a)Ta có:

\(A\left(x\right)=4x^2+4x-3x^2+1-x+3-x^2\)

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

          \(=3x+4\)

b) Thay \(x=2\) ta được:

\(A\left(2\right)=3.2+4=10\)

c) Ta có:

\(3x+4=0\)

\(\Rightarrow3x=-4\)

\(\Rightarrow x=-\dfrac{4}{3}\)

\(\Rightarrow\) Nghiệm của \(A\left(x\right)\) là \(-\dfrac{4}{3}\)

Ta có:

Vậy bậc của đa thức C(x) là 7.

Chọn đáp án D

\(a,x^2=5\Leftrightarrow x=\pm\sqrt{5}\)

Vậy \(S=\left\{\pm\sqrt{5}\right\}\)

\(b,3x^2-12=0\Leftrightarrow3x^2=12\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\)

Vậy \(S=\left\{\pm2\right\}\)

\(c,4x^2-3=-9\)

\(\Leftrightarrow4x^2=-6\)

\(\Leftrightarrow x^2=-\dfrac{3}{2}\) (loại)

Vậy pt vô nghiệm.

\(d,5x^2-3=-3\)

\(\Leftrightarrow5x^2=0\)

\(\Leftrightarrow x=0\)

Vậy \(S=\left\{0\right\}\)

a)

`x^2 =5`
`=>\(\left[{}\begin{matrix}x=\sqrt{5}\\x=-\sqrt{5}\end{matrix}\right.\)

b)

`3x^2 -12=0`

`<=>3x^2 =12`

`<=>x^2 =4`

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

c)

`4x^2 -3=-9`

`<=>4x^2 =-6`

`<=>x^2 =-3/2` (vô lí vì `x>=0AA x` )

d)

`5x^2 -3=3`

`<=>5x^2 =0`

`<=>x^2 =0`

`<=>x=0`

a,Thay k=0 vào pt ta có:
\(4x^2-25+0^2+4.0.x=0\\ \Leftrightarrow4x^2-25=0\\ \Leftrightarrow x^2=\dfrac{25}{4}\\ \Leftrightarrow x=\pm\dfrac{5}{2}\)

b, Thay x=-2 vào pt ta có:
\(4.\left(-2\right)^2-25+k^2+4k.\left(-2\right)=0\\ \Leftrightarrow4.4-25+k^2-8k=0\\ \Leftrightarrow k^2-8k+16-25=0\\ \Leftrightarrow k^2-8k-9=0\\ \Leftrightarrow\left(k^2+k\right)-\left(9k+9\right)=0\\ \Leftrightarrow k\left(k+1\right)-9\left(k+1\right)=0\\ \Leftrightarrow\left(k+1\right)\left(k-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}k=-1\\k=9\end{matrix}\right.\)

b.

\(B\left(x\right)=0\Rightarrow-18+2x^2=0\)

\(\Leftrightarrow2\left(x^2-9\right)=0\)

\(\Leftrightarrow2\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\)

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

c.

\(C\left(x\right)=0\Leftrightarrow x^3+4x^2-x-4=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x-1=0\\x+1=0\end{matrix}\right.\)

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

h(x)=5x+1

nghiệm_của_đa_thức_h(x)_là_-1/5

a)h(x)=f(x)-g(x)

        =(2x³ +3x² -2x +3)-(2x³ +3x² -7x +2)

        =2x³ + 3x² - 2x +3 - 2x³ -3x² + 7x -2

        =5x+1

b)h(x)=5x+1=0

=>5x=-1

    x=\(\frac{-1}{5}\)