Ta có:

\(x^2-4x=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

Ta có:

\(D=\dfrac{1!+2!+3!+4!+5!-5!}{1!+2!+3!+4!+5!}=1-\dfrac{1}{5!}=\dfrac{119}{120}\)

Ta có:

\(P=\dfrac{2n-4+3}{n-2}\)

\(P=2+\dfrac{3}{n-2}\)

Dể P nguyên thì \(\dfrac{3}{n-2}\)nguyên hay \(n-2\inƯ\left(3\right)\)

\(Ư\left(3\right)\in\left\{-3;-1;1;3\right\}\)

Ta có bảng:

Vậy n\(\in\left\{-1;1;3;5\right\}\)

\(A=2.\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+...+\dfrac{1}{95.98}\right)\)

\(A=\dfrac{2}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+....+\dfrac{3}{95.98}\right)\)

\(A=\dfrac{2}{3}\left(\dfrac{1}{2}-\dfrac{1}{98}\right)\)

\(A=\dfrac{2}{3}\dfrac{24}{49}=\dfrac{16}{49}\)

Hiệu số phần bằng nhau là:

2 - 1 = 1 (phần )

Chiều dài là:

15 : 1 x 2 = 30 (m)

Chiều rộng là:

30 - 15 = 15 (m)

Chu vi là:

(30 + 15) x 2 = 90 (m)

Đáp số: 90 m

\(a^2+b^2+c^2+14-2a-4b-6c=0\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b-2\right)^2+\left(c-3\right)^2=0\)

mà \(\left(a-1\right)^2\ge0;\left(b-2\right)^2\ge0;\left(c-3\right)^2\ge0\)nên

\(\left\{{}\begin{matrix}a-1=0\\b-2=0\\c-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=2\\c=3\end{matrix}\right.\)

tự vẽ hình nha

gọi các góc lần lượt là O₁ ;O₂;O₃;O4

Ta có:

O1+O2+O3+O4=180

2O₂+2O₃=180

=>O2+O3=90

\(x^4+x^3+2x^2+x+1\)

=\(x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

=\(\left(x^2+x+1\right)\left(x^2+1\right)\)

\(A=y^2-2y\left(x-1\right)+\left(x-1\right)^2-\left(x-1\right)^2+2x^2+4x+5\)

\(A=\left(y-x+1\right)^2-x^2+2x-1+2x^2+4x+5\)

\(A=\left(y-x+1\right)^2+x^2+6x+9-5\)

\(A=\left(y-x+1\right)^2+\left(x+3\right)^2-5\ge-5\)

Vậy Amin là -5 \(\Leftrightarrow\left\{{}\begin{matrix}\left(y-x+1\right)^2=0\\\left(x+3\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-4\end{matrix}\right.\)

-5;0