`@` `\text {Ans}`

`\downarrow`

`a,`

` F(x)=3x^2-7+5x-6x^2-4x^2+8`

`= (3x^2 - 6x^2 - 4x^2) + 5x + (-7 + 8)`

`= -7x^2 + 5x + 1`

Bậc của đa thức: `2`

`G(x)=x^4+2x-1+2x^4+3x^3+2-x`

`= (x^4 + 2x^4) + 3x^3 + (2x - x) + (-1+2)`

`= 3x^4 + 3x^3 + x + 1`

Bậc của đa thức: `4`

`b,`

`F(x) + G(x) = (-7x^2 + 5x + 1)+(3x^4 + 3x^3 + x + 1)`

`= -7x^2 + 5x + 1+3x^4 + 3x^3 + x + 1`

`= 3x^4 + 3x^3 - 7x^2 + (5x + x) + (1+1)`

`= 3x^4 + 3x^3 - 7x^2 + 6x + 2`

`F(x) - G(x) = (-7x^2 + 5x + 1) - (3x^4 + 3x^3 + x + 1)`

`= -7x^2 + 5x + 1 - 3x^4 - 3x^3 - x - 1`

`= -3x^4 - 3x^3 - 7x^2 + (5x - x) + (1-1)`

`= -3x^4 - 3x^3 - 7x^2 + 4x`

a/

\(F\left(x\right)=\left(3-6-4\right)x^2+5x+\left(-7+8\right)=-7x^2+5x+1\) -> Đa thức bậc 2

\(G\left(x\right)=\left(1+2\right)x^4+3x^3+\left(2-1\right)x+\left(-1+2\right)=3x^4+3x^3+x+1\) -> Đa thức bậc 4

b/

\(F\left(x\right)+G\left(x\right)=-7x^2+5x+1+3x^4+3x^3+x+1\\ =3x^4+3x^3-7x^2+6x+2\)

\(F\left(x\right)-G\left(x\right)=-7x^2+5x+1-3x^4-3x^3-x-1\\ =-3x^4-3x^3-7x^2+4x\)

a) Ta có: \(3x^2+2x-1=0\)

\(\Leftrightarrow3x^2+3x-x-1=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{-1;\dfrac{1}{3}\right\}\)

b) Ta có: \(x^2-5x+6=0\)

\(\Leftrightarrow x^2-2x-3x+6=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

Vậy: S={2;3}

c) Ta có: \(x^2-3x+2=0\)

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

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

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

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

Vậy: S={1;2}

d) Ta có: \(2x^2-6x+1=0\)

\(\Leftrightarrow2\left(x^2-3x+\dfrac{1}{3}\right)=0\)

mà \(2\ne0\)

nên \(x^2-3x+\dfrac{1}{3}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{23}{12}=0\)

\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=\dfrac{23}{12}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=\dfrac{\sqrt{69}}{6}\\x-\dfrac{3}{2}=\dfrac{-\sqrt{69}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9+\sqrt{69}}{6}\\x=\dfrac{9-\sqrt{69}}{6}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{9+\sqrt{69}}{6};\dfrac{9-\sqrt{69}}{6}\right\}\)

e) Ta có: \(4x^2-12x+5=0\)

\(\Leftrightarrow4x^2-10x-2x+5=0\)

\(\Leftrightarrow2x\left(2x-5\right)-\left(2x-5\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{5}{2};\dfrac{1}{2}\right\}\)

cho vào máy tính là ra hết

a: Ta có: \(F=\left(\dfrac{2\sqrt{x}}{2\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{3x}{x-2\sqrt{x}+1}\right)\)

\(=\dfrac{2x-2\sqrt{x}+1}{\sqrt{x}\left(2\sqrt{x}-1\right)}\cdot\dfrac{x-1+3x}{\left(\sqrt{x}-1\right)^2}\)

\(=\dfrac{2x-2\sqrt{x}+1}{\sqrt{x}\left(2\sqrt{x}-1\right)}\cdot\dfrac{4x-1}{\left(\sqrt{x}-1\right)^2}\)

\(=\dfrac{\left(2x-2\sqrt{x}+1\right)\left(2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)^2}\)

Câu a đã làm: F=(2√x/2√x-1     -    1/√x) ( √x+1/√x-1    +       3x/x-2√x+1) với x >0, x khác 1, x khác 1/4 a) rút gọn F - Hoc24

\(b,F=2\Leftrightarrow\dfrac{\left(2\sqrt{x}+1\right)\left(2x-2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)^2}=2\\ \Leftrightarrow2\sqrt{x}\left(x-2\sqrt{x}+1\right)=2x\sqrt{x}-4x+2\sqrt{x}+2x-2\sqrt{x}+1\\ \Leftrightarrow2x\sqrt{x}-4x+2\sqrt{x}=2x\sqrt{x}-2x+1\\ \Leftrightarrow2x-2\sqrt{x}+1=0\\ \Leftrightarrow2\left(x-\sqrt{x}+\dfrac{1}{4}\right)+\dfrac{1}{2}=0\\ \Leftrightarrow2\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{1}{2}=0\\ \Leftrightarrow x\in\varnothing\)

a: \(f\left(x\right)=\sqrt{x^2-6x+9}=\sqrt{\left(x-3\right)^2}=\left|x-3\right|\)

\(f\left(-1\right)=\left|-1-3\right|=4\)

\(f\left(5\right)=\left|5-3\right|=\left|2\right|=2\)

b: f(x)=10

=>\(\left[{}\begin{matrix}x-3=10\\x-3=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\\x=-7\end{matrix}\right.\)

c: \(A=\dfrac{f\left(x\right)}{x^2-9}=\dfrac{\left|x-3\right|}{\left(x-3\right)\left(x+3\right)}\)

TH1: x<3 và x<>-3

=>\(A=\dfrac{-\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{-1}{x+3}\)

TH2: x>3

\(A=\dfrac{\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{1}{x+3}\)