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\(A=\frac{3}{\left(1\cdot2\right)^2}+\frac{5}{\left(2\cdot3\right)^2}+\frac{7}{\left(3\cdot4\right)^2}+...+\frac{89}{\left(44\cdot45\right)^2}\)
\(=\frac{2^2-1^2}{1^2\cdot2^2}+\frac{3^2-2^2}{2^2\cdot3^2}+\frac{4^2-3^2}{3^2\cdot4^2}+...+\frac{45^2-44^2}{44^2\cdot45^2}\)
\(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+....+\frac{1}{44^2}-\frac{1}{45^2}\)
\(=1-\frac{1}{45^2}=1-\frac{1}{2025}=\frac{2024}{2025}\)
\(\frac{1}{n^2\left(n+1\right)^2}=\frac{1}{2n+1}.\left[\frac{1}{n^2}-\frac{1}{\left(n+1\right)^2}\right]\)
\(A_n=\frac{2n+1}{n^2\left(n+1\right)^2}=\frac{1}{n^2}-\frac{1}{\left(n+1\right)^2}\\ \)
\(A=1-\frac{1}{\left(45\right)^2}\)
Lời giải đây bn nhé :
\(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+...+\frac{89}{\left(44.45\right)^2}\)
=\(\frac{3}{1.4}+\frac{5}{4.9}+...+\frac{89}{1936.2025}\)
=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{1}{1936}-\frac{1}{2025}\)
=\(1-\frac{1}{2025}\)
=\(\frac{2024}{2025}\)
xong r nhé
a: \(\Leftrightarrow3x^3-2x^2+15x^2-10x+3x-2+7⋮3x-2\)
\(\Leftrightarrow3x-2\in\left\{1;-1;7;-7\right\}\)
hay \(x\in\left\{3;1\right\}\)
b: \(\Leftrightarrow2x^5-7x^3+4x^4-14x^2+14x^2-49x+49x-44⋮2x^2-7\)
\(\Leftrightarrow2401x^2-1936⋮2x^2-7\)
\(\Leftrightarrow4802x^2-3872⋮2x^2-7\)
\(\Leftrightarrow2x^2-7\inƯ\left(12935\right)\)
\(\Leftrightarrow2x^2-7\in\left\{1;5;13;65;199;995;2587;12935;-1;-5\right\}\)
\(\Leftrightarrow2x^2\in\left\{8;72;2\right\}\)
hay \(x\in\left\{2;-2;6;-6;1;-1\right\}\)
\(1.\)
\(-17-\left(x-3\right)^2\)
Ta có: \(\left(x-3\right)^2\ge0\)với \(\forall x\)
\(\Leftrightarrow-\left(x-3\right)^2\le0\)với \(\forall x\)
\(\Leftrightarrow17-\left(x-3\right)^2\le17\)với \(\forall x\)
Dấu '' = '' xảy ra khi:
\(\left(x-3\right)^2=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
Vậy \(Max=-17\)khi \(x=3\)
\(2.\)
\(A=x\left(x+1\right)+\frac{3}{2}\)
\(A=x^2+x+\frac{3}{2}\)
\(A=\left(x+\frac{1}{2}\right)^2+\frac{5}{4}\)
\(\left(x+\frac{1}{2}\right)^2+\frac{5}{4}\ge\frac{5}{4}\)với \(\forall x\)
\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2+\frac{5}{4}\ge\frac{5}{4}\)với \(\forall x\)
Vậy \(Max=\frac{5}{4}\)khi \(x=\frac{-1}{2}\)
