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Cho \(\frac{a}{b}=\frac{c}{d}\)CMR:
\(a,\frac{a.c}{b.d}=\frac{a^2+c^2}{b^2+d^2}\) \(b,\frac{a.c}{b.d}=\frac{a^2-c^2}{b^2-d^2}\)\(c,\frac{a.c}{b.d}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
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cho a,b,c là các số khác 0 thỏa mãn b2=a.c và c2 =b.d . CM :\(\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a}{d}\)
cho \(b^2=a.c;a^2=b.d\)
c/m \(\frac{a^3+b^3-c^3}{b^3+c^3-d^3}=\left(\frac{a+b-c}{b+c-d}\right)^2\)
Cho b2 = a.c ; c2 = b.d . Chứng minh :
a) \(\frac{a^3+b^3-c^3}{b^3+c^3-d^3}=\left(\frac{a+b-c}{b+c-d}\right)^3\)
b) \(\frac{a}{d}=\frac{a^3+8.b^3+125.c^3}{b^3+8.c^3+125.d^3}\)
cho \(b^2=a.c;c^2=b.d\) . với \(b,c,d\ne0;b+c\ne d;b^3+c^3\ne d^3\)
Chứng minh rằng
\(\frac{a^3+b^3-c^3}{b^3+c^3-d^3}=\left(\frac{a+b-c}{b+c-d}\right)^3\)
\(Cho\)\(\frac{a}{b}=\frac{c}{d}\)
Chứng minh rằng: \(a,\frac{a+2c}{a-c}=\frac{b+2d}{b-d}\)
\(b,\frac{a.c}{b.d}=\frac{\left(a+c\right)^2}{\left(b-d\right)^2}\)
Cho frac{a}{b}frac{c}{d}chứng minh rằnga)frac{a}{a-b}frac{c}{c-d}b)frac{a}{b}frac{a+c}{b+d}c)frac{a}{3a+b}frac{c}{3c+d}d)frac{a.c}{b.d}frac{a^2+c^2}{b^2+d^2}f)frac{a.b}{c.d}frac{left(a-bright)^2}{left(c-dright)^2}
Đọc tiếp
Cho \(\frac{a}{b}\)=\(\frac{c}{d}\)chứng minh rằng
a)\(\frac{a}{a-b}\)=\(\frac{c}{c-d}\)
b)\(\frac{a}{b}=\frac{a+c}{b+d}\)
c)\(\frac{a}{3a+b}=\frac{c}{3c+d}\)
d)\(\frac{a.c}{b.d}=\frac{a^2+c^2}{b^2+d^2}\)
f)\(\frac{a.b}{c.d}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
Chofrac{a}{b}frac{c}{d} chứng minh 1,frac{a^2+c^2}{b^2+d^2}frac{a.c}{b.d} 2,frac{a^2+c^2}{b^2+d^2}frac{a^2-c^2}{b^2-d^2} 3,left(a+cright).left(b-dright)left(a-cright).left(b+dright) 4,left(b+dright).cleft(c+cright).d 5,frac{4.a-12.b}{8.a+11.b}frac{4.c-12.d}{8.c+11.d} 6,frac{left(a+cright)^2}{left(b+dright)^2}frac{left(a+cright)^2}{left(b+dright)^2} 7,frac{a^{10}+b^{10}}{left(a+bright)^{10}}frac{c^{10}+d^{10}}{left(c+dright)^{10}}
Đọc tiếp
Cho\(\frac{a}{b}\)=\(\frac{c}{d}\) chứng minh
1,\(\frac{a^2+c^2}{b^2+d^2}\)=\(\frac{a.c}{b.d}\)
2,\(\frac{a^2+c^2}{b^2+d^2}\)=\(\frac{a^2-c^2}{b^2-d^2}\)
\(3,\left(a+c\right).\left(b-d\right)=\left(a-c\right).\left(b+d\right)\)
\(4,\left(b+d\right).c=\left(c+c\right).d\)
\(5,\frac{4.a-12.b}{8.a+11.b}=\frac{4.c-12.d}{8.c+11.d}\)
\(6,\frac{\left(a+c\right)^2}{\left(b+d\right)^2}=\frac{\left(a+c\right)^2}{\left(b+d\right)^2}\)
\(7,\frac{a^{10}+b^{10}}{\left(a+b\right)^{10}}=\frac{c^{10}+d^{10}}{\left(c+d\right)^{10}}\)
Cho b2=a.c và c2=b.d(với b;c;d khác 0;b+c không bằng d;b2017+c2017ko bằng d2017(ko bằng có nghĩa là lớn hơn hoặc nhỏ hơn một sô)). Chứng minh rằng \(\frac{a^{2017}+b^{2017}-c^{2017}}{b^{2017}+c^{2017}-d^{2017}}\)=\(\frac{\left(a+b-c\right)^{2017}}{\left(b+c-d\right)^{2017}}\)