\(A=\dfrac{x-1}{x+3}\\ \RightarrowĐKCĐ:x+3\ne0\\ \Rightarrow x\ne-3\\ \text{Ta có: }A=\dfrac{x-1}{x+3}\\ A=\dfrac{x+3-4}{x+3}\\ A=1-\dfrac{4}{x+3}\)

\(\RightarrowĐể\text{ }A\in Z\\ thì\Rightarrow\dfrac{4}{x+3}\in Z\\ \Rightarrow4⋮x+3\\ \Rightarrow x+3\inƯ_{\left(4\right)}\\ Mà\text{ }Ư_{\left(4\right)}=\left\{\pm1;\pm2;\pm4\right\}\)

Ta lập bảng giá trị:

\(\Rightarrow x\in\left\{-7;-5;-4;-2;-1;1\right\}\)

Vậy để \(A\in Z\)

thì \(x\in\left\{-7;-5;-4;-2;-1;1\right\}\)

Sao lại đăng ở đây. Giải trí nên vào mĩ thuật hoặc âm nhạc chứ bn.

\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\\ =\dfrac{2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)}{2}\\ =\dfrac{\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)}{2}\\ =\dfrac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)}{2}\\ =\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)}{2}\\ =\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)}{2}\\ =\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)}{2}\\ =\dfrac{\left(3^{32}-1\right)\left(3^{32}+1\right)}{2}\\ =\dfrac{3^{64}-1}{2}\)

65+33=98

5+124=129

Hai số liên tiếp có ƯCLN là 1

\(\text{a) }pthh:\text{ }2Cu+O_2\text{ }\text{ }\text{ }\underrightarrow{t^o}\text{ }\text{ }\text{ }2CuO\)

\(\text{b) Ta có công thức về khối lượng : }\\ m_{Cu}+m_{O_2}=m_{CuO}\)

\(\text{c) }m_{O_2}=m_{CuO}-m_{Cu}\\ =10-6\\ =4\left(g\right)\)

Vậy khối lượng của \(O_2\) phản ứng là \(4\left(g\right)\)

\(\text{Ta có công thức về khối lượng: }\\ m_{Cu}+m_{O_2}=m_{CuO}\\ \Rightarrow m_{CuO}=12,8+3,2\\ =16\left(g\right)\)

Vậy khối lượng \(Cu\) thu được là \(16\left(g\right)\)

theo hình trên ta có diện tích 1 ô vuông là:

432:3=144(cm²)

suy ra cạnh hình vuông hay chiều rộng hình chữ nhật là 12 cm

chiều dài hình chữ nhật là:

432:12=36(cm)

chu vi hình chữ nhật là:

(36+12)x2=96(cm)

đáp số: 96 cm

\(1.\text{ }\text{ }\text{ }\dfrac{\left(x^2+2\right)^2-4x^2}{y\left(x^2+2\right)-2xy-\left(x-1\right)^2-1}\\ =\dfrac{\left(x^2+2-2x\right)\left(x^2+2+2x\right)}{x^2y+2y-2xy-x^2+2x-1-1}\\ =\dfrac{\left(x^2+2-2x\right)\left(x^2+2+2x\right)}{\left(x^2y-x^2\right)-\left(2xy-2x\right)+\left(2y-2\right)}\\ =\dfrac{\left(x^2+2-2x\right)\left(x^2+2+2x\right)}{x^2\left(y-1\right)-2x\left(y-1\right)+2\left(y-1\right)}\\ =\dfrac{\left(x^2+2-2x\right)\left(x^2+2+2x\right)}{\left(x^2-2x+2\right)\left(y-1\right)}\\ =\dfrac{x^2+2x+2}{y-1}\)

\(2.\text{ }\text{ }\text{ }\text{ }\dfrac{x^2+5x+6}{x^2+3x+2}\\ =\dfrac{x^2+3x+2x+6}{x^2+2x+x+2}\\ =\dfrac{\left(x^2+3x\right)+\left(2x+6\right)}{\left(x^2+2x\right)+\left(x+2\right)}\\ =\dfrac{x\left(x+3\right)+2\left(x+3\right)}{x\left(x+2\right)+\left(x+2\right)}\\ =\dfrac{\left(x+2\right)\left(x+3\right)}{\left(x+2\right)\left(x+1\right)}\\ =\dfrac{x+3}{x+1}\)

\(3.\text{ }\text{ }\text{ }\dfrac{x^2+y^2-z^2-2zt+2xy-t^2}{x^2-y^2+z^2-2yt+2xz-t^2}\text{ ( Chữa đề ) }\\ =\dfrac{\left(x^2+2xy+y^2\right)-\left(z^2+2zt+t^2\right)}{\left(x^2+2xz+z^2\right)-\left(y^2+2yt+t^2\right)}\\ =\dfrac{\left(x+y\right)^2-\left(z+t\right)^2}{\left(x+z\right)^2-\left(y+t\right)^2}\\ =\dfrac{\left(x+y+z+t\right)\left(x+y-z-t\right)}{\left(x+z+y+t\right)\left(x+z-y-t\right)}\\ =\dfrac{x+y-z-t}{x+z-y-t}\)

\(4.\text{ }\text{ }\text{ }\dfrac{\left(n+1\right)!}{\left(n+1\right)!+\left(n+2\right)!}=\dfrac{\left(n+1\right)!}{\left(n+1\right)!\left(1+n+2\right)}=\dfrac{1}{n+3}\)

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

\(6.\text{ }\text{ }\text{ }\dfrac{x^2-3x}{2x^2-7x+3}\\ =\dfrac{x\left(x-3\right)}{2x^2-6x-x+3}\\ =\dfrac{x\left(x-3\right)}{\left(2x^2-6x\right)-\left(x-3\right)}\\ =\dfrac{x\left(x-3\right)}{2x\left(x-3\right)-\left(x-3\right)}\\ =\dfrac{x\left(x-3\right)}{\left(2x-1\right)\left(x-3\right)}\\ =\dfrac{x}{2x-1}\)

Gọi 2 số tự nhiên lẻ liên tiếp là \(a-1;a+1\left(a\in N\text{*}\right)\)

\(\text{Ta có : }\left(a+1\right)^2-\left(a-1\right)^2=40\\ \Leftrightarrow\left(a+1+a-1\right)\left(a+1-a+1\right)=40\\ \Leftrightarrow2a\cdot2=40\\ \Leftrightarrow4a=40\\ \Leftrightarrow a=10\\ \Rightarrow\left\{{}\begin{matrix}a-1=10-1\\a+1=10+1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a-1=9\\a+1=11\end{matrix}\right.\)

Vậy 2 số tự nhiên lẻ liên tiếp là \(9;11\)

Dấu của hạng tử bậc là dấu âm nên chỉ tìm được giá trị lớn nhất thôi nhé.

\(\text{a) }A=2x-x^2\\ A=2x-x^2+1-1\\ A=1-\left(x^2-2x+1\right)\\ A=1-\left(x-1\right)^2\\ Do\text{ }\left(x-1\right)^2\ge0\forall x\\ \Rightarrow A=1-\left(x-1\right)^2\le1\forall x\\ \text{ Dấu “=” xảy ra khi: }\\ \left(x-1\right)^2=0\\ \Leftrightarrow x-1=0\\ \Leftrightarrow x=1\\ Vậy\text{ }Max_A=1\text{ }khi\text{ }x=1\)

\(\text{b) }B=19-6x-9x^2\\ B=20-1-6x-9x^2\\ B=20-\left(1+6x+9x^2\right)\\ B=20-\left(1+3x\right)^2\\ Do\text{ }\left(1+3x\right)^2\ge0\forall x\\ \Rightarrow B=20-\left(1+3x\right)^2\le20\forall x\\ Dấu\text{ }"="\text{ }xảy\text{ }ra\text{ }khi:\\ \left(1+3x\right)^2=0\\ \Leftrightarrow1+3x=0\\ \Leftrightarrow3x=-1\\ \Leftrightarrow x=-\dfrac{1}{3}\\ Vậy\text{ }Max_B=20\text{ }khi\text{ }x=-\dfrac{1}{3}\)