Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(=\dfrac{\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc}{a^2+b^2+c^2-ab-ac-bc}\)
\(=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)}{a^2+b^2+c^2-ab-ac-bc}\)
=a+b+c
e: \(=\dfrac{a^2b-a^2c+b^2c-b^2a+c^2\left(a-b\right)}{a\left(b^2-c^2\right)-b\left(b^2-c^2\right)}\)
\(=\dfrac{ab\left(a-b\right)+c\left(b-a\right)\left(b+a\right)+c^2\left(a-b\right)}{\left(b-c\right)\left(b+c\right)\left(a-b\right)}\)
\(=\dfrac{\left(a-b\right)\left(ab-ac-bc+c^2\right)}{\left(b-c\right)\left(b+c\right)\left(a-b\right)}\)
\(=\dfrac{a\left(b-c\right)-c\left(b-c\right)}{\left(b-c\right)\left(b+c\right)}=\dfrac{a-c}{b+c}\)
a) \(A=\left(\frac{1}{4}x-y\right)\left(x^2+4xy+16y^2\right)+4\left(4y^3-\frac{1}{16}x^3+1\right)\)
\(\Leftrightarrow A=\frac{1}{4}\left(x-4y\right)\left(x^2+4xy+16y^2\right)+16y^3-\frac{1}{4}x^3+4\)
\(\Leftrightarrow A=\frac{1}{4}\left(x^3-64y^3\right)+16y^3-\frac{1}{4}x^3+4\)
\(\Leftrightarrow A=\frac{1}{4}x^3-16y^3+16y^3-\frac{1}{4}x^3+4\)
\(\Leftrightarrow A=4\)
b) \(B=2x\left(x-4\right)^2-\left(x+5\right)\left(x-2\right)\left(x+2\right)+2\left(x-5\right)^2-\left(x-1\right)^2\)
\(\Leftrightarrow B=2x\left(x^2-8x+16\right)-\left(x+5\right)\left(x^2-4\right)+2\left(x^2-10x+25\right)-\left(x^2-2x+1\right)\)
\(\Leftrightarrow B=2x^3-16x^2+32x-x^3-5x^2+4x+20+2x^2-20x+50-x^2+2x-1\)
\(\Leftrightarrow B=x^3-20x^2+18x+69\)
c) \(C=\frac{80x^3-125x}{3\left(x-3\right)-\left(x-3\right)\left(8-4x\right)}\)
\(\Leftrightarrow C=\frac{5x\left(16x^2-25\right)}{\left(x-3\right)\left(3-8+4x\right)}\)
\(\Leftrightarrow C=\frac{5x\left(4x-5\right)\left(4x+5\right)}{\left(x-3\right)\left(4x-5\right)}\)
\(\Leftrightarrow C=\frac{5x\left(4x+5\right)}{x-3}\)
\(\Leftrightarrow C=\frac{20x^2+25x}{x-3}\)
d) \(D=\frac{\left(a-b\right)\left(c-d\right)}{\left(b^2-a^2\right)\left(d^2-c^2\right)}\)
\(\Leftrightarrow D=\frac{\left(a-b\right)\left(c-d\right)}{\left(a^2-b^2\right)\left(c^2-d^2\right)}\)
\(\Leftrightarrow D=\frac{\left(a-b\right)\left(c-d\right)}{\left(a-b\right)\left(a+b\right)\left(c-d\right)\left(c+d\right)}\)
\(\Leftrightarrow D=\frac{1}{\left(a+b\right)\left(c+d\right)}\)
Chúc bạn học tốt !
Đặt \(b-c=x,c-a=y,a-b=z\)
\(\Rightarrow x+y+z=0\Rightarrow x^3+y^3+z^3=3xyz\)
\(\Rightarrow\left(b-c\right)^3+\left(c-a\right)^3+\left(a-b\right)^3=3\left(b-c\right)\left(c-a\right)\left(a-b\right)\)(1)
Ta có:
: \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)+b^2\left(c-b+b-a\right)+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)+b^2\left(c-b\right)+b^2\left(b-a\right)+c^2\left(a-b\right)\)
\(=\left(b-c\right)\left(a^2-b^2\right)+\left(a-b\right)\left(c^2-b^2\right)\)
\(=\left(b-c\right)\left(a-b\right)\left(a+b\right)+\left(a-b\right)\left(c-b\right)\left(c+b\right)\)
\(=\left(b-c\right)\left(a-b\right)\left(a+b-c-b\right)\)
\(=\left(b-c\right)\left(a-b\right)\left(a-c\right)\)(2)
Từ (1) và (2) giá trị biểu thức cần tìm là -3.
Chúc bạn học tốt
1a)
Đặt \(a^2+a+1=t\Rightarrow a^2+a+2=t+1\)
\(\Rightarrow A=t\left(t+1\right)-12=t^2+t-12=t^2-3t+4t-12=\left(t-3\right)\left(t+4\right)\)
\(=\left(a^2+a-2\right)\left(a^2+a+5\right)\)
Mà \(a>1\Rightarrow\hept{\begin{cases}a^2+a-2>0\\a^2+a+5>0\end{cases}}\forall a>1\)
Vậy A là hợp số
1b)
Ta có :
\(B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\cdot...\cdot\left(2^{1006}+1\right)+1\)
\(=\left(2^2-1\right)\left(2^2+1\right)\cdot...\cdot\left(2^{1006}+1\right)+1=....=\left(2^{1006}-1\right)\left(2^{1006}+1\right)+1\)
\(=2^{2012}-1+1=2^{2012}\)
a) \(\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{ab^2-ac^2-b^3+bc^2}\)
\(=\frac{a^2b-a^2c+b^2c-b^2a+c^2\left(a-b\right)}{ab^2-b^3-ac^2+bc^2}\)
\(=\frac{\left(a^2b-b^2a\right)+\left(b^2c-a^2c\right)+c^2\left(a-b\right)}{b^2\left(a-b\right)-c^2\left(a-b\right)}\)
\(=\frac{ab\left(a-b\right)+c\left(b^2-a^2\right)+c^2\left(a-b\right)}{\left(b^2-c^2\right)\left(a-b\right)}\)
\(=\frac{ab\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+c^2\left(a-b\right)}{\left(b-c\right)\left(b+c\right)\left(a-b\right)}\)
\(=\frac{ab-c\left(a+b\right)+c^2}{\left(b-c\right)\left(b+c\right)}\)
\(=\frac{ab-ac+c^2-bc}{\left(b-c\right)\left(b+c\right)}\)
\(=\frac{a\left(b-c\right)-c\left(b-c\right)}{\left(b-c\right)\left(b+c\right)}\)
\(=\frac{\left(b-c\right)\left(a-c\right)}{\left(b-c\right)\left(b+c\right)}\)
\(=\frac{a-b}{b+c}\)
Sửa lại: \(\frac{a-c}{b+c}\)