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1) Phân số đầu nhân 2.
_ Phân số thứ 2 nhân 3, p/s thứ 3 giữ nguyên.
_ Lấy phân số đầu + p/s thứ 2 - p/s thứ 3.
_ Dựa vào dãy tỉ số bằng nhau tìm x, y, z.
2) \(x-y-z=0\Rightarrow x=y+z\)
Khi đó thay vào B được:
\(B=\left(1-\dfrac{z}{y+z}\right)\left(1-\dfrac{y+z}{y}\right)\left(1+\dfrac{y}{z}\right)\)
\(=\dfrac{y}{y+z}.\dfrac{z}{y}.\dfrac{y+z}{z}\)
\(=1\)
Vậy B = 1.
\(\dfrac{x+y-z}{z}=\dfrac{y+z-x}{x}=\dfrac{x-y+z}{y}\)
\(\Rightarrow\dfrac{x+y-z}{z}+2=\dfrac{y+z-x}{x}+2=\dfrac{x-y+z}{y}+2\)
\(\Rightarrow\dfrac{x+y-z}{z}+\dfrac{2z}{z}=\dfrac{y+z-x}{x}+\dfrac{2x}{x}=\dfrac{x-y+z}{y}+\dfrac{2y}{y}\)
\(\Rightarrow\dfrac{x+y-z+2z}{z}=\dfrac{y+z-x+2x}{x}=\dfrac{x-y+z+2y}{y}\)
\(\Rightarrow\dfrac{x+y+z}{z}=\dfrac{y+z+x}{x}=\dfrac{x+z+y}{y}\)
Điều này xảy ra khi và chỉ khi: \(\left[{}\begin{matrix}x+y+z=0\\x=y=z\end{matrix}\right.\)
\(\circledast\)Với \(x+y+z=0\Leftrightarrow\left\{{}\begin{matrix}x+y=-z\\x+z=-y\\y+z=-x\end{matrix}\right.\)
Thay vào \(A\) ta có: \(A=\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{x}{z}\right)=\left(\dfrac{x+y}{y}\right)\left(\dfrac{y+z}{z}\right)\left(\dfrac{z+x}{z}\right)=\dfrac{-z.-x.-y}{xyz}=\dfrac{-xyz}{xyz}=-1\)
\(\circledast\) Với \(x=y=z\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{y}=1\\\dfrac{y}{z}=1\\\dfrac{x}{z}=1\end{matrix}\right.\)
Thay vào \(A\) ta có:
\(A=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
Áp dụng tích chất dãy tỉ số bằng nhau ta có :
\(\dfrac{y+z-x}{x}=\dfrac{z+x-y}{y}=\dfrac{x+y-z}{z}=\dfrac{x+y+z}{x+y+z}=1\\ \Rightarrow\left\{{}\begin{matrix}y+z-x=x\\z+x-y=y\\x+y-z=z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y+z=2x\\z+x=2y\\x+y=2z\end{matrix}\right.\)
\(\Rightarrow\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)=\dfrac{x+y}{y}.\dfrac{y+z}{z}.\dfrac{x+z}{x}=\dfrac{2z}{y}.\dfrac{2x}{z}.\dfrac{2y}{x}=8\)
Vào đây nhé: Câu hỏi của Vũ Ngọc Minh Anh - Toán lớp 7 | Học trực tuyến
Từ \(\dfrac{x+y-z}{x}=\dfrac{y+z-x}{y}=\dfrac{z+x-y}{z}\)
=> \(1+\dfrac{y-z}{x}=1+\dfrac{z-x}{y}=1+\dfrac{x-y}{z}\)
=> \(\dfrac{y-z}{x}=\dfrac{z-x}{y}=\dfrac{x-y}{z}\)
Áp dụng t/c dãy tỉ số bằng nhau:
\(\dfrac{y-z}{x}=\dfrac{z-x}{y}=\dfrac{x-y}{z}=\dfrac{y-z+z-x+x-y}{x+y+z}=\dfrac{0}{x+y+z}=0\)
Ta có : \(\dfrac{y-z}{x}=0\)
=> y - z = 0 ; Vì x # 0 => y = z
\(\dfrac{z-x}{y}=0\)
=> z - x = 0 . Vì y # 0 => z = x
=> y = z = x
Ta có: A = \(\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)\)
A = (1 + 1) (1 + 1) ( 1 + 1)
A = 2 . 2 . 2 = 8
ta có x-y-z=0
->x=y+z
y=x-z
z=x-y
B=\(\left(1-\dfrac{z}{x}\right)\left(1-\dfrac{x}{y}\right)\left(1-\dfrac{y}{z}\right)\)
B=\(\left(\dfrac{x-z}{x}\right)\left(\dfrac{y-x}{y}\right)\left(\dfrac{z+y}{z}\right)\)
B=\(\dfrac{y}{x}.\left(-\dfrac{z}{y}\right)\left(\dfrac{x}{z}\right)\)
B=\(\dfrac{-\left(xyz\right)}{xyz}\)
B=-1
Ta có: x-y-z=0 <=> x=y+z Thay vào A ta có:
A=\(\left(1-\dfrac{z}{y+z}\right)\left(1-\dfrac{y+z}{y}\right)\left(1+\dfrac{y}{z}\right)\)
=\(\dfrac{y}{y+z}\cdot\left(-\dfrac{z}{y}\right)\cdot\dfrac{y+z}{z}=\dfrac{y}{z}\cdot\left(-\dfrac{z}{y}\right)=-1\)
Vậy A=-1
theo bài ra táo:
\(A=\left(1-\dfrac{z}{x}\right)\left(1-\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\\ \Rightarrow A=\dfrac{x-z}{x}.\dfrac{y-x}{y}.\dfrac{z+y}{z}\left(1\right)\)
ta lại có:
\(x-y-z=0\\ \Rightarrow\left\{{}\begin{matrix}x-z=y\left(2\right)\\y-x=-z\left(3\right)\\z+y=x\left(4\right)\end{matrix}\right.\)
thay 2;3;4 vào 1 ta có:
\(A=\dfrac{y}{x}.\dfrac{-z}{y}.\dfrac{x}{z}=-1\)
vậy A = -1
\(\dfrac{y+z-x}{x}=\dfrac{z+x-y}{y}=\dfrac{x+y-z}{z}\)
\(\Rightarrow\dfrac{y+z-x}{x}+2=\dfrac{z+x-y}{y}+2=\dfrac{x+y-z}{z}+2\)
\(\Rightarrow\dfrac{x+y+z}{x}=\dfrac{x+y+z}{y}=\dfrac{x+y+z}{z}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x+y+z}{x}=\dfrac{x+y+z}{y}\\\dfrac{x+y+z}{y}=\dfrac{x+y+z}{z}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\left(x+y+z\right)=y\left(x+y+z\right)\\y\left(x+y+z\right)=z\left(x+y+z\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(x+y+z\right)=0\\\left(y-z\right)\left(x+y+z\right)=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=y\\x+y+z=0\end{matrix}\right.\\\left[{}\begin{matrix}y=z\\x+y+z=0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=y=z\\x+y+z=0\end{matrix}\right.\)
\(\circledast\) Với \(x=y=z\) thì \(A=\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
\(\circledast\) Với \(x+y+z=0\) thì\(\left\{{}\begin{matrix}x+y=-z\\x+z=-y\\y+z=-x\end{matrix}\right.\)
Khi đó \(A=\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)=\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz}=\dfrac{-xyz}{xyz}=-1\)
@ Mashiro Shiina
@Akai Haruma
@Nguyễn Thanh Hằng
@Đẹp Trai Không Bao Giờ Sai
Ta có :
\(\dfrac{x+y-z}{z}=\dfrac{y+z-x}{x}=\dfrac{z+x-y}{y}\\ \Leftrightarrow\dfrac{x+y+z}{z}=\dfrac{x+y+z}{x}=\dfrac{x+y+z}{y}\left(cùngcộngthêm2\right)\)
TH1: \(x+y+z\ne0\)
\(\Rightarrow x=y=z\)
\(\Rightarrow P=\left(1+1\right)\left(1+1\right)\left(1+1\right)\\ =2\cdot2\cdot2=8\)
TH2: \(x+y+z=0\Rightarrow\left\{{}\begin{matrix}x=-\left(y+z\right)\\y=-\left(x+z\right)\\z=-\left(y+x\right)\end{matrix}\right.\)(*)
\(\Rightarrow P=\left(1+\dfrac{-\left(y+z\right)}{y}\right)\left(1+\dfrac{-\left(z+x\right)}{z}\right)\left(1+\dfrac{-\left(x+y\right)}{z}\right)\\ =\left(1-1-\dfrac{z}{y}\right)\left(1-1-\dfrac{x}{z}\right)\left(1-1-\dfrac{y}{z}\right)\\ =\left(-\dfrac{z}{y}\right)\left(-\dfrac{x}{z}\right)\left(-\dfrac{y}{z}\right)\\ =-1\)
Vậy P=8 hoặc P=-1