 tháng 5 2018

\(\dfrac{1}{7}=\dfrac{8}{-x}\)=> \(-x=56\)

=> \(x=56\)

2) => 18x = 18

=> x = 1

3) \(\dfrac{-4}{3}+x=\dfrac{-11}{6}\)

=> \(x=\dfrac{-11}{6}+\dfrac{4}{3}\)

=> \(x=\dfrac{-1}{2}\)

4) 45%.x =\(\dfrac{3}{5}\)

=> \(x=\dfrac{3}{5}:\dfrac{9}{20}\)

=> \(x=\dfrac{4}{3}\)

NT

Nguyễn Thu Huyền

1 tháng 9 2023

Tìm...

#Toán lớp 7

3

T

Toru

1 tháng 9 2023

\(a,3-x=x+1,8\)

\(\Rightarrow-x-x=1,8-3\)

\(\Rightarrow-2x=-1,2\)

\(\Rightarrow x=0,6\)

\(b,2x-5=7x+35\)

\(\Rightarrow2x-7x=35+5\)

\(\Rightarrow-5x=40\)

\(\Rightarrow x=-8\)

\(c,2\left(x+10\right)=3\left(x-6\right)\)

\(\Rightarrow2x+20=3x-18\)

\(\Rightarrow2x-3x=-18-20\)

\(\Rightarrow-x=-38\)

\(\Rightarrow x=38\)

\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)

\(\Rightarrow8x-3+1=1+6x+x\)

\(\Rightarrow8x-3=7x\)

\(\Rightarrow8x-7x=3\)

\(\Rightarrow x=3\)

\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)

\(\Rightarrow-3x+x=\dfrac{4}{3}-\dfrac{2}{9}\)

\(\Rightarrow-2x=\dfrac{10}{9}\)

\(\Rightarrow x=-\dfrac{5}{9}\)

Đúng(3)

T

Toru

1 tháng 9 2023

\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)

\(\Rightarrow\dfrac{1}{2}x-\dfrac{3}{4}x=-\dfrac{1}{2}-\dfrac{5}{6}\)

\(\Rightarrow-\dfrac{1}{4}x=-\dfrac{4}{3}\)

\(\Rightarrow x=\dfrac{16}{3}\)

\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)

\(\Rightarrow x-4=5-x\)

\(\Rightarrow x+x=5+4\)

\(\Rightarrow2x=9\)

\(\Rightarrow x=\dfrac{9}{2}\)

\(k,7x^2-11=6x^2-2\)

\(\Rightarrow7x^2-6x^2=-2+11\)

\(\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

\(m,5\left(x+3\cdot2^3\right)=10^2\)

\(\Rightarrow5\left(x+24\right)=100\)

\(\Rightarrow x+24=20\)

\(\Rightarrow x=-4\)

\(n,\dfrac{4}{9}-\left(\dfrac{1}{6^2}\right)=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{4}{9}-\dfrac{1}{36}\)

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{5}{12}\)

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2=0\)

\(\Rightarrow x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)

#\(Urushi\text{☕}\)

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HQ

Hạ Quỳnh

20 tháng 12 2020

Bài 2:Tìm x...

#Toán lớp 6

1

NL

Nguyễn Lê Phước Thịnh

20 tháng 12 2020

a) Ta có: \(\dfrac{1}{7}+x=-\dfrac{2}{3}\)

\(\Leftrightarrow x=-\dfrac{2}{3}-\dfrac{1}{7}=\dfrac{-14}{21}-\dfrac{3}{21}\)

hay \(x=-\dfrac{17}{21}\)

Vậy: \(x=-\dfrac{17}{21}\)

b) Ta có: \(\dfrac{-2}{3}:x=\dfrac{-5}{6}\)

\(\Leftrightarrow x=\dfrac{-2}{3}:\dfrac{-5}{6}=\dfrac{-2}{3}\cdot\dfrac{6}{-5}=\dfrac{-12}{-15}=\dfrac{4}{5}\)

Vậy: \(x=\dfrac{4}{5}\)

c) Ta có: \(\left(\dfrac{3}{5}-2x\right)\cdot\dfrac{5}{8}=1\)

\(\Leftrightarrow\left(\dfrac{3}{5}-2x\right)=1:\dfrac{5}{8}=\dfrac{8}{5}\)

\(\Leftrightarrow-2x=\dfrac{8}{5}-\dfrac{3}{5}=1\)

hay \(x=-\dfrac{1}{2}\)

Vậy: \(x=-\dfrac{1}{2}\)

d) Ta có: \(\dfrac{3}{4}+\dfrac{2}{5}x=\dfrac{29}{60}\)

\(\Leftrightarrow x\cdot\dfrac{2}{5}=\dfrac{29}{60}-\dfrac{3}{4}=\dfrac{29}{60}-\dfrac{45}{60}=\dfrac{-16}{60}=\dfrac{-4}{15}\)

hay \(x=\dfrac{-4}{15}:\dfrac{2}{5}=\dfrac{-4}{15}\cdot\dfrac{5}{2}=\dfrac{-20}{30}=-\dfrac{2}{3}\)

Vậy: \(x=-\dfrac{2}{3}\)

e) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{8}{20}-\dfrac{15}{20}=\dfrac{-7}{20}\)

hay \(x=-\dfrac{1}{4}:\dfrac{7}{20}=\dfrac{-1}{4}\cdot\dfrac{20}{7}=\dfrac{-20}{28}=\dfrac{-5}{7}\)

Vậy: \(x=-\dfrac{5}{7}\)

f) Ta có: \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

\(\Leftrightarrow-x+\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}=0\)

\(\Leftrightarrow-x+\dfrac{55}{60}-\dfrac{24}{60}-\dfrac{40}{60}=0\)

\(\Leftrightarrow-x-\dfrac{9}{60}=0\)

\(\Leftrightarrow-x=\dfrac{9}{60}=\dfrac{3}{20}\)

hay \(x=-\dfrac{3}{20}\)

Vậy: \(x=-\dfrac{3}{20}\)

g) Ta có: \(\left|x+\dfrac{1}{3}\right|-4=\dfrac{-1}{2}\)

\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=\dfrac{-1}{2}+4=\dfrac{-1}{2}+\dfrac{8}{2}=\dfrac{7}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{7}{2}\\x+\dfrac{1}{3}=-\dfrac{7}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{21}{6}-\dfrac{2}{6}=\dfrac{19}{6}\\x=-\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{-21}{6}-\dfrac{2}{6}=\dfrac{-23}{6}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{19}{6};-\dfrac{23}{6}\right\}\)

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HQ

Hạ Quỳnh

20 tháng 12 2020

cảm ơn cậu cutee gì đó ơi nha