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Bài 8:
Ta có: \(A=-x^2+2x+4\)
\(=-\left(x^2-2x-4\right)\)
\(=-\left(x^2-2x+1-5\right)\)
\(=-\left(x-1\right)^2+5\le5\forall x\)
Dấu '=' xảy ra khi x=1
a: \(P\left(x\right)=-5x^3+3x^2+2x+5\)
\(Q\left(x\right)=-5x^3+6x^2+x+5\)
b: \(H\left(x\right)=Q\left(x\right)+P\left(x\right)=-10x^3+9x^2+3x+10\)
Khi x=1/2 thì \(H\left(x\right)=-10\cdot\dfrac{1}{8}+\dfrac{9}{4}+\dfrac{3}{2}+10=\dfrac{25}{2}\)
Ta có : x2 + 3x
= x2 + \(2.x.\frac{3}{2}+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2\)
\(=\left(x+\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2\)
này mình có vài câu không làm được, xin lỗi bạn nha
\(b,16x^2-8x+1=\left(4x-1\right)^2\\ c,4x^2+12xy+9y^2=\left(2x+3y\right)^2\\ e,=x^2+2x+1+y^2+2y+1+2\left(x+1\right)\left(y+1\right)\\ =\left(x+1\right)^2+2\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2\\ =\left[\left(x+1\right)+\left(y+1\right)\right]^2=\left(x+y+2\right)^2\\ g,=x^2-2x\left(y+2\right)+\left(x+2\right)^2=\left[x-\left(y+2\right)\right]^2=\left(x-y-2\right)^2\\ h,=\left[x+\left(y+1\right)\right]^2=\left(x+y+1\right)^2\)
a: \(P\left(x\right)=-5x^3+3x^2+2x+5\)
\(Q\left(x\right)=-5x^3+6x^2+2x+5\)
b: \(H\left(x\right)=P\left(x\right)+Q\left(x\right)=-10x^3+9x^2+4x+10\)
\(H\left(\dfrac{1}{2}\right)=-10\cdot\dfrac{1}{8}+\dfrac{9}{4}+2+10=13\)
c: Q(x)-P(x)=6
\(\Leftrightarrow3x^2=6\)
hay \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)
1, \(x^2+2xy+y^2=\left(x+y\right)^2\)
2, \(4x^2+12x+9=\left(2x\right)^2+2\cdot3\cdot2x+3^2=\left(2x+3\right)^2\)
3, \(x^2+5x+\dfrac{25}{4}=x^2+2\cdot\dfrac{5}{2}\cdot x+\left(\dfrac{5}{2}\right)^2=\left(x+\dfrac{5}{2}\right)^2\)
4, \(16x^2-8x+1=\left(4x\right)^2-2\cdot4x\cdot1+1^2=\left(4x-1\right)^2\)
5, \(x^2+x+\dfrac{1}{4}=x^2+2\cdot\dfrac{1}{2}\cdot x+\left(\dfrac{1}{2}\right)^2=\left(x+\dfrac{1}{2}\right)^2\)
1: =(x+y)^2
2: =(2x+3)^2
3: =(x+5/2)^2
4: =(4x-1)^2
5: =(x+1/2)^2
6: =(x-3/2)^2
7: =(x+1)^3
8: =(1/2x+1)^2
9: =(3y-1/3)^3
10: =(2x+y)^3
1,\(\dfrac{x^2-6x+9}{x^2-8x+15}=\dfrac{\left(x-3\right)^2}{\left(x-3\right).\left(x-5\right)}=\dfrac{x-3}{x-5}\)
2,\(\dfrac{x^2+5x}{2x+10}=\dfrac{x.\left(x+5\right)}{2.\left(x+5\right)}=\dfrac{x}{2}\)
3,\(\dfrac{25-10x+x^2}{xy-5y}=\dfrac{\left(x-5\right)^2}{y.\left(x-5\right)}=\dfrac{x-5}{y}\)
4,\(\dfrac{x^2+3x-y^2-3y}{x^2-y^2}\\ \\ =\dfrac{\left(x+y\right).\left(x-y\right)+3.\left(x-y\right)}{\left(x-y\right).\left(x+y\right)}\\ \\ =\dfrac{\left(x-y\right).\left(x+y+3\right)}{\left(x-y\right).\left(x+y\right)}\\ \\ =\dfrac{x+y+3}{x+y}\)5,\(\dfrac{x^3+2x^2-x-2}{x^3-3x+2}=\dfrac{x^2.\left(x+2\right)-\left(x+2\right)}{x.\left(x^2-1\right)-2.\left(x-1\right)}\\ \\ \dfrac{\left(x+2\right).\left(x^2-1\right)}{x.\left(x+1\right).\left(x-1\right)-2.\left(x-1\right)}\\ =\dfrac{\left(x+2\right).\left(x+1\right).\left(x-1\right)}{\left(x-1\right).\left[\left(x+1\right).x-2\right]}=\dfrac{\left(x+2\right).\left(x+1\right)}{\left(x+1\right).x-2}\)