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1: =(x+y)^2+x+y
=(x+y)(x+y+1)
2: =>3x-9+18<4x+2
=>4x+2>3x+9
=>x>7
![](https://rs.olm.vn/images/avt/0.png?1311)
Mik cần lời giải á, các bạn toàn cho mik đáp án hoặc là cho mỗi câu 123 (Q▪︎Q)
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Bài 3:
a) \(\left(2-3x\right)^2-\left(3-x\right)^2=\left[\left(2-3x\right)-\left(3-x\right)\right]\left[\left(2-3x\right)+\left(3-x\right)\right]\)
\(=\left(-1-2x\right)\left(5-4x\right)\)
b) \(49\left(x-3\right)^2-9\left(x+2\right)^2\)
\(=\left[7\left(x-3\right)\right]^2-\left[3\left(x+2\right)\right]^2\)
\(=\left[\left(7x-21\right)-\left(3x+6\right)\right]\left[\left(7x-21\right)+\left(3x+6\right)\right]\)
\(=\left(4x-27\right)\left(10x-15\right)\)
c) \(2xy-x^2-y^2+16=16-\left(x-y\right)^2=\left(16-x+y\right)\left(16+x-y\right)\)
d) \(2\left(x-3\right)+3\left(x^2-9\right)=2\left(x-3\right)+3\left(x-3\right)\left(x+3\right)\)
\(=\left(x-3\right)\left(3x+11\right)\)
e) \(16x^2-\left(x^2+4\right)^2=\left(4x-x^2-4\right)\left(4x+x^2+4\right)\)
\(=-\left(x-2\right)^2\left(x+2\right)^2\)
f) \(1-2x+2yz+x^2-y^2-z^2=\left(x-1\right)^2-\left(y-z\right)^2\)
\(=\left(x-1-y+z\right)\left(x-1+y-z\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 5:
a) \(x^2+4x-5=x^2-x+5x-5=x\left(x-1\right)+5\left(x-1\right)=\left(x+5\right)\left(x-1\right)\)
b) \(2x^2-14x+20=2x^2-4x-10x+20=2x\left(x-2\right)-10x\left(x-2\right)=2\left(x-5\right)\left(x-2\right)\)
c) \(3x^2+8x+5=3x^2+3x+5x+5=3x\left(x+1\right)+5\left(x+1\right)=\left(3x+5\right)\left(x+1\right)\)
d) \(6x^2-xy-7y^2=6x^2+6xy-7xy-7y^2=6x\left(x+y\right)-7y\left(x+y\right)\)
\(=\left(6x-7y\right)\left(x+y\right)\)
Bài 4:
a) \(x^3-6x^2+12x-8=x^3-2.3.x^2+3.2^2.x-2^3=\left(x-2\right)^3\)
b) \(\left(x-1\right)^3+\left(3-x\right)^3=\left(x-1+3-x\right)\left[\left(x-1\right)^2-\left(x-1\right)\left(3-x\right)+\left(3-x\right)^2\right]\)
\(=2\left(x^2-2x+1+x^2-4x+3+x^2-6x+9\right)\)
\(=2\left(3x^2-12x+13\right)\)
c) \(x^3+y^3+z^3-3xyz=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left(x+y+z\right)^3-3z\left(x+y\right)\left(x+y+z\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left[\left(x+y+z\right)^2-3xy-3yz-3zx\right]\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow6x^2-14x+4-6x^2-12x+18-7x+3=0\)
\(\Leftrightarrow-33x=-25\Rightarrow x=\frac{25}{33}\)
2( 3x - 1 )( x - 2 ) - 6( x - 1 )( x + 3 ) = 7x - 3
<=> 2( 3x2 - 7x + 2 ) - 6( x2 + 2x - 3 ) = 7x - 3
<=> 6x2 - 14x + 4 - 6x2 - 12x + 18 = 7x - 3
<=> -26x + 22 = 7x - 3
<=> -26x - 7x = -3 - 22
<=> -33x = -25
<=> x = 25/33
<=> -36x =
Bài 2 :
\(M=2\left(x+1\right)+\left(3x+2\right)\left(3x-2\right)-9x^2\)
\(=2x+2+9x^2-4-9x^2=2x-2\)
Thay x = 15 vào ta được : \(M=2.15-2=30-2=28\)
Bài 3 :
\(f\left(x\right)=x^2-4x+9=x^2-4x+4+5=\left(x-2\right)^2+5\ge5\forall x\)
Dấu ''='' xảy ra khi x = 2
Vậy GTNN của f(x) bằng 5 tại x = 2