\(\left(x-1\right)^2=\left(x-3\right)^4\)

\(\Leftrightarrow\left(x-1\right)^2-\left(x-3\right)^4=0\)

\(\Leftrightarrow\left(x-1\right)^2-\left[\left(x-3\right)^2\right]^2=0\)

\(\Leftrightarrow\left[\left(x-1\right)-\left(x-3\right)^2\right]\left[\left(x-1\right)+\left(x-3\right)^2\right]=0\)

\(\Leftrightarrow\left(x-1-x^2+6x-9\right)\left(x-1+x^2-6x+9\right)=0\)

\(\Leftrightarrow\left(-x^2+7x-10\right)\left(x^2-5x+8\right)=0\)

\(\Leftrightarrow-\left(x-5\right)\left(x-2\right)\left(x^2-5x+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)

Vậy: ... 

(x-1)^2 =(x-3)^4=\(\left\{{}\begin{matrix}1+1\\2+2\\3+3\\4+4\end{matrix}\right.=2+4+6+8=\sqrt[]{251234=\Sigma\dfrac{2}{2}22\dfrac{2}{2}}\max\limits_{212}=\dfrac{21}{23}2123=\sum\limits1^{ }_{ }\text{(x-1)^2 =x=}\sum1\)

Bổ sung cho @ Huỳnh Thanh Phong.

(- \(x^2\) + 7\(x\)  - 10).(\(x^2\) - 5\(x\) + 8) = 0

(- \(x^2\) + 5\(x\) + 2\(x\) - 10).(\(x^2\) - \(\dfrac{5}{2}\)\(x\) - \(\dfrac{5}{2}\)\(x\) + \(\dfrac{25}{4}\) + \(\dfrac{7}{4}\)) = 0

[(- \(x^2\) + 5\(x\)) + (2\(x\) - 10)].[(\(x^2\) - \(\dfrac{5}{2}\)\(x\)) - (\(\dfrac{5}{2}\)\(x\) - \(\dfrac{25}{4}\)) + \(\dfrac{7}{4}\)] = 0

[ -\(x\)(\(x\) - 5) + 2.(\(x\) - 5)]. [\(x\)(\(x\) - \(\dfrac{5}{2}\)) - \(\dfrac{5}{2}\).(\(x\) - \(\dfrac{5}{2}\)) + \(\dfrac{7}{4}\)] = 0

(\(x\) - 5).(-\(x\) + 2).[(\(x-\dfrac{5}{2}\)).(\(x\) - \(\dfrac{5}{2}\)) + \(\dfrac{7}{4}\)] = 0

(\(x\) - 5).(-\(x\) + 2).[(\(x\) - \(\dfrac{5}{2}\))² + \(\dfrac{7}{4}\)] = 0 (1)

Vì (\(x\) - \(\dfrac{5}{2}\))² ≥ 0 ⇒ (\(x\) - \(\dfrac{5}{2}\))² + \(\dfrac{7}{4}\) ≥ \(\dfrac{7}{4}\)  (2)

Kết hợp (1) và (2) ta có:

\(\left[{}\begin{matrix}x-5=0\\-x+2=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)

Vậy \(x\in\) {2; 5}

Có 2 Th  | x-2| , (x-y+1)^2 =0

| x-2| , (x-y+1)^2 là hai số đối ; lx-2/ nguyên dương => ( x - y + 1 )^2 là số nguyên âm 

TH1  | x-2| , (x-y+1)^2 =0

=> x = 2 để /x-2/ = 0 

thay vào bên kia ta có : ( 2  - y + 1 ) ^2 = 0 => 2 - y + 1 = 0 => 3 - y = 0 => y = 3 

TH2 : Tự xét nha bn 

x=y=z=0

VÌ \(\left|x\right|\ge0;\left|y\right|\ge0;\left|z\right|\ge0\)NÊN ĐỂ\(\left|x\right|+\left|y\right|+\left|z\right|=0\)\(\Leftrightarrow\hept{\begin{cases}\left|x\right|=0\\\left|y\right|=0\\\left|z\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\y=0\\z=0\end{cases}}}\)

CA BA SO DEU BANG 0

-2 tick nhé Tran Thi Xuan

a) (x-1).(x+2) < 0 

TH1: x - 1< 0

x < 1

TH2: x + 2 < 0

x < -2

b) ( x +3).(x-5) > 0

TH1: x + 3 > 0

x> -3

TH2: x - 5 > 0

x > 5

KL: x > 5

(x - 3).(x + 2) < 0

=> x - 3 và x + 2 trái dấu

Mà x + 2 > x - 3

=> x + 2 > 0 => x > -2

     x - 3 < 0 => x < 3

=> -2 < x < 3, mà x thuộc Z => x \(\in\) {-1;0;1;2}

x<3

x<3

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

\(\Rightarrow x-5=0\)

\(\Rightarrow x=0+5\)

\(\Rightarrow x=5\)

Vậy x=5

(x - 5)² = 0

=> (x - 5)² = 0²

=> x - 5 = 0

=> x = 0 + 5

=> x = 5

Vậy x = 5

(x - 5)² = 0

=> (x - 5)² = 0²

=> x - 5 = 0

=> x = 0 + 5

=> x = 5