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\(a,-x^2-x+2\)
\(=-x^2-2x+x+2\)
\(=-\left(x^2+2x\right)+\left(x+2\right)\)
\(=-x\left(x+2\right)+\left(x+2\right)\)
\(=\left(-x+1\right)\left(x+2\right)\)
\(b,x^3-x^2-4x^2-8x-4\)
\(=\left(x^3-x^2\right)-\left(4x^2+8x+4\right)\)
\(=\left(x^3-x^2\right)-\left[\left(2x\right)^2+2.2x.2+2^2\right]\)
\(=x^2\left(x-1\right)-\left(2x+2\right)^2\)
\(=-x^2\left(x+1\right)-\left[2\left(x+1\right)\right]^2\)
\(=-x^2\left(x+1\right)-4\left(x+1\right)^2\)
\(=\left(x+1\right)\left[-x^2-4\left(x+1\right)\right]\)
\(=\left(x+1\right)\left(-x^2-4x-4\right)\)
\(=-\left(x+1\right)\left(x^2+4x+4\right)\)
\(=-\left(x+1\right)\left(x+2\right)^2\)
Q=x2 - 2xy + y2 - 12x + 12y + 36 + 5y2 + 10y + 5 + 1976
Q=(x - y)2 - 2.(x - y).6 + 62 + 5(y2 + 2y + 1) + 1976
Q=(x - y - 6)2 +5.(y + 1)2 + 1976 (≥ 1976 > 0 ∀ x,y ∈ R)
Vậy biểu thức Q luôn nhận giá trị dương với mọi số thực x,y
Trả lời :
Anh có Củi
Em có Lửa
Hai chúng Mình cùng Đốt
12 + 26 = 38
\(\left(5x-4\right)\left(2x+3\right)=10x^2+15x-8x-12=10x^2+7x-12\)
\(b,\frac{x-4}{x-2}+\frac{5x-8}{x-2}=\frac{x-4+5x-8}{x-2}=\frac{6\left(x-2\right)}{x-2}=6\)
\(c,\frac{x-9}{x^2-9}-\frac{3}{x^2+3x}=\frac{x-9}{\left(x+3\right)\left(x-3\right)}-\frac{3}{x\left(x+3\right)}\)
\(=\frac{x^2-9x}{x\left(x+3\right)\left(x-3\right)}-\frac{3x-9}{x\left(x+3\right)\left(x-3\right)}=\frac{x^2-9x-3x+9}{x\left(x+3\right)\left(x-3\right)}=\frac{x^2-6x+9}{x\left(x+3\right)\left(x-3\right)}\)
\(=\frac{x-3}{x\left(x+3\right)}\)
CÂU 1 :
a, ( 5x-4 ) ( 2x + 3 )
= 10x + 15x -8x -12
= 17x - 12
b, \(\frac{x-4}{x-2}\)+ \(\frac{5x-8}{x-2}\)
= \(\frac{x-4+5x-8}{x-2}\)
= \(\frac{6x-12}{x-2}\)
= \(\frac{6\left(x-2\right)}{x-2}\)
= 6
c, \(\frac{x-9}{x^2-9}\)- \(\frac{3}{x^2+3x}\)
= \(\frac{x-9}{\left(x-3\right)\left(x+3\right)}\)- \(\frac{3}{x\left(x+3\right)}\)
= \(\frac{\left(x-9\right).x}{x\left(x-3\right).\left(x+3\right)}\)- \(\frac{3.\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}\)
= \(\frac{x^2-9x}{x\left(x-3\right)\left(x+3\right)}\)- \(\frac{3x-9}{x\left(x-3\right)\left(x+3\right)}\)
= \(\frac{x^2-9x-3x+9}{x\left(x-3\right)\left(x+3\right)}\)
= \(\frac{x^2-12x+9}{x\left(x-3\right)\left(x+3\right)}\)
