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Chỗ dấu bằng thứ hai sai nên bạn làm cũng chưa đúng
x^6 -y^6 = (x^2-y^2)(x^4 +x^2 .y^2 + y^4)
Bạn hiểu ra chỗ sai của mình chưa.Chúc bạn học tốt.
\(A=\left(x+y\right)^2-2xy=25-12=13\)
\(B=\left(x+y\right)\left(x^2+y^2-xy\right)=\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]=5\left(25-18\right)=35\)
\(C=x^2-y^2\Rightarrow C^2=x^4+y^4-2x^2y^2=\left(x^2+y^2\right)^2-4x^2y^2\)
\(C^2=\left[\left(x+y\right)^2-2xy\right]^2-4\left(xy\right)^2=\left(25-12\right)^2-4.36=25\Rightarrow C=\pm5\)
\(D=\frac{x^2+y^2}{xy}=\frac{\left(x+y\right)^2-2xy}{xy}=\frac{25-12}{6}=\frac{13}{6}\)
a)\(\dfrac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\dfrac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\dfrac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a-b+c\right)}=\dfrac{a+b-c}{a-b+c}\)Giá trị của biểu thức trên tại \(a=4;b=-5;c=6\) là:
\(\dfrac{4-5-6}{4-\left(-5\right)+6}=-\dfrac{7}{15}\)
b: \(=\dfrac{8x\left(2x-5y\right)}{8x\left(x-3y\right)}=\dfrac{2x-5y}{x-3y}\)
Đặt x/10=y/3=k
=>x=10k; y=3k
\(A=\dfrac{2\cdot10k-5\cdot3k}{10k-3\cdot3k}=\dfrac{5k}{k}=5\)
c: \(C=\left(\dfrac{x^3-y^3-x^3-y^3}{\left(x+y\right)\left(x-y\right)}\right):\dfrac{x^2-y^2-x^2}{x+y}\)
\(=\dfrac{-2y^3}{\left(x+y\right)\left(x-y\right)}\cdot\dfrac{x+y}{-y^2}=\dfrac{2y}{x-y}\)
\(=\dfrac{20}{9-10}=-20\)
Bài 1. Rút gọn:
\(a, x\left(1-x\right)+6\left(x+3\right)\left(x+3\right)\)
\(=x-x^2+6\left(x^2+6x+9\right)\)
\(=x-x^2+6x^2+36x+54\)
\(=5x^2+37x+54\)
\(b, \left(2-3x\right)\left(2+3x\right)-\left(x+5\right)\left(x-5\right)\)
\(=\left(4-9x^2\right)-\left(x^2-25\right)\)
\(=-10x^2+29\)
\(c, \left(3x+1\right)\left(x+5\right)-\left(x-1\right)\left(x+1\right)\)
\(=3x^2+15x+x+5-x^2+1\)
\(=2x^2+16x+6\)
\(d,\left(2-3x\right)\left(2x+3\right)+6\left(x-1\right)^2\)
\(=\left(4x+6-6x^2-9x\right)+6\left(x^2-2x+1\right)\)
\(=4x+6-6x^2-9x+6x^2-12x+6\)
\(=-17x+12\)
\(e, x\left(5-x\right)-\left(2x+2\right)\left(3x+2\right)-\left(x-2\right)\left(x+2\right)\)
\(=5x-x^2-\left(6x^2+4x+6x+4\right)-\left(x^2-4\right)\)
\(=5x-x^2-6x^2-4x-6x-4-x^2+4\)
\(=-8x^2-5x\)
Bài 2:
a: VT\(=x^3-xy+x^2y^2-y^3-x^3+y^3-x^2y^2\)
=-xy
b: \(VT=x^2+6xy+9y^2-x^2+9y^2-6xy=18y^2=VP\)
Lời giải:
\(B=x(x^2+xy+y^2)-y(y^2+xy+y^2)\)
\(=(x-y)(x^2+xy+y^2)=x^3-y^3=10^3-(-1)^3=1000-(-1)=1001\)
\(C=x^4+10x^3+10x^2+10\)
\(=x^4+9x^3+x^3+9x^2+x^2+10\)
\(=x^3(x+9)+x^2(x+9)+x^2+10\)
\(=(x+9)(x^3+x^2)+x^2+10\)
\(=(-9+9)[(-9)^3+(-9)^2]+(-9)^2+10\)
\(=0+(-9)^2+10=91\)
Thay $x=-1$ vào biểu thức:
\(D=x^2(x+y)-xy(x-y)-x(y^2+1)\)
\(=(-1)^2(x+y)-(-1)y(x-y)-(-1)(y^2+1)\)
\(=x+y+y(x-y)+(y^2+1)\)
\(=x+y+xy-y^2+y^2+1=x+y+xy+1\)
\(=(x+1)(y+1)=(-1+1)(y+1)=0\)
\(P=\left(x-y\right)^2+\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)-4x^2=\left(x-y-x-y\right)^2-\left(2x\right)^2=\left(-2y\right)^2-\left(2x\right)^2\)
\(=\left(2y-2x\right)\left(2y+2x\right)=2\left(y-x\right)2\left(y+x\right)=4\left(x+y\right)\left(y-x\right)\)
\(x^3-x^2y+3x-3y=x^2\left(x-y\right)+3\left(x-y\right)=\left(x-y\right)\left(x^2+3\right)\)
\(x^3-2x^2-4xy^2+x=x\left(x^2-2x+1-4y^2\right)=x\left[\left(x-1\right)^2-\left(2y\right)^2\right]=x\left(x+2y-1\right)\left(x-2y-1\right)\)
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\)
Đặt \(x^2+7x+10=t\), ta có:
\(t\left(t+2\right)-8=t^2+2t-8=t^2-2t+4t-8=t\left(t-2\right)+4\left(t-2\right)=\left(t-2\right)\left(t+4\right)\)
\(=\left(x^2+7x+10+4\right)\left(x^2+7x+10-2\right)=\left(x^2+7x+14\right)\left(x^2+7x-8\right)\)
