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\(=\frac{41^2-39}{41^2+39^2+2.41.39}\)
\(=\frac{41^2-39}{41^2+2.41.39+39^2}\)
\(=\frac{41^2-39}{\left(41+39\right)^2}\)
\(\)
a,\(=\left(\frac{3}{5}x+\frac{2}{7}y\right)^2=\left(\frac{3}{5}.5+\frac{2}{7}.\left(-7\right)\right)^2=0\)
\(b,=\left(\frac{5}{4}u^2v+\frac{2}{25}v^2\right)^2=\left(\frac{5}{4}.\left(\frac{2}{5}\right)^2.5+\frac{2}{25}.5^2\right)^2=3^2=9\)
Bài 1: \(\frac{a^2+c^2}{b^2+c^2}=\frac{a}{b}\) (1)
Từ \(\frac{a}{c}=\frac{c}{b}\Rightarrow ab=c^2\)
Thay vào (1) ta có:
\(\frac{a^2+ab}{b^2+ab}=\frac{a}{b}\Rightarrow\frac{a\left(a+b\right)}{b\left(a+b\right)}=\frac{a}{b}\) (luôn đúng)
Vậy ta có điều phải chứng minh
a) Ta có: \(25^{50}+3^{41}=\left(\left(25\right)^2\right)^{25}+\left(\left(3\right)^4\right)^{10}.3=625^{25}+81^{10}.3\)
\(2525^{25}+5^{31}=2525^{25}+\left(\left(5\right)^3\right)^{10}.5=2525^{25}+125^{10}.5\)
Vì \(625^{25}< 2525^{25}\),\(81^{10}.3< 125^{10}.5\)(\(81^{10}< 125^{10},3< 5\)) nên \(625^{25}+81^{10}.3< 2525^{25}+125^{10}.5\)
hay \(25^{50}+3^{41}< 2525^{25}+5^{31}\)
\(\)
1/\(\frac{84^2-16^2}{37^2-63^2}=\frac{\left(84-16\right)\left(84+16\right)}{\left(37-63\right)\left(37+63\right)}=\frac{68.100}{-26.100}=\frac{-68}{26}=\frac{-34}{13}\)
2/ \(199^2=\left(200-1\right)^2=40000-400+1=39601\)
3/ \(31^2=\left(30+1\right)^2=900+60+1=961\)
4/ \(45.55=\left(50-5\right)\left(50+5\right)=50^2-25=2500-25=2475\)
5/ \(78.82=\left(80-2\right)\left(80+2\right)=80^2-4=6400-4=6396\)
Pythagorean theorem: \(AD=\sqrt{BD^2-AB^2}=4\) (cm)
\(\Rightarrow BC=AD=4\left(cm\right)\)
\(CC'=\sqrt{BC'^2-BC^2}=4\sqrt{2}\)
The lateral surface area: \(2CC'.\left(BC+AB\right)=56\sqrt{2}\left(cm^2\right)\)
\(b.\frac{x+5}{2}+\frac{3-2x}{4}=x-\frac{7+x}{6}\\\Leftrightarrow \frac{6\left(x+5\right)}{12}+\frac{3\left(3-2x\right)}{12}=\frac{12x}{12}-\frac{2\left(7+x\right)}{12}\\ \Leftrightarrow6\left(x+5\right)+3\left(3-2x\right)=12x-2\left(7+x\right)\\ \Leftrightarrow6x+30+9-6x=12x-14-2x\\\Leftrightarrow 6x-6x-12x+2x=-30-9-14\\\Leftrightarrow -10x=-53\\ \Leftrightarrow x=\frac{53}{10}\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{\frac{53}{10}\right\}\)
\(25x^2y^4+30xy^2z+9z^2=\left(5xy^2\right)^2+2.5xy^2.3z+\left(3z\right)^2=\left(5xy^2+3z\right)^2\)
\(\frac{16}{9}x^2+4xyz^2+\frac{9}{4}y^2z^4=\left(\frac{4}{3}x\right)^2+2.\frac{4}{3}x.\frac{3}{2}yz^2+\left(\frac{3}{2}yz^2\right)^2=\left(\frac{4}{3}x+\frac{3}{2}yz^2\right)^2\)
\(\frac{9}{25}x^2+\frac{12}{35}xy+\frac{4}{49}y^2=\left(\frac{3}{5}x\right)^2+2.\frac{3}{5}x.\frac{2}{7}y+\left(\frac{2}{7}y\right)^2=\left(\frac{3}{5}x+\frac{2}{7}y\right)^2\)( tự thay vào tính nhé )
\(\frac{25}{16}u^4y^2+\frac{1}{5}u^2+y^3+\frac{4}{625}y^4=\left(\frac{5}{4}u^2y\right)^2+2.\frac{5}{4}u^2y.\frac{2}{25}.y^2+\left(\frac{2}{25}y^2\right)^2=\left(\frac{5}{4}u^2y+\frac{2}{25}y^2\right)^2\)( tự thay vào tính nhé )
Tham khảo nhé~
a) \(413.\left(413-26\right)+169=413^2-2.13.413+13^2=\left(413-13\right)^2=160000\)
b) \(\left(625^2+3\right).\left(25^4-3\right)-5^{16}+10\)
\(=\left(5^8+3\right)\left(5^8-3\right)-5^{16}+10\)
\(=5^{16}-9-5^{16}+10=1\)
c) \(\frac{41^2+39^2+8^2.39}{41^2-39^2}=\frac{\left(41+39\right)^2}{\left(41-39\right)\left(41+39\right)}=\frac{41+39}{41-39}=\frac{80}{2}=40\)