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

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

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

+) \(\left(x-1\right)^2=0\)

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

+) \(\left(x-1\right)^2-1=0\)

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

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

TH1: \(x-1=1\Rightarrow x=1+1=2\)

TH2: \(x-1=-1\Rightarrow x=-1+1=0\)

Vậy: \(x\in\left\{1;2;0\right\}\)

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

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

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

\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\1-\left(x-1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x-1=1\\x-1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)

Vậy \(x\in\left\{1;2;0\right\}\)

for you

700

Minh tính sai có gì bạn thông cảm ạ 

280$ ạ

=0,5176 m

=5176 cm

\(5176cm=51m76cm\)

chu vi mặt đáy là 

(8 + 6) x 2 = 28 (m)

diện tích cần sơn của căn phòng là 

28 x 3,5 - 28 - 20  = 50 (m²)

để sơn phòng khách thì hết số tiền là 

35000 x 50 = 1750000 (đồng)

Đáp số 

em cảm ơn cô!!!!!

Với \(0< x< \sqrt{3}\) ta có đánh giá sau:

\(\dfrac{1}{2-x}\ge\dfrac{x^2+1}{2}\)

Thực vậy, do \(x< \sqrt{3}\Rightarrow2-x>0\), BĐT tương đương:

\(2\ge\left(2-x\right)\left(x^2+1\right)\)

\(\Leftrightarrow x^3-2x^2+x\ge0\)

\(\Leftrightarrow x\left(x-1\right)^2\ge0\) (luôn đúng với \(x>0\))

Áp dụng cho bài toán:

\(\dfrac{1}{2-a}+\dfrac{1}{2-b}+\dfrac{1}{2-c}\ge\dfrac{a^2+1}{2}+\dfrac{b^2+1}{2}+\dfrac{c^2+1}{2}=\dfrac{a^2+b^2+c^2+3}{2}=3\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

h

A = \(\dfrac{2023^{2024^{2025}}-2017^{2024^{2023}}}{10}\)

A = \(\dfrac{2023^{2^{2025}.1012^{2025}}-2017^{2^{2023}.1012^{2023}}}{10}\)

A = \(\dfrac{2023^{2^2.2^{2023}.1012^{2025}}-2017^{2^2.2^{2021}1012^{2023}.}}{10}\)

A = \(\dfrac{2023^{4.2^{2023}.1012^{2025}}-2017^{4.2^{2021}.1012^{2023}}}{10}\)

A = \(\dfrac{\left(2023^4\right)^{2^{2023}.1012^{2025}}-\left(2017^4\right)^{2^{2021}.1012^{2023}}}{10}\)

A = \(\dfrac{\left(\overline{..1}\right)^{2^{2023}.1012^{2025}}-\left(\overline{..1}\right)^{2^{2021}.1012^{2023}}}{10}\)

A = \(\dfrac{\overline{..1}-\overline{..1}}{10}\)

A = \(\dfrac{\overline{..0}}{10}\) 

A  \(\in\) N (đpcm)

A = \(\dfrac{2023^{2024^{2025}}-2017^{2024^{2023}}}{10}\)

A = \(\dfrac{2023^{2^{2025}.1012^{2025}}-2017^{2^{2023}.1012^{2023}}}{10}\)

A = \(\dfrac{2023^{2^2.2^{2023}.1012^{2025}}-2017^{2^2.2^{2021}1012^{2023}.}}{10}\)

A = \(\dfrac{2023^{4.2^{2023}.1012^{2025}}-2017^{4.2^{2021}.1012^{2023}}}{10}\)

A = \(\dfrac{\left(2023^4\right)^{2^{2023}.1012^{2025}}-\left(2017^4\right)^{2^{2021}.1012^{2023}}}{10}\)

A = \(\dfrac{\left(\overline{..1}\right)^{2^{2023}.1012^{2025}}-\left(\overline{..1}\right)^{2^{2021}.1012^{2023}}}{10}\)

A = \(\dfrac{\overline{..1}-\overline{..1}}{10}\)

A = \(\dfrac{\overline{..0}}{10}\) 

A  \(\in\) N (đpcm)

\(5x+xy-4y=3\)

\(\Rightarrow x\left(y+5\right)-4y-20=3-20\)

\(\Rightarrow x\left(y+5\right)-4\left(y+5\right)=-17\)

\(\Rightarrow\left(y+5\right)\left(x-4\right)=-17\)

Bổ sung: \(x,y\in Z\) 

Ta có bảng:

Vậy: ... 

\(5x+xy-4y=3\)

\(\Rightarrow x\cdot\left(y+5\right)-4y-20=3-20\)

\(\Rightarrow x\cdot\left(y+5\right)-4\cdot\left(y+5\right)=-17\)

\(\Rightarrow\left(y+5\right)\cdot\left(x-4\right)=-17\)

\(\Leftrightarrow x,y\in Z\)

Lập bảng giá trị:

Vậy \(\left(x;y\right)\in\left\{\left(21;-6\right),\left(-13;-4\right),\left(3;12\right),\left(5;-22\right)\right\}\)