Ta có: \(A=\left(x-3\right)^2+\left(11-x\right)^2\)

\(=x^2-6x+9+x^2-22x+121\)

\(=2x^2-28x+130\)

\(=2\left(x^2-14x+49+16\right)\)

\(=2\left(x-7\right)^2+32\ge32\forall x\)

Dấu '=' xảy ra khi x=7

Cho C = 0

\(\Rightarrow x+\left(-\dfrac{5}{9}\right)x^2=0\)

\(x\left(1-\dfrac{5}{9}x\right)=0\)

\(x=0\) hoặc \(1-\dfrac{5}{9}x=0\)

*) \(1-\dfrac{5}{9}x=0\)

\(\dfrac{5}{9}x=1\)

\(x=\dfrac{1}{\dfrac{5}{9}}\)

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

Vậy nghiệm của đa thức C là \(x=0;x=\dfrac{9}{5}\)

Nhân − 5 19 với x 2 . c = x − 5 19 x 2 Rút gọn x − 5 19 x 2 . Rút gọn mỗi số hạng. Bấm để xem thêm các bước... c = x − 5 x 2 19 Sắp xếp lại x và − 5 x 2 19 . c = − 5 x 2 19 + x

1:

a: =>(|x|+4)(|x|-1)=0

=>|x|-1=0

=>x=1; x=-1

b: =>x^2-4>=0

=>x>=2 hoặc x<=-2

d: =>|2x+5|=2x-5

=>x>=5/2 và (2x+5-2x+5)(2x+5+2x-5)=0

=>x=0(loại)

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

Điều kiện: x≠ \(1\)

Ta có:

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

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

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

\(=3\left[\left(\dfrac{1}{x-1}\right)^2+2.\dfrac{1}{x-1}.\dfrac{1}{2}+\dfrac{1}{4}\right]+\dfrac{1}{4}\)

\(=3\left(\dfrac{1}{x-1}+\dfrac{1}{2}\right)^2+\dfrac{1}{4}\) ≥ \(\dfrac{1}{4}\) (Vì \(3\left(\dfrac{1}{x-1}+\dfrac{1}{2}\right)^2\text{≥}0\) )

Min P=\(\dfrac{1}{4}\) ⇔\(x=-1\)

cảm ơn nha!

x=2,5

thiếu r

Bài 2: 

a: =>x=0 hoặc x=-3

b: =>x-2=0 hoặc 5-x=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1

\(a,\Rightarrow2^{2x-5}=2^5\Rightarrow2x-5=5\Rightarrow x=5\\ b,\Rightarrow4^{2x-1}=4^3\Rightarrow2x-1=3\Rightarrow x=2\\ c,\Rightarrow\left[{}\begin{matrix}2x+2=11\\2x+2=-11\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{13}{2}\end{matrix}\right.\\ d,\Rightarrow2^{x^3}=256=2^8\Rightarrow x^3=8\Rightarrow x=2\)

a)
\(2^{2x-5}=2^5\)
2x-5=5
2x=10
x=5
b)
\(4^{2x-1}=4^3\)
2x-1=3
2x=4
x=2
c)
\(\left(2x+2\right)^2=11^2\)
2x+2=11
2x=9
x=9/2
 

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

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

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

\(=0\)

d) \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2\)

\(=\left(x-3\right)^3-\left(x-3\right)\left(x^2+x.3+3^2\right)+6\left(x+1\right)^2\)

\(=\left(x-3\right)^3-\left(x-3\right)^3+6\left(x+1\right)^2\)

\(=0+6\left(x+1\right)^2\)

\(=6\left(x+1\right)^2\)