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\(x^2+4x+5=x^2+4x+4+1\)
\(=\left(x+2\right)^2+1\)
Ta có:
\(\left(x+2\right)^2\text{≡}0,1\left(mod3\right)\)
\(1\text{≡}1\left(mod3\right)\)
\(\Rightarrow\left(x+2\right)^2+1\text{≡}1,2\left(mod3\right)\)
\(\Rightarrow\left(x+2\right)^2+1\) không chia hết cho 3
\(\Rightarrow x^2+4x+5\) không chia hết cho 3
Ta có:\(\left|x-1\right|\ge0;\forall x\)
\(\left|x+2\right|\ge0;\forall x\)
\(\left|x-3\right|\ge0;\forall x\)
\(\left|x+4\right|\ge0;\forall x\) ......
Cộng tất cả ta được:
\(\left|x-1\right|+\left|x+2\right|+\left|x-3\right|+\left|x+4\right|+...+\left|x-9\right|\ge0\)
\(\Rightarrow Min_T=0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}x=1\\x=-2\\x=3\\x=-4.....\end{matrix}\right.\)
a) Để y nguyên thì \(6x-4⋮2x+3\)
\(\Leftrightarrow-13⋮2x+3\)
\(\Leftrightarrow2x+3\in\left\{1;-1;13;-13\right\}\)
\(\Leftrightarrow2x\in\left\{-2;-4;10;-16\right\}\)
hay \(x\in\left\{-1;-2;5;-8\right\}\)
\(=\dfrac{\left|x-2020\right|+2022-1}{\left|x-2020\right|+2022}=1-\dfrac{1}{\left|x-2020\right|+2022}\\ mà\left|x-2020\right|\ge0\\ \Rightarrow\left|x-2022\right|+2022\ge2022\)
\(\Rightarrow\dfrac{1}{\left|x-2020\right|+2022}\le\dfrac{1}{2022}\\ =1-\dfrac{1}{\left|x-2020\right|+2022}\ge1-\dfrac{1}{2022}\\ =\dfrac{2021}{2022}\\ \Rightarrow B_{min}=\dfrac{2021}{2022}.tại.x-2020=0\Rightarrow x=2020\)
\(\dfrac{x}{8}-\dfrac{1}{4}=\dfrac{1}{y}\)
\(\Leftrightarrow\dfrac{x-2}{8}=\dfrac{1}{y}\)
\(\Leftrightarrow x-2=\dfrac{8}{y}\)
Do \(x-2\in Z\Rightarrow\dfrac{8}{y}\in Z\)
\(\Rightarrow y=Ư\left(8\right)\)
\(\Rightarrow y=\left\{-8;-4;-2;-1;1;2;4;8\right\}\)
\(\Rightarrow x=\left\{1;0;-2;-6;10;6;4;3\right\}\)
\(B=\dfrac{3}{4}xy^2-\dfrac{1}{3}x^2y-\dfrac{5}{6}xy^2+2x^2y=-\dfrac{1}{12}xy^2+\dfrac{5}{3}x^2y\)
Bậc:3
Thay x=-1, y=1 vào B ta có:
\(B=-\dfrac{1}{12}xy^2+\dfrac{5}{3}x^2y=-\dfrac{1}{12}.\left(-1\right).1^2+\dfrac{5}{3}.\left(-1\right)^2.1=\dfrac{1}{12}+\dfrac{5}{3}=\dfrac{7}{4}\)
b, Ta có : \(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{6}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}\)
Đặt \(x=15k;y=20k;z=24k\)
Thay vào A ta được : \(A=\dfrac{30k+60k+96k}{45k+80k+120k}=\dfrac{186k}{245k}=\dfrac{186}{245}\)
-Sửa đề: x,y nguyên.
\(x-\dfrac{1}{y}-\dfrac{4}{xy}=-1\left(x\ne0;y\ne0;x\ne-1\right)\)
\(\Rightarrow x-\dfrac{1}{y}-\dfrac{4}{xy}+1=0\)
\(\Rightarrow\dfrac{x^2y}{xy}-\dfrac{x}{xy}-\dfrac{4}{xy}+\dfrac{xy}{xy}=0\)
\(\Rightarrow x^2y-x-4+xy=0\)
\(\Rightarrow xy\left(x+1\right)=x+4\)
\(\Rightarrow y=\dfrac{x+4}{x\left(x+1\right)}\)
-Vì x,y nguyên:
\(\Rightarrow\left(x+4\right)⋮\left[x\left(x+1\right)\right]\)
\(\Rightarrow\left(x+4\right)⋮x\) và \(\left(x+4\right)⋮\left(x+1\right)\)
\(\Rightarrow4⋮x\) và \(\left(x+1+3\right)⋮\left(x+1\right)\)
\(\Rightarrow x\in\left\{1;-1;2;-2;4;-4\right\}\) và \(3⋮\left(x+1\right)\)
\(\Rightarrow x\in\left\{1;-1;2;-2;4;-4\right\}\) và \(x+1\in\left\{1;-1;3;-3\right\}\)
\(\Rightarrow x\in\left\{1;-1;2;-2;4;-4\right\}\) và \(x\in\left\{0;-2;2;-4\right\}\)
\(\Rightarrow x\in\left\{2;-2;-4\right\}\)
*\(x=2\Rightarrow y=\dfrac{2+4}{2.\left(2+1\right)}=1\)
\(x=-2\Rightarrow y=\dfrac{-2+4}{-2.\left(-2+1\right)}=1\)
\(x=-4\Rightarrow y=\dfrac{-4+4}{-4.\left(-4+1\right)}=0\left(loại\right)\)
-Vậy các cặp số (x,y) là: \(\left(2,1\right);\left(-2,1\right)\)