a) x = 8 3 .                            b) x = − 9 20 .  

c: \(=\left(x+1\right)^2+1>0\forall x\)

Trả lời:

a, \(x^2-6x+11=x^2-6x+9+2=\left(x-3\right)^2+2\ge2\forall x\)

Dấu "=" xảy ra khi x - 3 = 0 <=> x = 3

Vậy GTNN của biểu thức bằng 2 khi x = 3

b, \(-x^2+6x-11=-\left(x^2-6x+11\right)=-\left(x^2-6x+9+2\right)=-\left[\left(x-3\right)^2+2\right]\)

\(=-\left(x-3\right)^2-2\le-2\forall x\)

Dấu "=" xảy ra khi x - 3 = 0 <=> x = 3

Vậy GTLN của biểu thức bằng - 2 khi x = 3

c, \(x^2+2x+2=x^2+2x+1+1=\left(x+1\right)^2+1\ge1>0\forall x\inℤ\)  (đpcm)

Dấu "=" xảy ra khi x + 1 = 0 <=> x = - 1

1)(x²-4x+16)(x+4)-x(x+1)(x+2)+3x²=0

\(\Rightarrow\)(x³+64)-x(x²+2x+x+2)+3x²=0

\(\Rightarrow\)x³+64-x³-2x²-x²-2x+3x²=0

\(\Rightarrow\)-2x+64=0

\(\Rightarrow\)-2x=-64

\(\Rightarrow\)x=\(\dfrac{-64}{-2}\)

\(\Rightarrow x=32\)

2)(8x+2)(1-3x)+(6x-1)(4x-10)=-50

\(\Rightarrow\)8x-24x²+2-6x+24x²-60x-4x+10=50

\(\Rightarrow\)-62x+12=50

\(\Rightarrow\)-62x=50-12

\(\Rightarrow\)-62x=38

\(\Rightarrow\)x=\(-\dfrac{38}{62}=-\dfrac{19}{31}\)

3)(x²+2x+4)(2-x)+x(x-3)(x+4)-x²+24=0

\(\Rightarrow\)8-x³+x(x²+4x-3x-12)-x²+24=0

\(\Rightarrow\)8-x³+x³+4x²-3x²-12x-x²+21=0

\(\Rightarrow\)-12x+29=0

\(\Rightarrow\)-12x=-29

\(\Rightarrow\)x=\(\dfrac{-29}{-12}=\dfrac{29}{12}\)

\(2x^2-6x-1=0\)

Theo Vi - ét, ta có :

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{6}{2}=3\\x_1x_2=\dfrac{c}{a}=-\dfrac{1}{2}\end{matrix}\right.\)

Ta có :

\(A=\dfrac{x_1-2}{x_2-1}+\dfrac{x_2-2}{x_1-1}\)

\(=\dfrac{\left(x_1-2\right)\left(x_1-1\right)+\left(x_2-2\right)\left(x_2-1\right)}{\left(x_2-1\right)\left(x_1-1\right)}\)

\(=\dfrac{x_1^2-x_1-2x_1+2+x_2^2-x_2-2x_2+2}{x_1x_2-\left(x_1+x_2\right)+1}\)

\(=\dfrac{\left(x_1+x_2\right)^2-2x_1x_2-3\left(x_1+x_2\right)+4}{x_1x_2-\left(x_1+x_2\right)+1}\)

\(=\dfrac{3^2-2.\left(-\dfrac{1}{2}\right)-3.3+4}{-\dfrac{1}{2}-3+1}\)

\(=-2\)

\(1,\Leftrightarrow x^2+10x+25=x^2-4x-21\\ \Leftrightarrow14x=-46\\ \Leftrightarrow x=-\dfrac{23}{7}\\ 2,\Leftrightarrow x^3+8=15+x^3+2x\\ \Leftrightarrow2x=-7\Leftrightarrow x=-\dfrac{7}{2}\\ 3,\Leftrightarrow\left(x+3\right)^2=0\\ \Leftrightarrow x=-3\\ 4,\Leftrightarrow x^3-9x^2+27x-27=0\\ \Leftrightarrow\left(x-3\right)^3=0\\ \Leftrightarrow x-3=0\Leftrightarrow x=3\\ 5,\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\\ \Leftrightarrow-12x=24\Leftrightarrow x=-2\\ 6,\Leftrightarrow x^2-3x+5x-15=0\\ \Leftrightarrow\left(x-3\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

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

\(\Delta'=\left(-3\right)^2+3.2=15>0\)

⇒ Phương trình có hai nghiệm phân biệt với mọi m.

Theo hệ thức viét ta có : \(\left\{{}\begin{matrix}x_1+x_2=3\\x_1.x_2=-\dfrac{3}{2}\end{matrix}\right.\)

Ta có : \(B=3x_1x_2-x_1^2-x_2^2=-\left(x_1+x_2\right)^2+5x_1x_2=-9+5.\left(-\dfrac{3}{2}\right)=\dfrac{135}{2}\)

Vậy \(B=-\dfrac{135}{2}\) với hai nghiệm phân biệt thỏa mãn.

a: \(\Leftrightarrow x\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

c: \(\Leftrightarrow\left(x-1\right)\left(3x-1\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

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

b) \(\Leftrightarrow\left(3x-1-x-5\right)\left(3x-1+x+5\right)=0\\ \Leftrightarrow\left(2x-6\right)\left(4x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

c) \(9x^2\left(x-1\right)=x-1\\ \Leftrightarrow\left(9x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(3x-1\right)\left(3x+1\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\\x=1\end{matrix}\right.\)

d) \(x^2-4=\left(x-2\right)^2\\ \Leftrightarrow\left(x-2\right)\left(x+2-x+2\right)=0\\ \Leftrightarrow4\left(x-2\right)=0\\ \Leftrightarrow x=2\)

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

f) \(x^3-0,36=0\\ \Leftrightarrow x\left(x^2-0,36\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{3}{5}\\x=\dfrac{3}{5}\end{matrix}\right.\)

g) \(\Leftrightarrow\left(5x-1\right)\left(x-2018\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=2018\end{matrix}\right.\)

h) \(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

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

\(\Delta'=3^2+3.2=15>0\)

⇒ Phương trình có hai nghiệm phân biệt.

Theo hệ thức viét có : \(\left\{{}\begin{matrix}x_1+x_2=3\\x_1x_2=-\dfrac{3}{2}\end{matrix}\right.\)

Ta có : \(A=x_1^2x_2^2-2x_1-2x_2=\left(x_1x_2\right)^2-2\left(x_1+x_2\right)=\left(-\dfrac{3}{2}\right)^2-2.3=-\dfrac{15}{4}\)

Vậy \(A=-\dfrac{15}{4}\) thì thỏa mãn điều kiện bài ra.