Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
3b. Để A=\(\frac{4x^3-6x^2+8x}{2x-1}\) \(\in\)Z => 2x2-2x+3+\(\frac{3}{2x-1}\)\(\in\)Z =>\(\frac{3}{2x-1}\) \(\in\)Z
=> 2x-1 \(\in\)Ư(3)={\(\pm\)1,\(\pm\)3}
=> \(\left[{}\begin{matrix}2x-1=1\\2x-1=-1\\2x-1=3\\2x-1=-3\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=1\\x=0\\x=2\\x=-1\end{matrix}\right.\)(tm)
a) Ta có:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(\Rightarrow\frac{ab}{abc}+\frac{bc}{abc}+\frac{ac}{abc}=0\)
\(\Rightarrow\frac{ab+bc+ac}{abc}=0\)
\(\Rightarrow ab+bc+ac=0\)
Ta lại có:
\(a+b+c=1\)
\(\Rightarrow\left(a+b+c\right)^2=1\)
\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=1\)
=> Đpcm
Câu 1:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\) => ab + bc + ca = abc
=> (ab + bc + ca)(a+b+c) = abc (do a+b+c = 1)
=> \(a^2b+ac^2+a^2c+b^2c+ab^2+bc^2+2abc=0\)
=> ab(a+c) + ac(a+c) + \(b^2\left(a+c\right)\) + bc(c+a) = 0
=> (a+b)(b+c)(c+a) = 0
\(1.\)
\(a,\left(a+b\right)^2=a^2+2ab+b^2\)
\(\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab=a^2+2ab+b^2\)
\(\Rightarrow\left(a+b\right)^2=\left(a-b\right)^2+4ab\left(đpcm\right)\)
a) \(x^2+x+1=x^2+x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)(luôn dương)
b) \(x^2-x+\frac{1}{2}=x^2-x+\frac{1}{4}+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2+\frac{1}{4}>0\)(luôn dương)
Bài 3:
ĐKXĐ: $x\neq 0; x\neq 3$
\(\frac{x^2+1}{x^2-3x}+\frac{3}{x}-\frac{x}{x-3}=\frac{x^2+1}{x(x-3)}+\frac{3(x-3)}{x(x-3)}-\frac{x^2}{x(x-3)}\)
\(=\frac{x^2+1+3(x-3)-x^2}{x(x-3)}=\frac{3x-8}{x(x-3)}\)
Bài 2:
$(a-b)^2=a^2+b^2-2ab=(a^2+b^2+2ab)-4ab=(a+b)^2-4ab$
$=7^2-4.3=37$
1)\(4\left(a^4-1\right)x=5\left(a-1\right)\)
<=>x=\(\frac{5\left(a-1\right)}{a^4-1}\)
<=>x=\(\frac{5\left(a-1\right)}{\left(a-1\right)\left(a+1\right)\left(a^2+1\right)}=\frac{5}{\left(a+1\right)\left(a^2+1\right)}\)
Tương tự ta tính được y=\(\frac{4a^6+4}{5a^4-5a^2+5}\)
Suy ra x.y=\(\frac{5}{\left(a+1\right)\left(a^2+1\right)}.\frac{4\cdot\left(a^6+1\right)}{5\left(a^4-a^2+1\right)}\)=\(\frac{5}{\left(a+1\right)\left(a^2+1\right)}.\frac{4\left(a^2+1\right)\left(a^4-a^2+1\right)}{5\left(a^4-a^2+1\right)}\)
=\(\frac{5}{a+1}\)
Tương tự với x:y
\(A=\frac{4.6}{4.2}:\left(\frac{8.10}{6.8}.\frac{12.14}{10.12}.\frac{16.18}{14.16}...\frac{54.56}{54.53}\right)=\frac{6}{2}:\frac{56}{6}=\)
1) \(\left(a+b\right)^3=\left(a+b\right)\left(a+b\right)^2=\left(a+b\right)\left(a^2+2ab+b^2\right)\)
\(=a^3+2a^2b+ab^2+a^2b+2ab^2+b^3\)
\(=a^3+3a^2b+3ab^2+b^3\)
2) \(\left(a-b\right)^3=\left(a-b\right)\left(a-b\right)^2=\left(a-b\right)\left(a^2-2ab+b^2\right)\)\(=a^3-2a^2b+ab^2-a^2b+2ab^2-b^3\)
\(=a^3-3a^2b+3ab^2-b^3\)
a) b 3 + 3 b 2 + 2 b 3 + 1 . b) 0.