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a) \(5x^2-2x\left(3x+\frac{3}{2}\right)=-x^2-3x=-x\left(x+3\right)=-3\left(3+3\right)=-18\)
b) \(3x\left(x-4y\right)-\frac{12}{5}y\left(y-5x\right)=3x^2-\frac{12}{5}y^2=3\left(x^2-\frac{4}{5}y^2\right)\)
\(=3\left(4^2-\frac{4}{5}.5^2\right)=3.\left(-4\right)=-12\)
c) \(\left(x-2\right)^2-\left(x+7\right)\left(x-7\right)=x^2-4x+4-x^2+49=-4x+53=-4.3+53=41\)
d) \(x^2+12x+36=\left(x+6\right)^2=\left(64+6\right)^2=70^2=4900\)
e) \(\left(x-3\right)^2-\left(x-4\right)\left(x+4\right)=x^2-6x+9-x^2+16=-6x+25=-6\left(-1\right)+25\)
= 31
f) \(\left(3x+2y\right)^2-4y\left(3x+y\right)=9x^2+12xy+4y^2-12xy-4y^2=9x^2=9\left(-\frac{1}{3}\right)^2=1\)
Bài 2: \(a,\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\)
\(\frac{5-3x}{x^2-9}=\frac{5-3x}{\left(x-3\right)\left(x+3\right)}=\frac{\left(5-3x\right)2x}{2x\left(x-3\right)\left(x+3\right)}\)
\(b,\frac{x+1}{x-x^2}=\frac{x+1}{x\left(1-x\right)}=-\frac{x+1}{x\left(x+1\right)}=-\frac{2\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)^2}\)
\(\frac{x+2}{2-4x+2x^2}=\frac{x+2}{2\left(x-1\right)^2}=\frac{2x\left(x+2\right)}{2x\left(x-1\right)^2}\)
\(c,\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{2x}{x^2+x+1}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{6}{x-1}=\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(d,\frac{7}{5x}=\frac{7.2\left(2y-x\right)\left(2y+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{4}{x-2y}=-\frac{4}{2y-x}=-\frac{4.2.5x\left(2x+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{2.5x.\left(2y-x\right)\left(2y+x\right)}\)
\(P=4\left(\frac{3}{4}x-1\right)+\left(12x^2-3x\right):\left(-3x\right)-\left(2x+1\right)\)
\(=4.\frac{3}{4}x-4.1+12x^2:\left(-3x\right)+\left(-3x\right):\left(-3x\right)-2x-1\)
\(=3x-4-4x+1-2x-1=-3x-4\)
Thay \(x=\frac{-4}{3}\)vào P ta được \(P=-3.\frac{-4}{3}-4=4-4=0\)
a) Áp dụng hằng đẳng thức số 3 bạn nhé
b) (2x + 3)(4x^2 - 6x +9) = 8x^3 + 9
Thay x= 120:2 = 60 vào biểu thức.
8* 60^3 + 9 = 1728009
c) = (2x + 1)^3
Thay x= -0,5 vào biểu thức
[2*(-0,5)+1]^3 = 0
d) = x^2 - 49 - x^2 - 2x - 1 = -50 - 2x
Thay x=49 vào biểu thức.
-50 - 2* 49 = -148
a, \(x\left(x+y\right)-y\left(x-y\right)=x^2 +xy-xy+y^2=x^2+y^2\)
b, \(\left(x-\frac{1}{2}\right)^3=x^3-\frac{3}{2}x^2+\frac{3}{4}x-\frac{1}{8}\)
c, \(\left(x+3\right)\left(x^2+9x-3x\right)-\left(37+x^3\right)\)
\(=x^3+9x^2-3x^2+3x^2+27x-9x-37-x^3=9x^2+18x-37\)
Bài 2 :
a, \(4x^4-8x^2y-12x^3y=4x^2\left(x^2-2y-3xy\right)\)
b, \(x^2+3x-xy-3y=x\left(x+3\right)-y\left(x+3\right)=\left(x-y\right)\left(x+3\right)\)
Sửa đề c, \(x^2-1+4\left(x+1\right)=\left(x-1\right)\left(x+1\right)+4\left(x+1\right)=\left(x-5\right)\left(x+1\right)\)
a, \(A=\left(100+50\right)^2=22500\)
b, \(B=\left(127+73\right)^2=40000\)
c, \(C=-6x+25\)Thay x = 100 ta có :
\(C=-6.100+25=-600+25=-575\)
\(A=100^2+200.50+50^2\)
\(=100^2+2.100.5+50^2\)
\(=\left(100+50\right)^2=150^2\)
\(B=127^2+146.127+73^2\)
\(=127^2+2.73.127+73^2\)
\(=\left(127+73\right)^2=200^2\)
1) \(8x^3+12x^2+6x+1=\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1^2+1^3\)
\(=\left(2x+1\right)^3=\left(2.-2+1\right)^3=-27\)
2) \(8x^3-12x+6x-1=\left(2x\right)^3-3.\left(2x\right)^2.1+3.2x.1^2-1^3\)
\(=\left(2x-1\right)^3=\left(2.-\frac{1}{2}-1\right)^3=-8\)
3)\(\left(1-2x\right)^2-\left(3x+1\right)^2=\left(1-2x+3x+1\right)\left(1-2x-3x-1\right)\)
\(=\left(x+2\right)\left(-5x\right)=\left(-2+2\right).\left(-5.-2\right)=0\)
4) \(\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)=\left(2x-3y\right)\left[\left(2x\right)^2+2x.3y+\left(3y\right)^2\right]\)
\(=\left(2x\right)^3-\left(3y\right)^3=\left(2.-\frac{1}{2}\right)^3-\left(3.-\frac{1}{3}\right)^3=-1-\left(-1\right)=0\)
1) Ta có : \(8x^3+12x^2+6x+1\)
\(=\left(2x+1\right)^3=\left(2.-2+1\right)^3=\left(-3\right)^3=-27\)
b) \(8x^3-12x^2+6x-1\)
\(=\left(2x-1\right)^3=\left[2.\left(-\frac{1}{2}\right)-1\right]^3=-8\)