a) Có: y = f(x) = 4x² - 3

=> f(-2) = 4 . (-2) - 3

             = -11

Vậy f(-2) = -11

b) Có: f(x) = 4x² - 3

Mà f(x) = 1

=> 4x² - 3 = 1

<=> 4x² = 4

<=> x² = 1

<=> x = 1 hoặc x = -1

Vậy x = 1 hoặc x = -1 thì f(x) = 1.

c) Có: f(x) = 4x² - 3

Mà f(x) = x

=> 4x² - 3 = x

<=> 4x² - 3 - x = 0

<=> (4x² + 3x) - (4x + 3) = 0

<=> x(4x + 3) - (4x+ 3) = 0

<=> (x - 1)(4x + 3) = 0

<=> x - 1 = 0 hoặc 4x + 3 = 0

<=> x = 1 hoặc 4x = -3

<=> x = 1 hoặc x = \(-\frac{3}{4}\)

Vậy x = 1 hoặc x = \(-\frac{3}{4}\) thì f(x) = x.

Linz

a, \(f\left(-2\right)=4\left(-2\right)^2-3=16-3=13\)

b, \(f\left(x\right)=1\)hay \(f\left(x\right)=4x^2-3=1\)

\(\Leftrightarrow x^2=1\Leftrightarrow x=\pm1\)

c, \(f\left(x\right)=x\)hay \(4x^2-3=x\)

\(\Leftrightarrow4x^2-3-x=0\Leftrightarrow3x^2+x^2-3-x=0\)

\(\Leftrightarrow3\left(x^2-1\right)+x\left(x-1\right)=0\)

\(\Leftrightarrow3\left(x-1\right)\left(x+1\right)+x\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[3\left(x+1\right)+x\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(4x+3\right)=0\Leftrightarrow x=1;-\frac{3}{4}\)

x=7/27

\(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\)

\(\Leftrightarrow\left(x-\frac{2}{9}\right)^3=\left(\frac{4}{9}\right)^3\)

\(\Leftrightarrow x-\frac{2}{9}=\frac{4}{9}\)

\(\Leftrightarrow x=\frac{4}{9}+\frac{2}{9}\)

\(\Leftrightarrow x=\frac{2}{3}\)

Vậy\(x=\frac{2}{3}\)

Mink