\(\Leftrightarrow\frac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\frac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}+\frac{9x}{x^2-7x+10}=10\)

\(\Leftrightarrow\frac{3x^2-15x-x^2+2x+9x}{\left(x-2\right)\left(x-5\right)}=10\)

\(\Leftrightarrow2x^2-4x=10x^2-70x+100\)

\(\Leftrightarrow8x^2-66+100=0\)

\(\Leftrightarrow4x^2-33x+50=0\)

\(\Leftrightarrow4x\left(x-2\right)-25\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(4x-25\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=\frac{25}{4}\end{matrix}\right.\)

b) [(x-7)(x-2)][(x-4)(x-5)]=72

<=> (x²-9x+14)(x²-9x+20)=72

Đặt x²-9x+17=a

=> (a+3)(a-3)=72

<=> a²-9=72

<=> a²=81

=> a=\(\left\{9;-9\right\}\)

TH1: a=9

=> x²-9x+17=9

<=> x²-9x+8=0

<=> (x-1)(x-8)=0

=> x=\(\left\{1;8\right\}\)

TH2: a=-9

=> x²-9x+17=-9

<=> x²-9x+26=0

<=> x²-9x+20,25+5,75=0

<=> (x-4,5)²+5,75=0

=> x\(\in\varnothing\)

Vậy x=\(\left\{1;8\right\}\)

\(\frac{3x}{x-2}-\frac{x}{x-5}+\frac{9x}{x^2-7x+10}=10\)

\(\Rightarrow\frac{3x^2-15x-x^2+2x+9x}{x^2-7x+10}=10\)

\(\Rightarrow\frac{2x^2-4x}{x^2-7x+10}=10\)

\(\Rightarrow2x^2-4x=10x^2-70x+100\)

\(\Rightarrow8x^2-66x+100=0\)

Ta có \(\Delta=66^2-4.8.100=1156,\sqrt{\Delta}=34\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{66+34}{16}=\frac{25}{4}\\x=\frac{66-34}{16}=2\end{cases}}\)

a) \(\frac{3x}{x-2}-\frac{x}{x-5}+\frac{9x}{x^2-7x+10}=10\)

<=> \(\frac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\frac{x\left(x-2\right)}{\left(x-5\right)\left(x-2\right)}+\frac{9x}{\left(x-2\right)\left(x-5\right)}=10\)

<=> \(\frac{3x^2-15x-x^2+2x+9x}{\left(x-5\right)\left(x-2\right)}=10\)

<=> \(\frac{2x^2-4x}{\left(x-5\right)\left(x-2\right)}=10\)

<=> \(\frac{2x\left(x-2\right)}{\left(x-5\right)\left(x-2\right)}=10\)

<=> \(2x=10\left(x-5\right)\)

<=> 2x - 10x = -50

<=> -8x = -50

<=>x = 6,25

Vậy S = {6,25}

b) (x - 7)(x - 2)(x - 4)(x - 5) = 72

<=> (x² - 9x + 14)(x² - 9x + 20) = 72

Đặt x² - 9x + 14 = t <=> t(t + 6) = 72

<=> t² + 6t - 72 = 0

<=> t² + 12t - 6t - 72 = 0

<=> (t + 12)(t - 6) = 0

<=> \(\orbr{\begin{cases}t+12=0\\t-6=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x^2-9x+14+12=0\\x^2-9x+14-6=0\end{cases}}\)

<=> \(\orbr{\begin{cases}\left(x-9x+20,25\right)+5,75=0\\x^2-9x+8=0\end{cases}}\)

<=> \(\orbr{\begin{cases}\left(x-4,5\right)^2+5,75=0\left(vn\right)\\x^2-x-8x+8=0\end{cases}}\)

<=> (x - 1)(x - 8) = 0

<=> \(\orbr{\begin{cases}x-1=0\\x-8=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=1\\x=8\end{cases}}\)

Vậy S = {1; 8}

a) ( x - 1 )² - ( x - 1 ).( x + 1 ) = 3x - 5

\(\Leftrightarrow\) ( x - 1 ).( x - 1 ) - ( x - 1 ) .( x + 1 ) = 3x - 5

\(\Leftrightarrow\)( x - 1 ) .( x - 1 - x - 1 ) - 3x + 5 = 0

\(\Leftrightarrow\) ( x - 1 ) .( -2 ) - 3x + 5 = 0

\(\Leftrightarrow\) - 2x + 2 - 3x + 5 = 0

\(\Leftrightarrow\)- 5x + 7 = 0

\(\Leftrightarrow\) - 5x = - 7

\(\Leftrightarrow\) x = \(\frac{7}{5}\)

Vậy phương trình có nghiệm là : x = \(\frac{7}{5}\)

c) x³ - 6x² + 9x = 0

\(\Leftrightarrow\)x.( x² - 6x + 9 ) = 0

\(\Leftrightarrow\) x.( x - 3 )² = 0

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\\left(x-3\right)^2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-3=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

Vậy phương trình có nghiệm là : x = 0 , x = 3

\(a,\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)\(ĐKXĐ:x\ge-\frac{5}{7}\)

\(\Leftrightarrow9x-7=7x+5\)

\(\Leftrightarrow9x-7x=5+7\)

\(\Leftrightarrow2x=12\)

\(\Leftrightarrow x=6\)

\(b,\sqrt{4x-20}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=4\)

\(\Leftrightarrow\sqrt{4\left(x-5\right)}+3.\frac{\sqrt{x-5}}{\sqrt{9}}-\frac{1}{3}\sqrt{9\left(x-5\right)}=4\)

\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

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

\(\Leftrightarrow2\sqrt{x-5}=4\)

\(\Leftrightarrow\sqrt{x-5}=2\)

\(\Leftrightarrow x-5=4\)

\(\Leftrightarrow x=9\)

b) đặt x^2+2x+2=t => t>0

\(\frac{t-1}{t}+\frac{t}{t+1}=\frac{7}{6}\Leftrightarrow\frac{2t^2-1}{t^2+t}=\frac{7}{6}\Leftrightarrow12t^2-6=7t^2+7t\)

\(\Leftrightarrow5t^2-7t-6=0\Leftrightarrow5t\left(t-2\right)+3t-6=\left(t-2\right)\left(5t+3\right)\Rightarrow\left[\begin{matrix}t=2\\t=\frac{-3}{5}\left(loai\right)\end{matrix}\right.\)

với t=2

\(x^2+2x+2=2\Rightarrow x^2+2x=0\Rightarrow\left[\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

giup minh voi cac bạn

\(2+\frac{2x^2-8x}{2x^2+8x}+\frac{2x^2+7x+23}{2x^2+7x-4}=\frac{2x+5}{2x-1}\)

\(\Leftrightarrow2+\frac{2x\left(x-4\right)}{2x\left(x+4\right)}+\frac{2x^2+7x+23}{\left(2x-1\right)\left(x+4\right)}=\frac{2x+5}{2x-1}\)

\(\Leftrightarrow2+\frac{x-4}{x+4}+\frac{2x^2+7x+23}{\left(2x-1\right)\left(x+4\right)}-\frac{2x+5}{2x-1}=0\)

\(\Leftrightarrow\frac{2\left(x+4\right)\left(2x-1\right)}{\left(x+4\right)\left(2x-1\right)}+\frac{\left(x-4\right)\left(2x-1\right)}{\left(x+4\right)\left(2x-1\right)}+\frac{2x^2+7x+23}{\left(2x-1\right)\left(x+4\right)}-\frac{\left(2x+5\right)\left(x+4\right)}{\left(2x-1\right)\left(x+4\right)}=0\)

\(\Leftrightarrow\frac{2\left(x+4\right)\left(2x-1\right)+\left(x-4\right)\left(2x-1\right)+2x^2+7x+23-\left(2x+5\right)\left(x+4\right)}{\left(x+4\right)\left(2x-1\right)}=0\)

\(\Leftrightarrow2\left(x+4\right)\left(2x-1\right)+\left(x-4\right)\left(2x-1\right)+2x^2+7x+23-\left(2x+5\right)\left(x+4\right)=0\)

\(\Leftrightarrow2\left(2x^2+7x-4\right)+\left(2x^2-9x+4\right)+2x^2+7x+23-\left(2x^2+13x+20\right)=0\)

\(\Leftrightarrow4x^2+14x-8+2x^2-9x+4+2x^2+7x+23-2x^2-13x-20=0\)

\(\Leftrightarrow6x^2+7x-1=0\)

\(\Leftrightarrow6\left(x^2+2.\frac{7}{12}.x+\frac{49}{144}\right)-\frac{193}{144}=0\)

\(\Leftrightarrow\left(x+\frac{7}{12}\right)^2=\frac{\frac{193}{144}}{6}=\frac{193}{864}\)

Bạn tự làm nốt.

Tương tự với Cb.

Có chắc là đề ổn không bạn?

Hoặc là xem bài mình hộ với; ngộ nhỡ mình sai. Chứ kết quả lẻ quá; chẳng đẹp gì :>