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\(a,b,c\ne0\)
\(\dfrac{ac+bc-c^2}{abc}-\dfrac{ab+ac-a^2}{abc}-\dfrac{ab+bc-b^2}{abc}=0\)
\(\Leftrightarrow\dfrac{ac+bc-c^2-ab-ac+a^2-ab-bc+b^2}{abc}=0\)
\(\Leftrightarrow a^2+b^2-c^2-2ab=0\)
\(\Leftrightarrow\left(a-b\right)^2-c^2=0\)
\(\Leftrightarrow\left(a-b-c\right)\left(a-b+c\right)=0\)
\(\Leftrightarrow\left(b+c-a\right)\left(a+c-b\right)=0\) \(\Rightarrow\left[{}\begin{matrix}b+c-a=0\\a+c-b=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{b+c-a}{bc}=0\\\dfrac{a+c-b}{ac}=0\end{matrix}\right.\) (đpcm)
Lời giải:
\(\frac{x^2+y^2-z^2}{2xy}+\frac{y^2+z^2-x^2}{2yz}+\frac{x^2+z^2-y^2}{2xz}=1\)
\(\Leftrightarrow \frac{x^2+y^2-z^2}{2xy}+1+\frac{y^2+z^2-x^2}{2yz}-1+\frac{x^2+z^2-y^2}{2xz}-1=0\)
\(\Leftrightarrow \frac{(x+y-z)(x+y+z)}{2xy}+\frac{(y-z-x)(y-z+x)}{2yz}+\frac{(x-z-y)(x-z+y)}{2xz}=0\)
\(\Leftrightarrow (x+y-z)\left[\frac{x+y+z}{2xy}+\frac{y-z-x}{2yz}+\frac{x-z-y}{2xz}\right]=0\)
\(\Leftrightarrow (x+y-z)(xz+yz+z^2+xy-zx-x^2+xy-zy-y^2)=0\)
\(\Leftrightarrow (x+y-z)[z^2-(x-y)^2]=0\Leftrightarrow (x+y-z)(z-x+y)(x+z-y)=0\)
Nếu $x+y-z=0$ thì:
\(\frac{x^2+y^2-z^2}{2xy}=\frac{(x+y)^2-z^2-2xy}{2xy}=-1\); \(\frac{y^2+z^2-x^2}{2yz}=\frac{z(y-x)+z^2}{2yz}=\frac{y-x+z}{2y}=\frac{y-x+y+x}{2y}=1\)
\(\frac{x^2+z^2-y^2}{2xz}=1-(-1)-1=1\)
Ta có đpcm.
Các TH còn lại tương tự.
Vậy........
giải giúp mình với, mình gấp lắm,mai phải nộp rồi(1like nha)
Bài 3:
a: ĐKXĐ: x<>2
b: \(M=\dfrac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{3}{x-2}\)
c: Khi x=4001/2000 thì \(M=\dfrac{3}{\dfrac{4001}{2000}-2}=3:\dfrac{1}{2000}=6000\)
a. ĐK: a, b, c khác 0.
\(\frac{a^2+b^2-c^2}{2ab}+\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ac}=1\)
\(\Leftrightarrow\left[\frac{a^2+b^2-c^2}{2ab}-1\right]+\left[\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ca}\right]=0\)
\(\Leftrightarrow\frac{\left(a-b\right)^2-c^2}{2ab}+\frac{1}{2c}\left[\frac{c^2-\left(a^2-b^2\right)}{b}+\frac{c^2+\left(a^2-b^2\right)}{a}\right]=0\)
\(\Leftrightarrow\frac{\left(a-b\right)^2-c^2}{2ab}+\frac{1}{2c}\left[\frac{c^2\left(a+b\right)-\left(a^2-b^2\right)\left(a-b\right)}{ab}\right]=0\)
\(\Leftrightarrow\frac{\left(a-b\right)^2-c^2}{2ab}+\frac{\left(a+b\right)\left(c^2-\left(a-b\right)^2\right)}{2abc}=0\)
\(\Leftrightarrow\left[\left(a-b\right)^2-c^2\right]\left(1-\frac{a+b}{c}\right)=0\)
\(\Leftrightarrow\left(a-b-c\right)\left(a-b+c\right)\left(c-a-b\right)=0\)
\(\Leftrightarrow a=b+c\)hoặc \(b=a+c\)hoặc \(c=a+b\).
b) Không mất tính tổng quả. G/s: a = b + c
Khi đó ta có:
\(\frac{a^2+b^2-c^2}{2ab}=\frac{\left(b+c\right)^2+b^2-c^2}{2\left(b+c\right)b}=1\)
\(\frac{b^2+c^2-a^2}{2bc}=\frac{b^2+c^2-\left(b+c\right)^2}{2bc}=-1\)
\(\frac{c^2+a^2-b^2}{2ca}=\frac{c^2+\left(b+c\right)^2-b^2}{2\left(b+c\right)c}=1\)
=> Điều phải chứng minh.