Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bạn chỉ cần để ý điều này thôi: \(\left(x-\frac{1}{x}\right)^2=x^2-2.x.\frac{1}{x}+\frac{1}{x^2}=x^2-2+\frac{1}{x^2}\)
Do đó giả thiết viết lại thành:
\(\left(a^2-2+\frac{1}{a^2}\right)+\left(b^2-2+\frac{1}{b^2}\right)+\left(c^2-2+\frac{1}{c^2}\right)=0\)
\(\Leftrightarrow\left(a-\frac{1}{a}\right)^2+\left(b-\frac{1}{b}\right)^2+\left(c-\frac{1}{c}\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-\frac{1}{a}=0\\b-\frac{1}{b}=0\\c-\frac{1}{c}=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=\frac{1}{a}\\b=\frac{1}{b}\\c=\frac{1}{c}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2=1\\b^2=1\\c^2=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(a^2\right)^{1010}=1^{1010}\\\left(b^2\right)^{1010}=1^{1010}\\\left(c^2\right)^{1010}=1^{1010}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^{2020}=1\\b^{2020}=1\\c^{2010}=1\end{matrix}\right.\) \(\Leftrightarrow a^{2020}+b^{2020}+c^{2020}=3\)
Sai thì bỏ qua ( bạn bè mà ) !
Nếu \(a+b+c=0\Rightarrow\hept{\begin{cases}a+b=-c\\a+c=-b\\b+c=-a\end{cases}}\)
\(\Rightarrow\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=-1-1-1=-3\)(vô lí )
\(\Rightarrow a+b+c\ne0\)
Ta có :
\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=1\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)=a+b+c\)
Đặt a + b + c = H
\(\Rightarrow\frac{a^2}{b+c}+\frac{ab}{a+c}+\frac{ac}{a+b}+\frac{b^2}{a+c}+\frac{ab}{b+c}+\frac{bc}{a+b}+\frac{c^2}{b+a}+\frac{ac}{c+b}+\frac{bc}{a+c}=H\)
\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{b+a}+\left(\frac{ab}{a+c}+\frac{bc}{a+c}\right)+\left(\frac{ac}{a+b}+\frac{bc}{a+b}\right)+\left(\frac{ab}{b+c}+\frac{ac}{c+b}\right)=H\)
\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{b+a}+a+b+c=H\)( Chỗ này làm hơi tắt bỏ qua nha )
\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{b+a}=H-\left(a+b+c\right)\)
\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{b+a}=0\left(đpcm\right)\)
ĐK:....
\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)=a+b+c\)
\(\Leftrightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}+a+b+c=a+b+c\)(nhân vào rồi tách)
\(\Leftrightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0\)
Việt Hoàng _ TTH (*Yonko Team*): Mình chưa xem kỹ nhưng có lẽ hướng làm bạn là sai òi nhé!
Bài 2:
Ta có: \(f\left(a\right)=6a^5-10a^4-5a^3+23a^2-29a+2005\)
\(=\left(6a^5-10a^4-2a^3\right)-\left(3a^3-5a^2-a\right)+\left(18a^2-30a-6\right)+2011\)
\(=2a^3\left(3a^2-5a-1\right)-a\left(3a^2-5a-1\right)+6\left(3a^2-5a-1\right)+2011\)
\(=\left(2a^3-a+6\right)\left(3a^2-5a-1\right)+2011\)
Mà \(3a^2-5a-1=0\)
\(\Rightarrow f\left(a\right)=2011\)
Vậy...
Áp dụng BĐT Cauchy schwarz dạng phân thức ta có :
\(\dfrac{a^2}{1+b-a}+\dfrac{b^2}{1+c-b}+\dfrac{c^2}{1+a-c}\ge\dfrac{\left(a+b+c\right)^2}{3}\ge\dfrac{3\left(ab+bc+ca\right)}{3\left(ab+bc+ca\right)}=1\)
( vì \(a^2+b^2+c^2\ge ab+bc+ca\) )
Xảy ra đẳng thức khi và chỉ khi a=b=c= \(\sqrt{\dfrac{1}{3}}\)
a) \(\frac{2x+1}{x-1}\)=\(\frac{5\left(x-1\right)}{x+1}\):dkxd x\(\ne\)\(\pm\)1
\(\Rightarrow\)(2x+1)(x+1)=5(x-1)2
\(\Leftrightarrow\)2x2+2x+x+1=5(x2-2x+1)
\(\Leftrightarrow\)2x2+2x+x+1=5x2-10x+5
\(\Leftrightarrow\)2x2+2x+x+1-5x2+10x-5=0
\(\Leftrightarrow\)-3x2+13x-4=0
\(\Leftrightarrow\)-3x2+12x+1x-4=0
\(\Leftrightarrow\)-4x(x-4)+(x-4)=0
\(\Leftrightarrow\)(x-4)(-4x+1)=0
\(\Leftrightarrow\)x-4=0 hoac -4x+1=0
\(\Leftrightarrow\)x=4(tmdkxd) \(\Leftrightarrow\)x=1/4(tmdkxd)
vay s={4;1/4}
b)\(\frac{x}{x-1}\)-\(\frac{2x}{x^{ }2^{ }-1}\)=0 dkxd x\(\ne\)\(\pm\)1
\(\Leftrightarrow\)\(\frac{x\left(X+1\right)-2x^{ }}{\left(x-1\right)\left(x+1\right)}\)=0
\(\Rightarrow\)x2+x-2x=0
\(\Leftrightarrow\)x2-x=0
\(\Leftrightarrow\)x(x-1)=0
\(\Leftrightarrow\)x=0 hoac x-1=0
\(\Leftrightarrow\)x=0(tmdkxd)\(\Leftrightarrow\)x=1(ktmdkxd)
vay s={0}
c.\(\frac{1}{x-2}\)+3=\(\frac{x-3}{2-x}\) dkxd x\(\ne\)2
\(\Leftrightarrow\)\(\frac{1}{x-2}\)+3=\(\frac{-\left(x-3\right)}{x-2}\)
\(\Leftrightarrow\)\(\frac{1+3\left(x-2\right)}{x-2}\)=\(\frac{-x+3}{x-2}\)
\(\Rightarrow\)1+3x-6=-x+3
\(\Leftrightarrow\)4x=8
\(\Leftrightarrow\)x=2(ktmdkxd)
vay s=\(\varnothing\)
chuc ban hoc tot
a.\(\frac{2x+1}{x-1}\) = \(\frac{5\left(x-1\right)}{x+1}\)
\(\leftrightarrow\) 2x+1 = 5x - 5
\(\leftrightarrow\) 2x - 5= -1-5
\(\leftrightarrow\) -3x = -6
\(\leftrightarrow\) x =2
Vậy S=\(\left\{2\right\}\)
b.\(\frac{x}{x-1}\) - \(\frac{2x}{x^2-1}\) =0
\(\leftrightarrow\) \(\frac{x}{x-1}\) - \(\frac{2x}{\left(x-1\left(x+1\right)\right)}\)= 0 (ĐK : x\(_{\ne}\) -1 và 1)
\(\leftrightarrow\)\(\frac{x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\) - \(\frac{2x}{\left(x-1\left(x+1\right)\right)}\) =0
\(\leftrightarrow\) x2 + x -2x = 0
\(\leftrightarrow\)(x2 + x) -2x =0
\(\leftrightarrow\)x(x+1) -2x =0
\(\leftrightarrow\) x =0 -> x=0
x+1 =0 -> x = -1(Loại)
-2x = 0 -> x= 2(TM)
Vậy x =\(\left\{0,2\right\}\)
(BẠN NHỚ COI LẠI CÁI CÂU TRẢ LỜI Ở CUỐI MỖI BÀI NHA ,MÌNH KO CHẮC CÂU TRẢ LỜI ĐÓ )
Ta có : \(a^2+ab=c^2+bc\Leftrightarrow a^2-c^2+b\left(a-c\right)=0\)
\(\Leftrightarrow\left(a-c\right)\left(a+b+c\right)=0\Leftrightarrow a-c=0\) ( do a;b;c \(\ne0\Rightarrow a+b+c\ne0\) )
\(\Leftrightarrow a=c\)
Làm tương tự ; ta có : a = b . Suy ra : a = b = c
\(A=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=6\)
Vậy ...
Ta có : a2+ab=c2+bc⇔a2−c2+b(a−c)=0a2+ab=c2+bc⇔a2−c2+b(a−c)=0
⇔(a−c)(a+b+c)=0⇔a−c=0⇔(a−c)(a+b+c)=0⇔a−c=0 ( do a;b;c ≠0⇒a+b+c≠0≠0⇒a+b+c≠0 )
⇔a=c⇔a=c
Làm tương tự ; ta có : a = b . Suy ra : a = b = c
A=(1+ab)(1+bc)(1+ca)=(1+1)(1+1)(1+1)=6A=(1+ab)(1+bc)(1+ca)=(1+1)(1+1)(1+1)=6
Vậy ...