bỏ dấu ngoặc rồi viết gọn các biểu thức sau:
a) (a^2-ax+by)-(by-a^2-ax)
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Lời giải:
a)
$x(a-b)-x(a+b)=xa-xb-xa-xb=-2xb$
b)
$(a^2-ax+by)-(by-a^2-ax)=a^2-ax+by-by+a^2+ax=2a^2$
c)
$(a-b)(a+b)-(b-a)b=a^2-b^2-(b^2-ab)=a^2-b^2-b^2+ab=a^2-2b^2+ab$
\(a,x\left(a-b\right)-x\left(a+b\right)\)
\(=ax-bx-ax-bx=-2bx\)
\(b,\left(a^2-ax+by\right)-\left(by-a^2-ax\right)\)
\(=a^2-ax+by-by+a^2+ax=2a^2\)
\(c,\left(a-b\right)\left(a+b\right)-\left(b-a\right).b\)
\(=a^2-b^2-b^2+ab\)
\(=\left(a-b\right)^2-b^2=\left(a-b-b\right)\left(a-b+b\right)\)
\(=\left(a-2b\right).a\)
a) (m - n) (m + n) = m2 + mn - mn + n2 = m2 + n2
b) x(a - b) - x(a + b) = ax - bx - ax - bx = -2bx
c) (a2 - ax + by) - (by - a2 - ax) = a2 - ax + by - by + a2 + ax = 2a2
d) (a - b) (a + b) - (b - a)b = a2 + ab - ab - b2 - b2 + ab = a2 - 2b2 + ab
a)A = (a - b) – (a – b + c)
A=a-b-a+b-c
A=(a-a)+(b-b)-c
A=-c
b)B = (a + b + c) – (a + b - 5)
B=a+b+c-a-b+5
B=(a-a)+(b-b)+(c+5)
B=c+5
a) A = (a - b) - (a - b + c)
A = a - b - a + b - c
A = -c
b) B = (a + b + c) - (a + b - 5)
= a + b + c - a - b + 5
= c + 5
\(\left(a-b\right)\left(a+b\right)-\left(b-a\right)b\)
\(=a^2-ab+ab-b^2-b^2+ab\)
\(=a^2-b^2-b^2+ab\)
\(=a^2-2b^2+ab\)
\(=a\left(b+1\right)-2b^2\)
\(2x-2y=by+cz-cz-ax=by-ax\)
\(\Rightarrow2x-2y=by-ax\)
\(\Rightarrow2x+ax=2y+by\)
\(\Rightarrow x\left(a+2\right)=y\left(b+2\right)\)
\(\Rightarrow a+2=\dfrac{y\left(b+2\right)}{x}\)
\(2z-2y=ax+by-cz-ax=by-cz\)
\(\Rightarrow2z+cz=2y+by\)
\(\Rightarrow z\left(c+2\right)=y\left(b+2\right)\)
\(\Rightarrow c+2=\dfrac{y\left(b+2\right)}{z}\)
\(A=\dfrac{2}{a+2}+\dfrac{2}{b+2}+\dfrac{2}{c+2}=\dfrac{2}{\dfrac{y\left(b+2\right)}{x}}+\dfrac{2}{b+2}+\dfrac{2}{\dfrac{y\left(b+2\right)}{z}}=\dfrac{2x}{y\left(b+2\right)}+\dfrac{2}{b+2}+\dfrac{2z}{y\left(b+2\right)}=\dfrac{2x}{y\left(b+2\right)}+\dfrac{2y}{y\left(b+2\right)}+\dfrac{2z}{y\left(b+2\right)}=\dfrac{2x+2y+2z}{y\left(b+2\right)}=\dfrac{by+cz+cz+ax+ax+by}{by+2y}=\dfrac{2\left(ax+by+cz\right)}{by+cz+ax}=2\)
a,(a+b)-(a-b)+(a-c)-(a+c)
=a+b-a+b+a-c-a-c
=2b+2(-c)
b,(a+b-c)+(a-b+c)-(b+c-a)-(a-b-c)
=a+b-c+a-b+c-b-c+a-a+b+c
=2a
Bài 1:
a) Ta có: \(-ax\left(xy^3\right)^2\cdot\left(-by\right)^3\)
\(=-a\cdot x\cdot x^2\cdot y^6\cdot\left(-b\right)^3\cdot y^3\)
\(=abx^3y^9\)
b) Ta có: \(xy\cdot\left(-ax\right)^2\cdot\left(-by\right)^3\)
\(=xy\cdot a^2\cdot x^2\cdot b^3\cdot y^3\)
\(=a^2b^3x^3y^4\)
Bài 2:
Ta có: \(P\left(x\right)=5x-4x^4+x^6+3-2x^3-7x-x^7+1-2x^6+3x^3+x^7\)
\(=\left(-x^7+x^7\right)+\left(x^6-2x^6\right)-4x^4+\left(-2x^3+3x^3\right)+\left(5x-7x\right)+\left(3+1\right)\)
\(=-x^6-4x^4+x^3-2x+4\)
Phân tích mẫu :
\(M=bc\left(y-z\right)^2+ca\left(z-x\right)^2+ab\left(x-y\right)^2\)
Khai triển các bình phương và gom các nhân tử chung :
\(M=\left(ab+ac\right)x^2+\left(ab+bc\right)y^2+\left(bc+ac\right)z^2-2abxy-2bcxy-2acxy\)
\(=\left[\left(ab+ac\right)x^2+a^2x^2+\left(ab+bc\right)y^2+b^2y^2+\left(bc+ac\right)z^2+c^2z^2\right]-\)\(\left(a^2x^2+b^2y^2+c^2z^2+2ab+2aczx+2bcyz\right)\)
\(=\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)-\left(ax+by+cz\right)^2\)
\(=\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)\) ( vì \(ax+by+cz=0\) )
Kết quả : \(M=\frac{1}{a+b+c},a+b+c\ne0\)
Giải:
\(\left(a^2-ax+by\right)-\left(by-a^2-ax\right)\)
\(=a^2-ax+by-by+a^2+ax\)
\(=\left(a^2+a^2\right)+\left(-ax+ax\right)+\left(by-by\right)\)
\(=2a^2+0+0=2a^2\)
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