Làm lại cho you đây -_- vừa nãy bấm mt nhầm,đời t nhọ vãi

1)\(P=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{16}\left(1+2+3+....+16\right)\)

\(P=1+\dfrac{1+2}{2}+\dfrac{1+2+3}{3}+\dfrac{1+2+3+4}{4}+...+\dfrac{1+2+3+...+16}{16}\)

Xét thừa số tổng quát: \(\dfrac{1+2+3+...+t}{t}=\dfrac{\left[\left(t-1\right):1+1\right]:2.\left(t+1\right)}{t}=\dfrac{\dfrac{t}{2}\left(t+1\right)}{t}=\dfrac{\dfrac{t^2}{2}+\dfrac{t}{2}}{t}=\dfrac{t\left(\dfrac{t}{2}+\dfrac{1}{2}\right)}{t}=\dfrac{t}{2}+\dfrac{1}{2}\)

Như vậy: \(P=1+\left(\dfrac{2}{2}+\dfrac{1}{2}\right)+\left(\dfrac{3}{2}+\dfrac{1}{2}\right)+\left(\dfrac{4}{2}+\dfrac{1}{2}\right)+...+\left(\dfrac{16}{2}+\dfrac{1}{2}\right)\)

\(P=1+\dfrac{3}{2}+\dfrac{4}{2}+\dfrac{5}{2}+....+\dfrac{17}{2}\)

\(P=\dfrac{2+3+4+5+...+17}{2}\)

\(P=\dfrac{152}{2}=76\)

2) \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{3}\)

\(\Rightarrow2016\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)=\dfrac{2016}{3}\)

\(\Rightarrow\dfrac{2016}{a+b}+\dfrac{2016}{b+c}+\dfrac{2016}{c+a}=\dfrac{2016}{3}\)

\(\Rightarrow\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}=\dfrac{2016}{3}\)

\(\Rightarrow\dfrac{a+b}{a+b}+\dfrac{c}{a+b}+\dfrac{b+c}{b+c}+\dfrac{a}{b+c}+\dfrac{c+a}{c+a}+\dfrac{b}{c+a}=\dfrac{2016}{3}\)

\(\Rightarrow1+\dfrac{c}{a+b}+1+\dfrac{a}{b+c}+1+\dfrac{b}{c+a}=\dfrac{2016}{3}\)

\(\Rightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{2016}{3}-1-1-1=\dfrac{2007}{3}\)

\(P=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{16}\left(1+2+3+...+16\right)\)

Xét thừa số tổng quát: \(\dfrac{1+2+3+..+n}{n}=\dfrac{\left[\left(n-1\right):1+1\right]:2.\left(n+1\right)}{n}=\dfrac{\dfrac{n}{2}\left(n+1\right)}{n}=\dfrac{\dfrac{n^2}{2}+\dfrac{n}{2}}{n}=\dfrac{n\left(\dfrac{n}{2}+\dfrac{1}{2}\right)}{n}=\dfrac{n}{2}+\dfrac{1}{2}\)

Như vậy:

\(P=1+\left(\dfrac{2}{2}+\dfrac{1}{2}\right)+\left(\dfrac{3}{2}+\dfrac{1}{2}\right)+\left(\dfrac{4}{2}+\dfrac{1}{2}\right)+...+\left(\dfrac{16}{2}+\dfrac{1}{2}\right)\)

\(P=1+\dfrac{2}{2}+\dfrac{1}{2}+\dfrac{3}{2}+\dfrac{1}{2}+\dfrac{4}{2}+\dfrac{1}{2}+...+\dfrac{16}{2}+\dfrac{1}{2}\)

\(P=1+\left(\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{16}{2}\right)+\left(\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+...+\dfrac{1}{2}\right)\)

\(P=1+\dfrac{2+3+4+...+16}{2}+\dfrac{15}{2}\)

\(P=1+\dfrac{\left[\left(16-2\right):1+1\right]:2.\left(16+2\right)}{2}+\dfrac{15}{2}\)

\(P=1+210+\dfrac{15}{2}=218,5\)

A = (\(\dfrac{5}{6}\) - \(\dfrac{4}{5}\)) . 1\(\dfrac{1}{5}\) + \(\dfrac{3}{16}\) : (\(\dfrac{-1}{2}\))³

A = \(\dfrac{1}{30}\) . \(\dfrac{6}{5}\) + \(\dfrac{3}{16}\) : \(\dfrac{-1}{8}\)

A = \(\dfrac{1}{25}\) + \(\dfrac{3}{16}\) . \(\dfrac{-8}{1}\)

A = \(\dfrac{1}{25}\) + \(\dfrac{-3}{2}\)

A = \(\dfrac{-73}{50}\)

B = \(\dfrac{4}{17}\) . (7\(\dfrac{3}{4}\) - 6\(\dfrac{1}{3}\)) + (5\(\dfrac{3}{4}\) - 6.95) : (-1\(\dfrac{3}{5}\))

B = \(\dfrac{4}{17}\) . \(\dfrac{17}{12}\) + (\(\dfrac{23}{4}\) - \(\dfrac{139}{20}\)) : \(\dfrac{-8}{5}\)

B = \(\dfrac{1}{3}\) + \(\dfrac{-6}{5}\) . \(\dfrac{-5}{8}\)

B = \(\dfrac{13}{12}\)

a: \(\left(18\dfrac{1}{3}:\sqrt{225}+8\dfrac{2}{3}\cdot\sqrt{\dfrac{49}{4}}\right):\left[\left(12\dfrac{1}{3}+8\dfrac{6}{7}\right)-\dfrac{\left(\sqrt{7}\right)^2}{\left(3\sqrt{2}\right)^2}\right]:\dfrac{1704}{445}\)

\(=\left(\dfrac{55}{3}:15+\dfrac{26}{3}\cdot\dfrac{7}{4}\right):\left[\left(12+\dfrac{1}{3}+8+\dfrac{6}{7}\right)-\dfrac{7}{18}\right]\cdot\dfrac{445}{1704}\)

\(=\left(\dfrac{55}{45}+\dfrac{91}{6}\right):\left[20+\dfrac{101}{126}\right]\cdot\dfrac{445}{1704}\)

\(=\dfrac{295}{18}:\dfrac{2621}{126}\cdot\dfrac{445}{1704}\)

\(=\dfrac{295}{18}\cdot\dfrac{126}{2621}\cdot\dfrac{445}{1704}\simeq0,21\)

b: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)

c: \(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{n+1}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{n}{n+1}\)

\(=\dfrac{1}{n+1}\)

d: \(-66\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{11}\right)+124\cdot\left(-37\right)+63\cdot\left(-124\right)\)

\(=-66\cdot\dfrac{33-22+6}{66}+124\left(-37-63\right)\)

\(=-17-12400=-12417\)

e: \(\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{3333}{2020}+\dfrac{333333}{303030}+\dfrac{33333333}{42424242}\right)\)

\(=\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)

\(=\dfrac{7}{4}\cdot33\cdot\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\right)\)

\(=33\cdot\dfrac{7}{4}\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)

\(=33\cdot\dfrac{7}{4}\cdot\left(\dfrac{1}{3}-\dfrac{1}{7}\right)\)

\(=33\cdot\dfrac{7}{4}\cdot\dfrac{4}{21}=\dfrac{33\cdot1}{3}=11\)

ai giúp mìn vứi ❤

a)\(\dfrac{3}{4}-\dfrac{5}{2}-\dfrac{3}{5}=\dfrac{15}{20}-\dfrac{50}{20}-\dfrac{12}{20}=-\dfrac{47}{20}\)

b) \(\sqrt{7^2}+\sqrt{\dfrac{25}{16}-\dfrac{3}{2}}=7+\sqrt{\dfrac{1}{16}}=7+\dfrac{1}{4}=\dfrac{29}{4}\)

c) \(\dfrac{1}{2}.\sqrt{100}-\sqrt{\dfrac{1}{16}+\left(\dfrac{1}{3}\right)^0}=\dfrac{1}{2}.10-\sqrt{\dfrac{1}{16}+1}=5-\sqrt{\dfrac{17}{16}}\)

\(1,A=-\dfrac{3}{4}.\left(0,125-1\dfrac{1}{2}\right):\dfrac{33}{16}-25\%\)

\(A=-\dfrac{3}{4}.\left(0,125-\dfrac{3}{2}\right):\dfrac{33}{16}-\dfrac{1}{4}\)

\(A=-\dfrac{3}{4}.\left(-\dfrac{11}{8}\right):\dfrac{33}{16}-\dfrac{1}{4}\)

\(A=\dfrac{33}{32}:\dfrac{33}{16}-\dfrac{1}{4}\)

\(A=\dfrac{33}{32}.\dfrac{16}{33}-\dfrac{1}{4}\)

\(A=\dfrac{1}{2}-\dfrac{1}{4}\)

\(A=\dfrac{2}{4}-\dfrac{1}{4}\)

\(A=\dfrac{1}{4}\)

\(D=6\dfrac{5}{12}:2\dfrac{5}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\)

\(D=\dfrac{77}{12}:\dfrac{13}{4}+\dfrac{45}{4}.\dfrac{2}{15}\)

\(D=\dfrac{77}{39}+\dfrac{3}{2}\)

\(D=\dfrac{271}{78}\)

\(C=\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)

\(C=\dfrac{5}{16}:0,125-\left(\dfrac{9}{4}-0,6\right).\dfrac{10}{11}\)

\(C=\dfrac{5}{16}:0,125-\dfrac{33}{20}.\dfrac{10}{11}\)

\(C=\dfrac{5}{2}-\dfrac{3}{2}\)

\(C=1\)

a) \(\left(\dfrac{1}{16}\right)^{-\dfrac{3}{4}}+810000^{0.25}-\left(7\dfrac{19}{32}\right)^{\dfrac{1}{5}}\)

\(=\left(\dfrac{1}{2}\right)^{4.\left(-\dfrac{3}{4}\right)}+\left(30\right)^{4.0,25}-\left(\dfrac{243}{32}\right)^{\dfrac{1}{5}}\)

\(=\left(\dfrac{1}{2}\right)^{-3}+30-\left(\dfrac{3}{2}\right)^{5.\dfrac{1}{5}}\)

\(=2^3+30-\dfrac{3}{2}\)

\(=36,5\)

b) \(=\left(0,1\right)^{3.\left(-\dfrac{1}{3}\right)}-2^{-2}.2^{6.\dfrac{2}{3}}-\left[\left(2\right)^3\right]^{-\dfrac{4}{3}}\)

\(=0,1^{-1}-2^2-2^{-4}\)

\(=10-4-\dfrac{1}{16}\)

\(=\dfrac{95}{16}\)

c) \(=3^{3.\dfrac{2}{3}}-\dfrac{1}{\left(-2\right)^2}+\left(\dfrac{27}{8}\right)^{-\dfrac{1}{3}}\)

\(=9-\dfrac{1}{4}+\left(\dfrac{3}{2}\right)^{3.\dfrac{-1}{3}}\)

\(=9-\dfrac{1}{4}+\left(\dfrac{3}{2}\right)^{-1}\)

\(=9-\dfrac{1}{4}+\dfrac{2}{3}\)

\(=\dfrac{113}{12}\)

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

\(=4.\left(-\dfrac{1}{8}\right)-2.\dfrac{1}{4}+3.\left(-\dfrac{1}{2}\right)+1\)

\(=-\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{3}{2}+1\)

\(=-\dfrac{3}{2}\)

b) \(8.\sqrt{9}-\sqrt{64}\)

\(=8.3-8\)

\(=24-8\)

\(=16\)

c) \(\sqrt{\dfrac{9}{16}}+\dfrac{25}{46}:\dfrac{5}{23}-\dfrac{7}{4}\)

\(=\dfrac{3}{4}+\dfrac{5}{2}-\dfrac{7}{4}\)

\(=-1+\dfrac{5}{2}\)

\(=\dfrac{3}{2}\)

56:54=

Câu 1:

\(A=21\left(a+\frac{1}{b}\right)+3\left(b+\frac{1}{a}\right)=21a+\frac{21}{b}+3b+\frac{3}{a}\)

\(=(\frac{a}{3}+\frac{3}{a})+(\frac{7b}{3}+\frac{21}{b})+\frac{62}{3}a+\frac{2b}{3}\)

Áp dụng BĐT Cô-si:
\(\frac{a}{3}+\frac{3}{a}\geq 2\sqrt{\frac{a}{3}.\frac{3}{a}}=2\)

\(\frac{7b}{3}+\frac{21}{b}\geq 2\sqrt{\frac{7b}{3}.\frac{21}{b}}=14\)

Và do $a,b\geq 3$ nên:

\(\frac{62}{3}a\geq \frac{62}{3}.3=62\)

\(\frac{2b}{3}\geq \frac{2.3}{3}=2\)

Cộng tất cả những BĐT trên ta có:

\(A\geq 2+14+62+2=80\) (đpcm)

Dấu "=" xảy ra khi $a=b=3$

Câu 2:

Bình phương 2 vế ta thu được:

\((x^2+6x-1)^2=4(5x^3-3x^2+3x-2)\)

\(\Leftrightarrow x^4+12x^3+34x^2-12x+1=20x^3-12x^2+12x-8\)

\(\Leftrightarrow x^4-8x^3+46x^2-24x+9=0\)

\(\Leftrightarrow (x^2-4x)^2+6x^2+24(x-\frac{1}{2})^2+3=0\) (vô lý)

Do đó pt đã cho vô nghiệm.