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a) a2+b2-2ab=(a-b)2>=0
b) \(\frac{a^2+b^2}{2}\)\(\ge\)ab <=> \(\frac{a^2+b^2}{2}\)-ab\(\ge\)0 <=> \(\frac{\left(a-b\right)^2}{2}\)\(\ge\)0 (ĐPCM)
c) a2+2a < (a+1)2=a2+2a+1 (ĐPCM)
a) \(A=\left(\frac{1}{1-x}+\frac{2}{x+1}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}\)
\(\Leftrightarrow A=\left(\frac{1+x+2-2x-5+2}{1-x^2}\right):\frac{1-2x}{x^2-1}\)
\(\Leftrightarrow-\frac{2}{1-x^2}:\frac{1-2x}{x^2-1}\Leftrightarrow\frac{2}{x^2-1}:\frac{1-2x}{x^2-1}\)
\(\Leftrightarrow\frac{2}{x^2-1}.\frac{x^2-1}{1-2x}=\frac{2}{1-2x}\)
b) Ta có: \(\frac{2}{1-2x}>0\)( Vì 2 > 0 )
\(\Rightarrow1-2x>0\)
\(\Leftrightarrow-2x>-1\)
\(\Leftrightarrow x< \frac{1}{2}\)
Vậy.......................
a)
Ta có:
\(a^2+b^2=a^2+2ab+b^2-2ab=\left(a+b\right)^2-2ab\)
\(=7^2-2.5=49-25=24\)
Ta có:
\(a^3+b^3=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)\)
\(=7^3-3.5.7=238\)
b)
Ta có: \(a^2-4a+5=a^2-4a+4+1\)
\(=\left(a+2\right)^2+1>0\) với mọi a
c)
A = \(x^2+8x-1\)
A = \(x^2+2.x.4+16-17\)
A = \(\left(x+4\right)^2-17\ge-17\) với mọi x
Dấu " = " xảy ra khi và chỉ khi x = -4
Vậy MinA = -17 khi x = -4
a) \(ĐKXĐ:\) \(x\ne\pm1\)
\(A=\left(\frac{3x^2-4}{x^2-1}-\frac{2}{1-x}-\frac{2}{x+1}\right):\left(\frac{1-x}{x+1}\right)\)
\(=\left(\frac{3x^2-4}{\left(x-1\right)\left(x+1\right)}+\frac{2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\right).\frac{x+1}{1-x}\)
\(=\frac{3x^2-4+2x+2-2x+2}{\left(x-1\right)\left(x+1\right)}.\frac{x+1}{1-x}\)
\(=\frac{3x^2}{\left(x-1\right)\left(x+1\right)}.\frac{x+1}{1-x}\)
\(=-\frac{3x^2}{\left(x-1\right)^2}\)
Bài 1: A = \(\frac{\left(x-1\right)^2}{x^2-x+1}=\frac{x^2-x+1-x}{x^2-x+1}=1-\frac{x}{x^2-x+1}\)
Ta có \(\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\in R\\x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\in R\end{cases}\Rightarrow A}\ge0\forall x\in R\)
Bài 2: \(4\left(a^3+b^3\right)\ge\left(a+b\right)^3\Leftrightarrow3\left(a^3-a^2b-ab^2+b^3\right)\ge0\)\(\Leftrightarrow\left(a+b\right)\left(a-b\right)^2\ge0\)(đúng với mọi a; b > 0)
a: \(M=\left(a+1\right)\cdot\left(a-1\right)=a^2-1\)
b: Để M=a2 thì a2-1=a2
=>-1=0(vô lý)