Câu 2:

\(A=2014+\dfrac{2014}{1+2}+\dfrac{2014}{1+2+3}+...+\dfrac{2014}{1+2+3+...+2013}\)

\(=2014\left(1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+3+...+2013}\right)\)

\(=2014\left(1+\dfrac{1}{2\left(2+1\right)}.2+\dfrac{1}{3\left(3+1\right)}.2+...+\dfrac{1}{2013\left(2013+1\right)}.2\right)\)

\(=2014\left(\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{2013.2014}\right)\)

\(=4028\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2013.2014}\right)\)

Bạn tự tính nốt nhé

1)

\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2012^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2011\cdot2012}\left(1\right)\)\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2011\cdot2012}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2011}-\dfrac{1}{2012}\\ =\dfrac{1}{1}-\dfrac{1}{2012}< 1\left(2\right)\)

Từ (1) và (2) ta có: A < 1

2)

\(A=2014+\dfrac{2014}{1+2}+\dfrac{2014}{1+2+3}+...+\dfrac{2014}{1+2+3+...+2013}\\ =2014\cdot\left(\dfrac{1}{1}+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+3+...+2013}\right)\\ =2014\cdot\left(\dfrac{1}{\left(1\cdot2\right):2}+\dfrac{1}{\left(2\cdot3\right):2}+\dfrac{1}{\left(3\cdot4\right):2}+...+\dfrac{1}{\left(2013\cdot2014\right):2}\right)\\ =2014\cdot\left(\dfrac{2}{1\cdot2}+\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{2013\cdot2014}\right)\\ =2014\cdot2\cdot\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2013\cdot2014}\right)\\ =4028\cdot\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2013}-\dfrac{1}{2014}\right)\\ =4028\cdot\left(1-\dfrac{1}{2014}\right)\\ =4028\cdot\dfrac{2013}{2014}\\ =4026\)

3)

Để A là số nguyên thì \(6n+42⋮6n\Rightarrow42⋮6n\Rightarrow6n\inƯ\left(42\right)\)

\(Ư\left(42\right)=\left\{1;2;3;6;7;14;21;42\right\}\)

Vì n là số tự nhiên nên n = 1 hoặc n = 7

4)

\(A=\dfrac{17^{18}+1}{17^{19}+1}< \dfrac{17^{18}+1+16}{17^{19}+1+16}=\dfrac{17^{18}+17}{17^{19}+17}=\dfrac{17\cdot\left(17^{17}+1\right)}{17\cdot\left(17^{18}+1\right)}=\dfrac{17^{17}+1}{17^{18}+1}=B\)

Vậy A<B

vì U₁ = I * R₁

U₂ = I * R₂

​nên ta có : \(\dfrac{U_1}{U_2}=\dfrac{I\cdot R_1}{I\cdot R_2}=\dfrac{R_1}{R_2}\Rightarrow\dfrac{U_1}{U_2}=\dfrac{R_1}{R_2}\)

đây mà là toán lớp 2 à bạn?

đề sai

a) \(U=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(U=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(U=\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(U=\dfrac{15\sqrt{x}-11-\left(3x+9\sqrt{x}-2\sqrt{x}-6\right)-\left(2x-2\sqrt{x}+3\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(U=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(U=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\Leftrightarrow\dfrac{-5\left(\sqrt{x}-1\right)\left(\sqrt{x}-\dfrac{2}{5}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)

\(\Leftrightarrow\dfrac{-5\left(\sqrt{x}+\dfrac{2}{5}\right)}{\sqrt{x}+3}\Leftrightarrow\dfrac{-5\sqrt{x}-2}{\sqrt{x}+3}\)

b) ta có \(U=\dfrac{1}{2}\) \(\Leftrightarrow\dfrac{-5\sqrt{x}-2}{\sqrt{x}+3}=\dfrac{1}{2}\) \(\Leftrightarrow\sqrt{x}+3=2\left(-5\sqrt{x}-2\right)\)

\(\Leftrightarrow\sqrt{x}+3=-10\sqrt{x}-4\Leftrightarrow\sqrt{x}+3-\left(-10\sqrt{x}-4\right)\)

\(\Leftrightarrow\sqrt{x}+3+10\sqrt{x}+4=0\Leftrightarrow11\sqrt{x}+7=0\Leftrightarrow11\sqrt{x}=-7\)

\(\Leftrightarrow\sqrt{x}=\dfrac{-7}{11}\left(vôlí\right)\)

vậy không có giá trị nào để \(U=\dfrac{1}{2}\)

nguyen van tuan

Bài này là tớ đăg lên ! Nhưg hôm nay thầy tớ giải rồi! Tớ đăg lời giải lên đây cho mấy bạn tham khảo ạ! ko kiếm GP nhá!

Câu 1 :

Vì x > y \(\Rightarrow\) \(x-y>0\)

\(\Rightarrow x^2+y^2\ge2\sqrt{2}.\left(x-y\right)\)

\(\Leftrightarrow x^2+y^2-2\sqrt{2}x+2\sqrt{2}y\ge0\)

Vì \(xy=1\Rightarrow x^2+y^2+\left(\sqrt{2}\right)^2-2\sqrt{2}x+2\sqrt{2}y-2xy\ge0\)

\(\Leftrightarrow\left(x-y-2\sqrt{2}\right)^2\ge0\)

Đúng với mọi x; y

Câu 2:

\(a^3+b^3+ab\ge\dfrac{1}{2}\)

\(\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^3\right)+ab-\dfrac{1}{2}\ge0\)

\(\Leftrightarrow a^2-ab+b^2+ab-\dfrac{1}{2}\ge0\) ( vì a+b = 1 )

\(\Leftrightarrow a^2+b^2-\dfrac{1}{2}\ge0\)

Vì \(a+b=1\Rightarrow b=1-a\)

\(\Rightarrow a^2+\left(1-a\right)^2-\dfrac{1}{2}\ge0\)

\(\Leftrightarrow a^2+1-2a+a^2-\dfrac{1}{2}\ge0\)

\(\Leftrightarrow2a^2-2a+\dfrac{1}{2}\ge0\)

\(\Leftrightarrow4a^2-4a+1\ge0\)

\(\Leftrightarrow\left(2a-1\right)^2\ge0\)

Đúng với mọi a;b

Dấu "=" xảy ra khi

\(2a-1=0\Rightarrow a=\dfrac{1}{2}\Rightarrow b=\dfrac{1}{2}\)

Mysterious Person

bài hay đấy để mk thử giải

à bạn xem lại câu a hộ mk với

\(A=\dfrac{u-v}{\sqrt{u}+\sqrt{v}}-\dfrac{\sqrt{u^3}+\sqrt{v^3}}{u-v}\)

\(=\sqrt{u}-\sqrt{v}-\dfrac{u\sqrt{u}+v\sqrt{v}}{\left(\sqrt{u}-\sqrt{v}\right)\left(\sqrt{u}+\sqrt{v}\right)}\)

\(=\sqrt{u}-\sqrt{v}-\dfrac{u-\sqrt{uv}+v}{\left(\sqrt{u}-\sqrt{v}\right)\left(\sqrt{u}+\sqrt{v}\right)}\)

\(=\sqrt{u}-\sqrt{v}-\dfrac{u-\sqrt{uv}+v}{\sqrt{u}-\sqrt{v}}\)

\(=\dfrac{\left(\sqrt{u}-\sqrt{v}\right)\sqrt{u}-\left(\sqrt{u}-\sqrt[]{v}\right)\sqrt{v}-\left(u-\sqrt{uv}+v\right)}{\sqrt{u}-\sqrt{v}}\)

\(=\dfrac{u-\sqrt{uv}-\sqrt{uv}+v-u+\sqrt{uv}-v}{\sqrt{u}-\sqrt{v}}\)

\(\Leftrightarrow\)\(-\dfrac{\sqrt{uv}}{\sqrt{u}-\sqrt{v}}\)

để cả căn hơi phức tạp nhỉ? nếu tinh ý 1 chút thì sẽ đơn giản thôi :3

chú ý nhé ! nếu ta đăt như sau \(\sqrt{u}=a;\sqrt{v}=b\)

đến đấy thì dễ nhỉ<3;

\(A=\dfrac{a^2-b^2}{a+b}-\dfrac{a^3+b^3}{a^2-b^2}\)

xem nào ~~ để ý xem nó có phải hằng đẳng thức quen thuộc k nhỉ, thôi k quan tâm cứ trâu bò vào xem ra cái j k đã bạn ạ

\(A=\dfrac{\left(a+b\right)\left(a-b\right)}{a+b}-\dfrac{\left(a+b\right)\left(a^2-ab+b^2\right)}{\left(a-b\right)\left(a+b\right)}\)

\(A=a-b-\dfrac{a^2-ab+b^2}{a-b}\) có thể bạn nghĩ đến đây là khó, đùng ngại ta hãy cứ quy đồng chúng

\(A=\dfrac{\left(a-b\right)^2-a^2+ab-b^2}{a-b}=\dfrac{-ab}{a-b}\)