\(\left(x+5\right)\left(x-5\right)-\left(x-2\right)\left(x+7\right)=0\)
\(\left(x^2-5^2\right)-\left(x^2+7x-2x-14\right)=0\)
\(x^2-25-x^2-7x+2x+14=0\)
\(-5x=25-14\)
\(-5x=11\)
\(x=-\frac{11}{5}\)
***
\(9x^2-4-2\left(3x-2\right)^2=0\)
\(\left(3x\right)^2-2^2-2\left(3x-2\right)^2=0\)
\(\left(3x-2\right)\left(3x+2\right)-2\left(3x-2\right)^2=0\)
\(\left(3x-2\right)\left[\left(3x+2\right)-2\left(3x-2\right)\right]=0\)
\(\left(3x-2\right)\left(3x+2-6x+4\right)=0\)
\(\left(3x-2\right)\left(6-3x\right)=0\)
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 .
