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Cho ba số x , y , z thỏa mãn xyz = 2017
Tính tổng D = 2017x / xy + 2017x + 2017+ y/yz+y+2017+z/zx+z+1
thay xyz=2017, ta có:
\(D=\frac{xyzx}{xy+xyzx+xyz}+\frac{y}{yz+y+xzy}+\frac{z}{xz+z+1}\)
\(D=\frac{xz}{1+xz+z}+\frac{1}{x+1+xz}+\frac{z}{xz+x+1}=1\)
\(\text{Bài làm }\)
\(\text{ Gọi xyz = 2017}\)
\(\text{Ta có:}\) \(D=\frac{xyzx}{xy+xyzx+xyz}+\frac{y}{yz+y+xzy}+\frac{z}{xz+z+1}\)
\(D=\frac{xz}{1+xz+z}+\frac{1}{x+1+xz}+\frac{z}{xz+x+1}=1\)
\(\text{# Chúc bạn học tốt #}\)
Ta có : A = \(\dfrac{2017x}{xy+2017x+2017}+\dfrac{y}{yz+y+2017}+\dfrac{z}{xz+z+1}\)
A = \(\dfrac{x^2yz}{xy+x^2yz+xyz}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{xz+z+1}\) (Vì xyz = 2017)
A = \(\dfrac{xy\left(xz\right)}{xy\left(1+xz+z\right)}+\dfrac{y}{y\left(z+1+xz\right)}+\dfrac{z}{xz+z+1}\)
A = \(\dfrac{xz}{1+xz+z}+\dfrac{1}{z+1+xz}+\dfrac{z}{xz+z+1}\)
A = \(\dfrac{xz+1+z}{xz+1+z}\) = 1
Vậy A = 1
làm lần lượt nhá,dài dòng quá khó coi.ahihihi!
\(\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{7\left(\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)
\(=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}=\frac{1}{4}\)
\(D=\dfrac{2017x}{xy+2017x+2017}+\dfrac{y}{yz+y+2017}+\dfrac{z}{zx+z+1}\)
\(D=\dfrac{x^2yz}{xy+x^2yz+xyz}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{zx+z+1}\)
Vì \(xyz=2017\)
\(D=\dfrac{xy\left(xz\right)}{xy\left(1+xz+z\right)}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{zx+z+1}\)
\(D=\dfrac{xz}{1+xz+z}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{zx+z+1}\)
\(D=\dfrac{xz}{1+xz+z}+\dfrac{y}{y\left(z+1+xz\right)}+\dfrac{z}{zx+z+1}\)
\(D=\dfrac{xz+1+z}{1+xz+z}=1\)
Vậy D = 1
Tính giá trị của đa thức sau biết x=2018
N=x^6-2017x^5-2017x^4-2017x^3-2017x^2-2017x-2017
Help me :(((
Ta có : x - 1 = 2018 - 1 = 2017
N = x6 - 2017x5 - 2017x4 - 2017x3 - 2017x2 - 2017x - 2017
N = x6 - ( x - 1 ).x5 - ( x - 1 ).x4 - ( x - 1 ).x3 - ( x - 1 ).x2 - ( x - 1 ).x - ( x - 1 )
N = x6 - x6 + x5 - x5 + x4 - x4 + x3 - x3 + x2 - x2 + x - x + 1
N = 1
\(\dfrac{x+y-2017z}{z}=\dfrac{y+z-2017x}{x}=\dfrac{z+x-2017y}{y}\)
<=> \(\dfrac{x+y}{z}-2017=\dfrac{z+y}{x}-2017=\dfrac{z+x}{y}-2017\)
<=> \(\dfrac{x+y}{z}=\dfrac{z+y}{x}=\dfrac{z+x}{y}\)
đặt x+y+z = t
=> \(\dfrac{t-z}{z}=\dfrac{t-x}{x}=\dfrac{t-y}{y}< =>\dfrac{t}{z}-1=\dfrac{t}{x}-1=\dfrac{t}{y}-1\) \(< =>\dfrac{t}{z}=\dfrac{t}{y}=\dfrac{t}{x}\)
=> x=y=z
ta lại có
\(P=\left(1+\dfrac{y}{x}\right)\left(1+\dfrac{x}{z}\right)\left(1+\dfrac{z}{y}\right)\)
vì x=y=z => P = \(\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
