(3 + x).2 - 47 = -147

(3 + x).2 = -147 + 47 

(3 + x).2= - 100 

3 + x = -100 : 2 

3 + x = -50

x = -50 - 3 

x = -53

\(\left(3+x\right)\cdot2-47=-147\)

=>\(2\left(x+3\right)=-147+47=-100\)

=>x+3=-50

=>x=-53

10.000 đồng + 30.000 đồng = 40000 đồng

2.000 đồng + 12.000 đồng = 14000 đồng

30 triệu đồng - 10 triệu đồng =20 triệu đồng

40.000 đồng

14.000 đồng

20 triệu đồng

a: \(\dfrac{x-3}{3}=\dfrac{2x+1}{5}\)

=>\(3\left(2x+1\right)=5\left(x-3\right)\)

=>6x+3=5x-15

=>6x-5x=-3-15

=>x=-18

b: \(\dfrac{x+1}{22}=\dfrac{6}{x}\)(ĐKXĐ: \(x\ne0\))

=>\(x\left(x+1\right)=6\cdot22\)

=>\(x^2+x-132=0\)

=>(x+12)(x-11)=0

=>\(\left[{}\begin{matrix}x+12=0\\x-11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-12\left(nhận\right)\\x=11\left(nhận\right)\end{matrix}\right.\)

c: \(\dfrac{2x-1}{2}=\dfrac{5}{x}\)(ĐKXĐ: \(x\ne0\))

=>\(x\left(2x-1\right)=5\cdot2\)

=>\(2x^2-x-10=0\)

=>\(2x^2-5x+4x-10=0\)

=>x(2x-5)+2(2x-5)=0

=>(2x-5)(x+2)=0

=>\(\left[{}\begin{matrix}2x-5=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\left(nhận\right)\\x=-2\left(nhận\right)\end{matrix}\right.\)

a) \(\dfrac{x-3}{3}=\dfrac{2x+1}{5}\\ \Rightarrow5\left(x-3\right)=3\left(2x+1\right)\\ \Rightarrow5x-15=6x+3\\ \Rightarrow6x-5x=-15-3\\ \Rightarrow x=-18\)

b) \(\dfrac{x+1}{22}=\dfrac{6}{x}\left(x\ne0\right)\\ \Rightarrow x\left(x+1\right)=6.22\\ \Rightarrow x^2+x=132\\ \Rightarrow x^2+x-132=0\\ \Rightarrow\left(x^2+12x\right)-\left(11x+132\right)=0\\ \Rightarrow x\left(x+12\right)-11\left(x+12\right)=0\\ \Rightarrow\left(x+12\right)\left(x-11\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+12=0\\x-11=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-12\\x=11\end{matrix}\right.\left(TM\right)\)

c) \(\dfrac{2x-1}{2}=\dfrac{5}{x}\left(x\ne0\right)\\ \Rightarrow x\left(2x-1\right)=2.5\\ \Rightarrow2x^2-x-10=0\\ \Rightarrow\left(2x^2+4x\right)-\left(5x+10\right)=0\\ \Rightarrow2x\left(x+2\right)-5\left(x+2\right)=0\\ \Rightarrow\left(x+2\right)\left(2x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+2=0\\2x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{5}{2}\end{matrix}\right.\left(TM\right)\)

\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2023\cdot2024}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2023}-\dfrac{1}{2024}\)

\(=1-\dfrac{1}{2024}=\dfrac{2023}{2024}\)

\(B=\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{85\cdot89}\)

\(=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{85}-\dfrac{1}{89}\)

\(=1-\dfrac{1}{89}=\dfrac{88}{89}\)

\(C=\dfrac{7}{10\cdot11}+\dfrac{7}{11\cdot12}+...+\dfrac{7}{69\cdot70}\)

\(=7\left(\dfrac{1}{10\cdot11}+\dfrac{1}{11\cdot12}+...+\dfrac{1}{69\cdot70}\right)\)

\(=7\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)

\(=7\left(\dfrac{1}{10}-\dfrac{1}{70}\right)=7\cdot\dfrac{6}{70}=\dfrac{42}{70}=\dfrac{6}{10}=\dfrac{3}{5}\)

\(D=\dfrac{1}{18}+\dfrac{1}{54}+...+\dfrac{1}{990}\)

\(=\dfrac{1}{3}\left(\dfrac{3}{3\cdot6}+\dfrac{3}{6\cdot9}+...+\dfrac{3}{30\cdot33}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{33}\right)=\dfrac{1}{3}\cdot\dfrac{10}{33}=\dfrac{10}{99}\)

\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2023\cdot2024}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2023}-\dfrac{1}{2024}\)

\(=1-\dfrac{1}{2024}\)

\(=\dfrac{2023}{2024}\)

\(B=\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{85\cdot89}\)

\(=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{85}-\dfrac{1}{89}\)

\(=1-\dfrac{1}{89}\)

\(=\dfrac{88}{89}\)

\(C=\dfrac{7}{10\cdot11}+\dfrac{7}{11\cdot12}+...+\dfrac{7}{69\cdot70}\)

\(=7\left(\dfrac{1}{10\cdot11}+\dfrac{1}{11\cdot12}+...+\dfrac{1}{69\cdot70}\right)\)

\(=7\cdot\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)

\(=7\cdot\left(\dfrac{1}{10}-\dfrac{1}{70}\right)\)

\(=7\cdot\dfrac{6}{70}\)

\(=\dfrac{3}{5}\)

\(D=\dfrac{1}{18}+\dfrac{1}{54}+...+\dfrac{1}{990}\)

\(=\dfrac{1}{3\cdot6}+\dfrac{1}{6\cdot9}+\dfrac{1}{9\cdot12}+...+\dfrac{1}{30\cdot33}\)

\(=\dfrac{1}{3}\cdot\left(\dfrac{3}{3\cdot6}+\dfrac{3}{6\cdot9}+...+\dfrac{3}{30\cdot33}\right)\)

\(=\dfrac{1}{3}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{6}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)

\(=\dfrac{1}{3}\cdot\left(\dfrac{1}{3}-\dfrac{1}{33}\right)\\ =\dfrac{1}{3}\cdot\dfrac{10}{33}=\dfrac{10}{99}\)

a: x-y=2(x+y)

=>x-y=2x+2y

=>-x=3y

\(x-y=x:y\)

=>\(-3y-y=\dfrac{-3y}{y}=-3\)

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

=>\(x=-3y=-\dfrac{9}{4}\)

b) \(x+y=xy=\dfrac{x}{y}\) 

Ta có: \(xy=\dfrac{x}{y}\Rightarrow xy^2=x\Rightarrow y^2=1\Rightarrow y=\pm1\) 

\(y=1\Rightarrow x+1=1\cdot x\Rightarrow1=0\) (vô lý) 

\(y=-1\Rightarrow x+\left(-1\right)=\left(-1\right)\cdot x\)

\(\Rightarrow x-1=-x\Rightarrow2x=1\Rightarrow x=\dfrac{1}{2}\) 

Vậy: ... 

a: \(\dfrac{x+2}{98}+\dfrac{x+4}{96}=\dfrac{x+6}{94}+\dfrac{x+8}{92}\)

=>\(\dfrac{x+2}{98}+1+\dfrac{x+4}{96}+1=\dfrac{x+6}{94}+1+\dfrac{x+8}{92}+1\)

=>\(\dfrac{x+100}{98}+\dfrac{x+100}{96}-\dfrac{x+100}{94}-\dfrac{x+100}{92}=0\)

=>\(\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{96}-\dfrac{1}{94}-\dfrac{1}{92}\right)=0\)

=>x+100=0

=>x=-100

b: \(\dfrac{x-214}{86}+\dfrac{x-132}{84}+\dfrac{x-54}{82}=6\)

=>\(\left(\dfrac{x-214}{86}-1\right)+\left(\dfrac{x-132}{84}-2\right)+\left(\dfrac{x-54}{82}-3\right)=0\)

=>\(\dfrac{x-300}{86}+\dfrac{x-300}{84}+\dfrac{x-300}{82}=0\)

=>x-300=0

=>x=300

c: \(\dfrac{148-x}{25}+\dfrac{169-x}{23}+\dfrac{186-x}{21}+\dfrac{199-x}{19}=10\)

=>\(\left(\dfrac{148-x}{25}-1\right)+\left(\dfrac{169-x}{23}-2\right)+\left(\dfrac{186-x}{21}-3\right)+\left(\dfrac{199-x}{19}-4\right)=0\)

=>\(\dfrac{123-x}{25}+\dfrac{123-x}{23}+\dfrac{123-x}{21}+\dfrac{123-x}{19}=0\)

=>123-x=0

=>x=123

d: \(\dfrac{x-17}{1990}+\dfrac{x-21}{1986}+\dfrac{x+1}{1004}=4\)

=>\(\dfrac{x-17}{1990}-1+\dfrac{x-21}{1986}-1+\dfrac{x+1}{1004}-2=0\)

=>\(\dfrac{x-2007}{1990}+\dfrac{x-2007}{1986}+\dfrac{x-2007}{1004}=0\)

=>x-2007=0

=>x=2007

a) \(\dfrac{x+2}{98}+\dfrac{x+4}{96}=\dfrac{x+6}{94}+\dfrac{x+8}{92}\)

\(\Leftrightarrow\dfrac{x+2}{98}+\dfrac{x+4}{96}-\dfrac{x+6}{94}-\dfrac{x+8}{92}=0\)

\(\Leftrightarrow\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+4}{96}+1\right)-\left(\dfrac{x+6}{94}+1\right)-\left(\dfrac{x+8}{92}+1\right)=0\)

\(\Leftrightarrow\dfrac{x+100}{98}+\dfrac{x+100}{96}-\dfrac{x+100}{94}-\dfrac{x+100}{92}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{96}-\dfrac{1}{94}-\dfrac{1}{92}\right)=0\)

\(\Leftrightarrow x+100=0\) (vì \(\dfrac{1}{98}+\dfrac{1}{96}-\dfrac{1}{94}-\dfrac{1}{92}\ne0\))

\(\Leftrightarrow x=-100\)

b) \(\dfrac{x-214}{86}+\dfrac{x-132}{84}+\dfrac{x-54}{82}=6\)

\(\Leftrightarrow\left(\dfrac{x-214}{86}-1\right)+\left(\dfrac{x-132}{84}-2\right)+\left(\dfrac{x-54}{82}-3\right)=0\)

\(\Leftrightarrow\dfrac{x-300}{86}+\dfrac{x-300}{84}+\dfrac{x-300}{82}=0\)

\(\Leftrightarrow\left(x-300\right)\left(\dfrac{1}{86}+\dfrac{1}{84}+\dfrac{1}{82}\right)=0\)

\(\Leftrightarrow x-300=0\) (vì \(\dfrac{1}{86}+\dfrac{1}{84}+\dfrac{1}{82}\ne0\))

\(\Leftrightarrow x=300\)

c) \(\dfrac{148-x}{25}+\dfrac{169-x}{23}+\dfrac{186-x}{21}+\dfrac{199-x}{19}=10\)

\(\Leftrightarrow\left(\dfrac{148-x}{25}-1\right)+\left(\dfrac{169-x}{23}-2\right)+\left(\dfrac{186-x}{21}-3\right)+\left(\dfrac{199-x}{19}-4\right)=0\)

\(\Leftrightarrow\dfrac{123-x}{25}+\dfrac{123-x}{23}+\dfrac{123-x}{21}+\dfrac{123-x}{19}=0\)

\(\Leftrightarrow\left(123-x\right)\left(\dfrac{1}{25}+\dfrac{1}{23}+\dfrac{1}{21}+\dfrac{1}{19}\right)=0\)

\(\Leftrightarrow123-x=0\) (vì \(\dfrac{1}{25}+\dfrac{1}{23}+\dfrac{1}{21}+\dfrac{1}{19}\ne0\))

\(\Leftrightarrow x=123\)

d) \(\dfrac{x-17}{1990}+\dfrac{x-21}{1986}+\dfrac{x+1}{1004}=4\)

\(\Leftrightarrow\left(\dfrac{x-17}{1990}-1\right)+\left(\dfrac{x-21}{1986}-1\right)+\left(\dfrac{x+1}{1004}-2\right)=0\)

\(\Leftrightarrow\dfrac{x-2007}{1990}+\dfrac{x-2007}{1986}+\dfrac{x-2007}{1004}=0\)

\(\Leftrightarrow\left(x-2007\right)\left(\dfrac{1}{1990}+\dfrac{1}{1986}+\dfrac{1}{1004}\right)=0\)

\(\Leftrightarrow x-2007=0\) (vì \(\dfrac{1}{1990}+\dfrac{1}{1986}+\dfrac{1}{1004}\ne0\))

\(\Leftrightarrow x=2007\)

$Toru$

\(\dfrac{1}{5^2}< \dfrac{1}{4\cdot5}=\dfrac{1}{4}-\dfrac{1}{5}\)

\(\dfrac{1}{6^2}< \dfrac{1}{5\cdot6}=\dfrac{1}{5}-\dfrac{1}{6}\)

...

\(\dfrac{1}{100^2}< \dfrac{1}{99\cdot100}=\dfrac{1}{99}-\dfrac{1}{100}\)

Do đó: \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

=>\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}\)

\(\dfrac{1}{5^2}>\dfrac{1}{5\cdot6}=\dfrac{1}{5}-\dfrac{1}{6}\)

\(\dfrac{1}{6^2}>\dfrac{1}{6\cdot7}=\dfrac{1}{6}-\dfrac{1}{7}\)

...

\(\dfrac{1}{100^2}>\dfrac{1}{100\cdot101}=\dfrac{1}{100}-\dfrac{1}{101}\)

Do đó: \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}>\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{101}\)

=>\(\dfrac{1}{5^2}+...+\dfrac{1}{100^2}>\dfrac{1}{5}\)

mà 1/5>1/6

nên \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}>\dfrac{1}{6}\)

Do đó: \(\dfrac{1}{6}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}\)

Gọi độ dài quãng đường từ thư viện đến trường là x(m)

(Điều kiện: x>0)

Thời gian Qiqi đi từ thư viện đến trường là \(\dfrac{x}{60}\left(phút\right)\)

Thời gian Weiling đi từ thư viện đến trường là \(\dfrac{x}{72}\left(phút\right)\)

Weiling đến trường trước Qiqi 4 phút và xuất phát sau 2 phút nên ta có: \(\dfrac{x}{60}-\dfrac{x}{72}=4+2=6\)

=>\(\dfrac{x}{360}=6\)

=>\(x=6\cdot360=2160\left(nhận\right)\)

Vậy: độ dài quãng đường từ thư viện đến trường là 2160(m)

Nếu đi từ thư viện đến trường, thời gian Weiling hoàn thành nhanh hơn Qiqi là:

   4 + 2 = 6 (phút)

Tỉ lệ vận tốc Qiqi so với Weiling là: \(\dfrac{60}{72}=\dfrac{5}{6}\)

Cùng một quãng đường, vận tốc và thời gian là hai đại lượng tỉ lệ nghịch với nhau

Do đó tỉ lệ thời gian hoàn thành quãng đường Qiqi so với Weiling là: \(\dfrac{6}{5}\)

Coi thời gian Qiqi đi là 6 phần, Weiling đi là 5 phần

Hiệu số phần bằng nhau:

  6 - 5 = 1 (phần)

Thời gian Qiqi đi là:

  6 : 1 x 6 = 36 (phút)

Quãng đường từ thư viện đến trường là:

  36 x 60 = 2160 (m)

        Đáp số: 2160m

\(3x^3-14x^2+4x+3\)

\(=3x^3+x^2-15x^2-5x+9x+3\)

\(=x^2\left(3x+1\right)-5x\left(3x+1\right)+3\left(3x+1\right)\)

\(=\left(3x+1\right)\left(x^2-5x+3\right)\)