Cho 3 số a,b,c thỏa mãn:
\(2a^2+2b^2+c^2+2ab-2ac-2bc-2a+2b+2=0\)
Tính \(A=\left(x-y\right)^{2015}-\left(b+2\right)^{2015}+c^{2015}\)
Chuyên Hà Tĩnh 2015
Cho ba số a,b,c thỏa mãn: \(c^2+2\left(ab-bc-ac\right)=0,b\ne c\)và \(a+b\ne c\). Chứng minh rằng
\(\frac{2a^2-2ac+c^2}{2b^2-2bc+c^2}=\frac{a-c}{b-c}\)
\(\frac{2a^2-2ac+c^2}{2b^2-2bc+c^2}=\frac{a-c}{b-c}\)
\(\Leftrightarrow2a^2b-2a^2c+ac^2-bc^2-2ab^2+2b^2c=0\)
\(\Leftrightarrow2a\left(ab-ac+\frac{c^2}{2}\right)-bc^2-2ab^2+2bc^2=b\left(2ac-c^2-2ab+2bc\right)=0\)(đúng)
=> đpcm
Từ \(c^2+2\left(ab-bc-ac\right)=0.\)
\(\Rightarrow c^2+2ab-2bc-2ac=0\)
\(\Rightarrow\frac{c^2}{2}+ab-bc-ac=0\)
\(\Rightarrow bc=\frac{c^2}{2}+ab-ac\)
Có : \(2a\left(ab-ac+\frac{c^2}{2}\right)-bc^2-2ab^2+2bc^2\)
\(=2abc-bc^2-2ab^2+2bc^2\)
\(=-b\left(-2ac+c^2+2ab-2bc\right)\)
\(=-b\left[c^2+2\left(ab-bc-ac\right)\right]=-b.0=0\)\(\left(đpcm\right)\)
Bài 1.Cho \(x+y+z=0\)
Tính \(S=\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}\)
Bài 2. Cho \(a+b+c=1;a^2+b^2+c^2=1;\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)
CMR: \(xy+yz+zx=0\)
Bài 3. Cho \(3x-y=2z\)
\(2x+y=7z\)
Tính \(S=\frac{x^2-2xy}{x^2+y^2}\)với \(x,y\ne0\)
Bài 4. Cho \(a,b,c\ne0\)thỏa mãn \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
Tính \(E=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)
Bài 5. Cho \(abc\ne0\)thỏa mãn: \(2ab+6bc+2ac=0\)
Tính \(A=\frac{\left(a+2b\right)\left(2b+3c\right)\left(3c+a\right)}{6abc}\)
Bài 6. Cho \(a,b,c\ne0\)thỏa mãn \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
Tính \(Y=\frac{a^2b^2c^2}{a^2b^2+b^2c^2-c^2a^2}+\frac{a^2b^2c^2}{b^2c^2+c^2a^2-a^2b^2}+\frac{a^2b^2c^2}{c^2a^2+a^2b^2-b^2c^2}\)
Bài 7. Cho \(\hept{\begin{cases}10a^2-3b^2+5ab=0\\9a^2-b^2\ne0\end{cases}}\)
Tính \(B=\frac{2a-b}{3a-b}+\frac{5b-a}{3a+b}\)
1. Ta có : x + y + z = 0 \(\Rightarrow\)( x + y + z )2 = 0 \(\Rightarrow\)x2 + y2 + z2 = - 2 ( xy + yz + xz )\(S=\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}=\frac{-2\left(xy+yz+xz\right)}{2\left(x^2+y^2+z^2\right)-2\left(yz+xz+xy\right)}\)
\(S=\frac{-2\left(xy+yz+xz\right)}{-4\left(xy+yz+xz\right)-2\left(yz+xz+xy\right)}=\frac{-2\left(xy+yz+xz\right)}{-6\left(xy+yz+xz\right)}=\frac{1}{3}\)
2. a + b + c = 1 \(\Rightarrow\)( a + b + c )2 = 1 \(\Rightarrow\)a2 + b2 + c2 + 2 ( ab + bc + ac ) = 1 \(\Rightarrow\)ab + bc + ac = 0
Áp dụng tính chất dãy tỉ số bằng nhau, ta có : \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\frac{x+y+z}{a+b+c}=x+y+z\)
\(\Rightarrow\)x = a ( x + y + z ) ; y = b ( x + y + z ) ; z = c ( x + y + z )
Ta có : xy + yz + xz = ab ( x + y + z )2 + bc ( x + y + z )2 + ac ( x + y + z )2 = ( x + y + z )2 ( ab + bc + ac ) = 0
3. sửa đề : 3x - y = 3z
Ta có : \(\hept{\begin{cases}3x-y=3z\\2x+y=7z\end{cases}\Rightarrow\hept{\begin{cases}\left(3x-y\right)+\left(2x+y\right)=3z+7z\\2x+y=7z\end{cases}\Rightarrow}\hept{\begin{cases}5x=10z\\y=7z-2x\end{cases}\Rightarrow}\hept{\begin{cases}x=2z\\y=3z\end{cases}}}\)
\(\Rightarrow\)\(S=\frac{x^2-2xy}{x^2+y^2}=\frac{\left(2z\right)^2-2.2z.3z}{\left(2z\right)^2+\left(3z\right)^2}=\frac{4z^2-12z^2}{4z^2+9z^2}=\frac{-8z^2}{13z^2}=\frac{-8}{13}\)
cho a,b,c dương và a+b+c=1.CMR: \(\frac{\sqrt{\left(^{a^2+2ab}\right)}}{\sqrt{\left(b^2+2c^2\right)}}+\frac{\sqrt{\left(^{b^2+2bc}\right)}}{\sqrt{\left(c^2+2a^2\right)}}+\frac{\sqrt{\left(^{c^2+2ac}\right)}}{\sqrt{\left(a^2+2b^2\right)}}\ge\frac{1}{a^2+b^2+c^2}\)
Cho 3 số thực dương a;b;c thỏa mãn \(7\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=6\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)+2015\)
Tìm GTLN của \(P=\frac{1}{\sqrt{3\left(2a^2+b^2\right)}}+\frac{1}{\sqrt{3\left(2b^2+c^2\right)}}+\frac{1}{\sqrt{3\left(2c^2+a^2\right)}}\)
a)\(\left(b-c\right)^3+\left(c-a\right)^3+\left(a-b\right)^3\)
b)\(\left(x+y\right)^5-x^5-y^5\)
c)\(\left(x^2+y^2\right)^3+\left(z^2-x^2\right)^3-\left(y^2+z^2\right)^3\)
d)\(3abc+a^2\left(a-b-c\right)+b^2\left(b-a-c\right)+c^2\left(c-a-b\right)-c\left(b-c\right)\left(a-c\right)\)
e) 2bc(b+2c)+2ac(c-2a)-2ab(a+2b)-7abc
f)3bc(3b-c)-3ac(3c-a)-3ab(3a+b)+28abc
PTĐTTNT:\(3abc+a^2\left(a-b-c\right)+b^2\left(b-a-c\right)+c^2\left(c-b-a\right)-c\left(b-c\right)\left(a-c\right)\)
\(=3abc+a^3-a^2b-a^2c+b^3-b^2a-b^2c+c^3-c^2b-c^2a-\left(abc-bc^2-c^2a+c^3\right)\)
\(=2abc+a^3-a^2b-a^2c+b^3-b^2c-b^2a\)
\(=\left(a^3+a^2b-a^2c\right)-\left(2a^2b+2ab^2-2abc\right)+\left(ab^2+b^3-b^2c\right)\)
\(=a^2\left(a+b-c\right)-2ab\left(a+b-c\right)+b^2\left(a+b-c\right)\)
\(=\left(a+b-c\right)\left(a^2-2ab+b^2\right)\)
\(=\left(a+b-c\right)\left(a^2-2ab+b^2\right)\)
\(=\left(a+b-c\right)\left(a-b\right)^2\) nha !
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Trả lời
Ở phần kết quả bạn vẫn chưa thu gọn hết đâu nha
\(=\left(a+b+c\right).\left(a-b\right)^2\)
Mk góp ý thôi mong mọi người đừng có đáp gạch đáp đá nha
Study well
Cho a, b, c là các số tự nhiên thỏa mãn \(\left(2a^2-b\right)^2+\left(3b^2-a\right)^2+c^2-12a^2b^2-2ab-2=-4a^3-6b^3-\frac{1}{c^2}\)
Chứng minh rằng \(\left(a-b\right)\) và \(\left(2a+2b+1\right)\) đồng thời là các số chính phương
Cho 3 số thực dương a,b,c thỏa mãn : \(7\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=6\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)+2015.\)
Tìm \(GTLN\) của biểu thức sau: \(P=\frac{1}{\sqrt{3\left(2a^2+b^2\right)}}+\frac{1}{\sqrt{3\left(2b^2+c^2\right)}}+\frac{1}{\sqrt{3\left(2c^2+a^2\right)}}\)
Ta có:\(7\left(\frac{1}{a^2}+...\right)=6\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)+2015\)
Mà \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\le\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
=> \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\le2015\)=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\sqrt{6045}\)
\(P=\frac{1}{\sqrt{3\left(2a^2+b^2\right)}}+...\)
Mà \(\left(2+1\right)\left(2a^2+b^2\right)\ge\left(2a+b\right)^2\)(bất dẳng thức buniacoxki)
=> \(P\le\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}\)
Lại có \(\frac{1}{2a+b}=\frac{1}{a+a+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\right)\)
=> \(P\le\frac{1}{9}\left(\frac{3}{a}+\frac{3}{b}+\frac{3}{c}\right)\le\frac{\sqrt{6045}}{3}\)
Vậy \(MaxP=\frac{\sqrt{6045}}{3}\)khi \(a=b=c=\frac{\sqrt{6045}}{2015}\)
Cho 3 số x,y,z>0 thỏa mãn:
\(\frac{4a^2+\left(b-c\right)^2}{2a^2+b^2+c^2}+\frac{4b^2+\left(c-a\right)^2}{2b^2+c^2+a^2}+\frac{4c^2+\left(a-b\right)^2}{2c^2+a^2+b^2}\ge3\)