ai lam day du dau tien minh se k cho nha

minh can gap lam

Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

Ta có : \(\frac{1}{2^2}< \frac{1}{1.2}\)

            \(\frac{1}{3^2}< \frac{1}{2.3}\)

            \(\frac{1}{4^2}< \frac{1}{3.4}\)

             ...

            \(\frac{1}{100^2}< \frac{1}{99.100}\)

\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow A< 1-\frac{1}{100}< 1\)

\(\Rightarrow A< 1\)

Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\).

bạn viết thế ma cũng chẳng hiểu

tớ ko biết

k cho mình nhé

Toán lớp 0 đây sao 

đề yêu câu ftinhs hay tính nhanh Đoàn Đức Hiếu cho nhok cày GP nè

a) \(\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}^2\right).\left(-1\right)^5\)

\(=\dfrac{8}{27}.\dfrac{9}{16}.\left(-1\right)\)

\(=\dfrac{1}{6}.\left(-1\right)\)

=\(\dfrac{-1}{6}\)

b) \(\left(\dfrac{2}{5}\right)^2.\left(\dfrac{-5}{12}\right)^2\)

\(=\dfrac{4}{25}.\dfrac{25}{144}\)

\(=\dfrac{1}{36}\)

giup mk

Hình như sai đề bài đó

\(\left(2x-1\right)\left(1+2x\right)-3\left(x-3\right)^2-\left(2+x\right)^2\)

\(=\left(2x-1\right)\left(2x+1\right)-3\left(x^2-6x+9\right)-\left(4+4x+x^2\right)\)

\(=4x^2-1-3x^2+18x-27-4-4x-x^2\)

\(=14x-32\)

Phần b ,c giải phương trình??

\(\left(2x-3\right)^2+\left(3-x\right)^2+2\left(3-x\right)\left(2x-3\right)=5\)

\(\Leftrightarrow\left(2x-3\right)\left(2x-3+2\left(3-x\right)\right)+\left(3-x\right)^2=5\)

\(\Leftrightarrow\left(2x-3\right)\left(2x-3+6-2x\right)+\left(3-x\right)^2=5\)

\(\Leftrightarrow3\left(2x-3\right)+9-6x+x^2=5\)

\(\Leftrightarrow6x-9+9-6x+x^2=5\)

\(\Leftrightarrow x^2=5\)

\(\Leftrightarrow x=\pm\sqrt{5}\)

\(\left(x+5\right)\left(5-x\right)+\left(2x-1\right)^2-\left(3x-1\right)\left(x+2\right)-7=0\)

\(\Leftrightarrow\left(5-x\right)\left(5-x\right)+4x^2-4x+1-\left(3x^2+6x-x-2\right)-7=0\)

\(\Leftrightarrow25-x^2+4x^2-4x+1-3x^2-6x+x+2-7=0\)

\(\Leftrightarrow21-9x=0\)

​\(\Leftrightarrow9x=21\)​

\(\Leftrightarrow x=3\)

A = \(\dfrac{1}{5^2}\) + \(\dfrac{1}{6^2}\) + \(\dfrac{1}{7^2}\) +.................+ \(\dfrac{1}{2004^2}\)

A = \(\dfrac{1}{5.5}\) + \(\dfrac{1}{6.6}\) + \(\dfrac{1}{7.7}\)+..............+ \(\dfrac{1}{2004.2004}\)

Vì \(\dfrac{1}{5}>\dfrac{1}{6}>\dfrac{1}{7}>...........>\dfrac{1}{2004}\)

nên ta có : \(\dfrac{1}{5.5}>\dfrac{1}{5.6}>\dfrac{1}{6.6}>\dfrac{1}{6.7}>\dfrac{1}{7.7}>.....>\dfrac{1}{2004.2004}>\dfrac{1}{2004.2005}\)

\(\dfrac{1}{5.5}+\dfrac{1}{6.6}+\dfrac{1}{7.7}+...+\dfrac{1}{2004.2004}>\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+..+\dfrac{1}{2004.2005}\)

A > \(\dfrac{1}{5}\) \(-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+....+\dfrac{1}{2004}-\dfrac{1}{2005}\)

A > \(\dfrac{1}{5}\) - \(\dfrac{1}{2005}\) = \(\dfrac{1}{5}\) - \(\dfrac{12}{24060}\)

\(\dfrac{1}{65}\) = \(\dfrac{1}{5}\) - \(\dfrac{12}{65}\) 

Vì \(\dfrac{12}{65}\) > \(\dfrac{12}{24060}\) nên A>  \(\dfrac{1}{65}\) ( phân số nào có phần bù nhỏ hơn thì phân số đó lớn hơn)

Tương tự ta có :

A = \(\dfrac{1}{5.5}\) + \(\dfrac{1}{6.6}\)+ \(\dfrac{1}{7.7}\)+......+\(\dfrac{1}{2004.2004}\) >\(\dfrac{1}{4.5}\)+\(\dfrac{1}{5.6}\)+.....\(\dfrac{1}{2003.2004}\)

A < \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) +......+ \(\dfrac{1}{2003}\) - \(\dfrac{1}{2004}\)

A < \(\dfrac{1}{4}-\dfrac{1}{2004}\) < \(\dfrac{1}{4}\)

\(\dfrac{1}{65}< \)A < \(\dfrac{1}{4}\) (đpcm)