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a) (25x5 – 5x4 + 10x2) : 5x2 = (25x5 : 5x2 ) - (5x4 : 5x2 ) + (10x2 : 5x2 )
= 5x3 – x2 + 2
b) (15x3y2 – 6x2y – 3x2y2) : 6x2y
= (15x3y2 : 6x2y) + (– 6x2y : 6x2y) + (– 3x2y2 : 6x2y)
= \(\dfrac{15}{6}\)xy - 1 - \(\dfrac{3}{6}\)y = \(\dfrac{5}{2}\)xy - \(\dfrac{1}{2}\)y - 1.
a) \(25x^2-10x+3=25x^2-10x+1+2\)
\(=\left(5x-1\right)^2+2\)
Vì \(\left(5x-1\right)^2\ge0\forall x\)
Nên \(\left(5x-1\right)^2+2>0\forall x\)
Vậy biểu thức luôn lớn hơn 0 với mọi giá trị x.
b) \(y^2-y+2=y^2-y+\dfrac{1}{4}+\dfrac{7}{4}\)
\(=\left(y-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\)
Vì \(\left(y-\dfrac{1}{2}\right)^2\ge0\forall x\)
Nên \(\left(y-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>0\forall x\)
Vậy biểu thức luôn lớn hơn 0 với mọi giá trị x.
c) \(y^2-3y+5=y^2-3y+\dfrac{9}{4}+\dfrac{11}{4}\)
\(=\left(y-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\)
Vì \(\left(y-\dfrac{3}{2}\right)^2\ge0\forall x\)
Nên \(\left(y-\dfrac{3}{2}\right)^2+\dfrac{11}{4}>0\forall x\)
Vậy biểu thức luôn lớn hơn 0 với mọi giá trị x.
d) \(16y^2-6y+9=16y^2-6y+\dfrac{9}{16}+\dfrac{135}{16}\)
\(=\left(4x-\dfrac{3}{4}\right)^2+\dfrac{135}{16}\)
Vì \(\left(4x-\dfrac{3}{4}\right)^2\ge0\forall x\)
Nên \(\left(4x-\dfrac{3}{4}\right)^2+\dfrac{135}{16}>0\forall x\)
Vậy biểu thức luôn lớn hơn 0 với mọi giá trị x.
a,
\(25x^2-10x+3\\ =\left(5x\right)^2-10x+1+2\\ =\left(5x-1\right)^2+2\\ \left(5x-1\right)^2\ge0\forall x\\ \Rightarrow\left(5x-1\right)^2+2\ge2\forall x\\ \Rightarrow\left(5x-1\right)^2+2>0\forall x\)
b,
\(y^2-y+2\\ =y^2-y+\dfrac{1}{4}+\dfrac{7}{4}\\ =\left(y-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\\ \left(y-\dfrac{1}{2}\right)^2\ge0\forall y\\ \Rightarrow\left(y-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\forall y\\ \Rightarrow\left(y-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>0\forall y\)
c,
\(y^2-3y+5\\ =y^2-3y+\dfrac{9}{4}+\dfrac{11}{4}\\ =\left(y-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\\ \left(y-\dfrac{3}{2}\right)^2\ge0\forall y\\ \Rightarrow\left(y-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\forall y\\ \Rightarrow\left(y-\dfrac{3}{2}\right)^2+\dfrac{11}{4}>0\forall y\)
d,
\(16y^2-6y+9\\ =\left(4y\right)^2-6y+\dfrac{9}{16}+\dfrac{135}{16}\\ =\left(4y-\dfrac{3}{4}\right)^2+\dfrac{135}{16}\\ \left(4y-\dfrac{3}{4}\right)^2\ge0\forall y\\ \Rightarrow\left(4y-\dfrac{3}{4}\right)^2+\dfrac{135}{16}\ge\dfrac{135}{16}\forall y\\ \Rightarrow\left(4y-\dfrac{3}{4}\right)^2+\dfrac{135}{16}>0\forall y\)
c) x^3+3x^2-3x-1=(x^3-1)+(3x^2-3x)=(x-1)(x^2+x+1)+3x(x-1)=(x-1)(x^2+4x+1)
\(-25x^6-y^8+10x^3y^4=-\left(5x^3\right)^2-\left(y^4\right)^2+10x^3y^4=-[5x^3-10x^34^4+y^{^{ }4}]=-\left(5y^3-y^4\right)\)
Xét:
\(\left(3x-2y\right)\left(25x^2-9y^2\right)\)
\(=\left(3x-2y\right)\left(5x-3y\right)\left(5x+3y\right)\)
\(=\left(5x-3y\right)\left(15x^2+9xy-10xy-6y^2\right)\)
\(=\left(5x-3y\right)\left(15x^2-xy-6y^2\right)\)
Từ đó dễ dàng suy ra tích chéo = nhau => đpcm
ta có : \(VP=\dfrac{15x^2-xy-6y^2}{25x^2-9y^2}=\dfrac{\left(3x-2y\right)\left(5x+3y\right)}{\left(5x-3y\right)\left(5x+3y\right)}=\dfrac{3x-2y}{5x-3y}=VT\)
ta có: