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Bài 1. A=\(\frac{1}{1}\)x\(\frac{1}{2}\)x\(\frac{1}{2}\)x\(\frac{1}{3}\)x\(\frac{1}{3}\)x\(\frac{1}{4}\)x\(\frac{1}{4}\)x\(\frac{1}{5}\)x\(\frac{1}{5}\)x\(\frac{1}{6}\) Bài 2. B=\(\frac{1}{1x2}\)+\(\frac{1}{2x3}\)+\(\frac{1}{3x4}\)+\(\frac{1}{4x5}\)+\(\frac{1}{5x6}\) Bài 3. B=\(\frac{2}{1x2}\)+\(\frac{2}{2x3}\)+\(\frac{2}{3x4}\)+\(\frac{2}{4x5}\)+\(\frac{2}{5x6}\) Bài 4. C=\(\frac{2}{1x3}\)+\(\frac{2}{3x5}\)+\(\frac{2}{5x7}\)+\(\frac{2}{7x9}\)+\(\frac{2}{9x11}\) Bài...
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Bài 1.

A=\(\frac{1}{1}\)x\(\frac{1}{2}\)x\(\frac{1}{2}\)x\(\frac{1}{3}\)x\(\frac{1}{3}\)x\(\frac{1}{4}\)x\(\frac{1}{4}\)x\(\frac{1}{5}\)x\(\frac{1}{5}\)x\(\frac{1}{6}\)

Bài 2.

B=\(\frac{1}{1x2}\)+\(\frac{1}{2x3}\)+\(\frac{1}{3x4}\)+\(\frac{1}{4x5}\)+\(\frac{1}{5x6}\)

Bài 3.

B=\(\frac{2}{1x2}\)+\(\frac{2}{2x3}\)+\(\frac{2}{3x4}\)+\(\frac{2}{4x5}\)+\(\frac{2}{5x6}\)

Bài 4.

C=\(\frac{2}{1x3}\)+\(\frac{2}{3x5}\)+\(\frac{2}{5x7}\)+\(\frac{2}{7x9}\)+\(\frac{2}{9x11}\)

Bài 5.

C=\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{90}+\frac{1}{110}\)

Bài 6.Tính bằng cách thuận tiện nhất.

a.(792,81 x 025 + 792,81 x 0,75) x (11 x 9 - 900 x 0,1 - 9).

b.\(\frac{7,2:2x57,2+2,86x2x64}{4+4+8+12+20+....+220}\)

c.\(\frac{2003x14+1998+2001x2002}{2002+2002x503+504x2002}\)

d.\(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{28}\)

đ.3,54 x 73 + 0,23 x 25 + 3,54 x 27 + 0,17 x 25

e.\(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\)

g.\(\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x\left(1-\frac{1}{5}\right)\)

0
10 tháng 10 2019

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NV
27 tháng 10 2019

a/ ĐKXĐ: ...

\(\Leftrightarrow3\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)-7\)

Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow a^2=x+\frac{1}{4x}+1\)

\(\Rightarrow x+\frac{1}{4x}=a^2-1\)

Pt trở thành:

\(3a=2\left(a^2-1\right)-7\)

\(\Leftrightarrow2a^2-3a-9=9\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=3\)

\(\Leftrightarrow2x-6\sqrt{x}+1=0\)

\(\Rightarrow\sqrt{x}=\frac{3+\sqrt{7}}{2}\Rightarrow x=\frac{8+3\sqrt{7}}{2}\)

b/ ĐKXĐ:

\(\Leftrightarrow5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)

Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow x+\frac{1}{4x}=a^2-1\)

\(\Rightarrow5a=2\left(a^2-1\right)+4\Leftrightarrow2a^2-5a+2=0\)

\(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+\frac{1}{2\sqrt{x}}=2\\\sqrt{x}+\frac{1}{2\sqrt{x}}=\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x-4\sqrt{x}+1=0\\2x-\sqrt{x}+1=0\left(vn\right)\end{matrix}\right.\)

NV
27 tháng 10 2019

c/ ĐKXĐ: ...

\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)

\(\Leftrightarrow\frac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)

\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\frac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)

\(\Leftrightarrow2x^2-8x+5=0\)

d/ ĐKXĐ: ...

\(\Leftrightarrow x+1-\frac{15}{6}\sqrt{x}+\sqrt{x^2-4x+1}-\frac{1}{2}\sqrt{x}=0\)

\(\Leftrightarrow\frac{x^2-\frac{17}{4}x+1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{x^2-\frac{17}{4}x+1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}=0\)

\(\Leftrightarrow\left(x^2-\frac{17}{4}x+1\right)\left(\frac{1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}\right)=0\)

\(\Leftrightarrow x^2-\frac{17}{4}x+1=0\)

\(\Leftrightarrow4x^2-17x+4=0\)