\(a,f\left(1\right)=3\cdot1^2+1+1=5\\ f\left(-\dfrac{1}{3}\right)=3\cdot\left(-\dfrac{1}{3}\right)^2-\dfrac{1}{3}+1=\dfrac{1}{3}-\dfrac{1}{3}+1=1\\ f\left(\dfrac{2}{3}\right)=3\cdot\left(\dfrac{2}{3}\right)^2-\dfrac{2}{3}+1=\dfrac{4}{3}-\dfrac{2}{3}+1=\dfrac{5}{3}\\ f\left(-2\right)=3\cdot\left(-2\right)^2-2+1=11\\ f\left(-\dfrac{4}{3}\right)=3\cdot\left(-\dfrac{4}{3}\right)^2-\dfrac{4}{3}+1=\dfrac{16}{3}-\dfrac{4}{3}+1=5\)

\(b,f\left(\dfrac{2}{3}\right)=\left|2\cdot\dfrac{2}{3}-9\right|-3=\dfrac{23}{3}-3=\dfrac{14}{3}\\ f\left(-\dfrac{5}{4}\right)=\left|2\cdot\left(-\dfrac{5}{4}\right)-9\right|-3=\dfrac{23}{2}-3=\dfrac{17}{2}\\ f\left(-5\right)=\left|2\left(-5\right)-9\right|-3=19-3=16\\ f\left(4\right)=\left|2\cdot4-9\right|-3=1-3=-2\\ f\left(-\dfrac{3}{8}\right)=\left|2\cdot\left(-\dfrac{3}{8}\right)-9\right|-3=\dfrac{39}{4}-3=\dfrac{27}{4}\)

\(c,x=0\Rightarrow y=2\cdot0^2-7=-7\\ x=-3\Rightarrow y=2\cdot\left(-3\right)^2-7=11\\ x=-\dfrac{1}{2}\Rightarrow y=2\cdot\left(-\dfrac{1}{2}\right)^2-7=\dfrac{-13}{2}\\ x=\dfrac{2}{3}\Rightarrow y=2\cdot\left(\dfrac{2}{3}\right)^2-7=-\dfrac{55}{9}\)

thay x=2 và x=1/2 ta có 

\hept⎧⎪⎨⎪⎩f(2)+3f(12)=4f(12)+3f(2)=14⇒f(2)=−1332

Ta có: \(y=f\left(x\right)=2x-3\)

\(f\left(x\right)=0\Rightarrow2x-3=0\Rightarrow x=\dfrac{3}{2}\)

\(f\left(x\right)=1\Rightarrow2x-3=1\Rightarrow x=2\)

\(f\left(x\right)=-\dfrac{3}{2}\Rightarrow2x-3=-\dfrac{3}{2}\Rightarrow x=\dfrac{3}{4}\)

\(f\left(x\right)=2022\Rightarrow2x-3=2022\Rightarrow x=\dfrac{2025}{2}\)

T(x) = f(x) + g(x) = 5x² - 2x + 3 (1)

H(x) = f(x) - g(X) = x² - 2x + 5 (2)

Lấy (1) cộng (2) theo vế ta có 

f(x) + g(x) + f(x) - g(x) = 5x² - 2x + 3 + x² - 2x + 5

=> 2.f(x) = 6x² - 4x + 8

=> f(x) =  3x² - 2x + 4

Thay f(x) vào (1) ta có 

f(x) + g(x) = 5x² - 2x + 3

=> (3x² - 2x + 4) + g(x) = 5x² - 2x + 3

=> g(x) = 5x² - 2x + 3 - 3x² + 2x - 4

=> g(x) = 2x² - 1

Vậy f(x) = 3x² - 2x + 4 ; g(x) = 2x² - 1

f(6)=12, f(8)= 5 xem thêm lời giải ở đây nhé https://cunghocvui.com/danh-muc/toan-lop-10

a)  Cho hàm số y = f(x) = -2x + 3.

Ta có: f(-2)= -2.(-2)+3

                 = 4+3=7

Ta có: f(0)= -2.0+3

                = 0+3=3

Ta có: f(

Lời giải:

a.

$f(-2)=(-2)(-2)+3=7$

$f(0)=(-2).0+3=3$

$f(\frac{-1}{2})=(-2).\frac{-1}{2}+3=4$

b.

$f(x)=-2x+3=5$

$\Rightarrow -2x=2$

$\Rightarrow x=-1$

$f(x)=-2x+3=1$

$\Rightarrow -2x=1-3=-2$

$\Rightarrow x=1$

a: \(\Leftrightarrow x^3-x^2-x^2+x+3⋮x-1\)

\(\Leftrightarrow x-1\in\left\{-1;1;3;-3\right\}\)

hay \(x\in\left\{0;2;4;-2\right\}\)

Lời giải:

Đặt $f(x)=ax^2+bx+c$ với $a\neq 0; a,b,c\in\mathbb{R}$

Xét điều kiện $f(x)-f(x-1)=2x-6$

Cho $x=0\Rightarrow f(0)-f(-1)=-6\Rightarrow f(-1)=f(0)+6=8$

Cho $x=1\Rightarrow f(1)-f(0)=-4\Rightarrow f(1)=f(0)-4=-2$

Vậy $f(0)=2; f(1)=-2; f(-1)=8$

\(\Leftrightarrow \left\{\begin{matrix} c=2\\ a+b+c=-2\\ a-b+c=8\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} c=2\\ a=1\\ b=-5\end{matrix}\right.\)

Vậy đa thức cần tìm là $x^2-5x+2$

Lời giải:

Đặt $f(x)=ax^2+bx+c$ với $a\neq 0; a,b,c\in\mathbb{R}$

Xét điều kiện $f(x)-f(x-1)=2x-6$

Cho $x=0\Rightarrow f(0)-f(-1)=-6\Rightarrow f(-1)=f(0)+6=8$

Cho $x=1\Rightarrow f(1)-f(0)=-4\Rightarrow f(1)=f(0)-4=-2$

Vậy $f(0)=2; f(1)=-2; f(-1)=8$

\(\Leftrightarrow \left\{\begin{matrix} c=2\\ a+b+c=-2\\ a-b+c=8\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} c=2\\ a=1\\ b=-5\end{matrix}\right.\)

Vậy đa thức cần tìm là $x^2-5x+2$