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a) \(4x^2-8x+4-9\left(x-y\right)^2\)
\(=4\left(x^2-2x+1\right)-9\left(x-y\right)^2\)
\(=\left[2\left(x-1\right)\right]^2-\left[3\left(x-y\right)\right]^2\)
\(=\left(2x-2+3x-3y\right)\left(2x-2-3x+3y\right)\)
\(=\left(5x-3y-2\right)\left(3y-x-2\right)\)
b) \(x^3-4x^2+12x-27\)
\(=\left(x^3-27\right)-\left(4x^2-12x\right)\)
\(=\left(x-3\right)\left(x^2+3x+9\right)-4x\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-x+9\right)\)
1.a)\(x^2-ax+bx-ab=x\left(x-a\right)+b\left(x-a\right)=\left(x+b\right)\left(x-a\right)\)
b)\(x^2+ay-y^2-ax=\left(x-y\right)\left(x+y\right)-a\left(x-y\right)=\left(x+y-a\right)\left(x-y\right)\)
c)\(x^3-3x^2-4x+12=x^2\left(x-3\right)-4\left(x-3\right)=\left(x^2-4\right)\left(x-3\right)=\left(x-2\right)\left(x+2\right)\left(x-3\right)\)
2.a)\(2x^2-12x=-18=>2x^2-12x+18=0=>x^2-6x+9=0=>\left(x-3\right)^2=0=>x-3=0=>x=3\)b)\(\left(4x^2-4x+1\right)-x^2=0=>3x^2-3x-x+1=3x\left(x-1\right)-\left(x-1\right)=\left(3x-1\right)\left(x-1\right)=0\)
\(=>\orbr{\begin{cases}3x-1=0\\x-1=0\end{cases}=>\orbr{\begin{cases}x=\frac{1}{3}\\x=1\end{cases}}}\)
a) 2x2 - 12x = -18
<=> 2x2 - 12x + 18 = 0
<=> 2(x2 - 6x + 9) = 0
<=> 2(x2 - 2.x.3 + 9) = 0
<=> 2(x - 3)2 = 0
<=> x - 3 = 0
<=> x = 0 + 3
<=> x = 3
b) (4x2 - 4x + 1) - x2 = 0
<=> 4x2 - 4x + 1 - x2 = 0
<=> 3x2 - 4x + 1 = 0
<=> 3x2 - x - 3x + 1 = 0
<=> x(3x - 1) - (3x - 1) = 0
<=> \(\orbr{\begin{cases}\left(3x-1\right)=0\\\left(x-1\right)=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=\frac{1}{3}\\x=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{3}\\x=1\end{cases}}\)
a) \(x^2+4x+3\)
\(=x^2+3x+x+3\)
\(=x\left(x+3\right)+\left(x+3\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
\(\left(x+1\right)\left(x+2\right)-\left(x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+1-x-3\right)=0\)
\(\Leftrightarrow-2\left(x+2\right)=0\)
\(\Leftrightarrow x=-2\)
a
4x2--25=0
=> (2x)22 --52 =0
=> (2x-5)(2x+5)=0
\(\orbr{\begin{cases}2x-5=0\\2x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}X=\frac{5}{2}\\X=\frac{-5\:\:. \:\:\:\:\:\:\:\:\:\:TT}{2}\end{cases}Mình\:}\)
\(4x^2=25\Rightarrow x^2=\frac{25}{4}\Rightarrow x=\sqrt{\frac{25}{4}}\) \(=\frac{5}{2}\)
\(\left(x^3-x^2\right)^2-\left(4x^2-8x+4\right)=0\)
= \(\left(x^3-x^2\right)^2-\left(2x-2\right)^2=0\)
=(\(\left(x^3-x^2-2x+2\right)\left(x^3-x^2+2x-2\right)=0\)
=\(\left[x^2\left(x-1\right)-2\left(x-1\right)\right]\) \(\left[x^2\left(x-1\right)+2\left(x-1\right)\right]\)=0
=\(\left(x-1\right)\left(x^2-2\right)\left(x-1\right)\left(x^2+2\right)\) = 0
= \(\left(x-1\right)\left(x^2-2\right)\left(x^2+2\right)=0\)
=\(\left(x-1\right)\left(x^4-4\right)\) = 0
=> \(x-1=0\) hoặc \(x^4-4=0\)
=> \(x=1\) hoặc \(x=\pm\sqrt{2}\)
câu 2
a)\(\left(3x^2\right)^3-\left(2x\right)^3\)
= \(\left(3x^2-2x\right)\left(9x^4-54x^5+36x^4-4x^2\right)\)
= \(x\left(3x-2\right)\left(9x^4-54x^5+36x^4-4x^2\right)\)
may be wrong , but chawsc k nhiều , chỗ nào k hiểu ib hỏi mk sai nha <3
\(\left(x+5\right)\left(x-5\right)-\left(x-2\right)\left(x+7\right)=0\)
\(\left(x^2-5^2\right)-\left(x^2+7x-2x-14\right)=0\)
\(x^2-25-x^2-7x+2x+14=0\)
\(-5x=25-14\)
\(-5x=11\)
\(x=-\frac{11}{5}\)
***
\(9x^2-4-2\left(3x-2\right)^2=0\)
\(\left(3x\right)^2-2^2-2\left(3x-2\right)^2=0\)
\(\left(3x-2\right)\left(3x+2\right)-2\left(3x-2\right)^2=0\)
\(\left(3x-2\right)\left[\left(3x+2\right)-2\left(3x-2\right)\right]=0\)
\(\left(3x-2\right)\left(3x+2-6x+4\right)=0\)
\(\left(3x-2\right)\left(6-3x\right)=0\)
TH1:
\(3x-2=0\)
\(3x=2\)
\(x=\frac{2}{3}\)
TH2:
\(6-3x=0\)
\(3x=6\)
\(x=\frac{6}{3}\)
\(x=2\)
Vậy \(x=\frac{2}{3}\) hoặc \(x=2\)
***
\(12\left(3-4x\right)+7\left(4x-3\right)=0\)
\(12\left(3-4x\right)-7\left(3-4x\right)=0\)
\(\left(3-4x\right)\left(12-7\right)=0\)
\(5\left(3-4x\right)=0\)
\(3-4x=0\)
\(4x=3\)
\(x=\frac{3}{4}\)
***
\(x^2-4-2xy+y^2=\left(x-y\right)^2-2^2=\left(x-y-2\right)\left(x-y+2\right)\)
***
\(x^3-4x^2-12x+27=\left(x+3\right)\left(x^2-3x+9\right)-4x\left(x+3\right)=\left(x+3\right)\left(x^2-3x+9-4x\right)=\left(x+3\right)\left(x^2-7x+9\right)\)
***
\(3x^2-18x+27=3\left(x^2-2\times x\times3+3^2\right)=3\left(x-3\right)^2\)
***
\(A=-x^2+3x-4=-\left(x^2-2\times x\times\frac{3}{2}+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2+4\right)=-\left[\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\right]\)
\(\left(x-\frac{3}{2}\right)^2\ge0\)
\(\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\ge\frac{7}{4}\)
\(-\left[\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\right]\le-\frac{7}{4}< 0\)
Vậy A < 0 với mọi x (đpcm)
1a (x+5)(x-5)-(x-2)(x+7) = 0
=> x2-25-(x2+5x-14) = 0
=> x2-25-x2-5x+14 = 0
=> -11-5x = 0
=> -5x = -11-0
=> -5x = -11
=> x = -11:5
=> x = \(\frac{-11}{5}\)
bài 2:
1) (x-y)2-4
3) 3(x2-6x+9)
tìm x
1,(x+5)(x-5) - ( x - 2 ).( x+ 7) =0
=>( x2 - 25 ) - ( x2 + 7x -2x -14) =0
=> x2 - 25 - x2 -7x +2x +14 =0
=> -5x -11 =0
=> x= \(\frac{11}{5}\)
2, 9x2 -4 - 2.(3x - 2 )2 =0
=> 9x2 -4 - 2 . (9x2 - 12x +4 ) = 0
=> 9x2 - 4 - 18x2 + 24x - 8 =0
=>-9x2 - 12 +24x =0
=>- ( 9x2 - 24x + 16 -4)=0
=> -(3x -4)2 +4 =0
=>-(3x - 4)2=-4
=>3x -4 =2
=> x=6
3 , x2- \(\frac{1}{4}x\)=0
=> x( x -\(\frac{1}{4}\))=0
=> x=0;x=
4, 12 .( 3 - 4x ) + 7x(4x - 3) =0
=> 12 . ( 3-4x) - 7x ( 3 -4x) = 0
=> ( 3 - 4x ).(12 - 7x)=0
=> x=\(\frac{3}{4}\) ;x=\(\frac{12}{7}\)
x2- 4 -2xy + y2= ( x - y )2-4
3x2 - 18x +27=3(x2 -6 +9)
mk viết tắt hơi nhiều nếu kĩ thì bạn làm chi tiết ra nha .