a) 4x² - 12x + 9 = 0 <=> (2x - 3)² = 0 <=> 2x - 3 = 0 <=> x = 3/2.KL

b) ( 5 - 2x )( 2x + 7 ) + ( 25 - 4x² ) = 0 <=> ( 5 - 2x )( 2x + 7 ) + ( 5 + 2x )( 5 - 2x ) = 0 <=> ( 5 - 2x )( 2x + 7 + 5 + 2x ) = 0. KL

<=> ( 5 - 2x )( 4x + 12 ) = 0 <=>\(\orbr{\begin{cases}5-2x=0\\4x+12=0\end{cases}}\)

<=>\(\orbr{\begin{cases}x=2\frac{1}{2}\\x=-3\end{cases}}\)KL.

c) ( x + 3 )( x² - 3x + 9 ) + ( x + 3 )( x - 3 ) = 0 <=> ( x + 3 )( x² - 3x + 9 + x - 3 ) = 0 <=> ( x + 3 )( x² -2x + 6 ) = 0 <=> x + 3 = 0 (vi x² - 2x + 6 = ( x + 1 )² + 5 > 0 voi moi x) KL

<=>x=-3.KL

d) [ 2 ( 2x + 7 ) ]² - [ 3 ( x + 3 ) ]² = 0 <=> ( 4x + 14 )² - ( 3x + 9 )² = 0 <=> ( 4x + 14 + 3x + 9 )( 4x + 14 - 3x -9 ) = 0

<=> ( 7x + 23 )( x + 5 ) = 0 <=> 7x + 23 = 0 hoac x + 5 = 0 <=> x = -23/7 hoac x = -5.KL

Bài 3:

a) \(\left(x-6\right).\left(2x-5\right).\left(3x+9\right)=0\)

\(\Leftrightarrow\left(x-6\right).\left(2x-5\right).3.\left(x+3\right)=0\)

Vì \(3\ne0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\2x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\2x=5\\x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\frac{5}{2}\\x=-3\end{matrix}\right.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{6;\frac{5}{2};-3\right\}.\)

b) \(2x.\left(x-3\right)+5.\left(x-3\right)=0\)

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

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

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{3;-\frac{5}{2}\right\}.\)

c) \(\left(x^2-4\right)-\left(x-2\right).\left(3-2x\right)=0\)

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

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

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

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

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

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2;\frac{1}{3}\right\}.\)

Chúc bạn học tốt!

Bài 4 xem lại đề nhé bác

a: =>|x-7|=3-2x

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(-2x+3\right)^2-\left(x-7\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(2x-3-x+7\right)\left(2x-3+x-7\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(x+4\right)\left(3x-10\right)=0\end{matrix}\right.\Leftrightarrow x=-4\)

b: =>|2x-3|=4x+9

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(4x+9-2x+3\right)\left(4x+9+2x-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(2x+12\right)\left(6x+6\right)=0\end{matrix}\right.\Leftrightarrow x=-1\)

c: =>3x+5=2-5x hoặc 3x+5=5x-2

=>8x=-3 hoặc -2x=-7

=>x=-3/8 hoặc x=7/2

Bài 1:

a) \(\left(2x+3\right)\cdot\left(4x^2-6x+9\right)-2\left(4x^3-1\right)\)

\(=8x^3-12x^2+18x+12x^2-18x+27-8x^3-3=27-3=24\)

--> đpcm

b) Sửa đề: \(\left(x+3\right)^3-\left(x+9\right)\left(x^2+27\right)\)

\(=x^3+9x^2+27x+27-\left(x^3+27x+9x^2+243\right)\)

\(=x^3+9x^2+27x+27-x^3-27x-9x^2-243=27-243=-216\)

--> đpcm

c) \(\left(x+y\right)\left(x^2-xy+y^2\right)+\left(x-y\right)\left(x^2+xy+y^2\right)-2x^3\)

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

--> đpcm

B1: a) \(\left(2x+3\right)\left(4x^2-6x+9\right)-2\left(4x^3-1\right)\)

\(=8x^3-27-8x^3+2\)

\(=-25\)

b) c) Làm theo câu a áp dụng HĐT.

B2:

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

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

\(\Rightarrow\left(x+5\right)\left(x-1\right)=0\)

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

Vậy \(\left\{{}\begin{matrix}x=-5\\x=1\end{matrix}\right..\)

Mấy câu b,c,d bn chịu khó tạo HĐT nhé.

e) \(\Rightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)

\(\Rightarrow2x=-255\)

\(\Rightarrow x=-\dfrac{255}{2}\)

Vậy \(x=-\dfrac{255}{2}\)

bài 2

a) ( x+2)²-9=0

=>(x+2)²- 3² =0

=> (x+2-3)(x+2+3)=0

=>(x-1)(x+5)=0

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

\(B1\\ a,2x+10y=2\left(x+5y\right)\\ b,x^2+4x+4=x^2+2.2x+2^2=\left(x+2\right)^2\\ c,x^2-y^2+10y-25\\ =\left(x^2-y^2\right)+5\left(2y-5\right)\\ =\left(x-y\right)\left(x+y\right)+5\left(2y-5\right)\\ B2\)

\(a,x^2-3x+x-3=0\\ =>x\left(x-3\right)+\left(x-3\right)=0\\ =>\left(x+1\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+1=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\\ b,2x\left(x-3\right)-\dfrac{1}{2}\left(4x^2-3\right)=0\\ =>2x^2-6x-2x^2+\dfrac{3}{2}=0\\ =>-6x=-\dfrac{3}{2}\\ =>x=\left(-\dfrac{3}{2}\right):\left(-6\right)\\ =>x=\dfrac{1}{4}\\ c,x^2-\left(x-3\right)\left(2x-5\right)=9\\ =>x^2-2x^2+6x+5x-15=9\\ =>-x^2+11-15-9=0\\ =>-x^2+11x-24=0\\ =>-x^2+8x+3x-24=0\\ =>-x\left(x-8\right)+3\left(x-8\right)=0\\ =>\left(3-x\right)\left(x-8\right)=0\\ =>\left[{}\begin{matrix}3-x=0\\x-8=0\end{matrix}\right.=>\left[{}\begin{matrix}x=3\\x=8\end{matrix}\right.\)

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

b) \(\Rightarrow\left(x+5\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)

c) \(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

e) \(\Rightarrow\left(x+2\right)\left(x+2-x+2\right)=0\Rightarrow\left(x+2\right).4=0\Rightarrow x=-2\)

f) \(\Rightarrow x\left(2x-3\right)+2\left(2x-3\right)=0\Rightarrow\left(2x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)

g) \(\Rightarrow2\left(3x-2\right)^2-\left(3x-2\right)\left(3x+2\right)=0\Rightarrow\left(3x-2\right)\left(3x-6\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)

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

i) \(\Rightarrow4x\left(x+1\right)+5\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(4x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{5}{4}\end{matrix}\right.\)

a) 4         b) 0,6 hoặc 1,75     e) -3,5 hoặc -3