Tính giá trị biểu thức:
A= (b+a)+(c-d)-(c+a)-(b-d)
B=(a-d)-(d+a)-(c-d)+(c+b)
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a/b+c+d =b/c+d+a=c/d+a+b=d/a+b+c
=>a+b+c+d/3(a+b+c+d)=1/3
có thể P=4
Áp dụng t/c dttsbn:
\(\dfrac{a+b+c-2020d}{d}=\dfrac{b+c+d-2020a}{a}=\dfrac{c+d+a-2020b}{b}=\dfrac{d+a+b-2020c}{c}=\dfrac{3\left(a+b+c+d\right)-2020\left(a+b+c+d\right)}{a+b+c+d}=-2017\)
\(\Rightarrow\left\{{}\begin{matrix}a+b+c-2020d=-2017d\\b+c+d-2020a=-2017a\\c+d+a-2020b=-2017b\\d+a+b-2020c=-2017c\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}a+b+c=3d\\b+c+d=3a\\c+d+a=3b\\d+a+b=3c\end{matrix}\right.\Rightarrow a=b=c=d\)
\(F=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{a+d}{b+c}\\ F=\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}=4\)
a. A = (a + b)3 - (a - b)3
A = \(\left[\left(a+b\right)-\left(a-b\right)\right]\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
A = (a + b - a + b)\(\left[a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2\right]\)
A = 2b(a2 + a2 + a2 + 2ab - 2ab + b2 - b2 + b2)
A = 2b(3a2 + b2)
A = 6a2b + 2b3
\(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}=\dfrac{1}{3}\\ \Rightarrow\left\{{}\begin{matrix}b+c+d=3a\\a+c+d=3b\\a+b+d=3c\\a+b+c=3d\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b+c+d=2a\\a+b+c+d=2b\\a+b+c+d=2c\\a+b+c+d=2d\end{matrix}\right.\\ \Rightarrow2a=2b=2c=2d\\ \Rightarrow a=b=c=d\\ \Rightarrow A=\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}=1+1+1+1=4\)
anh đi anh nhớ quê nha
nhớ canh rau muống nhớ cà dầm tương
nhớ thằng đẩy bố xuống mương
bố mà bắt được bố tương vỡ mồm
Ta có : \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{b+c+d+a}=1\Rightarrow a=b=c=d\)
Khi đó P = \(\frac{2019a-b}{c+d}+\frac{2019b-c}{d+a}+\frac{2019c-d}{a+b}+\frac{2019d-a}{b+c}\)
\(=\frac{2019a-a}{2a}+\frac{2019b-b}{2b}+\frac{2019c-c}{2c}+\frac{2019d-d}{2d}\)
\(=1014+1014+1014+1014=1014.4=4056\)
Tham khảo
\(A=\left(b+a\right)+\left(c-d\right)-\left(c+a\right)-\left(b-d\right)\)
\(A=b+a+c-d-c-a-b+d\)
\(A=\left(b-b\right)+\left(a-a\right)+\left(c-c\right)+\left(-d+d\right)\)
\(A=0\)
\(B=\left(a-d\right)-\left(d+a\right)-\left(c-d\right)+\left(c+b\right)\)
\(B=a-d-d-a-c+d+c+b\)
\(B=\left(a-a\right)+\left(d-d+d\right)+\left(-c+c\right)+b\)
\(B=d+b\)
a) Ta có: \(A=\left(b+a\right)+\left(c-d\right)-\left(c+a\right)-\left(b-d\right)\)
\(=a+b+c-d-c-a-b+d\)
=0
b) Ta có: \(B=\left(a-d\right)-\left(a+d\right)-\left(c-d\right)+\left(c+b\right)\)
\(=a-d-a-d-c+d+c+b\)
=b-d