Question

Tìm các giá trị a khác 2 để pt a(x+a+1)=a^3+2x-2 có nghiệm đạt GTNN

Answer 1

a(x + a + 1) = a³ + 2x - 2

<=> ax + a² + a = a³ + 2x - 2

<=> ax - 2x = a³ - a² - a - 2

<=> (a - 2).x = (a - 2).(a² + a + 1) 

<=> x = a² + a + 1 (Vì a khác 2 nên a - 2 khác 0)

<=> x = a² + 2.a.1/2 + 1/4 + 3/4

<=> x = (a + 1/2)² + 3/4 \(\ge\)3/4

Dấu "=" xảy ra <=> a + 1/2 = 0 <=> a = -1/2

Vậy a = -1/2 thì x có GTNN.

Answer 2

\(a\left(x+a+1\right)=a^3+2x-2\) 2

\(\Leftrightarrow ax+a^2+a=a^3+2x-2\)

\(\Leftrightarrow ax-2x=a^3-a^2-a-2\)

\(\Leftrightarrow\left(a-2\right)\times x=\left(a-2\right)\times\left(a^2+a+1\right)\)

\(\Leftrightarrow x=a^2+a+1\). Vì \(a\ne2\)nên \(a-2\ne0\)

\(\Leftrightarrow x=a^2+2\times a\times\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)

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

Dấu \("="\) xảy ra 

\(\Leftrightarrow a+\frac{1}{2}=0\)

\(\Leftrightarrow a=-\frac{1}{2}\)

Vậy \(a=-\frac{1}{2}\)thì \(x\)có \(GTNN\)

Answer 3

khó quá mình mới học lớp 6 thôi

Answer 4

\(a\left(x+a+1\right)=a^3+2x-2\) 2

\(\Leftrightarrow ax+a^2+a=a^3+2x-2\)

\(\Leftrightarrow ax-2x=a^3-a^2-a-2\)

\(\Leftrightarrow\left(a-2\right)\times x=\left(a-2\right)\times\left(a^2+a+1\right)\)

\(\Leftrightarrow x=a^2+a+1\). Vì \(a\ne2\)nên \(a-2\ne0\)

\(\Leftrightarrow x=a^2+2\times a\times\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)

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

Dấu \("="\) xảy ra 

\(\Leftrightarrow a+\frac{1}{2}=0\)

\(\Leftrightarrow a=-\frac{1}{2}\)

Vậy \(a=-\frac{1}{2}\)thì \(x\)có \(GTNN\)