Các câu hỏi tương tự
cho \(\frac{a}{b}=\frac{c}{d}\) .CM \(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{a^2-b^2}{c^2-d^2}\) (b,c,d khác 0,c+d khác 0, c-d khác 0)
Cho \(\text{a,b,c \in R; a,b,c \ne0}\)thỏa mãn: b2 = a.c
Chứng minh rằng : \(\frac{a}{c}=\left(\frac{a+2018b}{b+2018c}\right)^2\)
1.Tính:left(frac{1}{4times9}+frac{1}{9times14}+frac{1}{14times19}+...+frac{1}{44times49}right)timesfrac{1-3-5-7-...-49}{89}2.Cho frac{a}{2b}frac{b}{2c}frac{c}{2d}frac{d}{2a}. Tính: Afrac{2019a-2018b}{c+d}+frac{2019b-2018c}{a+d}+frac{2019c-2018d}{a+b}+frac{2019d-2018a}{b+c}3.Tìm x biết:left(x-1right)left(x-3right) 0
Đọc tiếp
1.Tính:
\(\left(\frac{1}{4\times9}+\frac{1}{9\times14}+\frac{1}{14\times19}+...+\frac{1}{44\times49}\right)\times\frac{1-3-5-7-...-49}{89}\)
2.Cho \(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}\). Tính: \(A=\frac{2019a-2018b}{c+d}+\frac{2019b-2018c}{a+d}+\frac{2019c-2018d}{a+b}+\frac{2019d-2018a}{b+c}\)
3.Tìm x biết:\(\left(x-1\right)\left(x-3\right)< 0\)
\(Cho:\frac{a}{2b}+\frac{b}{2c}+\frac{c}{2d}+\frac{d}{2a}\)\(\left(a,b,c,d>0\right)\)Tính:\(\frac{2019a-2018b}{c+d}+\frac{2019b-2018c}{a+d}+\frac{2019c-2018d}{a+b}+\frac{2019d-2018a}{c+b}\)
Cho \(\frac{a}{b}=\frac{c}{d}\)và \(\left|a\right|#\left|b\right|;\:\left|k\right|#\left|d\right|\)và a, b, c, d # 0
Cm: \(\frac{a^2+ab}{a^2-b^2}=\frac{c^2+cd}{c^2-d^2}\)
\(\frac{a}{b}=\frac{c}{d}\)(a,b,c,d khác 0)
Chứng minh :\(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
cho 3 số đôi 1 khác nhau .CMR:
\(\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{2}{a-b}+\frac{2}{b-c}+\frac{2}{c-a}\)
cmr nếu \(\frac{a}{b}=\frac{c}{d}\)
thì: \(\frac{a^2+c^2}{b^2+d^2}=\frac{\left(a+c\right)^2}{\left(b+d\right)^2}\)(b+d khác 0)
1, Cho \(\frac{a}{b}=\frac{c}{d}\)( b,c,d khác 0; c+đ khác 0). CMR:
\(\frac{ab}{cd}=\frac{\left(a+b\right)^2}{\left(c+\text{d}\right)^2}\)