\(x=2\)

\(x=0,5\)

\(x=3,125\)

a. 

x = 2 ; 3

b. 

x = 0

c. 

x = 3,125

a, (3 - x) = (3 - x)² 
Suy ra 3-x = 1 hoặc 3-x = 0
--Nếu 3-x = 1
          x = 3-1
          x = 2
--Nếu 3-x = 0
          x = 3 - 0
          x = 3
Vậy x=2; x=3
b, x³ - 0,25x = 0
    x.(x² - 0,25) = 0
Suy ra x = 0   hoặc   x² - 0,25 = 0
                                 x² = 0 + 0,25
                                 x² = 0,25
                                 x² = (0,5)² 
                                 Suy ra x = 0,5
Vậy x = 0; x = 0,5
c, x2 - 10x = -25
    x.(2 - 10) = -25
    x.(-8) = -25
    x = -25 : (-8)
    x = 25/8
Vậy x = 25/8
Chúc bạn học tốt

\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)

\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)

\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a) \(x^3-0,25x=0\\ < =>x\left(x^2-0,25\right)=0\\ =>\left[{}\begin{matrix}x=0\\x^2-0,25=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=0\\x=\sqrt{0,25}\end{matrix}\right.\)

b) \(x^2-10x=-25\\ < =>x^2-10x+25=0\\ < =>\left(x-5\right)^2=0\\ < =>x-5=0\\=>x=5\)

a) \(x^3-0,25x=0\)

\(x\left(x^2-0,25\right)=0\)

\(\Leftrightarrow x=0\) hoặc \(x^2-0,25=0\)

\(\Leftrightarrow x=0\) hoặc \(x=0,25\) hoặc \(x=-0,25\)

b) \(x^2-10x=-25\)

\(\Leftrightarrow x\left(x-10\right)=-25\)

\(\Leftrightarrow x=-25\) hoặc \(\Leftrightarrow x-10=-25\)

\(\Leftrightarrow x=-25\) hoặc x=-15

a, \(x^3-0,25x=0\)

\(x\left(x^2-0,25\right)=0\)

\(x\left(x-0,5\right)\left(x+0,5\right)=0\)

=> \(\left[{}\begin{matrix}x=0\\x-0,5=0\\x+0,5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=0,5\\x=\left(-0,5\right)\end{matrix}\right.\)

b,\(x^2-10x=-25\)

x(x-10)=(-25)

\(\left[{}\begin{matrix}x=0\\x-10=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\)

a: Ta có: \(x\left(2-x\right)+\left(x^2+x\right)=7\)

\(\Leftrightarrow2x-x^2+x^2+x=7\)

\(\Leftrightarrow3x=7\)

hay \(x=\dfrac{7}{3}\)

b: Ta có: \(\left(2x+1\right)^2-x\left(4-5x\right)=17\)

\(\Leftrightarrow4x^2+4x+1-4x+5x^2=17\)

\(\Leftrightarrow9x^2=16\)

\(\Leftrightarrow x^2=\dfrac{16}{9}\)

hay \(x\in\left\{\dfrac{4}{3};-\dfrac{4}{3}\right\}\)

c: Ta có: \(\left(x-4\right)^2-\left(2x+1\right)^2=0\)

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

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

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

d: ta có: \(\dfrac{2x^3-8x^2+10x}{2x}=0\)

\(\Leftrightarrow x^2-4x+5=0\)

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

Mình cần gấp ạk

A và B

a) \(x^3-0,25x=0\)

\(\Rightarrow x^3=\dfrac{1}{4}x\)

\(\Rightarrow x^2=\dfrac{1}{4}\)

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

b) \(x^2-10x=-25\)

\(\Rightarrow x^2=-25+10x\)

\(\Rightarrow\left[{}\begin{matrix}x=-25+10x\\x=-\left(-25+10x\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}10x-x=-25\\-10x-x=25\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}9x=-25\\-11x=25\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-25}{9}\\x=-\dfrac{25}{11}\end{matrix}\right.\)

<=> (0,75+0,65+0,25+0,35)(0,25.2+0,25.3+0,25.5=2(0,25.10)=2.2,5=5

Đs: 5

[1.4+0.6]*[2.8+0.75+1.25]

     2      *             4.8      =9.6

minh chac chan lun do minh tinh may tinh lai rui

=9.6 do nha

x 2 - 10x = -25

⇔  x 2  –10x + 25 = 0

⇔  x 2  – 2.x.5 + 52 = 0

⇔ x - 5 2  = 0

⇔ x – 5 = 0 ⇔ x = 5