cho \(\left(a^2-bc\right)\left(b-abc\right)=\left(b^2-ac\right)\left(a-abc\right)\) ; \(abc\ne0\) và\(a\ne b\)
Chứng minh rằng: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=a+b+c\)
Những câu hỏi liên quan
cho a,b,c >0 CM \(\left(a^2+bc\right)\left(b^2+ac\right)\left(c^2+ab\right)>=abc\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
Cho a,b,c>0 thỏa mãn \(\left(ab\right)^2+\left(bc\right)^2+\left(ac\right)^2\ge\left(abc\right)^2\)
Chứng minh rằng \(\frac{\left(ab\right)^2}{\left(a^2+b^2\right)c^3}+\frac{\left(bc\right)^2}{\left(b^2+c^2\right)a^3}+\frac{\left(ac\right)^2}{\left(a^2+c^2\right)b^3}\ge\frac{\sqrt{3}}{2}\)
Cho \(\left(a^2-bc\right)\left(b-abc\right)=\left(b^2-ac\right)\left(a-abc\right);abc\ne0;a\ne b\)
CMR:\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=a+b+c\)
CMR nếu \(\left(a^2-bc\right).\left(b-abc\right)=\left(b^2-ac\right).\left(a-abc\right)\) và các số a, b, c, a-b khác 0 thì \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=a+b+c\)
\(\left(a^2-bc\right)\left(b-abc\right)=\left(b^2-ca\right)\left(a-abc\right)\)
\(\Leftrightarrow a^2b+ab^2c^2-a^3bc-b^2c=b^2a+a^2bc^2-ca^2-ab^3c\)
\(\Leftrightarrow a^2b-ab^2-b^2c+ca^2=a^2bc^2-ab^3c+a^3bc-ab^2c^2\)
\(\Leftrightarrow\left(a-b\right)\left(ab+bc+ca\right)=abc\left(a-b\right)\left(a+b+c\right)\)
\(\Leftrightarrow ab+bc+ca=abc\left(a+b+c\right)\Leftrightarrow a+b+c=\dfrac{ab+bc+ca}{abc}=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\left(đpcm\right)\)
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cho abc khác 0 , a khác b thõa mãn \(\left(a^2-bc\right)\left(b-abc\right)=\left(b^2-ac\right)\left(a-abc\right)\) cmr \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=a+b+c\)
cho a,b,c.>0 thoả mãn ab+bc+ac=1. CMR
\(\left(1+a\right)^2\left(1+b\right)^2\left(1+c\right)^2+\left(1-a\right)^2\left(1-b\right)^2\left(1-c\right)^2\ge8\sqrt{3}abc\)
phân tích đa thức thành nhân tử
a, \(ab\left(a+b\right)-bc\left(b+c\right)+ac\left(a-c\right)\)
b, \(\left(a+b\right)\left(a^2-b^2\right)+\left(b+c\right)\left(b^2-c^2\right)+\left(c+a\right)\left(c^2-a^2\right)\)
c, \(a^3\left(c-b^2\right)+b^3\left(a-c^2\right)+c^3\left(b-a^2\right)+abc\left(abc-1\right)\)
a)= ab (a + b) - bc [( a + b) - (a - c)] + ac (a - c)
= ab (a + b) - bc (a + b) + bc (a - c) +ac (a - c)
= b (a + b) (a - c) + c (a - c) (a + b)
= (a + b) (a - c) (b + c)
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b) \(=\left(a+b\right)\left(a^2-b^2\right)-\left(b+c\right)\left[\left(a^2-b^2\right)+\left(c^2-a^2\right)\right]+\left(c+a\right)\left(c^2-a^2\right)\)
\(=\left(a^2-b^2\right)\left[\left(a+b\right)-\left(b+c\right)\right]+\left(c^2-a^2\right)\left[\left(c+a\right)-\left(b+c\right)\right]\)
\(=\left(a-b\right)\left(a+b\right)\left(a-c\right)-\left(a-c\right)\left(a+c\right)\left(a-b\right)\)
\(=\left(a-b\right)\left(a-c\right)\left[\left(a+b\right)-\left(a+c\right)\right]\)
\(=\left(a-b\right)\left(a-c\right)\left(b-c\right)\)
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left(a+b+cright)^3-a^3-b^3-c^3left[left(a+bright)+cright]^3-a^3-b^3-c^3left(a+bright)^3+3left(a+bright)^2c+3left(a+bright)c^2+c^3-a^3-b^3-c^3a^3+3a^2b+3ab^2+b^3+3cleft(a^2+2ab+b^2right)+3ac^2+3bc^2-a^3-b^33a^2b+3ab^2+3a^2c+6abc+3b^2c+3ac^2+3bc^23left(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+2abcright)3left[left(a^2b+ab^2right)+left(a^2c+abcright)+left(ac^2+bc^2right)+left(b^2c+abcright)right]3left[ableft(a+bright)+acleft(a+bright)+c^2left(a+bright)+bcleft(a+bright)right]3left(a+bright)left(ab+ac+c^2+bcrigh...
Đọc tiếp
\(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3-a^3-b^3-c^3\)
\(=a^3+3a^2b+3ab^2+b^3+3c\left(a^2+2ab+b^2\right)+3ac^2+3bc^2-a^3-b^3\)
\(=3a^2b+3ab^2+3a^2c+6abc+3b^2c+3ac^2+3bc^2\)
\(=3\left(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+2abc\right)\)
\(=3\left[\left(a^2b+ab^2\right)+\left(a^2c+abc\right)+\left(ac^2+bc^2\right)+\left(b^2c+abc\right)\right]\)
\(=3\left[ab\left(a+b\right)+ac\left(a+b\right)+c^2\left(a+b\right)+bc\left(a+b\right)\right]\)
\(=3\left(a+b\right)\left(ab+ac+c^2+bc\right)\)
\(=3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+b\right)\)
Châu ơi!đăng làm j z
cho tam giác ABC, với AB=c, BC=a, AC=b, chứng minh rằng
\(\frac{a\left(b+c\right)\sqrt{bc\left(1-\frac{a^2}{b+c}\right)}+b\left(a+c\right)\sqrt{ac\left(1-\frac{b^2}{a+c}\right)}+c\left(a+b\right)\sqrt{ab\left(1-\frac{c^2}{a+b}\right)}}{a+b+c}\)
CHO TAM GIÁC ABC, ĐẶT ĐỘ DÀI 3 CẠNH BC=a, CA=b, AB=c
CHO BIẾT: \(\frac{ab}{b+c}+\frac{bc}{c+a}+\frac{ca}{a+b}=\frac{ca}{b+c}+\frac{ab}{c+a}+\frac{bc}{a+b}\)
A) CM TAM GIÁC ABC CÂN
B) NẾU CHO THÊM: \(c^4+abc\left(a+b\right)=c^2\left(a^2+b^2\right)+\left(c+b\right)\left(c-b\right)bc+\left(c-a\right)\left(c+a\right)ac\) .TÍNH CÁC GÓC CỦA TAM GIÁC ABC
