\(x^2+5x+6=0\\ \Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow x.\left(x+2\right)+3.\left(x+2\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)

\(x^2+5x+6=0\)

\(x^2+5x+6=0\)

\(x^2+3x+2x+6=0\)

\(\left(x+2\right)\left(x+3\right)=0\)

\(x+2=0\)

\(x=-2\)

\(x+3=0\)

\(x=-3\)

Vậy: \(x=-2;x=-3\)

\(\left(3x+2\right)\left(x-1\right)+\left(x+3\right)\left(x-7\right)+2x+23=0\\ \Leftrightarrow3x^2+2x-3x-2+x^2+3x-7x-21+2x+23=0\\ \Leftrightarrow3x^2-x^2+2x-3x+3x-7x+2x-2-21+23=0\\ \Leftrightarrow x^2-3x=0\\ \Leftrightarrow x.\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\)

<=> x=0 hoặc x=3

(3x+2)(x-1)+(x+3)(x-7)+2x+23=0

=>3x²+2x-3x-2+x²+3x-7x-21+2x=-23

=>(3x²+x²)+(2x-3x+3x-7x+2x) -(2+21)=-23

=>4x²-3x-23=-23

=>4x²-3x=-23+23=0

=>x(4x-3)=0

=>x=0 hoặc 4x-3=0

=>x=0 hoặc x=3/4.

\(A=\dfrac{6n-1}{3n-2}\)

\(\Rightarrow A=\dfrac{6n-4+3}{3n-2}\)

\(\Rightarrow A=\dfrac{2\left(3n-2\right)+3}{3n-2}\)

\(\Rightarrow A=2+\dfrac{3}{3n-2}\ge2+\dfrac{3}{3.1-2}=5\left(n=1\in Z\right)\)

\(\Rightarrow Min\left(A\right)=5\left(n=1\right)\)

mkmhkkkkkkkkkkkkkk

Bài 61: 

a, 5,5 : \(\dfrac{1}{2}\)  = \(\dfrac{55}{10}\) : \(\dfrac{1}{2}\) = \(\dfrac{11}{1}\)

    Vậy 5,5 : \(\dfrac{1}{2}\) = \(\dfrac{11}{1}\)

b, 1\(\dfrac{5}{7}\): (-1\(\dfrac{3}{9}\))  = \(\dfrac{12}{7}\): (-\(\dfrac{4}{3}\)) = - \(\dfrac{9}{7}\) = \(\dfrac{-9}{7}\)

    vậy 1\(\dfrac{5}{7}\) = \(\dfrac{-9}{7}\)

c, (-0,12) : 2\(\dfrac{3}{4}\) = - \(\dfrac{12}{100}\) : \(\dfrac{11}{4}\) = - \(\dfrac{12}{275}\)  = \(\dfrac{-12}{275}\)

    vậy (-0,12) : 2\(\dfrac{3}{4}\) = \(\dfrac{-12}{275}\) 

d, (-2,5) : 3,5 =  \(\dfrac{-25}{10}\) : \(\dfrac{35}{10}\) = \(\dfrac{-5}{7}\)

   vậy -2,5 : 3,5 = \(\dfrac{-5}{7}\) 

Bài 62: (-5) : 10 = - \(\dfrac{1}{2}\);     \(\dfrac{2}{5}\) : \(\dfrac{9}{7}\) = \(\dfrac{14}{45}\);

(-3,11)  : 12,5 = - \(\dfrac{311}{1250}\);       - \(\dfrac{14}{5}\) : 9 = - \(\dfrac{14}{45}\) 

(-1,5): 3 = \(-\dfrac{1}{2}\);                    - \(\dfrac{311}{50}\): 25  =    -  \(\dfrac{311}{1250}\)

\(\dfrac{-5}{10}\) = \(\dfrac{-1,5}{3}\);           \(\dfrac{-3,11}{12,5}\) = \(\dfrac{\dfrac{-311}{50}}{25}\)  

\(2^{91}=\left(2^{13}\right)^7=73728^7\)

\(5^{35}=\left(5^5\right)^7=3125^7\) nhỏ hơn \(73728^7\)

\(\Rightarrow2^{91}\) lớn hơn \(5^{35}\)

\(b,3^{400}=\left(3^4\right)^{100}=81^{100}\\ 4^{300}=\left(4^3\right)^{100}=64^{100}\\ Vì:81^{100}>64^{100}\left(Do:81>64\right)\\ \Rightarrow3^{400}>4^{300}\)

\(\left(x+\dfrac{1}{3}\right)^2=\dfrac{1}{25}\\ \Rightarrow\left(x+\dfrac{1}{3}\right)^2=\left(\pm\dfrac{1}{5}\right)^2\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{1}{5}\\x+\dfrac{1}{3}=-\dfrac{1}{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{15}\\x=-\dfrac{8}{15}\end{matrix}\right.\)

\(\left(x+\dfrac{1}{3}\right)^2=\dfrac{1}{25}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{1}{5}\\x+\dfrac{1}{3}=-\dfrac{1}{5}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{15}\\x=-\dfrac{8}{15}\end{matrix}\right.\)

Vậy phương trình có nghiệm \(S=\left\{-\dfrac{2}{15};-\dfrac{8}{15}\right\}\)

\(\left(x-\dfrac{3}{4}\right)^2=\dfrac{1}{16}\\ \Rightarrow\left(x-\dfrac{3}{4}\right)^2=\left(\pm\dfrac{1}{4}\right)^2\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{4}\\x-\dfrac{3}{4}=-\dfrac{1}{4}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2}\end{matrix}\right.\)

(x-3/4)²=1/16=(1/4)²

=>x-3/4=1/4 hoặc x-3/4=-1/4

=>x=1/4+3/4=1 hoặc x=-1/4+3/4=2/4=1/2

\(\left(2x-3\right)^2=\dfrac{4}{25}\\ \Rightarrow\left(2x-3\right)^2=\left(\pm\dfrac{2}{5}\right)^2\\ \Rightarrow\left[{}\begin{matrix}2x-3=\dfrac{2}{5}\\2x-3=-\dfrac{2}{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=\dfrac{17}{5}\\2x=\dfrac{13}{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{17}{10}\\x=\dfrac{13}{10}\end{matrix}\right.\)

\(\left(2x-3\right)^2=\dfrac{4}{25}\\ \Leftrightarrow\left(2x-3\right)^2=\left(\dfrac{2}{5}\right)^2\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=\dfrac{2}{5}\\-2x+3=\dfrac{2}{5}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{17}{5}\\-2x=-\dfrac{13}{5}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{10}\\x=\dfrac{13}{10}\end{matrix}\right.\)

Vậy...

Lời giải:

Gọi số bộ quần áo 2 phân xưởng may được lần lượt là $a,b$ (bộ)

Theo bài ra ta có:

$a+b=1125$

$a:b=0,8=4:5$

$\Rightarrow \frac{a}{4}=\frac{b}{5}=\frac{a+b}{4+5}=\frac{1125}{9}=125$ (áp dụng tính chất dtsbn)

$\Rightarrow a=125.4=500;  b=125.5=625$ (bộ)

cảm ơn bạn nha