a, \(\left(x^2-5x+7\right)^2-\left(2x-5\right)^2=0\)

\(\Leftrightarrow\left(x^2-5x+7-2x+5\right)\left(x^2-5x+7+2x-5\right)=0\)

\(\Leftrightarrow\left(x^2-7x+12\right)\left(x^2-3x+2\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-3\right)\left(x-1\right)\left(x-2\right)=0\Leftrightarrow x=1;x=2;x=3;x=4\)

Vậy tập nghiệm phương trình là S = { 1 ; 2 ; 3 ; 4 } 

b, \(\left|2x-1\right|=5\Leftrightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Vậy tập nghiệm của phương trình là S = { -2 ; 3 } 

c, \(\left|2x-1\right|=\left|x+5\right|\Leftrightarrow\left(2x-1\right)^2=\left(x+5\right)^2\)

\(\Leftrightarrow\left(2x-1\right)^2-\left(x+5\right)^2=0\Leftrightarrow\left(2x-1-x-5\right)\left(2x-1+x+5\right)=0\Leftrightarrow x=6;x=-\dfrac{4}{3}\)

Vậy tập nghiệm của phương trình là S = { -4/3 ; 6 } 

d, \(\left|3x+1\right|=x-2\)

TH1 : \(3x+1=x-2\Leftrightarrow2x=-3\Leftrightarrow x=-\dfrac{3}{2}\)

TH2 : \(3x+1=-x+2\Leftrightarrow4x=1\Leftrightarrow x=\dfrac{1}{4}\)

Vậy tập nghiệm của phương trình là S = { -3/2 ; 1/4 } 

các ý còn lại tương tự 

a) Ta có: \(\left(x^2-5x+7\right)^2-\left(2x-5\right)^2=0\)

\(\Leftrightarrow\left(x^2-5x+7-2x+5\right)\left(x^2-5x+7+2x-5\right)=0\)

\(\Leftrightarrow\left(x^2-7x+12\right)\left(x^2-3x+2\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-4\right)\left(x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-4=0\\x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\\x=1\\x=2\end{matrix}\right.\)

Vậy: S={3;4;1;2}

b) Ta có: |2x-1|=5

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Vậy: S={3;-2}

a)

\(\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x^2-1\right)\)

\(\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x+1\right)\left(x-1\right)\)

\(\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)-\left(3x+2\right)\left(x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left[9x^2-4-\left[\left(3x+2\right)\left(x-1\right)\right]\right]=0\)

\(\Leftrightarrow\left(x+1\right)\left[9x^2-4-\left(3x^2-3x+2x-2\right)\right]=0\)

\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-3x^2+3x-2x+2\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(6x^2+x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\6x^2+x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(2x-1\right)\left(3x+2\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-2}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{1;\dfrac{-2}{3};\dfrac{1}{2}\right\}\)

b)

\(\left(x-1\right)^2-1+x^2=\left(1-x\right)\left(x+3\right)\)

\(\Leftrightarrow x^2-2x+1-1+x^2=x+3-x^2-3x\)

\(\Leftrightarrow2x^2-2x=x^2-2x+3\)

\(\Leftrightarrow3x^2=3\)

\(\Leftrightarrow x^2=1\)

\(\Leftrightarrow x=\left(\pm1\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

Vậy \(x\in\left\{1;-1\right\}\)

Bài 2:

a: =>2x^2-4x+1=x^2+x+5

=>x^2-5x-4=0

=>\(x=\dfrac{5\pm\sqrt{41}}{2}\)

b: =>11x^2-14x-12=3x^2+4x-7

=>8x^2-18x-5=0

=>x=5/2 hoặc x=-1/4

a) \(\left(3x-2\right)\left(4x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\4x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{4}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{2}{3};-\dfrac{5}{4}\right\}\)

b) \(\left(2,3x-6,9\right)\left(0,1x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2,3x-6,9=0\\0,1x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-20\end{matrix}\right.\)

c) \(\left(4x+2\right)\left(x^2+1\right)=0\)

Vì \(x^2+1\ge1>0\forall x\)

\(\Rightarrow4x+2=0\)

\(\Leftrightarrow x=-\dfrac{1}{2}\)

Vậy: \(S=\left\{-\dfrac{1}{2}\right\}\)

d) \(\left(2x+7\right)\left(x-5\right)\left(5x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+7=0\\x-5=0\\5x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\\x=-\dfrac{1}{5}\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{7}{2};5;-\dfrac{1}{5}\right\}\)

e) \(\left(x-1\right)\left(2x+7\right)\left(x^2+2\right)=0\)

Vì \(x^2+2\ge2>0\forall x\)

\(\Rightarrow\left(x-1\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\)

f) \(\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)

\(\Leftrightarrow\left(3x+2\right)\left(x-1\right)\left(x+1\right)-\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[\left(3x+2\right)\left(x+1\right)\right].\left(x-1-3x+2\right)=0\)

\(\Leftrightarrow\left(3x^2+5x+2\right)\left(-2x+1\right)=0\)

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

\(\Leftrightarrow\left[3x\left(x+1\right)+2\left(x+1\right)\right]\left(-2x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x+2\right)\left(-2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x+2=0\\-2x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{2}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{-1;-\dfrac{2}{3};\dfrac{1}{2}\right\}\)

a) \(\left(x+1\right)\left(x-1\right)\left(3x-6\right)>0\)

Lập bảng xét dấu ta được kết quả :

\(Bpt\Leftrightarrow\left[{}\begin{matrix}-1< x< 1\\x>2\end{matrix}\right.\)

b) \(\dfrac{x+3}{x-2}\le0\)

Lập bảng xét dấu ta được kết quả :

\(Bpt\Leftrightarrow-3\le x< 2\)

d) \(\dfrac{2x-5}{3x+2}< \dfrac{3x+2}{2x-5}\)

\(\Leftrightarrow\dfrac{2x-5}{3x+2}-\dfrac{3x+2}{2x-5}< 0\)

\(\Leftrightarrow\dfrac{\left(2x-5\right)^2-\left(3x+2\right)^2}{\left(3x+2\right)\left(2x-5\right)}< 0\)

\(\Leftrightarrow\dfrac{\left(2x-5+3x+2\right)\left(2x-5-3x-2\right)}{\left(3x+2\right)\left(2x-5\right)}< 0\)

\(\Leftrightarrow\dfrac{-\left(5x-3\right)\left(x+7\right)}{\left(3x+2\right)\left(2x-5\right)}< 0\)

Lập bảng xét dấu ta được kết quả :

\(Bpt\Leftrightarrow\left[{}\begin{matrix}-7< x< -\dfrac{2}{3}\\\dfrac{5}{3}< x< \dfrac{5}{2}\end{matrix}\right.\)

a) Ta có: \(\left(x^2-2x\right)^2-6x^2+12x+9=0\)

\(\Leftrightarrow\left(x^2-2x\right)^2-6\left(x^2-2x\right)+9=0\)

\(\Leftrightarrow\left(x^2-2x-3\right)^2=0\)

\(\Leftrightarrow x^2-2x-3=0\)

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

\(\Leftrightarrow x\left(x-3\right)+\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

Vậy: S={3;-1}

b) Ta có: \(\left(x^2+x+1\right)\left(x^2+x+2\right)=12\)

\(\Leftrightarrow\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12=0\)

\(\Leftrightarrow\left(x^2+x\right)^2+5\left(x^2+x\right)-2\left(x^2+x\right)-10=0\)

\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x+5\right)-2\left(x^2+x+5\right)=0\)

\(\Leftrightarrow\left(x^2+x+5\right)\left(x^2+x-2\right)=0\)

\(\Leftrightarrow x^2+x-2=0\)(Vì \(x^2+x+5>0\forall x\))

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

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

\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)

Vậy: S={-2;1}

2 ý a và b anh CTV nãy đã làm rồi nha, còn câu c này thì làm dài dòng+không chắc :VVV

c)\(\left(2x^2-3x+1\right)\left(2x^2+5x+1\right)-9x^2=0\)

\(\Leftrightarrow\left(2x^2-3x+1\right)\left(2x^2-3x+1+8x\right)-9x^2=0\)

\(\Leftrightarrow\left(2x^2-3x+1\right)^2+8x\left(2x^2-3x+1\right)+16x^2-25x^2=0\)

\(\Leftrightarrow\left(2x^2-3x+1+4x\right)^2-25x^2=0\)

\(\Leftrightarrow\left(2x^2+x+1\right)^2-25x^2=0\)

\(\Leftrightarrow\left(2x^2+x+1-5x\right)\left(2x^2+x+1+5x\right)=0\)

\(\Leftrightarrow\left(2x^2-4x+1\right)\left(2x^2+6x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(2x^2-4x+1\right)=0\\\left(2x^2+6x+1\right)=0\end{matrix}\right.\)

Rồi đến đây tự giải nhé, không phân tích được thì bấm máy tính là ra nha:vv

Tất cả những bài này bạn đều có thể đặt ẩn phụ. Sau đó phân tích thành nhân tử để tìm nghiệm.

a) Đặt $x^2-2x=a$

b) Đặt $x^2+x+1=a$

c) Đặt $2x^2-3x+1=a$

( 9 x 2 – 4 ) ( x + 1 ) = ( 3 x + 2 ) ( x 2 - 1 )

⇔ (3x – 2)(3x + 2)(x + 1) - (3x + 2)(x - 1)(x + 1) = 0

⇔(3x+ 2)(x + 1)(3x – 2 – x + 1) = 0

⇔ (3x + 2)(x + 1)(2x – 1) = 0

Đề bài thiếu bạn ạ