cho cac bieu thuc: P=(a+1)^2+(b+1)^2+2(ab+ac+bc) Q=(a+b+c+1)^2. tinh P-Q
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Cho cac bieu thuc :
P = ( a + 1 )2 + ( b + 1 )2 + ( c + 1 )2 + 2 ( ab + ac + bc )
Q = ( a + b + c + 1 )2
Tinh P - Q
biet ab-ac+bc=c^2-1 tinh gia tri bieu thuc b=\(\frac{a}{b}\)
\(ab-ac+bc=c^2-1\)
\(ab-ac+bc-c^2=-1\)
\(a\left(b-c\right)+c\left(b-c\right)=-1\)
\(\Leftrightarrow\left(a+c\right)\left(b-c\right)=-1\)
=> a + c = 1 thì b - c = - 1; a + c = - 1 thì b - c = 1 => a + c và b - c đối nhau
\(\Rightarrow a+c=-\left(b-c\right)\)
\(a+c=-b+c\)
\(\Rightarrow a=-b\)
\(\Rightarrow B=\frac{a}{b}=-1\)
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cho biet a,b,c >0 dieu kien \(a^2+b^2+c^2=1\)Tinh GTNN cua bieu thuc A = \(\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\)
\(A^2=\frac{a^2b^2}{c^2}+\frac{b^2c^2}{a^2}+\frac{c^2a^2}{b^2}+2\left(b^2+c^2+a^2\right)=\frac{a^2b^2}{c^2}+\frac{b^2c^2}{a^2}+\frac{c^2a^2}{b^2}+2\)
Áp dụng Côsi: \(\frac{a^2b^2}{c^2}+\frac{b^2c^2}{a^2}\ge2\sqrt{\frac{a^2b^2}{c^2}.\frac{b^2c^2}{a^2}}=2\sqrt{b^4}=2b^2\)
Tương tự \(\frac{b^2c^2}{a^2}+\frac{c^2a^2}{b^2}\ge2c^2;\text{ }\frac{c^2a^2}{b^2}+\frac{a^2b^2}{c^2}\ge2a^2\)
\(\Rightarrow2\left(\frac{a^2b^2}{c^2}+\frac{b^2c^2}{a^2}+\frac{c^2a^2}{b^2}\right)\ge2\left(a^2+b^2+c^2\right)=2\)
\(\Rightarrow\frac{a^2b^2}{c^2}+\frac{b^2c^2}{a^2}+\frac{c^2a^2}{b^2}\ge1\)
\(\Rightarrow A^2\ge1+2=3\)
\(\Rightarrow A\ge\sqrt{3}\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=c=\frac{1}{\sqrt{3}}\)
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1) Cho bieu thuc A=\(3+\frac{2}{x-1}\). Tinh gia tri cua bieu thuc A khi |2x-3|=1
2) Rut gon bieu thuc B=\(\frac{x}{x-1}\)-\(\frac{x-5}{x+1}\)-\(\frac{3-x}{1-x^2}\)
3) Tim cac gia tri nguyen cua x de bieu thuc \(\frac{B}{A}\)co gia tri nguyen duong
cho cac so a,b,c va thoa man \(\frac{ab}{a+b}=\frac{1}{3},\frac{bc}{b+c}=\frac{1}{4},\frac{ca}{c+a}=\frac{1}{5}\)Tinh gia tri bieu thuc P=\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
Thêm đk \(a,b,c\ne0\)
Ta có: \(\frac{ab}{a+b}=\frac{1}{3}\Rightarrow\frac{a+b}{ab}=3\)
\(\frac{bc}{b+c}=\frac{1}{4}\Rightarrow\frac{bc}{b+c}=4\)
\(\frac{ca}{c+a}=\frac{1}{5}\Rightarrow\frac{c+a}{ca}=5\)
\(\Rightarrow\frac{a+b}{ab}+\frac{b+c}{bc}+\frac{c+a}{ca}=12\)
\(\Leftrightarrow\frac{1}{b}+\frac{1}{a}+\frac{1}{c}+\frac{1}{b}+\frac{1}{a}+\frac{1}{c}=12\)
\(\Leftrightarrow2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=12\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\)
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cho a,b,c khac nhau doi mot va 1/a+1/b+1/c=0.rut gon cac bieu thuc
N=bc/a^2+2bc+CA/B^2+2AC+AB/C^2+2AB
cho 1/a+1/b+1/c=0.tinh gia tri bieu thuc P=ab/c2+bc/a2+ca/b2
Câu hỏi của Conan Kudo - Toán lớp 8 - Học toán với OnlineMath
Bạn tham khảo link trên!
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Cho a+b=x va ab=y. Tinh cac bieu thuc sau theo x va y
1, \(a^3\)- \(b^3\)
2, \(a^4\)- \(b^4\)
Ta có : \(a+b=x\Rightarrow a^2+2ab+b^2=x^2\Rightarrow a^2+b^2=x^2-2y\)
\(\Rightarrow a^2+b^2-2ab=x^2-2y-2y=x^2-4y\Rightarrow\left(a-b\right)^2=x^2-4y\Rightarrow a-b=\sqrt{x^2-4y}\)
1 . \(a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)=\sqrt{x^2-4y}\left(x^2-2y+y\right)=\sqrt{x^2-4y}\left(x^2-y\right)\)
2 . \(a^4-b^4=\left(a^2-b^2\right)\left(a^2+b^2\right)=\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)=x\sqrt{x^2-4y}\left(x^2-2y\right)\)
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Cho a^2 + b^2 + c^2 = a^3 + b^3 + c^3 = 1. tinh gt cac bieu thuc : C = a^2 + b^9 + c^1945.