a, \(-4x+5+2x-1=3\Leftrightarrow-2x=-1\Leftrightarrow x=\dfrac{1}{2}\)

b, \(-2x+2=2\Leftrightarrow x=0\)

c, \(-2x-6=-8\Leftrightarrow x=1\)

P(x) = (5x³ - 4x² + 2x - 1) + (3 - x + 4x² - 5x³)

= 5x³ - 4x² + 2x - 1 + 3 - x + 4x² - 5x³

= x + 2

Để P (x) = -3 thì x + 2 = -3 <=> x = -5

Ta có : \(P\left(x\right)=\left(5x^3-4x^2+2x-1\right)+\left(3-x+4x^2-5x^3\right)\)

\(\Rightarrow P\left(x\right)=5x^3-4x^2+2x-1+3-x+4x^2-5x^3\)

\(\Rightarrow P\left(x\right)=\left(5x^3-5x^3\right)+\left(-4x^2+4x^2\right)+\left(2x-x\right)+\left(-1+3\right)\)

\(\Rightarrow P\left(x\right)=x+2\)

\(P\left(x\right)=-3\) \(\Leftrightarrow x+2=-3\)

\(\Leftrightarrow x=-5\)

Vậy : \(x=-5\) để \(P\left(x\right)=-3\)

Ta có : P(x)=(5x3−4x2+2x−1)+(3−x+4x2−5x3)

⇒P(x)=5x3−4x2+2x−1+3−x+4x2−5x3

⇒P(x)=(5x3−5x3)+(−4x2+4x2)+(2x−x)+(−1+3)

⇒P(x)=x+2

P(x)=−3 ⇔x+2=−3

⇔x=−5

Vậy : x=−5 để P(x)=−3

a) 3x(4x-3)-2x(5-6x)=0

\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)

\(\Leftrightarrow24x^2-19x=0\)

\(\Leftrightarrow x\left(24x-19\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{19}{24}\)

b) 5(2x-3)+4x(x-2)+2x(3-2x)=0

\(\Leftrightarrow\)10x-15+4x²-8x+6x-4x²=0

\(\Leftrightarrow8x-15=0\)

\(\Leftrightarrow8x=15\)

\(\Leftrightarrow x=\dfrac{15}{8}\)

vậy x=\(\dfrac{15}{8}\)

c)3x(2-x)+2x(x-1)=5x(x+3)

\(\Leftrightarrow6x-3x^2+2x^2-2x=5x^2+15x\\ \Leftrightarrow4x-x^2=5x^2+15x\\ \Leftrightarrow4x-x^2-5x^2-15x=0\\ \)

\(\Leftrightarrow-6x^2-11x=0\\ \Leftrightarrow-x\left(6x+11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\6x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\6x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-11}{6}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{-11}{6}\)

1, x= 2

2, x = 4

**** bạn mình trước nhé

trieu dang sai ket qua vi chua doi dau

a: \(P\left(x\right)=5x^5-4x^4-2x^3+4x^2+3x+6\)

Bậc là 5

\(Q\left(x\right)=-5x^5+4x^4+2x^3-4x^2+7x+\dfrac{1}{4}\)

Bậc là 5

b: H(x)=P(x)+Q(x)

\(=5x^5-4x^4-2x^3+4x^2+3x+6-5x^5+4x^4+2x^3-4x^2+7x+\dfrac{1}{4}\)

=10x+6,25

c: Để H(x)=0 thì 10x+6,25=0

hay x=-0,625

Thực hiện phép chia đơn thức ta có :

4x³ + 11x² + 5x + 5 : x + 2 dư 7

Để 4x³ + 11x² + 5x + 5 ⋮ x + 2 thì 7 ⋮ x + 2

=> x + 2 ∈ Ư(7) = { 1; 7; -1; -7 }

Ta có bảng:

Vậy để 4x³ + 11x² + 5x + 5 ⋮ x + 2 thì 7 ⋮ x + 2 thì x ∈ { -9; -3; -1; 5 }

thank you!