Đáp án: B hoặc C (Sao B và C lại giống nhau nhỉ ?!?)

\(3x^2-5x+2+3x^2+5x=\left(3x^2+3x^2\right)+\left(-5x+5x\right)+2=6x^2+2\)

(3x²-5x+2)+(3x²+5x)

=3x²-5x+2+3x²+5x

=6x²+2

=>Đáp án đúng là B và C

a: Ta có: \(40x^4+5x=0\)

\(\Leftrightarrow5x\left(8x^3+1\right)=0\)

\(\Leftrightarrow x\left(2x+1\right)=0\)

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

b: Ta có: \(8x^2-2x-1=0\)

\(\Leftrightarrow8x^2-4x+2x-1=0\)

\(\Leftrightarrow\left(2x-1\right)\left(4x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{4}\end{matrix}\right.\)

1. 2x(3x² - 5x + 3) = 6x³ - 10x² + 6x

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

3. -2x(x² + 5x - 3) = -2x³ - 10x² + 6x

4. x(3x² - 2x + 5) = 3x³ - 2x² + 5x

5. 3xy²(2x - 4y + 3xy) = 6x²y² - 12xy³ = 9x²y³

\(3x^2-5x-2\\ =3x^2-6x+x-2\\ =3x\left(x-2\right)+\left(x-2\right)\\ =\left(x-2\right)\left(3x+1\right)\)

\(3x^2-5x-2\)
\(=3x^2+x-6x-2\)
\(=x\left(3x+1\right)-2\left(3x+1\right)\)
\(=\left(x-2\right)\left(3x+1\right)\)

\(3x^2-5x-2\)

\(=\left(3x^2-6x\right)+\left(x-2\right)\)

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

\(=\left(x-2\right)\left(3x+1\right)\)

\(=\left[x\left(3x-1\right)+2\left(3x-1\right)\right]:\left(x+2\right)=\left[\left(3x-1\right)\left(x+2\right)\right]:\left(x+2\right)=3x-1\)

1) Ta có: \(2x\left(x-3\right)+5\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)

2) Ta có: \(\left(x^2-4\right)-\left(x-2\right)\left(3-2x\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)+\left(x-2\right)\left(2x-3\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

3) Ta có: \(\left(2x-1\right)^2-\left(2x+5\right)^2=11\)

\(\Leftrightarrow4x^2-4x-1-4x^2-20x-25=11\)

\(\Leftrightarrow-24x=11+1+25=37\)

hay \(x=-\dfrac{37}{24}\)

5) Ta có: \(3x^2-5x-8=0\)

\(\Leftrightarrow3x^2+3x-8x-8=0\)

\(\Leftrightarrow3x\left(x+1\right)-8\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{8}{3}\end{matrix}\right.\)

8) Ta có: \(\left|x-5\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)

10) Ta có: \(\left|2x+1\right|=\left|x-1\right|\)

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

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

a) \(3x^2-5x+2=0\)

Vì \(a+b+c=3-5+2=0\)

\(\Rightarrow\) pt co 2 ngiệm pb : \(x_1=1\) ; \(x_2=\frac{2}{3}\)

Vậy \(S=\left\{1;\frac{2}{3}\right\}\)

b) \(-3x^2+14x-8=0\)

\(\Delta'=7^2-\left(-3\right)\times\left(-8\right)=49-24=25\)

\(\Rightarrow\) pt có 2 nghiệm pb : \(x_1=4\) ; \(x_2=\frac{2}{3}\)

Vậy \(S=\left\{4;\frac{2}{3}\right\}\)

\(\left(3x-2\right)\left(3x^2-5x-4\right)\)=\(9x^3-15x^2-12x-6x^2+10x+8\)

=\(9x^3-21x^2-2x+8\)

Nhớ tick cho mình nha!

Ta có P(x) - Q(x) = (3x2 + 5x - 1) - (3x2 + 2x + 2) = 3x - 3

Vì 3x - 3 = 0 ⇒ x = 1 nên x = 1 là nghiệm cần tìm. Chọn A