trong 2 số sau đây số nào lớn hơn :a=\(\sqrt{1969}+\sqrt{1971}\);b=2 \(\sqrt{1970}\)tại sao nêu cách giải
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b) \(\left(3x^2-2x+1\right).\left(3x^2+2x+1\right)-\left(3x^2+1\right)^2\)=\(\left(3x^2\right)^2-\left(2x+1\right)^2-\left(3x^2+1\right)^2\)=\(\left(3x^2\right)^2-[\left(2x\right)^2+4x+1]-[\left(3x^2\right)^2+6x^2+1]\)=\(\left(2x\right)^2+4x+1+6x^2-1\)=\(4x^2+4x+6x^2\)=\(10x^2+4x\)
c)\(\left(x^2-5x+2\right)^2-2\left(x^2-5x+2\right)\left(5x-2\right)+\left(5x-2\right)^2\)=\([\left(x^2-5x+2\right)-\left(5x-2\right)]^2\)=\(x^2-5x+2-5x+2\)=\(x^2-10x+4\)=\(x^2-4x+2^2-6x\)=\(\left(x-2\right)^2-6x\)
a,\(=\left(\frac{3}{5}x+\frac{2}{7}y\right)^2=\left(\frac{3}{5}.5+\frac{2}{7}.\left(-7\right)\right)^2=0\)
\(b,=\left(\frac{5}{4}u^2v+\frac{2}{25}v^2\right)^2=\left(\frac{5}{4}.\left(\frac{2}{5}\right)^2.5+\frac{2}{25}.5^2\right)^2=3^2=9\)
\(\frac{a+3c}{a+b}+\frac{a+3b}{a+c}+\frac{2a}{b+c}\)
\(=\frac{a+c}{a+b}+\frac{2c}{a+b}+\frac{a+b}{a+c}+\frac{2b}{a+c}+\frac{2a}{b+c}\)
\(=2\left(\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c}\right)+\left(\frac{a+c}{a+b}+\frac{a+b}{a+c}\right)\)
Áp dụng BĐT Cauchy - Schwar:
\(\frac{a+c}{a+b}+\frac{a+b}{a+c}\ge2\sqrt{\frac{\left(a+c\right)\left(a+b\right)}{\left(a+b\right)\left(a+c\right)}}=2\)(1)
Áp dụng BĐT Nesbit:
\(\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c}\ge\frac{3}{2}\)
\(\Leftrightarrow2\left(\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c}\right)\ge3\)(2)
Từ (1) và (2) suy ra \(2\left(\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c}\right)+\left(\frac{a+c}{a+b}+\frac{a+b}{a+c}\right)\ge5\)
hay \(\frac{a+3c}{a+b}+\frac{a+3b}{a+c}+\frac{2a}{b+c}\ge\left(đpcm\right)\)
Ta có: \(\frac{a+3c}{a+b}+\frac{a+3b}{a+c}+\frac{2a}{b+c}-5\ge0\)
\(\Leftrightarrow\frac{a+3c}{a+b}-2+\frac{a+3b}{a+c}-2+\frac{2a}{b+c}-1\ge0\)
Giải bất phương trình
Cuối cùng ta được: \(\left(c-a\right)^2\left(\frac{1}{\left(a+b\right)\left(b+c\right)}\right)+2\left(b-c\right)^2\left(\frac{1}{\left(a+c\right)\left(a+b\right)}\right)+\left(a-b\right)^2\) \(\left(\frac{1}{\left(a+c\right)\left(b+c\right)}\right)\ge0\)
BĐT đúng <=> a = b = c
\(x^2-6x+9=x^2-2.3x+3^2=\left(x-3\right)^2\)
\(\frac{1}{4}a^2+2ab^2+4b^4=\left(\frac{1}{2}a\right)^2+2.\frac{1}{2}a.2b^2+\left(2b\right)^2=\left(\frac{1}{2}a+2b\right)^2\)
\(25+10x+x^2=5^2+2.5x+x^2=\left(5+x\right)^2\)
\(\frac{1}{9}-\frac{2}{3}y^4+y^8=\left(\frac{1}{3}\right)^2-2.\frac{1}{3}y^4+\left(y^4\right)^2=\left(\frac{1}{3}-y^4\right)^2\)
a)\(\left(x+1\right)\left(x+3\right)-x\left(x-2\right)=7\)
\(x\left(x+3\right)+x+3-x^2+2x=7\)
\(x^2+3x+x+3-x^2+2x=7\)
\(6x+3=7\)
\(6x=4\)
\(x=\frac{4}{6}=\frac{2}{3}\)
Vậy \(x=\frac{2}{3}\)
b) \(2x\left(3x+5\right)-x\left(6x-1\right)=33\)
\(6x^2+10x-6x^2-x=33\)
\(9x=33\)
\(x=\frac{33}{9}\)
Vậy \(x=\frac{33}{9}\)
a) \(\left(x+1\right)\left(x+2\right)\left(x-3\right)=\left(x^2+2x+x+2\right)\left(x-3\right)\)
\(=\left(x^2+3x+2\right)\left(x-3\right)\)
\(=x\left(x^2+3x+2\right)-3\left(x^2+3x+2\right)\)
\(=x^3+3x^2+2x-3x^2-9x-6\)
\(=x^3-7x-6\)
b) \(\left(2x-1\right)\left(x+2\right)\left(x+3\right)=\left(2x-1\right)\left(x^2+3x+2x+6\right)\)
\(=\left(2x-1\right)\left(x^2+5x+6\right)\)
\(=2x\left(x^2+5x+6\right)-\left(x^2+5x+6\right)\)
\(=2x^3+10x^2+12x-x^2-5x-6\)
\(=2x^3+9x^2+7x-6\)
Bình phương a và b lên để so sánh