tìm x,y,z biết \(\dfrac{xy}{ay+bx}=\dfrac{yz}{bz+cy}=\dfrac{xz}{cx+az}=\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}\)(a,b,c là hằng số)
Cho a, b, c là các số thực khác 0. Tìm các số thực x, y, z khác 0 thỏa mãn
\(\dfrac{xy}{ay+bx}=\dfrac{yz}{bz+cy}=\dfrac{zx}{cx+az}=\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}\)
\(\dfrac{xy}{ay+bx}=\dfrac{yz}{bz+cy}=\dfrac{zx}{cx+az}=\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}\left(1\right)\)
Ta có: \(\dfrac{xy}{ay+bx}=\dfrac{yz}{bz+cy}=\dfrac{zx}{cx+az}\)
\(\Rightarrow\dfrac{xyz}{ayz+bxz}=\dfrac{xyz}{bxz+cxy}=\dfrac{xyz}{cxy+ayz}\)
\(\Rightarrow ayz+bxz=bxz+cxy=cxy+ayz\)
\(\Rightarrow\left\{{}\begin{matrix}ayz+bxz=bxz+cxy\\ayz+bxz=cxy+ayz\\bxz+cxy=cxy+ayz\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}ayz=cxy\\bxz=cxy\\bxz=ayz\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}az=cx\\bz=cy\\bx=ay\end{matrix}\right.\left(2\right)\)
Thay (2) vào (1) ta có :
\(\dfrac{xy}{2ay}=\dfrac{yz}{2bz}=\dfrac{xz}{2cx}=\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}\)
\(\Rightarrow\dfrac{x}{2a}=\dfrac{y}{2b}=\dfrac{z}{2c}=\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}\left(3\right)\)
\(\Rightarrow\dfrac{x^2}{4a^2}=\dfrac{y^2}{4b^2}=\dfrac{z^2}{4c^2}=\dfrac{\left(x^2+y^2+z^2\right)^2}{\left(a^2+b^2+c^2\right)^2}=\)\(\dfrac{x^2+y^2+z^2}{4a^2+4b^2+4c^2}\)
\(\Rightarrow\dfrac{x^2+y^2+y^2}{a^2+b^2+c^2}=\dfrac{1}{4}\left(4\right).\)Thay (3) vào (2) ta có :
\(\dfrac{x}{2a}=\dfrac{y}{2b}=\dfrac{z}{2c}=\dfrac{1}{4}\Rightarrow\left\{{}\begin{matrix}x=\dfrac{a}{2}\\y=\dfrac{b}{2}\\z=\dfrac{c}{2}\end{matrix}\right.\)
Đâu lak Đạt hả Đẹp Trai Không Bao Giờ Sai???
Cho \(\dfrac{bz+cy}{x\left(-ax+by+cz\right)}=\dfrac{cx+az}{y\left(ax-by+cz\right)}=\dfrac{ay+bx}{z\left(ax+by-cz\right)}\)
CMR : \(\dfrac{ay+bx}{c}=\dfrac{bz+cy}{a}=\dfrac{cx+az}{b}\)
b) \(\dfrac{x}{a\left(b^2+c^2-a^2\right)}=\dfrac{y}{b\left(a^2+c^2-b^2\right)}=\dfrac{z}{c\left(a^2+b^2-c^2\right)}\)
Phương Ann Nhã Doanh đề bài khó wá Mashiro Shiina Đinh Đức Hùng
Nguyễn Huy Tú Lightning Farron Akai Haruma
\(\dfrac{bz-cy}{z}=\dfrac{cx-az}{b}=\dfrac{ay-bx}{c}\) (1) CMR: \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\) (*)
Lời giải:
Sửa đề: $z$ đầu tiên ở mẫu đổi thành $a$.
Ta có:
$\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}$
$=\frac{abz-cya}{a^2}=\frac{bcx-abz}{b^2}=\frac{acy-bcx}{c^2}$
$=\frac{abz-cya+bcx-abz+acy-bcx}{a^2+b^2+c^2}=\frac{0}{a^2+b^2+c^2}=0$
$\Rightarrow bz-cy=cx-az=ay-bx=0$
$\Rightarrow bz=cy; cx=az; ay=bx$
$\Rightarrow \frac{x}{a}=\frac{y}{b}=\frac{z}{c}$
Ta có đpcm.
Bai1:
1) Tìm x;y;z biết; (xy+1)/9=(xz+2)/15=(yz+3)/27 và xy+xz+yz=11
2) Biết (bz-cy)/a= (cx-az)/b=(ay-bx)/c (a,b,c khong bang 0). Chung minh rang x/a=y/b=z/c
cho \(\dfrac{cy-bz}{x}=\dfrac{az-cx}{y}=\dfrac{bx-ay}{z}\) chứng minh rằng :\(\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)
Ta có :
\(\dfrac{cy-bx}{x}=\dfrac{az-cx}{y}=\dfrac{bx-ay}{z}=\dfrac{bxz-cxy+cxy-ayz+ayz-bxz}{ax+by+cz}=0\)
\(\Rightarrow\dfrac{cy-bz}{x}=0\) \(\Rightarrow cy=bz\) \(\Rightarrow\) \(\dfrac{b}{y}=\dfrac{c}{z}\left(1\right)\)
\(\Rightarrow\dfrac{az-cx}{y}=0\) \(\Rightarrow az=cx\) \(\Rightarrow\dfrac{a}{x}=\dfrac{c}{z}\left(2\right)\)
Từ (1) và (2) suy ra : \(\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)
Cho các số a, b, c, x, y, z Thỏa mãn điều kiện: \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\).Chứng minh rằng:
\(\dfrac{bz-cy}{a}=\dfrac{cx-az}{b}=\dfrac{ay-bx}{c}\)
cho các số a, b, c khác 0, \(\dfrac{bz-cy}{a}=\dfrac{cx-az}{a}=\dfrac{ay-bx}{c}\)cmr: \(\dfrac{x}{a}=\dfrac{y}{b}\)=\(\dfrac{z}{c}\)
Cho biết : \(\dfrac{bz-cy}{a}=\dfrac{cx-az}{b}=\dfrac{ay-bx}{c}\) với a,b,c \(\ne\) 0
Chứng minh rằng \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\)
Ta có : \(\dfrac{bz-cy}{a}\text{=}\dfrac{cx-az}{b}\text{=}\dfrac{ay-bx}{c}\)
\(\Rightarrow\dfrac{a\left(bz-cy\right)}{a^2}\text{=}\dfrac{b\left(cx-az\right)}{b^2}\text{=}\dfrac{c\left(ay-bx\right)}{c^2}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{a\left(bz-cy\right)}{a^2}\text{=}\dfrac{b\left(cx-az\right)}{b^2}\text{=}\dfrac{c\left(ay-bx\right)}{c^2}\text{=}\dfrac{abz-acy+bcz-baz+cay-cbx}{a^2+b^2+c^2}\text{=}0\)
\(\Rightarrow\dfrac{bz-cy}{a}\text{=}0\Rightarrow bz\text{=}cy\)
\(\Rightarrow\dfrac{b}{c}\text{=}\dfrac{y}{z}\left(1\right)\)
\(\dfrac{cx-az}{b}\text{=}0\Rightarrow cx\text{=}az\)
\(\Rightarrow\dfrac{c}{a}\text{=}\dfrac{z}{x}\left(2\right)\)
Từ (1) và (2):
\(\Rightarrow dpcm\)
Tìm các số x,y,z khác 0 biết: \(\frac{xy}{ay+bx}=\frac{yz}{cy+bz}=\frac{xz}{az+cx}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}\left(a,b,c\ne0\right)\)