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\(a,\dfrac{2x+2y}{a^2+2ab+b^2}.\dfrac{ax-ay+bx-by}{2x^2-2y^2}\)
\(=\dfrac{2\left(x+y\right)}{\left(a+b\right)^2}.\dfrac{a\left(x-y\right)+b\left(x-y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2\left(x+y\right)}{\left(a+b\right)^2}.\dfrac{\left(x-y\right)\left(a+b\right)}{2\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{1}{a+b}\)
\(b,\dfrac{a+b-c}{a^2+2ab+b^2-c^2}.\dfrac{a^2+2ab+b^2+ac+bc}{a^2-b^2}\)
\(=\dfrac{a+b-c}{\left(a+b\right)^2-c^2}.\dfrac{\left(a+b\right)^2+c\left(a+b\right)}{\left(a-b\right)\left(a+b\right)}\)
\(=\dfrac{a+b-c}{\left(a+b-c\right)\left(a+b+c\right)}.\dfrac{\left(a+b\right)\left(a+b+c\right)}{\left(a-b\right)\left(a+b\right)}\)
\(=\dfrac{1}{a-b}\)
\(c,\dfrac{x^3+1}{x^2+2x+1}.\dfrac{x^2-1}{2x^2-2x+2}\)
\(=\dfrac{\left(x+1\right)\left(x^2-x+1\right)}{\left(x+1\right)^2}.\dfrac{\left(x-1\right)\left(x+1\right)}{2\left(x^2-x+1\right)}\) \(=\dfrac{x-1}{2}\) \(d,\dfrac{x^8-1}{x+1}.\dfrac{1}{\left(x^2+1\right)\left(x^4+1\right)}\) \(=\dfrac{\left(x^4\right)^2-1}{x+1}.\dfrac{1}{\left(x^2+1\right)\left(x^4+1\right)}\) \(=\dfrac{\left(x^4-1\right)\left(x^4+1\right)}{x+1}.\dfrac{1}{\left(x^2+1\right)\left(x^4+1\right)}\) \(=\dfrac{\left(x^2+1\right)\left(x^2-1\right)}{x+1}.\dfrac{1}{x^2+1}\) \(=\dfrac{\left(x-1\right)\left(x+1\right)}{x+1}\) \(=x-1\) \(e,\dfrac{x-y}{xy+y^2}-\dfrac{3x+y}{x^2-xy}.\dfrac{y-x}{x+y}\) \(=\dfrac{x-y}{y\left(x+y\right)}-\dfrac{3x+y}{x\left(x-y\right)}.\dfrac{-\left(x-y\right)}{x+y}\) \(=\dfrac{x-y}{y\left(x+y\right)}-\dfrac{3x+y}{x}.\dfrac{-1}{x+y}\) \(=\dfrac{x-y}{y\left(x+y\right)}-\dfrac{-3x-y}{x\left(x+y\right)}\) \(=\dfrac{x\left(x-y\right)+y\left(3x+y\right)}{xy\left(x+y\right)}\) \(=\dfrac{x^2-xy+3xy+y^2}{xy\left(x+y\right)}\) \(=\dfrac{x^2+2xy+y^2}{xy\left(x+y\right)}\) \(=\dfrac{\left(x+y\right)^2}{xy\left(x+y\right)}=\dfrac{x+y}{xy}\)tìm giá trị của m để pt 2x-m=1-x nhận giá trị x=-2 là nghiệm
giải hộ e với :)
\(a.\) \(\left(a-b\right)^2=\left(a+b\right)^2-4ab\)
\(\left(a-b\right)^2+2ab-2ab=\left(a+b\right)^2-4ab\)
\(\left(a-b\right)^2=a^2+2ab+b^2-4ab\)
\(\left(a-b\right)^2=a^2-2ab+b^2\)
\(\left(a-b\right)^2=\left(a-b\right)^2\)
Vậy \(\left(a-b\right)^2=\left(a+b\right)^2-4ab\)
Tương tự mấy câu kia
b: \(\left(a+b+c\right)^2+a^2+b^2+c^2\)
\(=2a^2+2b^2+2c^2+2ab+2bc+2ac\)
\(=\left(a^2+2ab+b^2\right)+\left(b^2+2bc+c^2\right)+\left(a^2+2ac+c^2\right)\)
\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(a+c\right)^2\)
c: \(x^4+y^4-2\left(x^2+xy+y^2\right)^2\)
\(=\left(x^2+y^2\right)^2-2x^2y^2-2\left[\left(x^2+y^2\right)^2+2xy\left(x^2+y^2\right)+x^2y^2\right]\)
\(=-\left(x^2+y^2\right)^2-4x^2y^2-4xy\left(x^2+y^2\right)\)
\(=-\left(x^2+2xy+y^2\right)^2=-\left(x+y\right)^4\)
=>\(x^4+y^4+\left(x+y\right)^4=2\left(x^2+xy+y^2\right)^2\)
\(a,VT=\left(a^2+b^2\right)\left(c^2+d^2\right)=a^2c^2+b^2c^2+a^2d^2+b^2d^2\)
\(VP=\left(ac+bd\right)^2+\left(ad-bc\right)^2=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2=a^2c^2+b^2c^2+a^2d^2+b^2d^2\)
\(\Rightarrow VT=a^2c^2+b^2c^2+a^2d^2+b^2d^2=VP\left(đpcm\right)\)
b, Tham khảo:Chứng minh hằng đẳng thức:(a+b+c)3= a3 + b3 + c3 + 3(a+b)(b+c)(c+a) - Hoc24
Lời giải:
\((5a-3b+8c)(5a-3b-8c)=(5a-3b)^2-(8c)^2\)
\(=25a^2+9b^2-30ab-(8c)^2\)
\(=(9a^2+25b^2-30ab)+(16a^2-16b^2)-64c^2\)
\(=(3a-5b)^2+16.4c^2-64c^2\)
\(=(3a-5b)^2\)
a. Biến đổi vế phải, ta có:
(a+b)2- 4ab
= a2+2ab+b2-4ab
=a2+2ab-4ab+b2
= a2-2ab+b2
= (a-b)2
Vậy: ( a - b )2 = ( a + b )2 - 4ab
Mik chỉ làm đc câu a thui àk