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b: Ta có: \(N=a^3+b^3+3ab\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\)
\(=1-3ab+3ab\)
=1
a: Thay x=-3 vào B, ta được:
\(B=\dfrac{2\cdot\left(-3\right)^2}{3\cdot\left(-3\right)+6}=\dfrac{2\cdot9}{-9+6}=\dfrac{18}{-3}=-6\)
b: \(A=\dfrac{2x^2+20+3x-6-7x-14}{\left(x+2\right)\left(x-2\right)}=\dfrac{2x^2-4x}{\left(x+2\right)\left(x-2\right)}=\dfrac{2x}{x+2}\)
Ta có:\(\left(a-b+c\right)^2+\left(a-b+c\right)^2-2\left(b-c\right)^2\)
\(=2\left(a-b+c\right)^2-2\left(b-c\right)^2\\ =2\left(\left(a-b+c\right)^2-\left(b-c\right)^2\right)\)
\(=2\left(a-b+c-b+c\right)\left(a-b+c+b-c\right)\\ =2\left(a-2b+2c\right)a \)
\(=2a^2-4ab+4ac\)
Q = (1 - 2x)(x - 3)
= x - 3 - 2x2 + 6x
= - 2x2 + 5x - 3
= \(-2\left(x^2-\frac{5}{2}x+3\right)=-2\left(x^2-2.\frac{5}{4}.x+\frac{25}{16}+\frac{23}{16}\right)=-2\left(x-\frac{5}{4}\right)^2-\frac{23}{8}\le-\frac{23}{8}\)
Dấu "=" xảy ra <=> x - 5/4 = 0
=> x = 1,25
Vậy Max Q = -23/8 <=> x = 1,25
Q = ( 1 - 2x )( x - 3 )
= x - 3 - 2x2 + 6x
= -2x2 + 7x - 3
= -2( x2 - 7/2x + 49/16 ) + 25/8
= -2( x - 7/4 )2 + 25/8 ≤ 25/8 ∀ x
Đẳng thức xảy ra <=> x - 7/4 = 0 => x = 7/4
=> MaxQ = 25/8 <=> x = 7/4
a: Ta có: \(\left(a^2-1\right)^3-\left(a^4+a^2+1\right)\left(a^2-1\right)\)
\(=a^6-3a^4+3a^2-1-\left(a^6-1\right)\)
\(=-3a^4+3a^2\)
b: Ta có: \(\left(a^4-3a^2+9\right)\left(a^2+3\right)-\left(a^2+3\right)^3\)
\(=a^6+27-a^6-9a^4-27a^2-27\)
\(=-9a^4-27a^2\)
\(a,=\left(x+8-x+2\right)^2=10^2=100\\ b,=x^2\left(x^2-16\right)-\left(x^4-1\right)=x^4-16x^2-x^4+1=1-16x^2\\ c,=x^3+1-x^3+1=2\)
\(\left(36-6x+x^2\right)\left(6+x\right)\)
= \(216+36x-36x-6x^2+6x^2+x^3\)
= \(x^3+216\)
1.\(\left(\frac{9}{x^3-9x}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3x+9}\right)=\left(\frac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}\right):\left(\frac{3x-9-x^2}{3x\left(x+3\right)}\right)=-\frac{1}{x-3}\)
2.\(\left(\frac{2\left(x+2\right)-2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\right).\frac{\left(x+2\right)^2}{8}=\frac{4}{\left(x-2\right)\left(x+2\right)}.\frac{\left(x+2\right)^2}{8}=\frac{x-2}{2}\)
3.\(\left(\frac{3\left(3x+1\right)+2x\left(1-3x\right)}{\left(1-3x\right)\left(3x+1\right)}\right):\frac{2x\left(x+5\right)}{\left(1-3x\right)^2}=\frac{-6x^2+11x+3}{\left(1-3x\right)\left(3x+1\right)}.\frac{\left(1-3x\right)^2}{2x\left(x+5\right)}=\frac{-6x^2+11x+3}{\left(3x+1\right)}.\frac{\left(1-3x\right)}{2x\left(x+5\right)}\)
4.\(\left(\frac{x^2-\left(x-5\right)^2}{x\left(x-5\right)\left(x+5\right)}\right):\frac{2x-5}{x\left(x+5\right)}+\frac{x}{5-x}=\frac{10x-25}{x\left(x-5\right)\left(x+5\right)}.\frac{x\left(x+5\right)}{2x-5}+\frac{x}{5-x}=\frac{5}{x-5}-\frac{x}{x-5}=-1\)