\(ac=bd\Rightarrow\frac{a}{d}=\frac{b}{c}\Rightarrow\frac{a}{b}=\frac{d}{c}=\frac{a^{2000}}{b^{2000}}=\frac{d^{2000}}{c^{2000}}=\left(\frac{a}{b}\right)^{2000}=\left(\frac{c}{d}\right)^{2000}=\frac{\left(a+d\right)^{2000}}{\left(b+c\right)^{2000}}\)
\(=\frac{a^{2000}+d^{2000}}{b^{2000}+c^{2000}}\)
Vậy...
Cho tỉ lệ thức\(\frac{a}{b}=\frac{c}{d}\). Chứng minh :
d) \(\frac{a+c}{b+d}=\frac{a-c}{b-d}\) f) \(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{ab}{cd}\) h) \(\frac{7a^2+ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
e) \(\frac{3a+5b}{2a-7b}=\frac{3c+5d}{2c-7d}\) g) \(\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\) i) \(\left(\frac{a-b}{c-d}\right)^{2000}=\frac{a^{2000}+b^{2000}}{c^{2000}+d^{2000}}\)
a) CHO a+b+c+d=2000 và
\(\frac{1}{a+b+c}+\frac{1}{b+c+d}+\frac{1}{c+d+a}+\frac{1}{d+a+b}=\frac{1}{40}\)
TÍNH GIÁ TRỊ CỦA S=\(\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{d+a+b}+\frac{d}{a+b+c}\)
b) XÁC ĐỊNH TỔNG CÁC HỆ SỐ CỦA ĐA THỨC f(x)=\(\left(5-6x+x^2\right).\left(5+6+x^2\right)^{2017}\)
Cho \(b^2=ac\) và \(c^2=bd\)(với \(b,c,d\ne0;b+d\ne d;b^{2017}+c^{2017}\ne d\))
CMR \(\frac{a^{2017}+b^{2017}-c^{2017}}{b^{2017}+c^{2017}+d^{2017}}=\frac{\left(a+b-c\right)^{2017}}{\left(b+c-d\right)^{2017}}\)
Cho \(b^2\)=ac,\(^{c^2}\)=bd , b,c,d \(\ne\)0, b+c \(\ne\)d ,\(b^3+c^3=d^3\)
CM \(\frac{a^3+b^3-c^3}{b^3+c^3-d^3}=\left(\frac{a+b-c}{b+c-d}\right)^3\)
tìm x và y
a) \(\left(x-1\right)^2+\left(y+3\right)^2=0\)
b) \(2\left(x-5\right)^4+5\left|2y-7\right|^5=0\)
c) \(3\left(x-2y\right)^{2004}+4\left|y+\frac{1}{2}\right|=0\)
d) \(\left|x+3y-1\right|+\left(2y-\frac{1}{2}\right)^{2000}=0\)
Cho 4 số a,b,c,d \(\ne\)0;\(b^2=ac;c^2=bd.\)Chứng minh a/b=\(\left(\frac{a+b+c}{b+c+d^{ }}\right)^3\)
Cho b^2=ac ; c^2= bd. Với b,c,d \(\ne\)0; 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 b2 = ac ; c2 = bd với b, c, d \(\ne\)0 ; b + c \(\ne\)d , b3 + c3 \(\ne\)d3
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 a,b,c\(\ne\)0,\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{a+a-b}{b}\)
Tính D=\(\left(1+\frac{b}{a}\right)\left(1+\frac{c}{b}\right)\left(1+\frac{a}{c}\right)\)
