x=6 hoặc -6

x=3 hoặc x=-3

9-\(X^2\)=0

9-\(X^2\)=0:2

9-\(X^2\)=0

X=0+9

X=9

9=3.3

=\(3^2\)

VẬY X=3

MIK LM KO LOGIC CHO LẮM ĐÂU

`5/(-x^2+5x-6)+(x+3)/(2-x)=0`

Đk:`x ne 2,x ne 3`

`pt<=>-5/(x^2-5x+6)-(x+3)/(x-2)=0`

`<=>-5-(x+3)(x-3)=0`

`<=>(x+3)(x-3)=-5`

`<=>x^2-9=-5`

`<=>x^2-4=0`

`<=>(x-2)(x+2)=0`

`x ne 2=>x-2 ne 0`

`<=>x+2=0`

`<=>x=-2`

Vậy `S={-2}`

\(a,\Leftrightarrow25x^2-70x+49-25x^2=32\\ \Leftrightarrow-70x=-17\Leftrightarrow x=\dfrac{17}{70}\\ b,\Leftrightarrow x^2-6x+9+x^2+2x+1-5=0\\ \Leftrightarrow2x^2-4x+5=0\\ \Leftrightarrow2\left(x^2-2x+1\right)+3=0\\ \Leftrightarrow2\left(x-1\right)^2=-3\Leftrightarrow\left(x-1\right)^2=-\dfrac{3}{2}\left(\text{vô lí}\right)\\ \Leftrightarrow x\in\varnothing\)

a, \(2x\left(x-3\right)-15+5x=0\\ \Rightarrow2x\left(x-3\right)-\left(15-5x\right)=0\\ \Rightarrow2x\left(x-3\right)-5\left(3-x\right)=0\\ \Rightarrow\left(2x+5\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=3\end{matrix}\right.\)

b, \(x^3-7x=0\\ \Rightarrow x\left(x^2-7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\pm7\end{matrix}\right.\)

c, \(\left(2x-3\right)^2-\left(x+5\right)^2=0\\ \Rightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\\ \Rightarrow\left(x-8\right)\left(3x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)

Xem lại đề câu d 

đặt \(t=x^2-5x+7\) pt thành \(t\ge0\)

\(t^2+t-2=0\) (t)

<=>\(\left(t-1\right)\left(t+2\right)=0\)

<=>\(\left[{}\begin{matrix}t=1\\t=-2\end{matrix}\right.\)

so với điều kiện =>t=1 thỏa 

=>\(x^2+-5x+7=1\)

<=> \(x^2-5x+6=0\)

<=>\(\left(x-2\right)\left(x-3\right)=0\)

<=>\(\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

KL vậy pt có 2 nghiệm là \(\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

\(\Leftrightarrow-\dfrac{2}{5}\left(4x-3\right)^2=-\dfrac{5}{18}\)

\(\Leftrightarrow\left(4x-3\right)^2=\dfrac{25}{36}\)

\(\Leftrightarrow4x-3\in\left\{\dfrac{5}{6};-\dfrac{5}{6}\right\}\)

hay \(x\in\left\{\dfrac{23}{24};\dfrac{13}{24}\right\}\)

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

\(\Leftrightarrow3\left(x-2\right)\left(3-x\right)\left(3+x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=2\end{matrix}\right.\)

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

@Lâm

\(f\left(x\right)=2x^2-5x+b\)

\(\Rightarrow f\left(1\right)=2.1^2-5.1+b\)

                \(=2-5+b\)

Mà \(f\left(1\right)=0\)

\(\Rightarrow2-5+b=0\)

\(\Rightarrow5+b=2\)

\(\Rightarrow b=-3\)

Vậy \(b=-3\)

Đa thức  \(f\left(x\right)=2x^2-5x+b\) có nghiệm tức là \(f\left(1\right)=0\)

Khi đó,ta có:

\(2\cdot1^2-5\cdot1+b=0\)

\(\Rightarrow2-5+b=0\)

\(\Rightarrow-5+b=0\)

\(\Rightarrow b=5\)

@@ CTV làm sai rồi 

a: \(=25x^4-10x^3+5x^2\)

c: \(=2x^3-3x-5x^3-x^2+x^2=-3x^3-3x\)