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a)a+b+c=9
=>(a+b+c)2=81
=>a2+b2+c2+2ab+2bc+2ca=81
Từ a2+b2+c2=141=>2ab+2bc+2ca=81-141=-60
=>2(ab+bc+ca)=-60=>ab+bc+ca=-30
b)x+y=1
=>(x+y)3=1
=>x3+3x2y+3xy2+y3=1
=>x3+y3+3xy(x+y)=1
=>x3+y3+3xy=1(Do x+y=1)
c)a3-3ab+2c=(x+y)3-3(x+y)(x2+y2)+2(x3+y3)
=x3+3x2y+3xy2+y3-3x3-3y3-3x2y-3xy2+2x3+2y3=0
d)đang tìm hướng giải
a, ĐKXĐ: \(x\ne1;x\ne-1\)
b, Với \(x\ne1;x\ne-1\)
\(B=\left[\dfrac{x+1}{2\left(x-1\right)}+\dfrac{3}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+3}{2\left(x+1\right)}\right]\cdot\dfrac{4\left(x^2-1\right)}{5}\\ =\left[\dfrac{x^2+2x+1+6-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}\right]\cdot\dfrac{4\left(x^2-1\right)}{5}\\ =\dfrac{5}{x^2-1}\cdot\dfrac{4\left(x^2-1\right)}{5}\\ =4\)
=> ĐPCM
bài 1 : a. x^3 +27 -54-x^3 =-27
b. 8x^3 +y^3 -8x^3 +y^3 =2y^3
c. (2x-1+2x+2)(2x-1-2x-2)=(4x+1).(-3)=-12x-3
d. a^3 +b^3 +3ab(a+b) -3ab(a+b)=a^3+b^3
\(ab\left(x-y\right)^3-8ab=ab\left[\left(x-y\right)^3-2^3\right]=ab\left(x-y-2\right)\left[\left(x-y\right)^2+2\left(x-y\right)+4\right]\)
\(36x^2-y^2+6y-9=36x^2-\left(y-3\right)^2=\left(6x-y+3\right)\left(6x+y-3\right)\)
\(8x^2+10x-3=0\)
\(8x^2-2x+12x-3=0\)
\(2x\left(4x-1\right)+3\left(4x-1\right)=0\)
\(\left(4x-1\right)\left(2x+3\right)=0\)
\(\left[\begin{array}{nghiempt}4x-1=0\\2x+3=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}4x=1\\2x=-3\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=\frac{1}{4}\\x=-\frac{3}{2}\end{array}\right.\)
\(\left(2x-5\right)^2-\left(x+4\right)^2=0\)
\(\left(2x-5+x+4\right)\left(2x-5-x-4\right)=0\)
\(\left(3x-1\right)\left(x-9\right)=0\)
\(\left[\begin{array}{nghiempt}3x-1=0\\x-9=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=\frac{1}{3}\\x=9\end{array}\right.\)
a: \(A=\left(x+1\right)\left(x-2\right)-x\left(2x-3\right)+2x^2+4\)
\(=x^2-x-2-2x^2+3x+2x^2+4\)
\(=x^2+2x+2\)
\(a,A=x^2-x-2-2x^2+3x+4+2x^2=x^2+2x+2\\ c,A=\left(x^2+2x+1\right)+1=\left(x+1\right)^2+1\ge1>0\)