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a: Thay x=-3 vào A, ta được:
\(A=\frac{-3+2}{-3}=\frac{-1}{-3}=\frac13\)
\(x=\sqrt{\left(-3\right)^2}=\sqrt9=3\)
Thay x=3 vào A, ta được:
\(A=\frac{3+2}{3}=\frac53\)
b: \(B=\frac{3}{x+5}+\frac{20-2x}{x^2-25}\)
\(=\frac{3}{x+5}+\frac{20-2x}{\left(x+5\right)\left(x-5\right)}\)
\(=\frac{3\left(x-5\right)+20-2x}{\left(x+5\right)\left(x-5\right)}=\frac{3x-15+20-2x}{\left(x+5\right)\left(x-5\right)}=\frac{x+5}{\left(x+5\right)\left(x-5\right)}\)
\(=\frac{1}{x-5}\)
c: \(A=B\cdot\left|x-4\right|\)
=>\(\frac{x+2}{x}:\frac{1}{x-5}=\left|x-4\right|\)
=>\(\frac{\left(x+2\right)\left(x-5\right)}{x}=\left|x-4\right|\)
=>\(\begin{cases}\frac{\left(x+2\right)\left(x-5\right)}{x}\ge0\\ \left(x+2\right)^2\cdot\frac{\left(x-5\right)^2}{x^2}=\left(x-4\right)^2\end{cases}\Rightarrow\begin{cases}\left[\begin{array}{l}-2\le x<0\\ x\ge5\end{array}\right.\\ \left(x+2\right)^2\cdot\left(x-5\right)^2=x^2\cdot\left(x-4\right)^2\end{cases}\)
Ta có: \(\left(x+2\right)^2\cdot\left(x-5\right)^2=x^2\cdot\left(x-4\right)^2\)
=>\(\left(x^2-3x-10\right)^2=\left(x^2-4x\right)^2\)
=>\(\left(x^2-4x-x^2+3x+10\right)\left(x^2-4x+x^2-3x-10\right)=0\)
=>(-x+10)\(\left(2x^2-7x-10\right)=0\)
TH1: -x+10=0
=>-x=-10
=>x=10(nhận)
TH2: \(2x^2-7x-10=0\)
=>\(x^2-\frac72x-5=0\)
=>\(x^2-2\cdot x\cdot\frac74+\frac{49}{16}-\frac{129}{16}=0\)
=>\(\left(x-\frac74\right)^2=\frac{129}{16}\)
=>\(\left[\begin{array}{l}x-\frac74=\frac{\sqrt{129}}{4}\\ x-\frac74=-\frac{\sqrt{129}}{4}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\sqrt{129}+7}{4}\left(loại\right)\\ x=\frac{-\sqrt{129}+7}{4}\left(nhận\right)\end{array}\right.\)
a = 2
b = 3
rồi tính ra nhé
ai k mình mình k lại cho
a: \(x^2-8x+5\)
\(=x^2-8x+16-11\)
\(=\left(x-4\right)^2-11\ge-11\forall x\)
Dấu '=' xảy ra khi x-4=0
=>x=4
b: \(a^3+b^3+c^3=3bac\)
=>\(\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
=>\(\left(a+b+c\right)\left\lbrack\left(a+b\right)^2-c\left(a+b\right)+c^2\right\rbrack-3ab\left(a+b+c\right)=0\)
=>\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)
=>\(a^2+b^2+c^2-ab-ac-bc=0\)
=>\(2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
=>a=b=c
\(N=\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}\)
\(=\frac{a^2+a^2+a^2}{\left(a+a+a\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{3}{3^2}=\frac13\)
Vì \(a+b=3\)
\(\Rightarrow\left(a+b\right)^2=9\)
\(\Leftrightarrow a^2+b^2+2ab=9\)
\(\Leftrightarrow a^2+b^2=7\)
Vì \(a+b=3\)
\(\Leftrightarrow\left(a+b\right)^3=27\)
\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=27\)
\(\Leftrightarrow a^3+b^3=18\)
Lời giải:
\(P=\frac{a^4-a-b^4+b}{(b^3-1)(a^3-1)}+\frac{2(a-b)}{a^2b^2+3}\)
\(=\frac{(a^4-b^4)-(a-b)}{a^3b^3-(a^3+b^3)+1}+\frac{2(a-b)}{a^2b^2+3}=\frac{(a-b)[(a+b)(a^2+b^2)-1]}{a^3b^3-[(a+b)^3-3ab(a+b)]+1}+\frac{2(a-b)}{a^2b^2+3}\)
\(=\frac{(a-b)[(a^2+b^2)-(a+b)^2]}{a^3b^3-[1-3ab]+1}+\frac{2(a-b)}{a^2b^2+3}=\frac{-2ab(a-b)}{a^3b^3+3ab}+\frac{2(a-b)}{a^2b^2+3}\)
\(=\frac{-2(a-b)}{a^2b^2+3}+\frac{2(a-b)}{a^2b^2+3}=0\)
Lời giải:
\(P=\frac{a^4-a-b^4+b}{(b^3-1)(a^3-1)}+\frac{2(a-b)}{a^2b^2+3}\)
\(=\frac{(a^4-b^4)-(a-b)}{a^3b^3-(a^3+b^3)+1}+\frac{2(a-b)}{a^2b^2+3}=\frac{(a-b)[(a+b)(a^2+b^2)-1]}{a^3b^3-[(a+b)^3-3ab(a+b)]+1}+\frac{2(a-b)}{a^2b^2+3}\)
\(=\frac{(a-b)[(a^2+b^2)-(a+b)^2]}{a^3b^3-[1-3ab]+1}+\frac{2(a-b)}{a^2b^2+3}=\frac{-2ab(a-b)}{a^3b^3+3ab}+\frac{2(a-b)}{a^2b^2+3}\)
\(=\frac{-2(a-b)}{a^2b^2+3}+\frac{2(a-b)}{a^2b^2+3}=0\)