\(a^{\left(x-2\right)\left(x-3\right)}=1\)

\(\Rightarrow a^{\left(x-2\right)\left(x-3\right)}=a^0\)

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

\(\Rightarrow x-2=0\) hoặc \(x-3=0\)

+) \(x-2=0\Rightarrow x=2\)

+) \(x-3=0\Rightarrow x=3\)

Vậy \(x\in\left\{2;3\right\}\)

a) 5^x + 5^x+2 = 650

   5^x + 5^x+2 = 5⁴ + 5²

   5^x +  5^x+2 = 5⁶

 ^{\(\Rightarrow\)}x+x+2 = 6

         2x       = 6-2 

          x        = 2

b) 3x - ( 2x +1) = 2

   3x - 2x -1     = 2 

    3x - 2x         =2+1 

   (3 - 2).x         = 3

     1.x             = 3

       x              = 3

sao ko k  cho mình ?

a) Ta có : 2017 - |x - 2017| = x

=> |x - 2017| = 2017 - x (1)

Điều kiện xác định : \(2017-x\ge0\Rightarrow2017\ge x\Rightarrow x\le2017\)

Khi đó (1) <=> \(\orbr{\begin{cases}x-2017=2017-x\\x-2017=-\left(2017-x\right)\end{cases}\Rightarrow\orbr{\begin{cases}2x=2017+2017\\x-2017=-2017+x\end{cases}\Rightarrow}\orbr{\begin{cases}2x=4034\\0x=0\end{cases}}}\)

\(\Rightarrow\orbr{\begin{cases}x=2017\\x\text{ thỏa mãn }\Leftrightarrow x\le2017\end{cases}}\Rightarrow x\le2017\)

b) Ta có : \(\hept{\begin{cases}\left(2x-1\right)^{2016}\ge0\forall x\\\left(y-\frac{2}{5}\right)^{2016}\ge\\\left|x+y+z\right|\ge0\forall x;y;z\end{cases}0\forall y}\Rightarrow\left(2x-1\right)^{2016}+\left(y-\frac{2}{5}\right)^{2016}+\left|x+y+z\right|\ge0\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}2x-1=0\\y-\frac{2}{5}=0\\x+y+z=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{2}{5}\\\frac{1}{2}+\frac{2}{5}+z=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{2}{5}\\z=-\frac{9}{10}\end{cases}}}\)

A) Có \(f\left(x\right)=3x^2-2x-1\)

\(\Rightarrow f\left(1\right)=3.1^2-2.1-1\)

     \(f\left(1\right)=3-2-1=0\)

\(\Rightarrow f\left(-\frac{1}{3}\right)=3.\left(-\frac{1}{3}\right)^2-2.\frac{1}{3}-1\)

\(f\left(-\frac{1}{3}\right)=3.\frac{1}{9}-\frac{2}{3}-1\)

\(f\left(-\frac{1}{3}\right)=\frac{1}{3}-\frac{2}{3}-1=-\frac{4}{3}\)

Có \(f\left(x\right)=3x^2-2x-1\)

Để \(f\left(x\right)=0\)

Thì \(3x^2-2x-1=0\)

\(\Rightarrow x\left(3x-2\right)-1=0\)

\(\Rightarrow x\left(3x-2\right)=1\)

TH1 : \(\hept{\begin{cases}x=1\\3x-2=1\end{cases}\Rightarrow\hept{\begin{cases}x=1\\x=1\end{cases}}}\)

TH2 : \(\hept{\begin{cases}x=-1\\3x-2=-1\end{cases}\Rightarrow}\hept{\begin{cases}x=-1\\x=\frac{1}{3}\end{cases}}\)( loại )

Vậy x=1 thì f(x) = 0

A, ta có f(x) =3x²-2x-1

=> f(1)=3.1²-2.1-1=3-2-1=0

\(f\left(\frac{-1}{3}\right)=3.\left(-\frac{1}{3}\right)^2-2.\left(-\frac{1}{3}\right)-1\)

= \(\frac{1}{3}+\frac{2}{3}-1=1-1=0\)

B,  theo a ta có \(f\left(1\right)=f\left(\frac{-1}{3}\right)=0\)

Vậy x=1 và x=\(\frac{-1}{3}\)thì đa thức f(x)=0

tk mk nha bạn ,mk xong đầu tiên

*****Chúc bạn học giỏi*****

a,x^2-7x=0

<=>x(x-7)=0

<=>th1 x=0

th2 x-7=0=>x=7

vậy x=0 hoặc 7

\(a^2-7a=0\)

\(\Rightarrow a\left(a-7\right)=0\)

\(\Rightarrow\hept{\begin{cases}a=0\\a-7=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}a=0\\a=7\end{cases}}\)