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\(A=\left(b+c\right)^2+b^2+c^2=2b^2+2c^2+2bc=2\left(b^2+bc+c^2\right)\) (tự hiểu nhé)
Mà \(a^2=2\left(a+c+1\right)\left(a+b-1\right)=2a^2+2\left(ab+bc+ca\right)+2\left(b-c\right)-2\)
\(\Leftrightarrow a^2+2a\left(b+c\right)+2bc-2=0\) (*)
\(\Leftrightarrow2bc=2-a^2-2a\left(b+c\right)=2-\left(b+c\right)^2+2\left(b+c\right)^2\) (mấy cái này là từ a + b + c =0 suy ra a = -(b+c) suy ra a2 = [-(b+c)]2 = (b+c)2 thôi!)
\(\Leftrightarrow\left(b+c\right)^2-2bc=-2\)
hay c2 + b2 = -2?? hay là mình làm sai nhì?
\(a^2=2\left(a+c+1\right)\left(a+b-1\right)\)
\(\Leftrightarrow\left(b+c\right)^2=\left(b-1\right)\left(c+1\right)\)
\(\Leftrightarrow\left(b-1\right)^2+\left(c+1\right)^2=0\)
\(\Rightarrow a=0,b=1,c=-1\)
\(\Rightarrow A=2\)
Ta có: \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+a+b+c=2+2018\)
\(\Leftrightarrow\frac{a+ab+bc}{b+c}+\frac{b+bc+ab}{c+a}+\frac{c+ac+bc}{a+b}=2020\)
\(\Leftrightarrow a\left(\frac{1+b+c}{b+c}\right)+b\left(\frac{1+a+c}{a+c}\right)+c\left(\frac{1+a+b}{a+b}\right)=2020\left(1\right)\)
Vì \(a+b+c=2018\Rightarrow\hept{\begin{cases}a+b=2018-c\\b+c=2018-a\\c+a=2018-b\end{cases}\left(2\right)}\)
Thay (2) vào (1) ta được:
\(a\left(\frac{2019-a}{b+c}\right)+b\left(\frac{2019-b}{a+c}\right)+c\left(\frac{2019-c}{a+b}\right)=2020\)
\(\Leftrightarrow\frac{2019a-a^2}{b+c}+\frac{2019b-b^2}{a+c}+\frac{2019c-c^2}{a+b}=2020\)
\(\Leftrightarrow\frac{2019a}{b+c}-\frac{a^2}{b+c}+\frac{2019b}{a+c}-\frac{b^2}{a+c}+\frac{2019c}{a+b}-\frac{c^2}{a+b}=2020\)
\(\Leftrightarrow2019\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)-\left(\frac{a^2}{c+b}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)=2020\)
\(\Leftrightarrow4038-\left(\frac{a^2}{c+b}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)=2020\)( vì \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=2\))
\(\Leftrightarrow\frac{a^2}{c+b}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=2018\)
\(\Leftrightarrow\frac{a^2}{c+b}+\frac{b^2}{c+a}+\frac{c^2}{a+b}+1=2019\)
\(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
\(\Rightarrow2+2\left(ab+bc+ca\right)=0\Rightarrow ab+bc+ca=-1\Rightarrow\left(ab+bc+ca\right)^2=1\)
\(\Rightarrow a^2b^2+b^2c^2+c^2a^2+2ab^2c+2abc^2+2a^2bc=1\)
\(\Rightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=1\)\(\Rightarrow a^2b^2+b^2c^2+c^2a^2+2abc.0=1\)
\(\Rightarrow a^2b^2+b^2c^2+c^2a^2+0=1\Rightarrow a^2b^2+b^2c^2+c^2a^2=1\)
Mặt khác:
\(a^2+b^2+c^2=2\Rightarrow\left(a^2+b^2+c^2\right)^2=4\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\)
=>\(\Rightarrow a^4+b^4+c^4+2.1=4\Rightarrow a^4+b^4+c^4+2=4\Rightarrow a^4+b^4+c^4=2\)
tính tương tự câu kia
a, Xét : 196 = 14^2 = (a^2+b^2+c^2) = a^4+b^4+c^4+2.(a^2b^2+b^2c^2+c^2a^2)
<=> a^4+b^4+c^4 = 196 - 2.(a^2b^2+b^2c^2+c^2a^2)
Xét : 0 = (a+b+c)^2 = a^2+b^2+c^2+2.(ab+bc+ca)
Mà a^2+b^2+c^2 = 14
<=> 2.(ab+bc+ca) = -14
<=> ab+bc+ca = -7
<=> a^2b^2+b^2c^2+c^2a^2+2abc.(a+b+c) = 49
Lại có : a+b+c = 0
<=> a^2b^2+b^2c^2+c^2a^2 = 49
<=> A = a^4+b^4+c^4 = 196 - 2.49 = 98
Tk mk nha
b) \(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)
\(\Leftrightarrow\)\(\frac{x^2}{a^2}-\frac{x^2}{a^2+b^2+c^2}+\frac{y^2}{b^2}-\frac{y^2}{a^2+b^2+c^2}+\frac{z^2}{c^2}-\frac{z^2}{a^2+b^2+c^2}=0\)
\(\Leftrightarrow\)\(x^2\left(\frac{1}{a^2}-\frac{1}{a^2+b^2+c^2}\right)+y^2\left(\frac{1}{b^2}-\frac{1}{a^2+b^2+c^2}\right)+z^2\left(\frac{1}{c^2}-\frac{1}{a^2+b^2+c^2}\right)=0\)
\(\Leftrightarrow\)\(x^2=y^2=z^2=0\)
\(\Leftrightarrow\)\(x=y=z=0\)
Vậy \(D=0\)
4
ta có : \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)\(\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{-1}{z}\)
Ta có: \(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=\dfrac{1}{x^3}+3\times\dfrac{1}{x^2}\times\dfrac{1}{y}+3\times\dfrac{1}{x}\times\dfrac{1}{y^2}+\dfrac{1}{y^3}-3\times\dfrac{1}{x^2}\times\dfrac{1}{y}-3\times\dfrac{1}{x}\times\dfrac{1}{y^2}+\dfrac{1}{z^3}\) \(\Leftrightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^3-3\times\dfrac{1}{xy}\times\left(\dfrac{1}{x}+\dfrac{1}{y}\right)+\dfrac{1}{z^3}\)
\(\Leftrightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=\left(\dfrac{-1}{z}\right)^3-3\times\dfrac{1}{xy}\times\left(\dfrac{-1}{z}\right)+\dfrac{1}{z^3}\)
\(\Leftrightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=-\dfrac{1}{z^3}+3\times\dfrac{1}{xyz}+\dfrac{1}{z^3}\)
\(\Leftrightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=\dfrac{3}{xyz}\Leftrightarrow xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=3\)(ĐPCM)
\(a+b+c=0\)
\(\Rightarrow\left(a+b+c\right)^2=0\)
\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
\(\Rightarrow ab+bc+ca=\frac{-1}{2}\)
\(\Rightarrow\left(ab+bc+ca\right)^2=\frac{1}{4}\)
\(\Rightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=\frac{1}{4}\)
\(\Rightarrow a^2b^2+b^2c^2+c^2a^2=\frac{1}{4}\)( 1 )
\(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=1\)
Mà theo ( 1 ) nên có \(a^2+b^4+c^4=\frac{1}{2}\)
P/S:Hướng lm là như vầy nhé !
Cho a + b + c = 0 và a2 + b2 +c2= 1 Tính giá trị của biểu thức M = a4+b4+c4 Giúp mk vs nha!!
Tham khảo
\(2T=\frac{a^2-2ac+c^2+c^2-2bc+b^2+a^2-2ab+b^2}{\left(a-c\right)\left(a+c\right)-2b\left(a-c\right)}\)
\(2T=\frac{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}{\left(a-c\right)\left(a-b+c-b\right)}\)
Theo đề bài ta có:\(\hept{\begin{cases}a-b=4\\b-c=2\end{cases}\Rightarrow}a-c=6\)
\(\Rightarrow2T=\frac{4^2+2^2+6^2}{6\cdot\left(4-2\right)}=\frac{14}{3}\)
Ta có:
\(a+b+c=0\)
\(\Leftrightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=0\)
\(\Leftrightarrow2+2ab+2bc+2ca=0\)(theo bài ra a^2 + b^2 + c^2 = 2)
\(\Leftrightarrow ab+bc+ca=-1\)
\(\Leftrightarrow\left(ab+bc+ca\right)^2=-1\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=1\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=1\)
Vậy:\(a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2\left(a^2b^2+b^2c^2+c^2a^2\right)=4-2-2\)