Cho\(x,y,z\ne0\), biết:
\(\frac{y+z-x}{z}=\frac{z+x-y}{y}=\frac{x+y-z}{z}\)
Tính:\(B=\left(1+\frac{x}{y}\right).\left(1+\frac{y}{z}\right).\left(1+\frac{z}{x}\right)\)
Tính B: \(B=\left(1-\frac{z}{x}\right).\left(1-\frac{x}{y}\right).\left(1-\frac{y}{z}\right)cho\left(x,y,z\ne0,x-y-z=0\right).\)
thực hiện phép tính
a,\(x^3+\left[\frac{x\left(2y^3-x^3\right)}{x^3+y^3}\right]^3-\left[\frac{y\left(2x^3-y^3\right)}{x^3+y^3}\right]^3\)
b,\(\frac{\frac{x\left(x+y\right)}{x-y}+\frac{x\left(x+z\right)}{x-z}}{1+\frac{\left(y-z\right)^2}{\left(x-y\right)\left(x-z\right)}}+\frac{\frac{y\left(y+z\right)}{y-z}+\frac{y\left(y+x\right)}{y-x}}{1+\frac{\left(z-x\right)^2}{\left(y-z\right)\left(y-x\right)}}+\frac{\frac{z\left(z+x\right)}{z-x}+\frac{z\left(z+y\right)}{z-y}}{1+\frac{\left(x-y\right)^2}{\left(z-x\right)\left(z-y\right)}}\)
c,\(\left[\frac{y+z-2x}{\frac{\left(y-z\right)^3}{y^3-z^3}+\frac{\left(x-y\right)\left(x-z\right)}{y^2+yz+z^2}}+\frac{z+x-2y}{\frac{\left(z-x\right)^3}{z^3-x^3}+\frac{\left(y-z\right)\left(y-x\right)}{z^2+xz+x^2}}+\frac{x+y-2z}{\frac{\left(x-y\right)^3}{x^3-y^3}+\frac{\left(z-x\right)\left(z-y\right)}{x^2+xy+y^2}}\right]:\frac{1}{x+y+z}\)
Tính B: \(B=\left(1-\frac{z}{x}\right).\left(1-\frac{x}{y}\right).\left(1-\frac{y}{z}\right)cho\left(x,y,z\ne0,x-y-z=0\right)\)
Đề học kì đấy mọi người, giải giúp với.
Ta có: \(x-y-z=0\Rightarrow x-z=y,z-y=x,y-x=-z\)
\(B=\left(1-\frac{z}{x}\right)\cdot\left(1-\frac{x}{y}\right)\cdot\left(1-\frac{y}{z}\right)\)
\(\Rightarrow B=\frac{x-z}{x}\cdot\frac{y-x}{y}\cdot\frac{z-y}{z}=\frac{y}{x}\cdot\frac{-z}{y}\cdot\frac{x}{z}=\frac{-xyz}{xyz}=-1\)
x - y - z = 0
=> x = y + z
y = x - z
-z = x - y
Thay vào B ta được :
\(B=\left(1-\frac{z}{x}\right)\left(1-\frac{x}{y}\right)\left(1-\frac{y}{z}\right)\)
\(=\left(1-\frac{x-y}{x}\right)\left(1-\frac{y+z}{y}\right)\left(1-\frac{x-z}{z}\right)\)
\(=\left(\frac{-y}{x}\right)\left(\frac{z}{y}\right)\left(\frac{-x}{z}\right)\)
\(=\frac{-yz\left(-x\right)}{xyz}\)
\(=\frac{xyz}{xyz}=1\)
Mình k dám chắc nhá
@Phạm Trà Giang : \(z-y=x????\)phải biến đối là: \(x=y+z\)
@Không Tên : \(-z=x-y???\)chuyển vế không đối dấu hả man:)) \(x-y=z\)mà :D
Đề:
Cho các số thực x, y, z thoả mãn x + y + z = 1 và \(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=1\)
\(\left(x\ne-y;y\ne-z;z\ne-x\right)\)
Giá trị của biểu thức \(P=\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\) là . . .
Giải:
x + y + z = 1
=> x = 1 - (y + z)
y = 1 - (x + z)
z = 1 - (x + y)
Thay x = 1 - (y + z); y = 1 - (x + z) và z = 1 - (x + y) vào P, ta có:
\(P=\frac{x\left[1-\left(y+z\right)\right]}{y+z}+\frac{y\left[1-\left(x+z\right)\right]}{x+z}+\frac{z\left[1-\left(x+y\right)\right]}{x+y}\)
\(=\frac{x-x\left(y+z\right)}{y+z}+\frac{y-y\left(x+z\right)}{x+z}+\frac{z-z\left(x+y\right)}{x+y}\)
\(=\frac{x}{y+z}-\frac{x\left(y+z\right)}{y+z}+\frac{y}{x+z}-\frac{y\left(x+z\right)}{x+z}+\frac{z}{x+y}-\frac{z\left(x+y\right)}{x+y}\)
\(=\left(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\right)-\left(x+y+z\right)\)
\(=1-1\)
\(=0\)
ĐS: 0
Trịnh Trân Trân <3
Hay quớ ak! Mơn m nhìu nha ný! <3 <3 <3 (not thả thính =))))
Tính:a) \(A=\frac{2}{x-y}+\frac{2}{y-z}+\frac{2}{z-x}+\frac{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
b) Cho \(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=1\) . Tính \(A=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\)
a) \(A=\frac{2}{x-y}+\frac{2}{y-z}+\frac{2}{z-x}+\frac{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{2\left(y-z\right)\left(z-x\right)+2\left(x-y\right)\left(z-x\right)+2\left(x-y\right)\left(y-z\right)+\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{\left[\left(x-y\right)+\left(y-z\right)+\left(z-x\right)\right]^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=\frac{\left(x-y+y-z+z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=0\)
Áp dụng: \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
b)Ta có: \(\frac{x^2}{y+z}+x=\frac{x^2+x\left(y+z\right)}{y+z}=\frac{x^2+xy+xz}{y+z}=\frac{x\left(x+y+z\right)}{y+z}\)
Tương tự: \(\frac{y^2}{x+z}+y=\frac{y^2+xy+zy}{x+z}=\frac{y\left(x+y+z\right)}{x+z}\)
\(\frac{z^2}{x+y}+z=\frac{z^2+xz+zy}{x+y}=\frac{z\left(x+y+z\right)}{x+y}\)
Suy ra: \(A+\left(x+y+z\right)\)
\(=\frac{x\left(x+y+z\right)}{y+z}+\frac{y\left(x+y+z\right)}{z+x}+\frac{z\left(x+y+z\right)}{x+y}+\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}+1\right)\)
\(=2.\left(x+y+z\right)\)
Nên \(A=2.\left(x+y+z\right)-\left(x+y+z\right)=x+y+z\)
Mình có sai chỗ nào không nhỉ?
Cho \(x,y,z\ne0\) và \(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}\)
Tính giá trị biểu thức \(A=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)\)
Ez
ta có \(A=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{x}{z}\right)\)
\(\Leftrightarrow A=\left(\frac{y}{y}+\frac{x}{y}\right)\left(\frac{z}{z}+\frac{y}{z}\right)\left(\frac{x}{x}+\frac{z}{x}\right)\)
\(\Leftrightarrow A=\frac{x+y}{y}.\frac{y+z}{z}.\frac{z+x}{x}\left(1\right)\)
theo giả thiết \(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}\)
\(\Leftrightarrow\frac{y+z}{x}-\frac{x}{x}=\frac{z+x}{y}-\frac{y}{y}=\frac{x+y}{z}-\frac{z}{z}\)
\(\Leftrightarrow\frac{y+z}{x}-1=\frac{z+x}{y}-1=\frac{x+y}{z}-1\)
\(\Leftrightarrow\frac{y+z}{x}=\frac{z+x}{y}=\frac{x+y}{z}\)
theo tính chất dãy tỉ số bằng nhau
\(\frac{y+z}{x}=\frac{z+x}{y}=\frac{x+y}{z}=\frac{2x+2y+2z}{x+y+z}=\frac{2\left(x+y+z\right)}{\left(x+y+z\right)}=2\)
\(\left\{{}\begin{matrix}\frac{y+z}{x}=2\Leftrightarrow y+z=2x\left(2\right)\\\frac{z+x}{y}=2\Leftrightarrow z+x=2y\left(3\right)\\\frac{x+y}{z}=2\Leftrightarrow x+y=2z\left(4\right)\end{matrix}\right.\)
thay (2); (3); (4) vào (1)
\(\Leftrightarrow A=\frac{2z}{y}.\frac{2x}{z}.\frac{2y}{x}=\frac{2z.2x.2y}{xyz}=\frac{2^3\left(xyz\right)}{\left(xyz\right)}=2^3=8\)
Cho x,y,z\(\ne0\)và x-y-z=0, tính giá trị của biểu thức
B=\(\left(1-\frac{z}{x}\right)\left(1-\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\)
cho x;y;z khac 0 thỏa mãn
\(\frac{y+z-x}{x}=\frac{z+x-y}{y}x=\frac{x+y-z}{z}x\)
tính
\(B=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)\)
Cho \(x,y,z\ne0\); đôi một cùng dấu thỏa mãn: \(\left(1+\frac{y}{x}\right)\left(1+\frac{z}{y}\right)\left(1+\frac{x}{z}\right)=8\)
Tính \(M=\frac{x^2}{y^2+z^2}+\frac{y^2}{z^2+x^2}+\frac{z^2}{x^2+y^2}\)
Áp dụng bđt côsi cho 2 số dương lần lượt ta có :
\(1+\frac{y}{x}\ge2\sqrt{\frac{y}{x}}\)
\(1+\frac{z}{y}\ge2\sqrt{\frac{z}{y}}\)
\(1+\frac{x}{z}\ge2\sqrt{\frac{x}{z}}\)
Nhân vế theo vế ta đc : \(\left(1+\frac{y}{x}\right)\left(1+\frac{z}{y}\right)\left(1+\frac{x}{z}\right)\ge8\sqrt{\frac{xyz}{xyz}}=8\)
Dấu = xảy ra khi : \(1=\frac{y}{x}\)=> x=y và \(1=\frac{z}{y}\) => z=y và \(1=\frac{x}{z}\) => x=z
=> x=y=z
Thay vào M ta được : \(M=\frac{x^2}{2x^2}+\frac{y^2}{2y^2}+\frac{z^2}{2z^2}=\frac{3}{2}\).