a) Check lại đề

b) Cho \(g\left(x\right)=x^2-4=0\)

\(\Rightarrow x^2=0+4=4\)

\(\Rightarrow x=\pm2\)

Vậy g (x) có 2 nghiệm là x = 2 và x = -2

c) Cho \(h\left(x\right)=x^2-16x=0\)

\(\Rightarrow x\left(x-16\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-16=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=0+16=16\end{matrix}\right.\)

Vây g (x) có 2 nghiệm là x = 0 và x = 16

d) Cho \(t\left(x\right)=x^2+8x=0\)

\(\Rightarrow x\left(x+8\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x+8=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=0-8=-8\end{matrix}\right.\)

Vậy t (x) có 2 nghiêm là x = 0 và x = -8

a) \(f\left(x\right)=6+12=18=0\)(vô lý)

Nên đa thức trên vô nghiệm

\(b,g\left(x\right)=x^2-4=0\\ \Leftrightarrow x^2=4\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy đa thức trên có 2 nghiệm là x=2 ; x= -2

\(c,h\left(x\right)=x^2-16x=0\\ \Leftrightarrow x\left(x-16\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=16\end{matrix}\right.\)

Vậy...

\(d,t\left(x\right)=x^2+8x=0\\ \Leftrightarrow x\left(x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\)

Vậy...

\(A\left(x\right)=43x-\left(52x^2+34x^2-8x^4\right)-\left(8x^4+16x^3-42x^2+43x\right)+19\)

\(\Leftrightarrow A\left(x\right)=43x-86x^2+8x^4-16x^3+42x^2-43x+19\)

\(\Leftrightarrow A\left(x\right)=-16x^3-44x^2+19\)

Bậc là: 3

\(16x^2-9=\left(4x-3\right)\left(4x+3\right)\)

\(16x^2-8x+1=\left(4x-1\right)^2\)

16x^2-9=(4x-3)(4x+3)

16x^2-8x+1=(4x-1)^2 

a) Ta có: \(x^2-8x+7=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=7\end{matrix}\right.\)

b) Ta có: \(x^2+x-20=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=4\end{matrix}\right.\)

c) Ta có: \(3x^2+4x-4=0\)

\(\Leftrightarrow3x^2+6x-2x-4=0\)

\(\Leftrightarrow3x\left(x+2\right)-2\left(x+2\right)=0\)

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

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

d) Ta có: \(3x^2-4x-7=0\)

\(\Leftrightarrow3x^2-7x+3x-7=0\)

\(\Leftrightarrow\left(3x-7\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-1\end{matrix}\right.\)

e) Ta có: \(5x^2-16x+3=0\)

\(\Leftrightarrow5x^2-15x-x+3=0\)

\(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)

f) Ta có: \(x^2+3x-10=0\)

\(\Leftrightarrow x^2+5x-2x-10=0\)

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

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

a)

\(x^2-8x+7=0\text{⇔}\text{⇔}x^2-7x-x-7=\left(x-7\right)\left(x-1\right)=0\text{⇔}\left[{}\begin{matrix}x=1\\x=7\end{matrix}\right.\)

Vậy nghiệm của đa thức : \(S=\left\{1;7\right\}\)

c)

\(3x^2+4x-4=0\text{⇔}3x^2+6x-2x-4=\left(3x-2\right)\left(x+2\right)=0\text{⇔}\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)

Vậy nghiệm của đa thức : \(S=\left\{\dfrac{2}{3};-2\right\}\)

b)

\(x^2+x-20=0⇔\left(x-4\right)\left(x+5\right)=0\text{⇔}\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)

d)

\(3x^2-4x-7=0\text{⇔}\left(3x-7\right)\left(x+1\right)=0\text{⇔}\left[{}\begin{matrix}x=-1\\x=\dfrac{7}{3}\end{matrix}\right.\)

e)

\(5x^2-16x+3\text{⇔}\left(x-3\right)\left(5x-1\right)=0\text{⇔}\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)

f)

\(x^2+3x-10=0\text{⇔}\left(x-2\right)\left(x+5\right)=0\text{⇔}\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)

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