a) \(\sqrt{2x}=12\left(đk:x\ge0\right)\)

\(2x=144\)

\(x=72\)

b) \(\sqrt{9x^2-6x}+1=10\)\(\left(Đk:x\le0;x\ge\dfrac{2}{3}\right)\)

\(\sqrt{9x^2-6x}=9\)

\(9x^2-6x=81\)

\(\left(3x-1\right)^2=82\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{82}+1}{3}\\x=\dfrac{1-\sqrt{82}}{3}\end{matrix}\right.\)

c) \(x^2\sqrt{5}-\sqrt{125}=0\)

\(x^2\sqrt{5}=5\sqrt{5}\)

\(x^2=5\)

\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{5}\\x=-\sqrt{5}\end{matrix}\right.\)

các thầy cô giúp e vs ạ

M = 1 + 3 + (32  + 33  + 34 )  +..... + (398   + 399  + 3100 )

M = 4 + 32 .13 + .....  +398 .13

= 13.(32  + 35  + ... + 398 ) + 4

=> M chia 13 dư 4 

M = 1 + 3 + 3² + ... + 3⁹⁹ + 3¹⁰⁰

M = 1 + ( 3 + 3² + 3³ ) + .. + ( 3⁹⁸ + 3⁹⁹ + 3¹⁰⁰ )

M = 1 + 3 . ( 1 + 3 + 3²  ) + ... + 3⁹⁸ . ( 1 + 3 + 3² )

M = 1 + 3 . 13 + ... + 3⁹⁸ . 13 = 1 + 13 . ( 3 + ... + 3⁹⁸ ) : 13 dư 1

The Pagoda of the Celestial Lady (Vietnamese: Chùa Thiên Mụ; also called Linh Mụ Pagoda) is a historic temple in the city of Huế in Vietnam. Its iconic seven-story pagoda is regarded as the unofficial symbol of the city,^[1] and the temple has often been the subject of folk rhymes and ca dao about Huế.^[2]

The pagoda sits on the Hà Khê hill, in the ward of Hương Long in Huế. It is around 3 kilometres (1.9 mi) from the Citadel of Huế constructed by the Nguyễn Dynasty and sits on the northern bank of the Perfume River.^[2][1]

Built in 1601 on the order of the first Nguyễn lords, Nguyễn Hoàng, who at that time was the governor of Thuận Hóa (now known as Huế). The Nguyen Lords were in name, officials of the ruling Lê Dynasty in Hanoi, but was the de factoindependent ruler of central Vietnam. According to the royal annals, Hoang while touring the vicinity, was told of the local legend in which an old lady, known as Thiên Mụ (literally "celestial lady"), dressed in red and blue sat at the site, rubbing her cheeks. She foretold that a lord would come and erect a pagoda on the hill to pray for the country's prosperity. She then vanished after making her prophecy. Upon hearing this, Hoang ordered the construction of a temple at the site, thus the beginning of Thiên Mụ Tự.^[2][1]

The original temple was simply constructed, then later expanded and refurbished. In 1665, major construction was undertaken by the Nguyễn Lord Nguyễn Phúc Tần.^[2]

Hơi dài bn ạ!Nhưng dù sao cx thanks bn nhoa

During the summer of 1963, Thien Mu Pagoda, many in South Vietnam, became a hotbed of anti-government protest. South Vietnam's Buddhist majority had long been discontented with the rule of President Ngo Dinh Diem since his rise to power in 1955. Diem had shown strong favouritism towards Catholics and discrimination against Buddhists in the army, public service and distribution of government aid. In the countryside, Catholics were de facto exempt from performing corvée labour and in some rural areas, Catholic priests led private armies against Buddhist villages. Discontent with Diem exploded into mass protest in Huế during the summer of 1963 when nine Buddhists died at the hand of Diem's army and police on Vesak, the birthday of Gautama Buddha. In May 1963, a law against the flying of religious flags was selectively invoked; the Buddhist flag was banned from display on Vesak while the Vaan flag was displayed to celebrate the anniversary of the consecration of Archbishop Ngo Dinh Thuc, Diem's brother. The Buddhists defied the ban and a protest that began with a march starting from Từ Đàm Pagoda to the government broadcasting station was ended when government forces opened fire. As a result, Buddhist protests were held across the country and steadily grew in size, asking for the signing of a Joint Communique to end religious inequality. Thien Mu Pagoda was a major organising point for the Buddhist movement and was often the location of hunger strikes, barricades and protests.^[1][3][4]

A) (15-2x)(4x+1)-(13-4x)(2x-3)-(x-1)(x+2)+x^2=52

..............bn phân rồi gộp lại để ra kq như dòng dưới nha....

=>19x + 56 = 52

=> 19x = -4

=> x = ‐ 4 / 1 9

NHỚ TK MK ĐÓ

Bài 2 : 

a,\(\frac{x-1}{3}=2-\frac{x}{-2}\)

\(\Leftrightarrow\frac{x-1}{3}=\frac{-4-x}{-2}\Leftrightarrow-2x+2=-12-3x\Leftrightarrow x=-14\)

b, \(\frac{x-1}{x+5}=\frac{6}{7}\Leftrightarrow7x-7=6x+30\Leftrightarrow x=37\)

c, \(\frac{2x-1}{4}=\frac{4}{2x-1}\Leftrightarrow\left(2x-1\right)^2=16\)

\(\Leftrightarrow\left(2x-1\right)^2-4^2=0\Leftrightarrow\left(2x-5\right)\left(2x+3\right)=0\Leftrightarrow x=\frac{5}{2};-\frac{3}{2}\)

Bạn iair thích ý a cho mk một xíu được không bạn

Cho mk hỏi trước dấu trị tuyện đối là dấu j z ?

ko có dấu j cả. Thôi ko cần giải đâu thầy mk giải rùi

Ta có: \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\Leftrightarrow\orbr{\begin{cases}5x-1=0\\2x-\frac{1}{3}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{6}\end{cases}}\)

Vậy.......

tích của hai số bằng 0 thì 1 trong 2 số bằng 0. bạn dựa vào mà tính

Bài 1 : 

\(4\left(x-1\right)^{100}-3^{100}=3^{101}\)

\(\Leftrightarrow\)\(4\left(x-1\right)^{100}=3^{101}+3^{100}\)

\(\Leftrightarrow\)\(4\left(x-1\right)^{100}=3^{100}\left(3+1\right)\)

\(\Leftrightarrow\)\(4\left(x-1\right)^{100}=3^{100}.4\)

\(\Leftrightarrow\)\(\left(x-1\right)^{100}=3^{100}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}\left(x-1\right)^{100}=3^{100}\\\left(x-1\right)^{100}=\left(-3\right)^{100}\end{cases}\Leftrightarrow\orbr{\begin{cases}x-1=3\\x-1=-3\end{cases}}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=3+1\\x=-3+1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-2\end{cases}}}\)

Vậy \(x=-2\) hoặc \(x=4\)

Chúc bạn học tốt ~ 

ai giúp mk bài 2 với