Cho các số x,y,z,t thoả mãn điều kiện xyzt = 1
Tính tổng : P = \(\dfrac{1}{1+x+xy+xyz}+\dfrac{1}{1+y+yz+yzt}+\dfrac{1}{1+z+zt+ztx}+\dfrac{1}{1+t+tx+txy}\)
Cho các số x, y, z, t thỏa mãn xyzt=1. Tính P= \(\dfrac{1}{1+x+xy+xyz}+\dfrac{1}{1+y+yz+yzt}+\dfrac{1}{1+z+zt+ztx}+\dfrac{1}{1+t+tx+txy}\)
Cho các số x, y, z, t thỏa mãn xyzt=1. Tính P=\(\dfrac{x}{xyz+xy+x+1}+\dfrac{y}{yzt+yz+y+1}+\dfrac{z}{xzt+zt+z+1}+\dfrac{t}{xyt+tx+t+1}\)
Thay xyzt = 1 vào P, có:
P= \(\frac{x}{xyz+xy+x+xyzt\ }\) + \(\frac{y}{yzt+yz+y+1}+\frac{z}{xzt+zt+z+xyzt}+\frac{t}{xyt+tx+t+1}\)
\(P=\frac{x}{x.\left(yz+y+1+yzt\right)}+\frac{y}{yzt+yz+y+1}+\frac{z}{z.\left(xt+t+1+xyt\right)}+\frac{t}{xyt+tx+t+1}\)
\(P=\frac{1\ +y}{yz+y+yzt+1}\) \(+\frac{1+t}{xyt+tx+t+1}\)
\(P=\frac{1+y}{yz+y+yzt+xyzt\ }+\frac{1+t}{xyt+tx+t+1}\)
\(P=\frac{1+y}{y.z.\left(xyt+tx+t+1\right)}+\frac{yz+tyz}{yz.\left(xyt+tx+t+1\right)}\)
\(P=\frac{1+y+yz+tyz}{yz.\left(xyt+tx+t+1\right)}=\frac{1+y+yz+tyz}{xyzt.\left(1+y+yz+tyz\right)}=\frac{1}{xyzt}=1\)
KL: P = 1 tại xyzt=1
\(\dfrac{x}{xyz+xy+x+1}+\dfrac{y}{yzt+yz+y+1}+\dfrac{z}{xzt+zt+z+1}+\dfrac{t}{xyt+tx+t+1}\)
= \(\dfrac{x}{xyz+xy+x+1}+\dfrac{xy}{xyzt+xyz+xy+x}+\dfrac{xyz}{x^2yzt+xyzt+xyz+xy}+\dfrac{xyzt}{x^{2^{ }}y^2zt+x^2yzt+xyzt+xyz}\)
= \(\dfrac{x}{xyz+xy+x+1}+\dfrac{xy}{1+xyz+xy+x}+\dfrac{xyz}{x+1+xyz+xy}+\dfrac{1}{xy+x+1+xyz}\)
= \(\dfrac{x+xy+xyz+1}{x+xy+xyz+1}\)
= 1
Tìm x,y,z,t biết xyzt=1
Tính P=1/(1+x+xy+xyz)+1/(1+y+yz+yzt)+1/(1+z+zt+ztx)+1/(1+t+tx+txy)
mọi người giúp mik với sắp đi hok r mà vẫn chưa xong hết bài =(((
Answer:
\(P=\frac{1}{1+x+xy+xyz}+\frac{1}{1+y+yz+yzt}+\frac{1}{1+z+zt+ztx}+\frac{1}{1+t+tx+txy}\)
\(=\frac{1}{1+x+xy+xyz}+\frac{x}{x+xy+xyz+xyzt}+\frac{xy}{xy+xyz+xyzt+xyzt.x}+\frac{xyz}{xyz+xyzt+xyzt.x+xyzt.xy}\)
\(=\frac{1}{1+x+xy+xyz}+\frac{x}{x+xy+xyz+1}+\frac{xy}{xy+xyz+1+x}+\frac{xyz}{xyz+1+x+xy}\)
\(=\frac{1+x+xy+xyz}{1+x+xy+xyz}\)
\(=1\)
Cho \(x,y,z,t>0\) thỏa mãn \(xyzt=1\)
Chứng minh \(\dfrac{1}{x^3\left(yz+zt+ty\right)}+\dfrac{1}{y^3\left(xz+zt+tx\right)}+\dfrac{1}{z^3\left(xy+yt+tx\right)}+\dfrac{1}{t^3\left(xy+yz+zx\right)}\ge\dfrac{1}{3}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{t}\right)\)
Từ \(xyzt=1\) ta có: \(\dfrac{1}{x^3\left(yz+zt+ty\right)}=\dfrac{xyzt}{x^3\left(yz+zt+ty\right)}=\dfrac{yzt}{x^2\left(yz+zt+ty\right)}\)
Đánh giá tương tự ta có:
\(pt\Leftrightarrow\dfrac{yzt}{x^2\left(yz+zt+ty\right)}+\dfrac{xzt}{y^2\left(xz+zt+tx\right)}+\dfrac{xyt}{z^2\left(xy+yt+tx\right)}+\dfrac{xyz}{t^2\left(xy+yz+zx\right)}\ge3\left(yzt+xzt+xyt+xyz\right)=3yzt+3xzt+3xyt+3xyz\)
Ta sẽ chứng minh:
\(\dfrac{yzt}{x^2\left(yz+zt+ty\right)}\ge3yzt\). Cộng theo vế rồi suy ra đpcm
T gần đi học r,có gì tối về giải full cho
Áp dụng cauchy-schwarz:
\(VT=\sum\dfrac{\dfrac{1}{x^2}}{\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{t}}\ge\dfrac{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{t}\right)^2}{3\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{t}\right)}=VF\)
@Neet
\(VT=\dfrac{1}{x^3\left(yz+zt+ty\right)}+\dfrac{1}{y^3\left(xz+zt+tx\right)}+\dfrac{1}{z^3\left(xy+yt+tx\right)}+\dfrac{1}{t^3\left(xy+yz+xz\right)}\)
\(=\dfrac{\dfrac{1}{x^2}}{xyz+xzt+xyt}+\dfrac{\dfrac{1}{y^2}}{xyz+yzt+txy}+\dfrac{\dfrac{1}{z^2}}{xyz+yzt+ztx}+\dfrac{\dfrac{1}{t^2}}{xyt+yzt+txz}\)
\(=\dfrac{\dfrac{1}{x^2}}{\dfrac{xyz}{xyzt}+\dfrac{xzt}{xyzt}+\dfrac{xyt}{xyzt}}+\dfrac{\dfrac{1}{y^2}}{\dfrac{xyz}{xyzt}+\dfrac{yzt}{xyzt}+\dfrac{txy}{xyzt}}+\dfrac{\dfrac{1}{z^2}}{\dfrac{xyz}{xyzt}+\dfrac{yzt}{xyzt}+\dfrac{ztx}{xyzt}}+\dfrac{\dfrac{1}{t^2}}{\dfrac{xyt}{xyzt}+\dfrac{yzt}{xyzt}+\dfrac{txz}{xyzt}}\)
\(=\dfrac{\dfrac{1}{x^2}}{\dfrac{1}{t}+\dfrac{1}{y}+\dfrac{1}{z}}+\dfrac{\dfrac{1}{y^2}}{\dfrac{1}{t}+\dfrac{1}{x}+\dfrac{1}{z}}+\dfrac{\dfrac{1}{z^2}}{\dfrac{1}{t}+\dfrac{1}{x}+\dfrac{1}{y}}+\dfrac{\dfrac{1}{t^2}}{\dfrac{1}{z}+\dfrac{1}{x}+\dfrac{1}{y}}\)
\(\ge\dfrac{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{t}\right)^2}{3\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{t}\right)}=\dfrac{1}{3}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{t}\right)=VP\)
Cho x;y;z;t thỏa mãn: \(xyzt=1\) Chứng minh rằng: \(\dfrac{1}{x^2\left(yz+zt+ty\right)}+\dfrac{1}{y^2\left(xz+zt+tx\right)}+\dfrac{1}{z^2\left(xy+xt+tz\right)}+\dfrac{1}{t^2\left(xy+yz+xz\right)}\ge\dfrac{4}{3}\)
đặt x/y=a hay xy/z=a hay j đó là ra nói chung là 4 biế
n lười nháp
Cho xyzt=1 tính tổng (x/xyz+xy+x+1)+(y/yzt+yz+yt+1)+(z/zxt+zt+z+1)+(t/xyt+t+t+1)
Cho các số thực x, y, z đồng thời thoả mãn các điều kiện: x+y+z=18 và xyz=-1. Tính giá trị của \(S=\dfrac{1}{xy+z-1}+\dfrac{1}{yz+x-1}+\dfrac{1}{xz-y+1}\)
Cho x, y, z thoả mãn xyz = 2023.
Chứng minh: \(\dfrac{2023x}{xy+2023x+2023}+\dfrac{y}{yz+y+2023}+\dfrac{z}{xz+z+1}=1\)
Có `xyz=2023=>2023=xyz`
Thay vào ta có :
\(\dfrac{xyz\cdot x}{xy+xyz\cdot x+xyz}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{xz+z+1}=1\\ \dfrac{x^2yz}{xy\left(1+xz+z\right)}+\dfrac{y}{y\left(z+1+xz\right)}+\dfrac{z}{xz+z+1}=1\\ \dfrac{xz}{1+xz+z}+\dfrac{1}{z+1+xz}+\dfrac{z}{xz+z+1}=1\\ \dfrac{xz+1+z}{1+xz+z}=1\left(dpcm\right)\)
Cho x, y, z, t thoả mãn điều kiện \(x^2+y^2+z^2+t^2=1\)và xy+yz+zt+tx=1. Tính giá trị của biểu thức \(A=\dfrac{2x-3y+4z}{5x-6y+7z}\)