Cộng vế với vế giả thiết:
\(a^2+4b+4+b^2+4c+4+c^2+4a+4=0\)
\(\Leftrightarrow\left(a^2+4a+4\right)+\left(b^2+4b+4\right)+\left(c^2+4c+4\right)=0\)
\(\Leftrightarrow\left(a+2\right)^2+\left(b+2\right)^2+\left(c+2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+2=0\\b+2=0\\c+2=0\end{matrix}\right.\) \(\Leftrightarrow a=b=c=-2\)
\(\Rightarrow P=1+1+1=3\)
giải các phương trình sau :
a. (x-3)(x-4)-2.(3x-2)=\(\left(4-x\right)^2\)
b. \(\left(x+2\right)\left(x-2\right)+5x^2=\left(3x+1\right)-3x^2\)
c. \(\left(x+2\right)^3-\left(x-1\right)^3=\left(3x+1\right).\left(3x-1\right)\)
d.\(\frac{3-x}{2018}+\frac{x-1}{2020}=\frac{-x}{2021}+1\)
cho: \(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\) . CMR: \(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=0\)
Bài 1. Cho a+b+c=0. Đặt P=\(\frac{a-b}{b}+\frac{b-c}{a}+\frac{c-a}{b}\); Q=\(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\).Tính P.Q
b) Rút gọn rồi tính giá trị biểu thức E=\(\frac{\left(a-x\right)^2}{a\left(b-a\right)\left(c-a\right)}+\frac{\left(b-x\right)^2}{b\left(a-b\right)\left(c-b\right)}+\frac{\left(c-x\right)^2}{c\left(a-c\right)\left(b-c\right)}\)biết \(1-\frac{x^2}{abc}=0\)
Cho 3 số thực a,b,c thỏa mãn \(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\) = 0. CMR
\(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}\) = 0
ta có
\(\frac{\left(2018-x\right)^2+\left(2018-x\right)\left(x-2019\right)+\left(x-2019\right)^2}{\left(2018-x\right)^2-\left(2018-x\right)\left(x-2019\right)-\left(x-2019\right)^2}=\frac{19}{49}\) ( điều kiện : x khác : 2018;2019 )
đặt a = x - 2019 ( a khác 0 )
ta có hệ thức :
\(\frac{\left(a+1\right)^2-\left(a+1\right)a+a^2}{\left(a+1\right)^2+\left(a+1\right)a+a^2}=\frac{19}{49}\\ \Leftrightarrow\frac{a^2+a+1}{3a^2+3a+1}=\frac{19}{49}\)
\(\Leftrightarrow49\left(a^2+a+1\right)=19\left(3a^2+3a+1\right)\)
\(\Leftrightarrow49a^2+49a+49=57a^2+57a+19\)
\(\Leftrightarrow8a^2+8a-30=0\\ \left(2a+1\right)^2-4^2=0\\ \Leftrightarrow\left(2a+3\right)\left(2a+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=\frac{3}{2}\\a=-\frac{5}{2}\end{matrix}\right.\)( thỏa mãn điều kiện )
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{4041}{2}\\x=\frac{4033}{2}\end{matrix}\right.\)( thỏa mãn điều kiện )
vậy \(x\in\left\{\frac{4041}{2};\frac{4033}{2}\right\}\)
Cho a,b,c #0 giải phân thức:\(\frac{x}{\left(a-b\right).\left(a-c\right)}+\frac{x}{\left(b-a\right).\left(b-c\right)}+\frac{x}{\left(c-a\right).\left(c-b\right)}=2\)
Cho \(f\left(x\right)=x^3+ax^2+bx+c\) (a, b thuộc R). Biết f(x) chia cho x+1 dư -4, chia cho x-2 dư 5. Tính: \(A=\left(a^{2019}+b^{2019}\right).\left(b^{2020}-c^{2020}\right).\left(c^{2021}+a^{2021}\right)\)
( Càng dễ hiểu càng tốt nhé , chương trình lớp 8 thôi ) .Cho a,b,c > 0 và 3 + \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=12.\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)
Chứng minh rằng \(\frac{1}{4a+b+c}+\frac{1}{4b+a+c}+\frac{1}{4c+a+b}\)≤ \(\frac{1}{6}\)
Cho a,b,c > 0. CMR P = \(\frac{a^2}{b\left(b+2c\right)}+\frac{b^2}{c\left(c+2a\right)}+\frac{c^2}{a\left(a+2b\right)}\) ≥ 1
