(x³-2x+1).(x-2)

=>x⁴-2x²+1x-x³+4x-2x

=> x⁴-x³-2x²+3x

2(x−1).(x+2)=1

=> 2^(x-1)(x+2)=2⁰

=> (x-1)(x+2)=0

=> \(\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

vậy x=1 hoặc x=-2

a, 2 (x+5) - x² - 5x = 0

=> 2(x+5)-(x2+5x)=0

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

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

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

vậy x=-5 và x=2

b, 2x² + 3x - 5 = 0

=> 2x²+5x-2x-5=0

=> x(2x+5)-1(2x+5)=0

=> (x-1)(2x+5)=0

=> \(\left[{}\begin{matrix}x-1=0\\2x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\2x=-5\Rightarrow x=\dfrac{-5}{2}\end{matrix}\right.\)

506766

tick nhé Nguyễn Khắc Vinh

M= x² - x+1

a) M = x²-x\(+\dfrac{1}{4}+\dfrac{3}{4}\)

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

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

d0 \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)

=> \(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

=>M \(\ge\dfrac{3}{4}\Rightarrow M\ge0\left(đpcm\right)\)

b)vì \(M\ge\dfrac{3}{4}\left(a\right)\)

GTNN M =\(\dfrac{3}{4}khi\) x-\(\dfrac{1}{2}=0\)

=> x=\(\dfrac{1}{2}\)

506766
nếu đúng thì tick cho mình nha

a) A=2x²+y²+2xy-8x+2028

=(x²+2xy+y²)+(x²-8x+16)+2012

=(x+y)²+(x-4)²+2012

do (x+y) ²≥ 0 ∀x;y

(x-4)²≥ 0 ∀x

=> (x+y)²+(x-4)² ≥ 0

=> (x+y)²+(x-4)²+2012 ≥ 2012

=> A≥2012

vậy GTNN A=2012 khi \(\left[{}\begin{matrix}x+y=0\\x-4=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}y=-4\\x=4\end{matrix}\right.\)