Ta có : x² - 10x = -25

=> x² - 10x + 25 = 0

=> (x - 5)² = 0

=> x - 5 = 0

=> x = 5

Ta có \(x^2-10x=-25\)

=>\(x^2-10x+25=0\)

=>\(\left(x-5\right)^2=0\)

=>x-5=0

=>x=5

Vậy x=5

        x²- 10x + 25= 0

<=> x² - 2.5x + 5²= 0 ( Bình phương một hiệu )

<=>  (x - 5)² = 0

  =>     x = 5

Bạn nhớ k cho mình nhé ^-^!

Học tốt!!!

\(\Rightarrow x^2-10x+25=0\)

\(\Rightarrow\left(x-5\right)^2=0\)

\(\Rightarrow x-5=0\)

\(\Rightarrow x=5\)

10x=15y=6z

=> \(\frac{10x}{60}=\frac{15y}{60}=\frac{6z}{60}\)

=> \(\frac{x}{6}=\frac{y}{4}=\frac{z}{10}\)

= \(\frac{10x}{60}=\frac{5y}{20}=\frac{z}{10}=\frac{10x-5y+z}{60-20+10}=\frac{25}{50}=\frac{1}{2}\)

=> x =3 ; y=2 ; z=5

\(x^5+x^4+x+1=0\)

\(\Leftrightarrow x+1=0\)

hay x=-1

x:x*15-3,7=2,3

x:2*15-3,7=2,3

x:2*15=2,3+3,7

x:2*15=6

x:2=6:15

x*2=0,4

x=0,4:2

x=0,2

         \(x^2-5x-4\left(x-5\right)=0\)

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

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-5=0\\x-4=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=5\\x=4\end{cases}}\)

Vậy....

\(2x\left(x+6\right)=7x+42\)

\(\Leftrightarrow\)\(2x\left(x+6\right)-7x-42=0\)

\(\Leftrightarrow\)\(2x\left(x+6\right)-7\left(x+6\right)=0\)

\(\Leftrightarrow\)\(\left(x+6\right)\left(2x-7\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x+6=0\\2x-7=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-6\\x=\frac{7}{2}\end{cases}}\)

Vậy......

\(x^3-5x^2+x-5=0\)

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

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

\(\Leftrightarrow\)\(x-5=0\)

\(\Leftrightarrow\)\(x=5\)

\(x^4-2x^3+10x^2-20x=0\)

\(\Leftrightarrow\)\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

Vậy...