Cho 4 số a,b,c,d sao cho a+b+c+d \(\ne\)0
Biết \(\dfrac{b+c+d}{a}=\dfrac{c+d+a}{a}=\dfrac{d+a+b}{c}=\dfrac{a+b+c}{d}=k\)
Tính giá trị của k
Cho 4 số a,b,c,d sao cho a + b +c + d ≠ 0
Biết : \(\dfrac{b+c+d}{a}=\dfrac{c+d+a}{c}=\dfrac{a+b+c}{a}=k\)
Tính k
Cho 4 số a,b,c,d sao cho a,b,c,d khác 0
Biết \(\dfrac{b+c+d}{a}=\dfrac{c+d+a}{b}=\dfrac{d+a+b}{c}=\dfrac{a+b+c}{d}=k\)
Tính k
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{b+c+d}{a}=\dfrac{c+d+a}{b}=\dfrac{d+a+b}{c}=\dfrac{a+b+c}{d}=\dfrac{b+c+d+c+d+a+d+a+b+a+b+c}{a+b+c+d}=\dfrac{\left(a+a+a\right)+\left(b+b+b\right)+\left(c+c+c\right)+\left(d+d+d\right)}{a+b+c+d}=\dfrac{3a+3b+3c+3d}{a+b+c+d}=\dfrac{3\left(a+b+c+d\right)}{a+b+c+d}=3\)
Vậy \(k=3\)
Cho bốn số a;b;c;d sao cho a+b+c+d khác 0. Biết \(\dfrac{b+c+d}{a}=\dfrac{c+d+a}{b}=\dfrac{d+a+b}{c}=\dfrac{a+b+c}{d}=k\)
Vậy k =..........
theo bài ra ta có:
\(\frac{b+c+d}{a}=\frac{c+d+a}{b}=\frac{d+a+b}{c}=\frac{a+b+c}{d}=k\)
\(\Rightarrow\frac{b+c+d}{a}+1=\frac{c+d+a}{b}+1=\frac{d+a+b}{c}+1=\frac{a+b+c}{d}+1=k+1\) \(\Rightarrow\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}=k+1\)
vì a + b + c + d khác 0 => a = b = c = d
ta có:
\(\Rightarrow\frac{4a}{a}=\frac{4b}{b}=\frac{4c}{c}=\frac{4d}{d}=k+1\)
=> 4 = 4 = 4 = 4 = k + 1
=> k + 1 = 4
=> k = 3
vật k = 3
theo đầu bài
=>\(\dfrac{b+c+d}{a}\)=\(\dfrac{c+d+a}{b}\)=\(\dfrac{d+a+b}{c}\)=\(\dfrac{a+b+c}{d}\)=\(\dfrac{b+c+d+c+d+a+d+a+b+a+b+c}{a+b+c+d}\)=\(\dfrac{3\left[a+b+c+d\right]}{a+b+c+d}\)=>=3
=>k=3
Cho a + b + c ≠ 0 và \(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}\) tính giá trị của biểu thức \(A=\dfrac{a+b}{c+d}+\dfrac{b+c}{a+d}+\dfrac{c+d}{a+b}+\dfrac{d+a}{b+c}\)
\(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}=\dfrac{1}{3}\\ \Rightarrow\left\{{}\begin{matrix}b+c+d=3a\\a+c+d=3b\\a+b+d=3c\\a+b+c=3d\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b+c+d=2a\\a+b+c+d=2b\\a+b+c+d=2c\\a+b+c+d=2d\end{matrix}\right.\\ \Rightarrow2a=2b=2c=2d\\ \Rightarrow a=b=c=d\\ \Rightarrow A=\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}=1+1+1+1=4\)
ab+c+d=ba+c+d=ca+b+d=da+b+c=a+b+c+d3(a+b+c+d)=13⇒⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩b+c+d=3aa+c+d=3ba+b+d=3ca+b+c=3d⇒⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩a+b+c+d=2aa+b+c+d=2ba+b+c+d=2ca+b+c+d=2d⇒2a=2b=2c=2d⇒a=b=c=d⇒A=a+aa+a+a+aa+a+a+aa+a+a+aa+a=1+1+1+1=4
Cho dãy tỉ số bằng nhau \(\dfrac{a}{a+b+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}\) Tính giá trị của biểu thức M=\(\dfrac{a+b}{c+d}=\dfrac{b+c}{d+a}-\dfrac{c+d}{a+b}+\dfrac{d+a}{b+c}\)
cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\)(b\(\ne\)0;d\(\ne\)0)
a) \(\dfrac{a-b}{b}=\dfrac{c-d}{d}\)
b)\(\dfrac{a+b}{a}=\dfrac{c+d}{d}\)
a: \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Leftrightarrow\dfrac{a}{b}-1=\dfrac{c}{d}-1\)
hay \(\dfrac{a-b}{b}=\dfrac{c-d}{d}\)
cho dãy tỉ số bằng nhau\(\dfrac{2a+b+c+d}{a}\) =\(\dfrac{a+2b+c+d}{b}\) =\(\dfrac{a+b+2c+d}{c}\)=\(\dfrac{a+b+c+2d}{d}\)
tính giá trị của biểu thức M= \(\dfrac{a+b}{c+d}=\dfrac{b+c}{d+a}=\dfrac{c+d}{a+b}=\dfrac{d+a}{b+c}\)
Ta có:
\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\)
⇔ \(\dfrac{2a+b+c+d}{a}-1=\dfrac{a+2b+c+d}{b}-1=\dfrac{a+b+2c+d}{c}-1\)
\(=\dfrac{a+b+c+2d}{d}-1\)
⇔ \(\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}\)
Nếu a+b+c+d=0
⇒a+b=−(c+d);c+b=−(a+d);c+d=−(a+b);a+d=−(c+b)
Thay vào M, ta có:
\(M=\dfrac{a+b}{-\left(a+b\right)}=\dfrac{b+c}{-\left(b+c\right)}=\dfrac{c+d}{-\left(c+d\right)}=\dfrac{a+d}{-\left(a+d\right)}=-1\)
Nếu a+b+c+d ≠0
⇒ \(a=b=c=d\)
Thay vào M, ta có
\(M=\dfrac{a+b}{a+b}=\dfrac{b+c}{b+c}=\dfrac{c+d}{c+d}=\dfrac{d+a}{d+a}=1\)
\(\text{Cùng trừ mỗi tỉ số trên 1 đơn vị ta được:}\)
\(\dfrac{2a+b+c+d}{a}-1=\dfrac{a+2b+c+d}{b}-1=\dfrac{a+b+2c+d}{c}-1=\dfrac{a+b+c+2d}{d}-1\) \(\Rightarrow\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}\)
\(\text{Từ đây ta suy ra 2 trường hợp:}\)
\(\text{Trường hợp 1:}\)
\(\text{Nếu }a+b+c+d\notin0\Rightarrow a=b=c=d\)
\(\Rightarrow M=1+1+1+1=1.4=4\)
\(\text{Trường hợp 2:}\)
\(\text{Nếu }a+b+c+d=0\text{ thì:}\)
\(a+b=-\left(c+d\right);b+c=-\left(d+a\right)\)
\(c+d=-\left(a+b\right);d+a=-\left(b+c\right)\)
\(\text{Do đó }M=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)
Cho a+b+c+d\(\ne\)0 và \(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}\)
Tìm giá trị của A=\(\dfrac{a+b}{c+d}=\dfrac{b+c}{a+d}=\dfrac{c+d}{a+b}=\dfrac{d+a}{b+c}\)
\(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}\dfrac{1}{3}\)(vìa+b+c+d\(\ne\)0)
=>3a=b+c+d: 3b=a+c+d=>3a-3b=b-a
=>3(a-b)=-(a-b)=>4(a-b)=0=>a=b
Tương tự => a=b=c=d=> A=4
Ta có: \(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}=\dfrac{1}{3}\)
Ta có: \(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{a+b}{a+b+2\left(c+d\right)}=\dfrac{1}{3}\)
\(\Rightarrow3\left(a+b\right)=\left(a+b\right)+2\left(c+d\right)\)
\(\Rightarrow2\left(a+b\right)=2\left(c+d\right)\)
\(\Rightarrow a+b=c+d\)
\(\Rightarrow\dfrac{a+b}{c+d}=1\)
Tương tự:\(\dfrac{b+c}{a+d}=1;\dfrac{c+d}{a+b}=1;\dfrac{d+a}{b+c}=1\)
Vậy A=4.
a) Cho \(a+b+c+d=2000\) và \(\dfrac{1}{a+b+c}+\dfrac{1}{b+c+d}+\dfrac{1}{c+d+a}+\dfrac{1}{d+a+b}=\dfrac{1}{40}\)
Tính giá trị của: \(S=\dfrac{a}{b+c+d}+\dfrac{b}{c+d+a}+\dfrac{c}{d+a+b}+\dfrac{d}{a+b+c}\)
b) Xác định tổng các hệ số của đa thức \(f\left(x\right)=\left(5-6x+x^2\right)^{2016}\cdot\left(5-6x+x^2\right)^{2017}\)