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

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

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

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

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

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

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

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

\(\Leftrightarrow10\left(x^2+\dfrac{1}{x^2}+2\right)+5\left(x^2+\dfrac{1}{x^2}\right)^2-5\left(x^2+\dfrac{1}{x^2}\right)\left(x^2+\dfrac{1}{x^2}+2\right)=\left(x-5\right)^2-5\)

\(\Leftrightarrow10\left(x^2+\dfrac{1}{x^2}\right)+20+5\left(x^2+\dfrac{1}{x^2}\right)^2-5\left(x^2+\dfrac{1}{x^2}\right)^2-10\left(x^2+\dfrac{1}{x^2}\right)=\left(x-5\right)^2-5\)

\(\Leftrightarrow\left(x-5\right)^2=25\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\)

a, - Đặt \(x^2+x=a\) ta được phương trình :\(a^2+4a-12=0\)

=> \(a^2-2a+6a-12=0\)

=> \(a\left(a-2\right)+6\left(a-2\right)=0\)

=> \(\left(a+6\right)\left(a-2\right)=0\)

=> \(\left[{}\begin{matrix}a+6=0\\a-2=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}a=2\\a=-6\end{matrix}\right.\)

- Thay lại \(x^2+x=a\) vào phương trình trên ta được :\(\left[{}\begin{matrix}x^2+x=2\\x^2+x=-6\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x^2+x-2=0\\x^2+x+6=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}\left(x+\frac{1}{2}\right)^2-\frac{9}{4}=0\\\left(x+\frac{1}{2}\right)^2+\frac{23}{4}=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}\left(x+\frac{1}{2}\right)^2=\frac{9}{4}\\\left(x+\frac{1}{2}\right)^2=-\frac{23}{4}\left(VL\right)\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x+\frac{1}{2}=\sqrt{\frac{9}{4}}\\x+\frac{1}{2}=-\sqrt{\frac{9}{4}}\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\sqrt{\frac{9}{4}}-\frac{1}{2}=1\\x=-\sqrt{\frac{9}{4}}-\frac{1}{2}=-2\end{matrix}\right.\)

Vậy phương trình trên có nghiệm là \(S=\left\{1,-2\right\}\)

b, Đặt \(x^2+2x+3=a\) -> làm tương tự câu a .

c, Ta có : \(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)=72\)

=> \(\left(x^2-4\right)\left(x^2-10\right)=72\)

- Đặt \(x^2-4=a\) và \(x^2-10=a-6\) ta được phương trình :

\(a\left(a-6\right)=72\)

=> \(a^2-6a-72=0\)

=> \(a^2+6a-12a-72=0\)

=> \(a\left(a+6\right)-12\left(a+6\right)=0\)

=> \(\left(a+6\right)\left(a-12\right)=0\)

=> \(\left[{}\begin{matrix}a+6=0\\a-12=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}a=-6\\a=12\end{matrix}\right.\)

- Thay lại \(x^2-4=a\) vào phương trình trên ta được :\(\left[{}\begin{matrix}x^2-4=-6\\x^2-4=12\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x^2=-2\left(VL\right)\\x^2=16\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\sqrt{16}=4\\x=-\sqrt{16}=-4\end{matrix}\right.\)

Vậy phương trình trên có nghiệm là \(S=\left\{4,-4\right\}\)

d, Ta có : \(x\left(x+1\right)\left(x^2+x+1\right)=42\)

=> \(\left(x^2+x\right)\left(x^2+x+1\right)=42\)

- Đặt \(x^2+x=a\) ta được phương trình : \(a\left(a+1\right)=42\)

=> \(a^2+a-42=0\)

=> \(a^2+7a-6a-42=0\)

=> \(a\left(a+7\right)-6\left(a+7\right)=0\)

=> \(\left(a-6\right)\left(a+7\right)=0\)

=> \(\left[{}\begin{matrix}a=6\\a=-7\end{matrix}\right.\)

- Thay \(a=x^2+x\) vào phương trình ta được : \(\left[{}\begin{matrix}x^2+x=6\\x^2+x=-7\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x^2+x-6=0\\x^2+x+7=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}\left(x+\frac{1}{2}\right)^2-\frac{25}{4}=0\\\left(x+\frac{1}{2}\right)^2+\frac{27}{4}=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}\left(x+\frac{1}{2}\right)^2=\frac{25}{4}\\\left(x+\frac{1}{2}\right)^2=-\frac{27}{4}\left(VL\right)\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x+\frac{1}{2}=\sqrt{\frac{25}{4}}\\x+\frac{1}{2}=-\sqrt{\frac{25}{4}}\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\sqrt{\frac{25}{4}}-\frac{1}{2}=2\\x=-\sqrt{\frac{25}{4}}-\frac{1}{2}=-3\end{matrix}\right.\)

Vậy phương trình trên có tập nghiệm là \(S=\left\{2;-3\right\}\)

a: \(\Leftrightarrow\left(1-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{17}+...+\dfrac{1}{49}-\dfrac{1}{57}\right)+2x-2=\dfrac{2}{3}x+\dfrac{7}{3}+\dfrac{5}{4}x-2\)

\(\Leftrightarrow\dfrac{56}{57}+2x-2=\dfrac{23}{12}x+\dfrac{1}{3}\)

=>1/12x=77/57

=>x=308/19

b: =>(x^2-4)(x^2-10)=72

=>x^4-14x^2+40-72=0

=>x^4-14x^2-32=0

=>(x^2-16)(x^2+2)=0

=>x^2-16=0

=>x^2=16

=>x=4 hoặc x=-4

Bài làm:

Ta có: \(\left(x+2\right)\left(x-2\right)\left(x^2-10\right)=72\)

\(\Leftrightarrow\left(x^2-4\right)\left(x^2-10\right)=72\)

\(\Leftrightarrow x^4-14x^2+40-72=0\)

\(\Leftrightarrow x^4-14x^2-32=0\)

\(\Leftrightarrow\left(x^4-16x^2\right)+\left(2x^2-32\right)=0\)

\(\Leftrightarrow x^2\left(x^2-16\right)+2\left(x^2-16\right)=0\)

\(\Leftrightarrow\left(x^2+2\right)\left(x^2-16\right)=0\)

Mà \(x^2+2\ge2>0\left(\forall x\right)\)

\(\Rightarrow x^2-16=0\Leftrightarrow\left(x-4\right)\left(x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x+4=0\end{cases}}\Rightarrow x=\pm4\)

( x + 2 )( x - 2 )( x² - 10 ) = 72

<=> ( x² - 4 )( x² - 10 ) = 72

<=> x⁴ - 14x² + 40 - 72 = 0

<=> x⁴ - 14x² - 32 = 0

Đặt t = x² ( \(t\ge0\))

Pt <=> t² - 14t - 32 = 0

     <=> t² + 2t - 16t - 32 = 0

     <=> t( t + 2 ) - 16( t + 2 ) = 0

     <=> ( t - 16 )( t + 2 ) = 0

     <=> \(\orbr{\begin{cases}t-16=0\\t+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}t=16\\t=-2\end{cases}}\)

\(t\ge0\Rightarrow t=16\)

=> x² = 16

=> \(x=\pm4\)

\(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)=72\)

\(\Leftrightarrow\left(x^2-4\right)\left(x^2-10\right)=72\)

\(\Leftrightarrow\left(x^2-4\right)\left(x^2-4-6\right)=72\\\) ( * )

Đặt \(t=x^2-4\)

Khi đó phương trình ( * ) trở thành:

\(t.\left(t-6\right)=72\)

\(\Leftrightarrow t^2-6t=72\)

\(\Leftrightarrow t^2-6t-72=0\)

\(\Leftrightarrow\left(t-12\right)\left(t+6\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}t-12=0\\t+6=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}t=12\\t=-6\end{matrix}\right.\)

* Với \(t=12\)\(\Rightarrow x^2-4=12\Rightarrow x^2=16\Rightarrow x=+-4\)

* Với \(t=-6\Rightarrow x^2-4=-6\Leftrightarrow x^2=-2\) ( vô lý )

Vậy.............................................

\(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)=72\)

\(\Leftrightarrow\left(x^2-4\right)\left(x^2-10\right)=72\)

\(\Leftrightarrow x^4-10x^2-4x^2+40=72\)

\(\Leftrightarrow x^4-14x^2+40=72\)

\(\Leftrightarrow\left(x^2-7\right)^2=81\)

\(\Leftrightarrow x^2-7=9\)

\(\Leftrightarrow x^2=16\)

\(\Leftrightarrow x=\pm\sqrt{16}=\pm4\).

Vậy .........

( x-2) ( x+2 ) ( x² - 10 ) = 72

<=> ( x²-4 ) (  x² - 10 ) = 72

<=> ( x²-7+3) ( x²-7-3)=72

<=> ( x^2-7)^2 -9 = 72 

<=> ( x^2 -7)^2 = 81

<=> x^2-7 = -9 hoặc 9

mà x^2-7 luôn lớn hơn hoặc bằng -7

<=> x^2-7 = 9

<=> x^2 = 16

<=> x = 4 hoặc -4

Đúng thì nhấn đáng hộ nhé 

pt này có kết quả

x=4 hoặc -4

Đặt pt là (1)

Ta có :

(1) <=> \(\left[\left(x-3\right)\left(x-10\right)\right]\left[\left(x-5\right)\left(x-6\right)\right]-24x^2=0\)

\(\Leftrightarrow\left(x^2-13x+30\right)\left(x^2-11x+30\right)-24x^2=0\)

Đặt \(x^2-12x+30=t\) (*)

Phương trình trở thành \(\left(t-x\right)\left(t+x\right)-24x^2=0\)

\(\Leftrightarrow t^2-x^2-24x^2=0\)

\(\Leftrightarrow t^2-25x^2=0\)

\(\Leftrightarrow\left(t-5x\right)\left(t+5x\right)=0\)

Thay (*) vào ta có :

\(\left(x^2-17x+30\right)\left(x^2+7x+30\right)=0\)

Để ý thấy \(x^2-7x+30\ne0\)

\(\Rightarrow x^2-17x+30=0\)

\(\Leftrightarrow x^2-15x-2x+30=0\)

\(\Leftrightarrow x\left(x-15\right)-2\left(x-15\right)=0\)

\(\Leftrightarrow\left(x-15\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=15\end{matrix}\right.\)

Vậy S={1 ; 15 }

cảm ơn bạn nhìu

\(\left(x+2\right)\left(x-3\right)+3=\left(x-4\right)\left(x+2\right)-7\)

\(\Leftrightarrow x^2-x-6+3=x^2-2x-8-7\)

\(\Leftrightarrow x^2-x-x^2+2x=6-3-8-7\)

\(\Leftrightarrow x=-12\)

Vậy: Phương trình có tập nghiệm \(S=\left\{-12\right\}\)

\(x^2-3x+2x-6+3-x^2-2x+4x+8+7=0\)

\(x+12=0\)

\(x=-12\)