Bài 1: Khái niệm về biểu thức đại số

1: \(3a+2b^3\)

2: \(2x-5\left(x+y\right)\)

\(2x-xy\)

2x(x-y)

f(x) = ax^2 + bx + cf(1) = a + b + cf(-1) = a - b - cVì f(1) = f(-1) => a + b + c = a - b - c=> b = -b=> b = 0Vậy f(x) =  ax^2 + bx + c = ax^2 + cf(-x) = a(-x)^2 + 0 + c = ax^2 + c=> f(x) = f(-x)

Có : \(f\left(-1\right)=f\left(1\right)\)

\(\Rightarrow a.\left(-1\right)^2+b.\left(-1\right)+c=a.1^2+b.1+c\)

\(\Leftrightarrow a-b+c=a+b+c\)

\(\Leftrightarrow b=0\)

Khi đó \(f\left(x\right)=\) \(a.x^2+c\) và \(f\left(-x\right)=a.\left(-x\right)^2+c=a.x^2+c\)

Do vậy \(f\left(x\right)=f\left(-x\right)\)

\(\left(x+y\right)^2\)

\(x^2+y^2\)

\(S=x.y\)

\(a^2+3b^2\)

1, (x + y)²

2, x² + y²

3, xy

4, a² + 3b²

a) \(A=x^{15}+3x^{14}+5\)

\(=x^{14}\left(x+3\right)+5\)

\(=x^{14}.0+5\)

= 5

b) x = -3 => x + 3 = 0

\(B=\left(x^{2007}+3x^{2006}+1\right)^{2007}\)

\(=\left[x^{2006}\left(x+3\right)+1\right]^{2007}\)

\(=\left(x^{2006}.0+1\right)^{2007}\)

\(=1^{2007}=1\)

\(A=x^{15}+3.x^{14}+5\text{ biết x+3=0}\)

\(A=x^{14}.\left(x+3\right)+5\)

\(\text{Do x+3=0}\Rightarrow A=x^{14}.0+5\)

\(A=0+5\)

\(A=5\)        \(\text{Vậy }A=5\text{ với x+3=0}\)

\(B=\left(x^{2007}+3.x^{2006}+1\right)^{2007}\text{ biết x=-3}\)

\(B=\left[x^{2006}.\left(x+3\right)+1\right]^{2007}\)

\(\text{Do x=-3}\Rightarrow B=\left[x^{2006}.\left(-3+3\right)+1\right]^{2007}\)

\(B=\left(x^{2006}.0+1\right)^{2007}\)

\(B=\left(0+1\right)^{2007}\)

\(B=1^{2007}\)

\(B=1\)           \(\text{Vậy }B=1\text{ với x=-3}\)

Ngu thì câm mồm đi

a/ \(\Leftrightarrow9x^2=36\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=6\\3x=-6\end{matrix}\right.\)

\(\Leftrightarrow x=\pm2\)

b/ \(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\) (do \(x^2+\dfrac{1}{2}>0\))

\(\Leftrightarrow x=\pm1\)

c/ Có \(\left|x+4\right|\ge0\forall x\)

=> \(\left|x+4\right|+5\ge5>0\forall x\)

\(\Rightarrow\left|x+4\right|+5=0\left(vô-lí\right)\)

\(\Rightarrow x\in\varnothing\)

d/ \(\sqrt{2x}-3-1=0\)

\(\Leftrightarrow\sqrt{2x}=4\)

\(\Leftrightarrow2x=16\)

\(\Leftrightarrow x=8\)

Bài 1 :

a) Xét P(x) = 0, ta có :

-3x + 8 = 8 - 3x = 0

⇒ 3x = 8 ⇒ x = 8/3

b) Xét Q(x) = 0, ta có :

x² - 1 = 0 ⇒ x² = 1

⇒ x = 1

c) Xét M(x) = 0, ta có :

(2x - 1)² - 16 = 0 ⇒ (2x - 1)² = 16

⇒ 2x -1 = 4 ⇒ x = 2,5

d) Xét N(x) = 0, ta có :

x³ - 9x = x(x² - 9) = 0

\(\Rightarrow\left\{{}\begin{matrix}x=0\\x^2-9=0\Rightarrow x=3\end{matrix}\right.\)

Vậy ..........

a) Xét P(x) = 0, ta có :

-3x + 8 = 8 - 3x = 0

⇒ 3x = 8 ⇒ x = 8/3

b) Xét Q(x) = 0, ta có :

x2 - 1 = 0 ⇒ x2 = 1

⇒ x = 1

c) Xét M(x) = 0, ta có :

(2x - 1)2 - 16 = 0 ⇒ (2x - 1)2 = 16

⇒ 2x -1 = 4 ⇒ x = 2,5

d) Xét N(x) = 0, ta có :

x3 - 9x = x(x2 - 9) = 0

Vậy ..........

tick mk cái

Mk làm 1 cách thôi..

f (x) + g (x) - h(x) =\(\left(5x^3-2x^2+x-3\right)+\left(2x^3-5x^2+4\right)-\left(4x^3+5x\right)\)

= \(5x^3-2x^2+x-3+2x^3-5x^2+4-4x^3-5x\)

= \(5x^3+2x^3-4x^3-2x^2-5x^2+x-5x-3+4\)

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

#Yiin - girl ><