Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có a-4ab=b. suy ra: \(a=b+4ab\)
Suy ra: \(P=\frac{b+4ab+b}{b+4ab-b}=\frac{2b+4ab}{4ab}=1+\frac{1}{2a}\)
a+b+c = 0 <=> (a+b+c)^2 = 0
<=> 2(ab+bc+ca) = 0 - (a^2+b^2+c^2) = 0 - 1 = -1
<=> ab+bc+ca = -1/2
<=> (ab+bc+ca)^2 = 1/4
<=> a^2b^2+b^2c^2+c^2a^2 = 1/4 - 2abc.(a+b+c) = 1/4 - 0 = 1/4
Có : a^2+b^2+c^2 = 1
<=> (a^2+b^2+c^2) = 1
<=> A = a^4+b^4+c^4 = 1 - 2.(a^2b^2+b^2c^2+c^2a^2) = 1 - 2.1/4 = 1/2
Vậy A = 1/2
k mk nha
\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ac\right)=2+2\left(ab+bc+ac\right)\)
=> \(0=2+2\left(ab+bc+ac\right)\)=> \(ab+bc+ca=-1\)
=> \(\left(ab+bc+ac\right)^2=1\)
Mà \(\left(ab+bc+ac\right)^2=a^2b^2+b^2c^2+a^2c^2+2\left(ab^2c+a^2bc+abc^2\right)\)
\(=a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=a^2b^2+b^2c^2+a^2c^2\)
=> \(a^2b^2+b^2c^2+c^2a^2=1\)
Mặt khác : \(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
=> \(a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
\(=4-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
=> \(a^4+b^4+c^4=4-2=2\)
a:
ĐKXĐ: x<>2
|2x-3|=1
=>\(\left[{}\begin{matrix}2x-3=1\\2x-3=-1\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=2\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)
Thay x=1 vào A, ta được:
\(A=\dfrac{1+1^2}{2-1}=\dfrac{2}{1}=2\)
b: ĐKXĐ: \(x\notin\left\{-1;2\right\}\)
\(B=\dfrac{2x}{x+1}+\dfrac{3}{x-2}-\dfrac{2x^2+1}{x^2-x-2}\)
\(=\dfrac{2x}{x+1}+\dfrac{3}{x-2}-\dfrac{2x^2+1}{\left(x-2\right)\left(x+1\right)}\)
\(=\dfrac{2x\left(x-2\right)+3\left(x+1\right)-2x^2-1}{\left(x+1\right)\left(x-2\right)}\)
\(=\dfrac{2x^2-4x+3x+3-2x^2-1}{\left(x+1\right)\left(x-2\right)}\)
\(=\dfrac{-x+2}{\left(x+1\right)\left(x-2\right)}=-\dfrac{1}{x+1}\)
c: \(P=A\cdot B=\dfrac{-1}{x+1}\cdot\dfrac{x\left(x+1\right)}{2-x}=\dfrac{x}{x-2}\)
\(=\dfrac{x-2+2}{x-2}=1+\dfrac{2}{x-2}\)
Để P lớn nhất thì \(\dfrac{2}{x-2}\) max
=>x-2=1
=>x=3(nhận)
a) Theo đầu bài ta có:
\(x+y=2\Rightarrow x=2-y\)
\(x^2+y^2=10\)
\(\Rightarrow\left(2-y\right)^2+y^2=10\)
\(\Rightarrow4+y^2-4y+y^2=10\)
\(\Rightarrow2y^2-4y=6\)
\(\Rightarrow2\left(y^2-2y\right)=6\)
\(\Rightarrow y\left(y-2\right)=3\)
Mà \(\hept{\begin{cases}y-\left(y-2\right)=2\\y+\left(y-2\right)=k\end{cases}\Rightarrow\hept{\begin{cases}y=\frac{k+2}{2}\\y-2=\frac{k-2}{2}\end{cases}}}\)( với k là hằng số )
\(\Rightarrow y\left(y-2\right)=\frac{k+2}{2}\cdot\frac{k-2}{2}\)
\(\Rightarrow\frac{\left(k+2\right)\left(k-2\right)}{4}=3\)
\(\Rightarrow k^2-4=12\)
\(\Rightarrow k^2=16\)
\(\Rightarrow k=4;-4\)
- Nếu k = 4 thì:
\(\Rightarrow\hept{\begin{cases}y=\frac{k+2}{2}=3\\x=2-y=-1\end{cases}\Rightarrow x^3+y^3=-1+27=26}\)
- Nếu k = -4 thì:
\(\Rightarrow\hept{\begin{cases}y=\frac{k+2}{2}=-1\\x=2-y=3\end{cases}\Rightarrow x^3+y^3=27+-1=26}\)
Vậy x3 + y3 = 26
a, \(x+y=2\Rightarrow\left(x+y\right)^2=4\Rightarrow x^2+2xy+y^2=4\Rightarrow10+2xy=4\Rightarrow xy=-3\)
\(\Rightarrow x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=2.13=26\)
vậy............
b, \(x+y=a\Rightarrow\left(x+y\right)^2=a^2\)
\(\Rightarrow x^2+2xy+y^2=a^2\)
\(\Rightarrow xy=\frac{a^2-b}{2}\)
\(\Rightarrow x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=a\left(b-\frac{a^2-b}{2}\right)=ab-\frac{a^3-ab}{2}\)
Vậy....
Ta có: a3+b3+c3=3abc <=> a3+b3+c3-3abc=0
<=>\(a^3+3a^2b+3ab^2+b^3+c^3-3ab\left(a+b\right)-3abc=0\)
<=>\(\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
<=>\(\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
<=>\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)
<=>\(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
Mà a+b+c khác 0
=>\(a^2+b^2+c^2-ab-bc-ca=0\)
<=>\(2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
<=>\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
<=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
<=>\(\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow}}a=b=c}\)
=>\(N=\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)
Ta có\(\left(a-b\right)^2=a^2+2ab+b^2\)
\(=\left(a^2+2ab+b^2\right)-4ab\)
\(=\left(a+b\right)^2-4ab\)
\(=81-80=1\)
Mà \(a< b\Rightarrow a-b=-1\)
\(\Rightarrow\left(a-b\right)^{2017}=-1\)
bài này sai