\(a,A=\left(x^2-x\right)\left(x^2-x-12\right)\\ A=\left(x^2-x\right)^2-12\left(x^2-x\right)\\ A=\left(x^2-x\right)^2-12\left(x^2-x\right)+36-36\\ A=\left(x^2-x+6\right)^2-36\ge-36\\ A_{min}=-36\Leftrightarrow x^2-x+6=0\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\\ b,B=4x^4+4x^3+5x^2+4x+3\\ B=\left(4x^4+4x^3+x^2\right)+\left(x^2+4x+4\right)-1\\ B=x^2\left(2x+1\right)^2+\left(x+2\right)^2-1\ge-1\\ B_{min}=-1\Leftrightarrow\left\{{}\begin{matrix}x\left(2x+1\right)=0\\x+2=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

Vậy dấu \("="\) không xảy ra

a)

(x+4)(3x-5) = 0

=> x + 4 = 0 hoặc 3x-5 = 0

     x = -4                 x = 5/3

b)

  2x² + 7x + 3 = 0

  2x² + 6x + x + 3= 0

  (2x+1)(x+3) = 0

=> 2x+1 = 0 hoặc x + 3 = 0

    x = -1/2              x = -3

a) Ta có: \(\left|2.5-x\right|=1.3\)

\(\Leftrightarrow\left|x-2.5\right|=1.3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2.5=1.3\\x-2.5=-1.3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3.8\\x=-1.3+2.5=1.2\end{matrix}\right.\)

Vậy: \(x\in\left\{3.8;1.2\right\}\)

b) Ta có: \(1.6-\left|x-0.2\right|=0\)

\(\Leftrightarrow\left|x-0.2\right|=1.6\)

\(\Leftrightarrow\left[{}\begin{matrix}x-0.2=1.6\\x-0.2=-1.6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1.8\\x=-1.4\end{matrix}\right.\)

Vậy: \(x\in\left\{1.8;-1.4\right\}\)

c) Ta có: \(13^x=169\)

\(\Leftrightarrow13^x=13^2\)

\(\Leftrightarrow x=2\)

Vậy: x=2

d) Ta có: \(\dfrac{-2}{x}=\dfrac{-x}{\dfrac{8}{25}}\)

\(\Leftrightarrow\dfrac{2}{x}=\dfrac{x}{\dfrac{8}{25}}\)

\(\Leftrightarrow x^2=\dfrac{16}{25}\)

hay \(x\in\left\{\dfrac{4}{5};-\dfrac{4}{5}\right\}\)

Vậy: \(x\in\left\{\dfrac{4}{5};-\dfrac{4}{5}\right\}\)

\(|2,5-x|=1,3\)

\(\Rightarrow2,5-x=1,3\)    hoặc      -(2,5-x)=1,3

=>x=2,5-1,3                          x-2,5=1,3

=>x=1,2                                x=1,3+2,5=3,8

Vậy \(x\in\left\{1,2;3,8\right\}\)

\(1,6-|x-0,2|=0\Rightarrow|x-0,2|=1,6\)

=> x-0,2=1,6       hoặc       -(x-0,2)=1,6

=>x=1,6+0,2                      0,2-x=1,6

=>x=3,8                              x=0,2-1,6=-1,4

Vậy \(x\in\left\{3,8;-1,4\right\}\)

\(13^x=169\Rightarrow13^x=13^2\Rightarrow x=2\)

\(\dfrac{-2}{x}=\dfrac{-x}{\dfrac{8}{25}}\Rightarrow-2\times\dfrac{8}{25}=-x\times x\Rightarrow\dfrac{-16}{25}=-x^2\Rightarrow\dfrac{16}{25}=x^2\Rightarrow x^2=\left(\dfrac{4}{5}\right)^2=\left(-\dfrac{4}{5}\right)^2\)

\(\Rightarrow x=\dfrac{4}{5}\) hoặc \(x=-\dfrac{4}{5}\)

\(a.13.\left(78+22\right)=13.100=1300\)

\(b.25.70+28.25+25.2=25.\left(70+28+2\right)=25.100=2500\)

\(c.\dfrac{15.3.14.2.19}{19.14.15}=6\)

\(d.\dfrac{132.70.60}{12.35.20}=\dfrac{11.12.35.2.20.3}{12.35.20}=66\)

\(a,-3x^2+7x-9+\left(x-1\right)\left(x+2\right)\\ =-3x^2+7x-9+x^2-x+2x-2\\ =\left(-3x^2+x^2\right)+\left(7x-x+2x\right)-\left(9+2\right)\\ =-2x^2+8x-11\\ b,x\left(x-5\right)-2x\left(x+1\right)\\ =x^2-5x-2x^2-2x\\ =\left(x^2-2x^2\right)-\left(5x+2x\right)\\ =-3x^2-7x\\ c,4x\left(x^2-x+1\right)-\left(x-1\right)\left(x^2-x\right)\\ =4x^3-4x^2+4x-x\left(x^2-x\right)+x^2-x\\ =4x^3-4x^2+4x-x^3+x^2+x^2-x\\ =\left(4x^3-x^3\right)+\left(-4x^2+x^2+x^2\right)+\left(4x-x\right)\\ =3x^3-2x^2+3x\\ =x\left(3x^2-2x+3\right)\)

\(d,-5x\left(x-5\right)+\left(x-3\right)\left(x^2-7\right)\\ =-5x^2+25x+x\left(x^2-7\right)-3\left(x^2-7\right)\\ =-5x^2+25x+x^3-7x-3x^2+21\\ =\left(-5x^2-3x^2\right)+\left(25x-7x\right)+x^3+21\\ =-8x^2+x^3+18x+21\)

\(A=\left(7x-1\right)^2-4\left|1-7x\right|+5\)

\(\Rightarrow MinA=5\)khi và  chỉ khi x=1/7

nguyen hoang nhờ bạn giải cụ thể ra giùm mình được k ạ?

a) Đặt \(\left|1-7x\right|=t\left(t\ge0\right)\Rightarrow\left(7x-1\right)^2=t^2\)

Ta có \(A=t^2-4t+5=\left(t^2-4t+4\right)+1=\left(t-2\right)^2+1\ge1\)

Vậy min A = 1 khi t = 2 hay \(\left|1-7x\right|=2\Leftrightarrow\orbr{\begin{cases}1-7x=2\\1-7x=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{7}\\x=\frac{3}{7}\end{cases}}\)

b) \(B=\left(x^2+x+1\right)^4\)

Ta thấy \(x^2+x+1=\left(x^2+x+\frac{1}{4}\right)+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Vậy thì \(B\ge\left(\frac{3}{4}\right)^4=\frac{81}{256}\)

Vậy \(minB=\frac{81}{256}\) khi \(x=-\frac{1}{2}.\)

a) \(\frac{x^2+5x}{5x^2+x^3}\)

\(=\frac{x\left(x+5\right)}{x^2\left(x+5\right)}=\frac{1}{x}\)

b) \(\frac{x^4+x^2+1}{x^3+1}\)

\(=\frac{\left(x^2+x+1\right)\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\frac{x^2+x+1}{x+1}\)

\(a)\frac{x^2+5x}{5x^2+x^3}=\frac{x\left(x+5\right)}{x^2\left(5+x\right)}=\frac{1}{x}\)

\(a,\frac{x^2+5x}{5x^2+x^3}=\frac{x\left(x+5\right)}{x^2\left(x+5\right)}=\frac{1}{x}\)

\(b,\frac{x^4+x^2+1}{x^3+1}=\frac{x^4+x^3+x^2-\left(x^3-1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{x^2\left(x^2+x+1\right)-\left(x-1\right)\left(x^2+x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\frac{\left(x^2+x+1\right)\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{x^2+x+1}{x+1}\)

hok tốt nha