cho \(A=\dfrac{1}{b^2+c^2-a^2}+\dfrac{1}{c^2+a^2-b^2}+\dfrac{1}{a^2+b^2-c^2}\)Rút gọn biểu thức A biết a+b+c=0
Cho a, b, c khác nhau đôi một và \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\). Rút gọn các biểu thức:
a) M= \(\dfrac{1}{a^2+2bc}+\dfrac{1}{b^2+2ac}+\dfrac{1}{c^2+2ab}\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Leftrightarrow ab+bc+ca=0\Rightarrow\left\{{}\begin{matrix}bc=-ab-ac\\ab=-bc-ac\\ac=-ab-bc\end{matrix}\right.\)
\(M=\dfrac{1}{a^2+bc-ab-ac}+\dfrac{1}{b^2+ac-ab-bc}+\dfrac{1}{c^2+ab-bc-ac}\)
\(=\dfrac{1}{a\left(a-b\right)-c\left(a-b\right)}+\dfrac{1}{b\left(b-c\right)-a\left(b-c\right)}+\dfrac{1}{c\left(c-a\right)-b\left(c-a\right)}\)
\(=\dfrac{1}{\left(a-b\right)\left(a-c\right)}-\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(a-c\right)\left(b-c\right)}\)
\(=\dfrac{b-c-\left(a-c\right)+a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=0\)
Cho A=\(\dfrac{1}{b^2+c^2-a^2}+\dfrac{1}{c^2+a^2-b^2}+\dfrac{1}{a^2+b^2-c^2}\)
Rút gọn A, biết a+b+c=0
Ta có: a+b+c=0\(\Leftrightarrow\)b+c=-a
Bình phương hai vế có: (b+c)2=a2
⇔ b2+2bc+c2=a2\(\Leftrightarrow\) b2+c2-a2=-2bc
Tương tự, ta có: c2+a2-b2=-2ca
a2+b2-c2=-2ab
→ A=\(-\dfrac{1}{2bc}-\dfrac{1}{2ca}-\dfrac{1}{2ab}=\dfrac{-\left(a+b+c\right)}{2abc}=0\)(vì a+b+c=0)
Vậy A=0
Cho 3 số khác 0 a, b, c và a+b+c=0. Rút gọn biểu thức: \(M=\dfrac{a^2}{a^2-b^2-c^2}+\dfrac{b^2}{b^2-c^2-a^2}+\dfrac{c^2}{c^2-a^2-b^2}\)
\(\dfrac{a^2}{a^2-b^2-c^2}=\dfrac{a^2}{\left(a-b\right)\left(a+b\right)-c^2}=\dfrac{a^2}{\left(a-b\right)\left(-c\right)-c^2}=\dfrac{a^2}{c\left(b-a-c\right)}=\dfrac{a^2}{2bc}\\ \Leftrightarrow M=\sum\dfrac{a^2}{a^2-b^2-c^2}=\sum\dfrac{a^2}{2bc}=\dfrac{a^3+b^3+c^3}{2abc}\\ \Leftrightarrow M=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{2abc}=0\)
Cho biểu thức A = \(\dfrac{x-1}{x^2}\) và B = \(\dfrac{1}{x}-\dfrac{x}{2x+1}\)+\(\dfrac{2x^2-3x-1}{x\left(2x+1\right)}\) với x ≠ 0; x ≠ \(\dfrac{-1}{2}\); x ≠ 1
1. Rút gọn biểu thức B.
2. Đặt C = A : B. Chứng minh: C ≥ -1.
Giúp mình với ạ
1: \(B=\dfrac{2x+1-x^2+2x^2-3x-1}{x\left(2x+1\right)}=\dfrac{x^2-x}{x\left(2x+1\right)}=\dfrac{x-1}{2x+1}\)
2: \(C=A:B\)
\(=\dfrac{x-1}{x^2}:\dfrac{x-1}{2x+1}=\dfrac{2x+1}{x^2}\)
\(C+1=\dfrac{2x+1+x^2}{x^2}=\dfrac{\left(x+1\right)^2}{x^2}>=0\)
=>C>=-1
Cho a+b+c=0 (a khác 0, b khác 0, c khác 0). Rút gọn các biểu thức: \(A=\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab}\)
\(a+b=-c\Leftrightarrow\left(a+b\right)^3=-c^3\)
\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)
\(\Leftrightarrow a^3+b^3+c^3=-3ab\left(a+b\right)=3abc\)
\(A=\dfrac{a^3+b^3+c^3}{abc}=\dfrac{3abc}{abc}=3\)
1. Cho a,b,c không đồng thời bằng 0 và a+b+c=0. Rút gọn:
\(N=\dfrac{a^2}{a^2-b^2-c^2}+\dfrac{b^2}{b^2-c^2-a^2}+\dfrac{c^2}{c^2-a^2-b^2}\)
2. CMR: Nếu a+b+c=2x thì:
\(\dfrac{1}{x-a}+\dfrac{1}{x-b}+\dfrac{1}{x-c}-\dfrac{1}{x}=\dfrac{abc}{x\left(x-a\right)\left(x-b\right)\left(x-c\right)}\)
\(1,a+b+c=0\Leftrightarrow a=-b-c\Leftrightarrow a^2=b^2+2bc+c^2\Leftrightarrow b^2+c^2=a^2-2bc\)
Tương tự: \(\left\{{}\begin{matrix}a^2+b^2=c^2-2ab\\c^2+a^2=b^2-2ac\end{matrix}\right.\)
\(\Leftrightarrow N=\dfrac{a^2}{a^2-a^2+2bc}+\dfrac{b^2}{b^2-b^2+2ca}+\dfrac{c^2}{c^2-c^2+2ac}\\ \Leftrightarrow N=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2bc}=\dfrac{a^3+b^3+c^3}{2abc}=\dfrac{a^3+b^3+c^3-3abc+3abc}{2abc}\\ \Leftrightarrow N=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc}{2abc}\\ \Leftrightarrow N=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)
bài 1 :a) Cho abc=2. Rút gọn biểu thức : M=\(\dfrac{a}{ab+a+2}\)+\(\dfrac{b}{bc+b+1}\)+\(\dfrac{2c}{ac+2c+2}\)
b) Cho abc=1. Rút gọn biểu thức : N= \(\dfrac{a}{ab+a+1}\)+\(\dfrac{b}{bc+b+1}\)+\(\dfrac{c}{ac+c+1}\)
bài 2: Cho a+b+c=0 ( a,b,c cùng khác 0) . Rút gọn các biểu thức:
a) A=\(\dfrac{a^2}{bc}\)+\(\dfrac{b^2}{ca}\)+\(\dfrac{c^2}{ab}\)
b)B= \(\dfrac{a^2}{a^2-b^2-c^2}\)+\(\dfrac{b^2}{b^2-c^2-a^2}\)+\(\dfrac{c^2}{c^2-a^2-b^2}\)
Bài 2:
a) \(A=\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab}\)
\(A=\dfrac{a^3}{abc}+\dfrac{b^3}{abc}+\dfrac{c^3}{abc}\)
\(A=\dfrac{1}{abc}\left(a^3+b^3+c^3\right)\)
\(A=\dfrac{1}{abc}\left[\left(a+b\right)^3-3ab\left(a+b\right)+c^3\right]\)
Vì \(a+b+c=0\)
Nên a + b = -c (1)
Thay (1) vào A, ta được:
\(A=\dfrac{1}{abc}\left[\left(-c\right)^3-3ab\left(-c\right)+c^3\right]\)
\(A=\dfrac{1}{abc}.3abc\)
\(A=3\)
b) \(B=\dfrac{a^2}{a^2-b^2-c^2}+\dfrac{b^2}{b^2-c^2-a^2}+\dfrac{c^2}{c^2-a^2-b^2}\)
\(B=\dfrac{a^2}{a^2-\left(b^2+c^2\right)}+\dfrac{b^2}{b^2-\left(c^2+a^2\right)}+\dfrac{c^2}{c^2-\left(a^2+b^2\right)}\)
Vì \(a+b+c=0\)
Nên b + c = -a
=> ( b + c )2 = (-a)2
=> b2 + c2 + 2bc = a2
=> b2 + c2 = a2 - 2bc (1)
Tương tự ta có: c2 + a2 = b2 - 2ac (2)
a2 + b2 = c - 2ab (3)
Thay (1), (2) và (3) vào B, ta được:
\(B=\dfrac{a^2}{a^2-\left(a^2-2bc\right)}+\dfrac{b^2}{b^2-\left(b^2-2ac\right)}+\dfrac{c^2}{c^2-\left(c^2-2ab\right)}\)
\(B=\dfrac{a^2}{a^2-a^2+2bc}+\dfrac{b^2}{b^2-b^2+2ac}+\dfrac{c^2}{c^2-c^2+2ab}\)
\(B=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2ab}\)
\(B=\dfrac{a^3}{2abc}+\dfrac{b^3}{2abc}+\dfrac{c^3}{2abc}\)
\(B=\dfrac{1}{2abc}\left(a^3+b^3+c^3\right)\)
Mà \(a^3+b^3+c^3=3abc\) ( câu a )
\(\Rightarrow B=\dfrac{1}{2abc}.3abc\)
\(\Rightarrow B=\dfrac{3}{2}\)
Bài 1:
a) GT: abc = 2
\(M=\dfrac{a}{ab+a+2}+\dfrac{b}{bc+b+1}+\dfrac{2c}{ac+2c+2}\)
\(M=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{2cb}{abc+2cb+2b}\)
\(M=\dfrac{a}{a\left(b+1+bc\right)}+\dfrac{b}{bc+b+1}+\dfrac{2cb}{2+2cb+2b}\)
\(M=\dfrac{1}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{2cb}{2\left(1+cb+b\right)}\)
\(M=\dfrac{1}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{bc}{bc+b+1}\)
\(M=\dfrac{1+b+bc}{bc+b+1}\)
\(M=1\)
b) GT: abc = 1
\(N=\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}\)
\(N=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{cb}{b\left(ac+c+1\right)}\)
\(N=\dfrac{a}{a\left(b+1+bc\right)}+\dfrac{b}{bc+b+1}+\dfrac{bc}{abc+bc+b}\)
\(N=\dfrac{1}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{bc}{bc+b+1}\)
\(N=\dfrac{1+b+bc}{bc+b+1}\)
\(N=1\)
Rút gọn biểu thức: \(B=\left(ab+bc+ca\right).\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)-abc.\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\)
\(B=\left(ab+bc+ca\right)\left(\dfrac{ab+bc+ca}{abc}\right)-abc\left(\dfrac{a^2b^2+b^2c^2+c^2a^2}{a^2b^2c^2}\right)\)
\(=\dfrac{\left(ab+bc+ca\right)^2-\left(a^2b^2+b^2c^2+c^2a^2\right)}{abc}\)
\(=\dfrac{a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)-\left(a^2b^2+b^2c^2+c^2a^2\right)}{abc}\)
\(=2\left(a+b+c\right)\)
2 a. rút gọn biểu C = \(\dfrac{2x^{\text{2}}-x}{\text{x }-1}+\dfrac{x+1}{1-x}+\dfrac{2-x^2}{x-1}\)
b. Rút gọn biểu thức D = \(\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{\text{a}}-1}\right):\dfrac{\sqrt{\text{a}}+1}{a-2\sqrt{a}+1}\)
Vậy khi rút gọn một biểu thức hửu tỉ và một biểu thức chứa căn có tìm điều kiện xác định không?
\(a,C=\dfrac{2x^2-x-x-1+2-x^2}{x-1}\left(x\ne1\right)\\ C=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\\ b,D=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\left(a>0;a\ne1\right)\\ D=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
Có