Nhiều bài quá e, nên làm những bài dễ rùi bài khó hãng hỏi nhé
150.
\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\Rightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\)
\(\Rightarrow ayz+bxz+cxy=0\)
\(x+y+z=0\Rightarrow\left\{{}\begin{matrix}-x=y+z\\-y=x+z\\-z=x+y\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=\left(y+z\right)^2\\y^2=\left(x+z\right)^2\\z^2=\left(x+y\right)^2\end{matrix}\right.\)
Do đó:
\(ax^2+by^2+cz^2=a\left(y+z\right)^2+b\left(x+z\right)^2+c\left(x+y\right)^2\)
\(=a\left(y^2+z^2+2yz\right)+b\left(x^2+z^2+2xz\right)+c\left(x^2+y^2+2xy\right)\)
\(=x^2\left(b+c\right)+y^2\left(a+c\right)+z^2\left(a+b\right)+2\left(ayz+bxz+cxy\right)\)
\(=x^2.\left(-a\right)+y^2\left(-b\right)+z^2.\left(-c\right)+2.0\)
\(=-\left(ax^2+by^2+cz^2\right)\)
\(\Rightarrow2\left(ax^2+by^2+cz^2\right)=0\)
\(\Rightarrow ax^2+by^2+cz^2=0\)
151.
Biểu thức xác định khi \(xyz\ne0\)
Giả thiết tương đương:
\(x+\dfrac{1}{y}=y+\dfrac{1}{z}=z+\dfrac{1}{x}\)
Từ \(x+\dfrac{1}{y}=y+\dfrac{1}{z}\Rightarrow x-y=\dfrac{1}{z}-\dfrac{1}{y}\)
\(\Rightarrow yz\left(x-y\right)=y-z\) (1)
Tương tự: \(y+\dfrac{1}{z}=z+\dfrac{1}{x}\Rightarrow xz\left(y-z\right)=z-x\) (2)
\(x+\dfrac{1}{y}=z+\dfrac{1}{x}\Rightarrow xy\left(z-x\right)=x-y\) (3)
- Nếu tồn tại 2 trong 3 số x;y;z bằng nhau, giả sử \(x=y\)
Thay vào (3) \(\Rightarrow xy\left(z-x\right)=0\Rightarrow z=x\)
\(\Rightarrow x=y=z\)
- Nếu x;y;z đôi một phân biệt \(\Rightarrow\left(x-y\right)\left(y-z\right)\left(z-x\right)\ne0\)
Nhân vế (1),(2) và (3):
\(\Rightarrow x^2y^2z^2\left(x-y\right)\left(y-z\right)\left(z-x\right)=\left(x-y\right)\left(y-z\right)\left(z-x\right)\)
\(\Rightarrow x^2y^2z^2=1\)
152.
Ta có:
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1\)
\(\Rightarrow\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)\left(a+b+c\right)=a+b+c\)
\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{a}{b+c}\left(b+c\right)+\dfrac{b^2}{c+a}+\dfrac{b}{c+a}\left(c+a\right)+\dfrac{c^2}{a+b}+\dfrac{c}{a+b}\left(a+b\right)=a+b+c\)
\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+a+b+c=a+b+c\)
\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=0\)
153.
Ta có:
\(\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}=0\)
\(\Rightarrow\left(\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}\right)\left(\dfrac{1}{b-c}+\dfrac{1}{c-a}+\dfrac{1}{a-b}\right)=0\)
\(\Rightarrow\dfrac{a}{\left(b-c\right)^2}+\dfrac{a}{b-c}\left(\dfrac{1}{c-a}+\dfrac{1}{a-b}\right)+...=0\) (phần đằng sau hơi dài em tự ghi)
\(\Rightarrow\dfrac{a}{\left(b-c\right)^2}+\dfrac{a\left(c-b\right)}{\left(b-c\right)\left(c-a\right)\left(a-b\right)}+...=0\)
\(\Rightarrow\dfrac{a}{\left(b-c\right)^2}-\dfrac{a}{\left(a-b\right)\left(c-a\right)}+...=0\)
\(\Rightarrow\dfrac{a}{\left(b-c\right)^2}+\dfrac{b}{\left(c-a\right)^2}+\dfrac{c}{\left(a-b\right)^2}-\dfrac{a\left(b-c\right)+b\left(c-a\right)+c\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
\(\Rightarrow\dfrac{a}{\left(b-c\right)^2}+\dfrac{b}{\left(c-a\right)^2}+\dfrac{c}{\left(a-b\right)^2}+0=0\)
154.
\(x^2+\dfrac{1}{x^2}=\left(x+\dfrac{1}{x}\right)^2-2=a^2-2\)
\(x^3+\dfrac{1}{x^3}=\left(x+\dfrac{1}{x}\right)^3-3x.\dfrac{1}{x}\left(x+\dfrac{1}{x}\right)=a^3-3a\)
\(x^4+\dfrac{1}{x^4}=\left(x^2+\dfrac{1}{x^2}\right)^2-2=\left(a^2-2\right)^2-2=a^2-4a+2\)
\(x^5+\dfrac{1}{x^5}=\left(x^2+\dfrac{1}{x^2}\right)\left(x^3+\dfrac{1}{x^3}\right)-\left(\dfrac{x^2}{x^3}+\dfrac{x^3}{x^2}\right)\)
\(=\left(x^2+\dfrac{1}{x^2}\right)\left(x^3+\dfrac{1}{x^3}\right)-\left(x+\dfrac{1}{x}\right)\)
\(=\left(a^2-2\right)\left(a^3-3a\right)-a\)
155.
\(\left(x^2-\dfrac{1}{x^2}\right):\left(x^2+\dfrac{1}{x^2}\right)=a\)
\(\Rightarrow\left(\dfrac{x^4-1}{x^2}\right):\left(\dfrac{x^4+1}{x^2}\right)=a\)
\(\Rightarrow\dfrac{x^4-1}{x^4+1}=a\)
\(\Rightarrow x^4-1=a\left(x^4+1\right)\)
\(\Rightarrow x^4-1=ax^4+a\)
\(\Rightarrow x^4\left(1-a\right)=a+1\)
\(\Rightarrow x^4=\dfrac{a+1}{1-a}\)
Do đó:
\(M=\left(\dfrac{x^8-1}{x^4}\right):\left(\dfrac{x^8+1}{x^4}\right)=\dfrac{x^8-1}{x^8+1}=\dfrac{\left(\dfrac{a+1}{1-a}\right)^2-1}{\left(\dfrac{a+1}{1-a}\right)^2+1}\)
\(=\dfrac{\left(a+1\right)^2-\left(1-a\right)^2}{\left(a+1\right)^2+\left(1-a\right)^2}=\dfrac{4a}{2a^2+2}=\dfrac{2a}{a^2+1}\)
156.
\(x^2-4x+1=0\Rightarrow x^2+1=4x\)
\(\Rightarrow x+\dfrac{1}{x}=4\)
\(\Rightarrow\left(x+\dfrac{1}{x}\right)^2=16\)
\(\Rightarrow x^2+\dfrac{1}{x^2}+2=16\)
\(\Rightarrow x^2+\dfrac{1}{x^2}=14\)
Do đó:
\(A=\dfrac{x^4+x^2+1}{x^2}=x^2+\dfrac{1}{x^2}+1=14+1=15\)
157.
Với \(x=0\Rightarrow a=0\Rightarrow M=0\)
Với \(x\ne0\)
\(\dfrac{x}{x^2-x+1}=a\Rightarrow\dfrac{x^2-x+1}{x}=\dfrac{1}{a}\)
\(\Rightarrow x+\dfrac{1}{x}-1=\dfrac{1}{a}\)
\(\Rightarrow x+\dfrac{1}{x}=\dfrac{1}{a}+1\)
\(\Rightarrow\left(x+\dfrac{1}{x}\right)^2=\left(\dfrac{1}{a}+1\right)^2\)
\(\Rightarrow x^2+\dfrac{1}{x^2}+1=\dfrac{1}{a^2}+\dfrac{2}{a}\)
Do đó:
\(M=\dfrac{1}{x^2+1+\dfrac{1}{x^2}}=\dfrac{1}{\dfrac{1}{a^2}+\dfrac{2}{a}}\)
\(=\dfrac{a^2}{2a+1}\)
158.
\(x+1=\dfrac{b^2+c^2-a^2}{2bc}+1=\dfrac{\left(b+c\right)^2-a^2}{2bc}\)
\(y+1=\dfrac{a^2-\left(b-c\right)^2}{\left(b+c\right)^2-a^2}+1=\dfrac{\left(b+c\right)^2-\left(b-c\right)^2}{\left(b+c\right)^2-a^2}=\dfrac{4bc}{\left(b+c\right)^2-a^2}\)
Từ đó ta có:
\(x+y+xy=\left(xy+x+y+1\right)-1\)
\(=\left(x+1\right)\left(y+1\right)-1\)
\(=\dfrac{\left(b+c\right)^2-a^2}{2bc}.\dfrac{4bc}{\left(b+c\right)^2-a^2}-1\)
\(=\dfrac{4}{2}-1=1\)
159.
a. ĐKXĐ: \(b\ne0\)
Đặt \(a-b=\dfrac{a}{b}=k\in N\)
\(\Rightarrow\left\{{}\begin{matrix}a-b=k\\a=kb\end{matrix}\right.\)
\(\Rightarrow kb-b=k\)
\(\Rightarrow kb-b-k+1=1\)
\(\Rightarrow b\left(k-1\right)-\left(k-1\right)=1\)
\(\Rightarrow\left(b-1\right)\left(k-1\right)=1\)
\(\Rightarrow\left[{}\begin{matrix}b=k=0\left(loại\right)\\b=k=2\end{matrix}\right.\)
\(\Rightarrow a=kb=4\)
Vậy \(\left(a;b\right)=\left(4;2\right)\)
b.
Tương tự câu a, đặt \(a-b=\dfrac{a}{2b}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a-b=k\\a=2kb\end{matrix}\right.\)
\(\Rightarrow2kb-b=k\)
\(\Rightarrow2b\left(2k-1\right)-\left(2k-1\right)=1\)
\(\Rightarrow\left(2b-1\right)\left(2k-1\right)=1\)
\(\Rightarrow\left[{}\begin{matrix}b=k=0\left(loại\right)\\b=k=1\end{matrix}\right.\)
\(\Rightarrow a=2kb=2\)
Vậy \(\left(a;b\right)=\left(2;1\right)\)