đỡ hơn chưa??? mong các bn giúp mình vs

Vê trái: 

\(=\frac{2}{\left(x-1\right)\left(x+1\right)}+\frac{4}{\left(x-2\right)\left(x+2\right)}+...+\frac{20}{\left(x-10\right)\left(x+10\right)}\)

\(=\frac{\left(x+1\right)-\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{\left(x+2\right)-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+...+\frac{\left(x+10\right)-\left(x-10\right)}{\left(x+10\right)\left(x-10\right)}\)

\(=\frac{1}{x-1}-\frac{1}{x+1}+\frac{1}{x-2}-\frac{1}{x+2}+...+\frac{1}{x-10}-\frac{1}{x+10}\)

\(=\left(\frac{1}{x-1}+\frac{1}{x-2}+...+\frac{1}{x-10}\right)-\left(\frac{1}{x+1}+\frac{1}{x+2}+...+\frac{1}{x+10}\right)\)

Vế phải:

\(=\frac{\left(x+1\right)-\left(x-10\right)}{\left(x-10\right)\left(x+1\right)}+\frac{\left(x+2\right)-\left(x-9\right)}{\left(x-9\right)\left(x+2\right)}+...+\frac{\left(x+10\right)-\left(x-1\right)}{\left(x-1\right)\left(x+10\right)}\)

\(=\frac{1}{x-10}-\frac{1}{x+1}+\frac{1}{x-9}-\frac{1}{x+2}+...+\frac{1}{x-1}-\frac{1}{x+10}\)

\(=\left(\frac{1}{x-1}+\frac{1}{x-2}+...+\frac{1}{x-10}\right)-\left(\frac{1}{x+1}+\frac{1}{x+2}+...+\frac{1}{x+10}\right)\) = vế phải

=> đpcm

\(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\left(x\ne1\right)\)

\(\Leftrightarrow\frac{1}{x-1}-\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x}{x^2+x+1}=0\)

\(\Leftrightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=0\)

\(\Leftrightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x^2-2x}{\left(x-1\right)\left(x^2+x+1\right)}=0\)

\(\Leftrightarrow\frac{1}{\left(x-1\right)\left(x^2+x+1\right)}\left(x^2+x+1-3x^2-2x^2+2x\right)=0\)

\(\Leftrightarrow-4x^2+3x+1=0\left(\frac{1}{\left(x-1\right)\left(x^2+x+1\right)}\ne0\right)\)

\(\Leftrightarrow-4x^2+4x-x+1=0\)

\(\Leftrightarrow-4x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(-4x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\-4x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\-4x=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\left(loại\right)\\x=\frac{-1}{4}\end{cases}}}\)

Vậy \(x=\frac{-1}{4}\)

Tìm x, biết:

3(x+2)(x+5) +5(x+5)(x+10) +7(x+10)(x+17) =x(x+2)(x+17) (x∉−2;−5;−10;−17)

2(x−1)(x−3) +5(x−3)(x−8) +12(x−8)(x−20) −1x−20 =−34 (x∉1;3;8;20)

x+110 +2+111 x+112 =x+113 +x+114 

x−1030 +x−1443 +x−595 +x−1488 =0

Trả lời luôn à bạn

1.

\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)

\(MC:12\)

Quy đồng :

\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)

\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)

\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)

\(\Leftrightarrow6x+9-3x=-4-9+16\)

\(\Leftrightarrow-7x=3\)

\(\Leftrightarrow x=\frac{-3}{7}\)

2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)

\(MC:20\)

Quy đồng :

\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)

\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)

\(\Leftrightarrow30x+15-20=15x-2\)

\(\Leftrightarrow15x=3\)

\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)

bạn xem lại xem có nhầm đề ko nhé

c) \(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)

\(\Leftrightarrow\left(\frac{x-1}{2009}-1\right)+\left(\frac{x-2}{2008}-1\right)=\left(\frac{x-3}{2007}-1\right)+\left(\frac{x-4}{2006}-1\right)\)

\(\Leftrightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)

\(\Leftrightarrow\left(x-2010\right).\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)

\(\Leftrightarrow x-2010=0\)

\(\Leftrightarrow x=0+2010\)

\(\Rightarrow x=2010\)

Vậy \(x=2010.\)

Mình chỉ làm câu c) thôi nhé.

Chúc bạn học tốt!

\(1\)) \(70:\frac{4x+720}{x}=\frac{1}{2}\)

\(\Leftrightarrow\frac{4x+720}{x}=70:\frac{1}{2}\)

\(\Leftrightarrow\frac{4x+720}{x}=140\)

\(\Leftrightarrow\left(4x+720\right):x=140\)

\(\Leftrightarrow4x+720=140.x\)

\(\Leftrightarrow4x-140x=-720\)

\(\Leftrightarrow x.\left(-136\right)=-720\)

\(\Leftrightarrow x=-720:\left(-136\right)\)

\(\Leftrightarrow x=\frac{90}{17}\)

\(2\)) Mình đang nghĩ

1) \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)

<=> \(\frac{21x}{24}-\frac{100\left(x-9\right)}{24}=\frac{80x+6}{24}\)

<=> 21x - 100x + 900 = 80x + 6

<=> -79x - 80x = 6 - 900

<=> -159x = -894

<=> x = 258/53

Vậy S = {258/53}

2) \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x+1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)

<=> \(\frac{3\left(4x^2+4x+1\right)}{15}-\frac{5\left(x^2+2x+1\right)}{15}=\frac{7x^2-14x-5}{15}\)

<=> 12x² + 12x + 3 - 5x² - 10x - 5 = 7x² - 14x - 5

<=> 7x² + 2x - 7x² + 14x = -5 + 2

<=> 16x = 3

<=> x = 3/16

Vậy S  = {3/16}

3) 4(3x - 2) - 3(x - 4) = 7x+  10

<=> 12x - 8 - 3x + 12 = 7x + 10

<=> 9x - 7x = 10 - 4

<=> 2x = 6

<=> x = 3

Vậy S = {3}

4) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)

<=> \(\frac{x^2+14x+40}{12}+\frac{3\left(x^2+2x-8\right)}{12}=\frac{4\left(x^2+8x-20\right)}{12}\)

<=> x² + 14x + 40 + 3x² + 6x - 24 = 4x² + 32x - 80

<=> 4x² + 20x - 4x² - 32x = -80 - 16

<=> -12x = -96

<=> x = 8

Vậy S = {8}