Áp dụng BĐT Cô-si:

\(x^3+x^3+8\ge3\sqrt[3]{8x^6}=6x^2\)

\(y^6+y^6+1+1+1+1\ge6\sqrt[6]{y^{12}}=6y^2\)

Cộng vế:

\(2\left(x^3+y^6\right)+12\ge6\left(x^2+y^2\right)\ge30\)

\(\Rightarrow x^3+y^6\ge9\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(2;1\right)\)

a \ |x-2|+|x-5|=5x

==> x - 2 + x - 5=5x

      x + x - 2 - 5 =5x

    2x - 7 =5x

   2x : x -7=5

   x-7=5

  x=5+7

x=12[22222222222222222222222222222222222222222 =]]]

hoac -(x-2)-(x-5)=5

-x+2-x+5=5

-x-x+2+5=5x

-2x+7=5x

-2x:x+7=5

-x+7=5

-x=5-7

-x=-2

==> x=2

vay x=12 hoac x=2

hinh nhu t sai cho nao do ;{

\(x^2-2x-11=\left(x^2-2x+1\right)-12=y^2\Leftrightarrow\left(x-1\right)^2-y^2=12\Leftrightarrow\left(x-1-y\right)\left(x+1+y\right)=12\)

\(\text{Điều kiện x;y nguyên thì đến đây easy rồi}\)

\(A=x^2-12x+7=x^2-12x+36-29\)

\(=\left(x-6\right)^2-29\ge-29\)

Vậy \(A_{min}=-29\Leftrightarrow x=6\)

\(C=x-x^2-4=-\left(x^2-x+4\right)\)

\(=-\left(x^2-x+\frac{1}{4}+\frac{3}{4}\right)\)

\(=-\left[\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\right]\)

\(=-\left[\left(x-\frac{1}{2}\right)^2\right]-\frac{3}{4}\le-\frac{3}{4}\)

Vậy \(C_{min}=\frac{-3}{4}\Leftrightarrow x=\frac{1}{2}\)

\(\left(x-\dfrac{1}{5}\right)^2=\dfrac{4}{25}\Leftrightarrow\left(x-\dfrac{1}{5}\right)^2=\left(\dfrac{2}{5}\right)^2\)

\(\Leftrightarrow x-\dfrac{1}{5}=\dfrac{2}{5}\Leftrightarrow x=\dfrac{3}{5}\)

pt \(\Leftrightarrow\)\(\left[{}\begin{matrix}x-\dfrac{1}{5}=\dfrac{2}{5}\\x-\dfrac{1}{5}=-\dfrac{2}{5}\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)

\(\left(x-\dfrac{1}{5}\right)^2=\dfrac{4}{25}\)

\(\Rightarrow\left(x-\dfrac{1}{5}\right)^2=\left(\pm\dfrac{2}{5}\right)^5\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{5}=\dfrac{2}{5}\\x-\dfrac{1}{5}=-\dfrac{2}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}+\dfrac{1}{5}=\dfrac{3}{5}\\x=-\dfrac{2}{5}+\dfrac{1}{5}=\dfrac{-1}{5}\end{matrix}\right.\)

Vậy \(x=\dfrac{3}{5}\) hoặc \(x=\dfrac{-1}{5}\)

helpppp meeeee nowwww

x = (-1) + (-99)

x = -100

x = (-105) + (-15)

x = -120

k mình nha

Chúc bạn học giỏi

Mình cảm ơn bạn nhiều

x = ( -1 ) + ( - 99 )

x = - 100

x = ( -105 ) + ( -15 )

x = -120

4 . (2x)² - 7² = 0

=> (2x + 7 ).(2x+7 )= 0

=> th1 : 2x - 7 = 0 => x = 7/2

=> th2 : 2x + 7 = 0 => x = -7/2

5 . x(x -1 ) - 2( 1- x) = 0

=> x(x - 1) + 2 (x- 1 )= 0

=> (x - 2) .(x - 1 )= 0

=> th1 : x-2 = 0 => x=2

th2 : x-1 =0 => x= 1

6. (x-3)²-(x - 3 ) = 0

=> ( x- 3 ) ( x-4 ) = 0

=> th1 : x-3 = 0 => x=3

th2 : x-4= 0 => x =4

7. x³ = x⁵ => x = 1 . x= -1

ok nhé !!!

1 . x²-2x+1 = 0

=> (x-1)² = 0 => x-1 = 0 => x = 1

2. x(x-3) -(x-3) = 0

=>(x-1).(x-3)=0

=> th1 : x-1 = 0 => x= 1

=> th2 : x-3=0 => x= 3

3. x² + 36 = 12x

=> x² + 36 - 12= 0

=> x² - 6x -6x + 36 = 0

=> x(x - 6) - 6(x-6) = 0

=> (x-6)² = 0

=> x = 6

Đề là j vậy ạ ??

Nếu là tìm x thì sao lại có z ở câu 2 thế ??

Lời giải:
\(|x-2019|-|x-1|=0\)

\(\Leftrightarrow |x-2019|=|x-1|\)

\(\Rightarrow \left[\begin{matrix} x-2019=x-1\\ x-2019=-(x-1)=1-x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} 2019=1(\text{vô lý})\\ x=1010\end{matrix}\right.\)

Vậy $x=1010$