rút gọn: A= (a+1)^2+(b+1)^2+(c+1)^2 + 2(ab+bc+ca)-(a+b+c+1)^2
Những câu hỏi liên quan
cho (a+b+c)^2 = a^2 + b^2 +c^2 và abc khác 0
cmr bc/a^2 + ac/b^2 +ab/c^2 = 3
cho abc=1. rút gọn
a/ab+a+1 + b/bc+b+1 + c/ca+c+1
Rút gọn biểu thức: \(B=\left(ab+bc+ca\right).\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)-abc.\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\)
\(B=\left(ab+bc+ca\right)\left(\dfrac{ab+bc+ca}{abc}\right)-abc\left(\dfrac{a^2b^2+b^2c^2+c^2a^2}{a^2b^2c^2}\right)\)
\(=\dfrac{\left(ab+bc+ca\right)^2-\left(a^2b^2+b^2c^2+c^2a^2\right)}{abc}\)
\(=\dfrac{a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)-\left(a^2b^2+b^2c^2+c^2a^2\right)}{abc}\)
\(=2\left(a+b+c\right)\)
Đúng 1
Bình luận (0)
Bài1:Cho a+b=1.Tính \(A=a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2.\left(a+b\right)\)
Bài 2: Cho a,b,c thuộc R t/m: ab+bc+ca=abc và a+b+c=1.CMR:(a-1)(b-1)(c-1)=0
Bài 3: Cho x-y=12.Tính A=x^3-y^3-36xy
Bài 4: Rút gọn A=(ab+bc+ca)(1/a+1/b+1/c)-abc(1/a^2 + 1/b^2 +1/c^2)
Ta có A=\(\left(ab+bc+ca\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-abc\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)
=\(2\left(a+b+c\right)+\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}-\frac{ab}{c}-\frac{bc}{a}-\frac{ca}{b}=2\left(a+b+c\right)\)
Đúng 0
Bình luận (0)
\(A=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2=a^2-ab+b^2+3ab\left(1-2ab\right)+6a^2b^2\)
=\(\left(a+b\right)^2-3ab+3ab-6a^2b^2+6a^2b^2=1\)
2) Ta có \(A=\left(a-1\right)\left(b-1\right)\left(c-1\right)=abc-ab-bc-ca+a+b+c-1=0\)
Đúng 0
Bình luận (0)
bài 3 : Ta có \(A=\left(x-y\right)\left(x^2+xy+y^2\right)-36xy=12\left(x^2+xy+y^2\right)-36xy=12\left(x^2-2xy+y^2\right)\)
\(=12\left(x-y\right)^2=12.12^2=1728\)
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
Cho ab + bc + ca = 1
Rút gọn: P =\(\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}-\frac{2\left(a+b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
cho a, b, c thỏa mãn a khác +-1 và abc=1
Rút gọn biểu thức M=ab+bc+ca-a-b-c /a^2b -a^2-b+1
Rút gọn
a) x(x+1)(x-1)-(x-1)(x2+x+1)
b) (9x-1)2+(1-5x)2+2(9x-1)(1-5x)
c) (a+b+c)(a2+b2+c2-ab-bc-ca)
a) x (x+1) (x-1) - (x-1) (x2+x+1)= x3 - x2 + x2 - x - x3 + 13
= 1- x
Đúng 0
Bình luận (0)
Rút gọn
\(A=\left(ab+bc+ca\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-abc\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)
\(A=\left(ab+bc+ca\right).\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-abc.\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right).\)
\(A=\frac{1}{b}+\frac{1}{a}+\frac{ab}{c}+\frac{bc}{a}+\frac{1}{c}+\frac{1}{b}+\frac{1}{c}+\frac{ca}{b}+\frac{1}{a}-\frac{bc}{a}-\frac{ac}{b}-\frac{ab}{c}\)
\(A=2\cdot\frac{1}{b}+2\cdot\frac{1}{a}+2\cdot\frac{1}{c}\)
\(A=2.\left(\frac{1}{b}+\frac{1}{a}+\frac{1}{c}\right)\)
Đúng 0
Bình luận (0)
Đặt;\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=m\Rightarrow mabc=ab+bc+ca\)
\(\Rightarrow m^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)
\(\Rightarrow m^2-2\left(\frac{a+b+c}{abc}\right)=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
Thay vào A=\(mabc.m-abc.\left(m^2-2\left(\frac{a+b+c}{abc}\right)\right)=m^2abc-abcm^2+2\left(a+b+c\right)\)
\(=2a+2b+2c\)
Đúng 0
Bình luận (0)
\(A=\left(ab+bc+ca\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-abc\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)
\(A=\left(ab+bc+ca\right).\frac{ab+bc+ca}{abc}-\frac{abc\left(a^2b^2+b^2c^2+c^2a^2\right)}{a^2b^2c^2}\)
\(A=\frac{\left(ab+bc+ca\right)^2}{abc}-\frac{a^2b^2+b^2c^2+c^2a^2}{abc}\)
\(A=\frac{2abc\left(a+b+c\right)}{abc}\)
\(A=2\left(a+b+c\right)\)
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
Cho \(ab+bc+ca=1\)
Rút gọn
\(\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}\)
Thay \(ab+bc+ca=1\) ta có:
\(1+a^2=ab+bc+ca+a^2=b\left(c+a\right)+a\left(c+a\right)=\left(c+a\right)\left(a+b\right)\)
Tương tự: \(1+b^2=\left(b+c\right)\left(a+b\right);\) \(1+c^2=\left(c+a\right)\left(b+c\right)\)
\(\Rightarrow\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)=\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\)
\(\Rightarrow\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}=1\). Vậy biểu thức đó rút gọn lại bằng 1.
Đúng 0
Bình luận (0)
Bài 1: Cho a+b+c=0; rút gọn biểu thức A= a^2/(a^2-b^2-c^2) + b^2/(b^2-c^2-a^2) + c^2/(c^2-b^2-a^2)
Bài 2: Cho abc=2; rút gọn A= a/(ab+a+2) + b/(bc+b+1) + 2c/(ac+2c+2)