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

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

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

Bài 1:

a: \(x\left(x+y\right)+5y-x^2\)

\(=x^2+xy+5y-x^2\)

=xy+5y

b: \(\left(x-2\right)\left(y+1\right)-xy+4\)

\(=xy+x-2y-2-xy+4\)

=-2y+x+2

c: \(\dfrac{\left(4x^2y+12xy^2-8xy\right)}{2xy}\)

\(=\dfrac{2xy\cdot2x+2xy\cdot6y-2xy\cdot4}{2xy}\)

=2x+6y-4

d: \(\left(x-4\right)^2+8x-7\)

\(=x^2-8x+16+8x-7\)

\(=x^2+9\)

ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

a) Ta có: \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)

\(=\dfrac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)

\(=x-\sqrt{x}-2x-\sqrt{x}+2\sqrt{x}+2\)

\(=2-x\)

b) Để P=3 thì 2-x=3

hay x=-1(Không thỏa mãn ĐKXĐ)

Vậy: Không có giá trị nào của x để P=3

c) Thay \(x=7+2\sqrt{3}\) vào P, ta được:

\(P=2-7-2\sqrt{3}=-5-2\sqrt{3}\)

Vậy: Khi \(x=7+2\sqrt{3}\) thì \(P=-5-2\sqrt{3}\)

\(Q=23x^3y^3+17x^3y^3-50x^3y^3+(-2xy)^3\)

\(Q=23x^3y^3+17x^3y^3-50x^3y^3+(-8)x^3y^3\)

\(Q=(23+17-50-8)x^3y^3\)

\(Q=-18x^3y^3\)

 ---

\(|x-1|=1\)

\(TH1:\) \(x-1=1\)

⇒ \(x=1+1=2\)

\(TH2: x-1=-1\)

⇒ \(x=(-1)+1=0\)

---

Tính giá trị của \(Q\) tại \(|x-1|=1\) và \(y=\dfrac{-1}{2}\)

\(TH1: x=2; y=\dfrac{-1}{2}\)

\(Q=-18.2^3.(\dfrac{-1}{2})^3\)

\(Q=-18.8.(\dfrac{-1}{8})^3\)

\(Q=36\)

\(TH1: x=0; y=\dfrac{-1}{2}\)

\(Q=-18.0^3.(\dfrac{-1}{2})^3\)

\(Q=0\)

Vậy \(Q\) ∈ {\({36;0}\)}

Ta có: \(Q=23x^2y^3+17x^3y^3-50x^3y^3+\left(-2xy\right)^3\)

\(=-10x^3y^3-8x^3y^3\)

\(=-18x^3y^3\)

Ta có: |x-1|=1

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

Thay x=2 và y=-1/2 vào Q, ta được:

\(Q=-18\cdot2^3\cdot\left(-\dfrac{1}{2}\right)^3=-18\cdot8\cdot\dfrac{-1}{8}=18\)

Thay x=0 và y=-1/2 vào Q, ta được:

\(Q=-18\cdot0^3\cdot\left(-\dfrac{1}{2}\right)^3=0\)

Nhầm, \(Q\) ∈ {\(18;0\)} mới đúng ạ ;-;

thay x=2 và x=1/2 ta có 

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

2) là gì vậy bạn , 17x hay là x¹⁷

x¹⁷ đó bạn nhé

`1,`

`f(x)+g(x)=(5+4-2+7)+(4x^4-2x^3+3x^2+4x-1)`

`= 5+4-2+7+4x^4-2x^3+3x^2+4x-1`

`=(5x^4+4x^4)-2x^3+(4x^2+4x^2)+(-2x+4x)+(7-1)`

`= 9x^4-2x^3+8x^2+2x+6`

Đề phải là `f(x)-g(x)` chứ nhỉ :v?

`f(x)-g(x)=(5+4-2+7)-(4x^4-2x^3+3x^2+4x-1)`

`= 5+4-2+7-4x^4+2x^3-3x^2-4x+1`

`= (5x^4-4x^4)+2x^3+(-2x-4x)+(4x^2-3x^2)+(7+1)`

`= x^4+2x^3-6x+x^2+8`

`2,`

`a, (x+3)(x-1)`

`= x(x-1)+3(x-1)`

`= x*x+x*(-1)+3*x+3*(-1)`

`=x^2-x+3x-3`

`= x^2+2x-3`

`b, (4x+3)(x-2)`

`= 4x(x-2)+3(x-2)`

`= 4x*x+4x*(-2)+3*x+3*(-2)`

`= 4x^2-8x+3x-6`

`c, (2x+3)(x+1)`

`= 2x(x+1)+3(x+1)`

`= 2x*x+2x*1+3*x+3*1`

`= 2x^2+2x+3x+3`

`= 2x^2+5x+3`

`d, (5x-2)(x^2-3x+1)`

`= 5x(x^2-3x+1)+(-2)(x^2-3x+1)`

`= 5x*x^2+5x*(-3x)+5x*1+(-2)*x^2+(-2)*(-3x)+(-2)*1`

`= 5x^3-15x^2+5x-2x^2+6x-2`

`= 5x^3-17x^2+11x-2`

A nhé

Chọn A