\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{91.93}+\frac{1}{x}=5\)

\(\Rightarrow\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{91.93}\right)+\frac{1}{x}=5\)

\(\Rightarrow\left[\frac{1}{2}\times\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{91.93}\right)\right]+\frac{1}{x}=5\)

\(\Rightarrow\left[\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{91}-\frac{1}{93}\right)\right]+\frac{1}{x}=5\)

\(\Rightarrow\left[\frac{1}{2}\times\left(1-\frac{1}{93}\right)\right]+\frac{1}{x}=5\)

\(\Rightarrow\left[\frac{1}{2}\times\frac{92}{93}\right]+\frac{1}{x}=5\)

\(\Rightarrow\frac{46}{93}+\frac{1}{x}=5\)

\(\Rightarrow\frac{1}{x}=5-\frac{46}{93}\)

\(\Rightarrow\frac{1}{x}=\frac{419}{93}\)

\(\Rightarrow x=1:\frac{419}{93}\)

\(\Rightarrow x=\frac{93}{419}\)

Vậy  \(x=\frac{93}{419}\)

_Chúc bạn học tốt_

mình thấy đề hơi sai sai sao á bạn đoạn \(\dfrac{x-4}{93}\) á bạn

a) \(85^2-15^2=\left(85-15\right)\left(85+15\right)=70.100=7000\)

b) \(93^3+21.93^2+3.49.93+343\)

\(=93^3+3.7.93^2+3.7^2.93+7^3\)

\(=\left(93+7\right)^3\)

\(=100^3=1000000\)

c) \(73^2-13^2-10^2+20.13\)

\(=73^2-\left(13^2+10^2-20.13\right)\)

\(=73^2-\left(13-10\right)^2\)

\(=73^2-3^2\)

\(=\left(73+3\right)\left(73-3\right)\)

\(=76.70=5320\)

d) Viết = Latex hộ mình

a) Ta có: \(85^2-15^2\)

\(=\left(85-15\right)\left(85+15\right)\)

\(=70\cdot100=7000\)

b) Ta có: \(93^3+21\cdot93^2+3\cdot49\cdot93+343\)

\(=93^3+3\cdot93^2\cdot7+3\cdot93+7^2+7^3\)

\(=\left(93+7\right)^3\)

\(=100^3=1000000\)

c) Ta có: \(73^2-13^2-10^2+20\cdot13\)

\(=73^2-\left(13^2+10^2-20\cdot13\right)\)

\(=73^2-\left(13^2-2\cdot13\cdot10+10^2\right)\)

\(=73^2-\left(13-10\right)^2\)

\(=73^2-3^2=\left(73-3\right)\left(73+3\right)\)

\(=70\cdot76=5320\)

các bạn giúp mình với a. Mình cảm ơn trước

\(\frac{x+1}{99}+\frac{x+3}{97}+\frac{x+5}{95}=\frac{x+7}{93}+\frac{x+9}{91}+\frac{x+11}{89}\)

\(\Rightarrow\frac{x+1}{99}+1+\frac{x+3}{97}+1+\frac{x+5}{95}+1\)\(=\frac{x+7}{93}+1+\frac{x+9}{91}+1+\frac{x+11}{89}+1\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}\)\(=\frac{x+100}{93}+\frac{x+100}{91}+\frac{x+100}{89}\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}\)\(-\frac{x+100}{93}-\frac{x+100}{91}-\frac{x+100}{89}=0\)

\(\Rightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{93}-\frac{1}{91}-\frac{1}{89}\right)=0\)

Mà \(\left(\frac{1}{99}< \frac{1}{97}< \frac{1}{95}< \frac{1}{93}< \frac{1}{91}< \frac{1}{89}\right)\)nên \(\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{93}-\frac{1}{91}-\frac{1}{89}\right)< 0\)

\(\Rightarrow x+100=0\Leftrightarrow x=-100\)

Vậy x = -100

1.

$\sqrt{3x^2}-\sqrt{12}=0$

$\Leftrightarrow \sqrt{3x^2}=\sqrt{12}$

$\Leftrightarrow 3x^2=12$

$\Leftrightarrow x^2=4$

$\Leftrightarrow (x-2)(x+2)=0\Leftrightarrow x=\pm 2$

2. 

$\sqrt{(x-3)^2}=9$

$\Leftrightarrow |x-3|=9$

$\Leftrightarrow x-3=9$ hoặc $x-3=-9$

$\Leftrightarrow x=12$ hoặc $x=-6$

3.

$\sqrt{4x^2+4x+1}=6$

$\Leftrightarrow \sqrt{(2x+1)^2}=6$

$\Leftrightarrow |2x+1|=6$

$\Leftrightarrow 2x+1=6$ hoặc $2x+1=-6$

$\Leftrightarrow x=\frac{5}{2}$ hoặc $x=\frac{-7}{2}$

a) \(85^2-15^2=\left(85-15\right)\left(85+15\right)=70.100=7000\)

c) \(73^2-13^2-10^2+20.13\)

\(=73^2-\left(13^2+10^2-20.13\right)\)

\(=73^2-\left(13^2-2.13.10+10^2\right)\)

\(=73^2-\left(13-10\right)^2\)

\(=73^2-3^2\)

\(=\left(73-3\right)\left(73+3\right)\)

\(=70.76\)

\(=5320\)

d)Viết đề = công thức trực quan hộ mình