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Có: 3x³ - 2x + 7 - x = 3x³ - 3x + 7

moi hok lop 6

`Answer:`

\(\frac{xy}{ay+bx}=\frac{yz}{bz+cy}=\frac{zx}{cx+ax}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}\left(1\right)\)

Theo đề ra, có: \(\frac{xy}{ay+bx}=\frac{yz}{bz+cy}=\frac{zx}{cx+az}\)

\(\Rightarrow\frac{xyz}{ayz+bxz}=\frac{xyz}{bxz+cxy}=\frac{xyz}{cxy+ayz}\)

\(\Rightarrow ayz+bxz=bxz+cxy=cxy+ayz\)

\(\Rightarrow\hept{\begin{cases}ayz+bxz=bxz+cxy\\ayz+bxz=cxy+ayz\\bxz+cxy=cxy+ayz\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}ayz=cxy\\bxz=cxy\\bxz=ayz\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}az=cx\\bz=cy\\bx=ay\end{cases}}\left(2\right)\)

Thế (2) và (1): \(\frac{xy}{2ay}=\frac{yz}{2bz}=\frac{xz}{2cx}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}\)

\(\Rightarrow\frac{x}{2a}=\frac{y}{2b}=\frac{z}{2c}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}\left(3\right)\)

\(\Rightarrow\frac{x^2}{4a^2}=\frac{y^2}{4b^2}=\frac{z^2}{4c^2}=\frac{\left(x^2+y^2+z^2\right)^2}{\left(a^2+b^2+c^2\right)^2}=\frac{x^2+y^2+z^2}{4a^2+4b^2+4c^2}\)

\(\Rightarrow\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{1}{4}\)

Thế (3) vào (2): \(\frac{x}{2a}=\frac{y}{2b}=\frac{z}{2c}=\frac{1}{4}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{a}{2}\\y=\frac{b}{2}\\z=\frac{c}{2}\end{cases}}\)

x=2. Chắc chắn 100%

minh moi hok lop 6

minh moi hok lop ? nen khong biet lam

mình mới học lớp 6 thôi !

minh moi hok lop 6

a) sai đề rồi phải là tam giác MHB=tam giác MKC chứ!!! happy new year ^_^

a/ \(\frac{x-1}{x+5}=\frac{6}{7}\Rightarrow7\left(x-1\right)=6\left(x+5\right)\Rightarrow7x-7=6x+30\Rightarrow x=37\)

b/ \(\frac{x-2}{x-1}=\frac{x+4}{x+7}\Rightarrow\left(x-2\right)\left(x+7\right)=\left(x+4\right)\left(x-1\right)\Rightarrow x^2+5x-14=x^2+3x-4\)

     \(\Rightarrow2x=10\Rightarrow x=5\)