Nếu\(x=\frac{a}{b},a\ne b,a;b\ne0\) thì \(\frac{a+b}{a-b}\) bằng bao nhiêu?
CMR: Nếu a(y+z)=b(z+x)=c(x+y)\(\left(a\ne b\ne c\ne0\right)\)thì \(\frac{y-z}{a\left(b-c\right)}=\frac{z-x}{b\left(c-a\right)}=\frac{x-y}{c\left(a-b\right)}\)
CMR: Nếu a(y + z) = b(z + x) = c(x + y) \(\left(a\ne b\ne c\ne0\right)\)
Thì \(\frac{y-z}{a\left(b-c\right)}=\frac{z-x}{b\left(c-a\right)}=\frac{x-y}{c\left(a-b\right)}\)
Từ giả thiết ta suy ra \(\frac{a\left(y+z\right)}{abc}=\frac{b\left(z+x\right)}{abc}=\frac{c\left(x+y\right)}{abc}\to\frac{y+z}{bc}=\frac{z+x}{ca}=\frac{x+y}{ab}\).
Áp dụng tính chất của dãy tỉ số bằng nhau ta được từ
\(\frac{y+z}{bc}=\frac{z+x}{ca}=\frac{x+y}{ab}\to\frac{y+z}{bc}=\frac{z+x}{ca}=\frac{x+y}{ab}=\frac{\left(z+x\right)-\left(x+y\right)}{ca-ab}=\frac{z-y}{a\left(c-b\right)}=\frac{y-z}{a\left(b-c\right)}.\) (1)
Tương tự, \(\frac{y+z}{bc}=\frac{z+x}{ca}=\frac{x+y}{ab}\to\frac{y+z}{bc}=\frac{z+x}{ca}=\frac{x+y}{ab}=\frac{\left(y+z\right)-\left(x+y\right)}{bc-ab}=\frac{z-x}{b\left(c-a\right)},\) (2)
và
\(\frac{y+z}{bc}=\frac{z+x}{ca}=\frac{x+y}{ab}\to\frac{y+z}{bc}=\frac{z+x}{ca}=\frac{x+y}{ab}=\frac{\left(y+z\right)-\left(z+x\right)}{bc-ca}=\frac{y-x}{c\left(b-a\right)}=\frac{x-y}{c\left(a-b\right)}.\) (3)
Từ (1), (2), (3) ta suy ra \(\frac{y-z}{a\left(b-c\right)}=\frac{z-x}{b\left(c-a\right)}=\frac{x-y}{c\left(a-b\right)}.\) (ĐPCM)
Nếu \(x=\frac{a}{b},a\ne b,a;b\ne0\) thì\(\frac{a+b}{a-b}\) bằng bao nhiêu?
Tìm x: \(\frac{x-ab}{a+b}+\frac{x-ac}{a+c}+\frac{x-bc}{b+c}=a+b+c\) với \(a\ne-b;b\ne-c;c\ne-a\)
\(\frac{x-ab}{a+b}+\frac{x-ac}{a+c}+\frac{x-bc}{b+c}=a+b+c\)
\(\frac{x-ab}{a+b}-c+\frac{x-ac}{a+c}-b+\frac{x-bc}{b+c}-a=0\)
\(\frac{x-ab-ac-bc}{a+b}+\frac{x-ac-ba-bc}{a+c}+\frac{x-bc-ab-ac}{b+c}=0\)
\(\left(x-ab-ac-bc\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)=0\)
\(x-ab-ac-bc=0\)
\(x=ab+ac+bc\)
Tìm x biết : \(\frac{x-ab}{a+b}+\frac{x-ac}{a+c}+\frac{x-bc}{b+c}=a+b+c\) với \(a\ne-b;b\ne-c;c\ne-a\)
<=> \(\left(\frac{x-ab}{a+b}-c\right)+\left(\frac{x-ac}{a+c}-b\right)+\left(\frac{x-bc}{b+c}-a\right)=0\)
<=>\(\frac{x-ab-ac-bc}{a+b}+\frac{x-ab-ac-bc}{a+c}+\frac{x-ab-ac-bc}{b+c}=0\)
<=>\(\left(x-ab-ac-bc\right)\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)=0\)
Vì \(a\ne-b;b\ne-c;c\ne-a\) nên tổng 3 phân số kia khác 0
=> (x-ab-ac-ca)=0
=>x=ab+ac+ca
\(CMR,nếu\)
\(a\left(y+z\right)=b\left(z+x\right)=c\left(x+y\right)\)\(a\ne b\ne c\)\(a,b,c\ne0\)
\(Thì\)\(\frac{y-x}{a\left(b-c\right)}=\frac{z-x}{b\left(c-a\right)}=\frac{x-y}{c\left(a-b\right)}\)
\(a\left(y+z\right)=b\left(z+x\right)=c\left(x+y\right)\Leftrightarrow\frac{y+z}{\frac{1}{a}}=\frac{z+x}{\frac{1}{b}}=\frac{x+y}{\frac{1}{c}}=\)
\(=\frac{y+z-\left(z+x\right)}{\frac{1}{a}-\frac{1}{b}}=\frac{z+x-\left(x+y\right)}{\frac{1}{b}-\frac{1}{c}}=\frac{x+y-\left(y+z\right)}{\frac{1}{c}-\frac{1}{a}}=\frac{y-x}{\frac{b-a}{ab}}=\frac{z-y}{\frac{c-b}{bc}}=\frac{x-z}{\frac{a-c}{ac}}\)
Chia các vế của 3 tỷ lệ thức cuối cho abc ta có:
\(\frac{y-x}{\frac{b-a}{ab}\cdot abc}=\frac{z-y}{\frac{c-b}{bc}\cdot abc}=\frac{x-z}{\frac{a-c}{ac}\cdot abc}=\frac{y-x}{c\left(b-a\right)}=\frac{z-y}{a\left(c-b\right)}=\frac{x-z}{b\left(a-c\right)}\)
Hay: \(\frac{x-y}{c\left(a-b\right)}=\frac{y-z}{a\left(b-c\right)}=\frac{z-x}{b\left(c-a\right)}\)đpcm
Giải pt: \(\frac{x-a}{b}+\frac{x-b}{a}=\frac{b}{x-a}+\frac{a}{x-b}\)với a,b \(\ne\) 0; a\(\ne\)+-b
Theo đầu bài ta có:
\(\frac{x-a}{b}+\frac{x-b}{a}=\frac{b}{x-a}+\frac{a}{x-b}\)
\(\Rightarrow\frac{a\left(x-a\right)+b\left(x-b\right)}{ab}=\frac{b\left(x-b\right)+a\left(x-a\right)}{\left(x-a\right)\left(x-b\right)}\)
\(\Rightarrow ab=\left(x^2-xb\right)-\left(xa-ab\right)\)
\(\Rightarrow x\left(x-b-a\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=0\\x-\left(a+b\right)=0\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\x=a+b\end{cases}}\)
Rút gọn:
a, A = \(\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}+\frac{x}{36-x}\) (đk: x ≥ 0 và x ≠ 36)
b, B = \(\frac{9-x}{\sqrt{x}+3}-\frac{x-6\sqrt{x}+9}{\sqrt{x}-3}-6\) (đk: x ≥ 0 và x ≠ 9)
c, C = \(\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}-\frac{2}{\sqrt{ab}}:\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\right)^2\) (đk: a > 0, b > 0 và a ≠ b)
d, D = \(\left(\frac{2-a\sqrt{a}}{2-\sqrt{a}}+\sqrt{a}\right)\left(\frac{2-\sqrt{a}}{2-a}\right)\) (đk: a ≥ 0, a ≠ 2, a ≠ 4)
\(B=\frac{9-x}{\sqrt{x}+3}-\frac{x-6\sqrt{x}+9}{\sqrt{x}-3}-6\)(đk: x ≥ 0 và x ≠ 9)
\(B=\frac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}{\sqrt{x}+3}-\frac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}-3}-6\)
\(B=\left(3-\sqrt{x}\right)-\left(\sqrt{x}-3\right)-6\)
\(B=3-\sqrt{x}-\sqrt{x}+3-6\)
\(B=-2\sqrt{x}\)
\(A=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}+\frac{x}{36-x}\)(đk: x ≥ 0 và x ≠ 36)
\(=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}-\frac{x}{x-36}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}-\frac{x}{x-36}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+6\right)-3\left(\sqrt{x-6}\right)-x}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{x+6\sqrt{x}-3\sqrt{x}+18-x}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{3\sqrt{x}+18}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{3(\sqrt{x}+6)}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{3}{\sqrt{x}-6}\)
1.cho x+y+z=xyz và xy+yz+zx≠3
cmr: x(y^2+z^2)+y(x^2+z^2)+z(x^2+y^2)/xy+yz+zx=xyz
2.cmr nếu c^2+2(ab-ac-bc)=0và b≠c,a+b≠c thì \(\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\frac{a-c}{b-c}\)
3. cho a,b,c thỏa mãn abc≠0 và ab+bc+ca=0
tính :P=\(\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)