Cho \(a+b+c=a^2+b^2+c^2=1\) và \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\left(a\ne0,b\ne0,c\ne0\right)\)
CMR \(\left(x+y+z\right)^2=x^2+y^2+z^2\)
Cho : \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1;\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\left(a,b,c,x,y,z\ne0\right)\)
CMR : \(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\)
Đặt : x/a = m ; y/b = n ; z/c = p
=> m+n+p = 1 ; 1/m+1/n+1/p=0
1/m+1/n+1/p=0
<=> mn+np+pm/mnp=0
<=> mn+np+pm=0
<=> 2mn+2np+2pm=0
Xét : 1 = (m+n+p)^2 = m^2+n^2+p^2+2mn+2np+2pm = m^2+n^2+p^2
=> x^2/a^2+y^2/b^2+z^2/c^2 = 1
=> ĐPCM
Tk mk nha
Ta có: \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\)
\(\Rightarrow\frac{ayz}{xyz}+\frac{bxz}{xyz}+\frac{cxy}{xyz}=0\)
\(\Rightarrow\frac{ayz+bxz+cxy}{xyz}=0\)
\(\Rightarrow ayz+bxz+cxy=0\left(1\right)\)
Mặt khác: \(\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\)
\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)=1\)
\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{cxy+ayz+bxz}{abc}\right)=1\left(2\right)\)
Thay (1) vào (2) \(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2.\frac{0}{abc}=1\)
\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\left(đpcm\right)\)
Cho \(a+b+c=a^2+b^2+c^2=1\)và \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)với \(a\ne0,b\ne0,c\ne0\)
Chứng minh \(\left(x+y+z\right)^2=x^2+y^2+z^2\)
Ta có : \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\frac{x+y+z}{a+b+c}=x+y+z\) vì a + b + c = 1
Do đó \((x+y+z)^2=\frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac{z^2}{c^2}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=x^2+y^2+z^2\)vì \(a^2+b^2+c^2=1\)
Vậy :
Cho \(a+b+c=a^2+b^2+c^2=1\) và \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\) \(\left(a\ne0,b\ne0,c\ne0\right)\)
Chứng minh rằng: \(\left(x+y+z\right)^2=x^2+y^2+z^2\)
Lời giải:
Đặt $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=t$
$\Rightarrow x=at; y=bt; z=ct$. Ta có:
$(x+y+z)^2=(at+bt+ct)^2=t^2(a+b+c)^2=t^2(*)$
Mặt khác:
$x^2+y^2+z^2=(at)^2+(bt)^2+(ct)^2=t^2(a^2+b^2+c^2)=t^2(**)$
Từ $(*); (**)\Rightarrow (x+y+z)^2=x^2+y^2+z^2$ (đpcm)
CMR: Nếu:
a) \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)\(\forall x,y\ne0\) thì \(\frac{a}{x}=\frac{b}{y}\)
b) \(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)=\left(ax+by+cz\right)^2\forall x,y,z\ne0\) thì\(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)
c)\(\left(a+b\right)^2=2\left(a^2+b^2\right)\) thì \(a=b\)
a) \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)
\(\Leftrightarrow a^2x^2+b^2x^2+a^2y^2+b^2y^2=a^2x^2+b^2y^2+2abxy\)
\(\Leftrightarrow b^2x^2-2abxy+a^2y^2=0\)
\(\Leftrightarrow\left(bx\right)^2-2\cdot bx\cdot ay+\left(ay\right)^2=0\)
\(\Leftrightarrow\left(bx-ay\right)^2=0\Rightarrow bx=ay\Rightarrow\left(\frac{a}{x}=\frac{b}{y}\right)\)
b) \(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)=\left(ax+by+cz\right)^2\)
\(\Leftrightarrow a^2x^2+b^2x^2+c^2x^2+a^2y^2+b^2y^2+c^2y^2+a^2z^2+b^2z^2+c^2z^2\)
\(=a^2x^2+b^2y^2+c^2z^2+2abxy+2bcyz+2acxz\)
\(\Leftrightarrow b^2x^2-2bxay+a^2y^2+b^2z^2-2bzcy+c^2y^2+a^2z^2-2azcx+c^2x^2=0\)
\(\Leftrightarrow\left(bx-ay\right)^2+\left(bz-cy\right)^2+\left(az-cx\right)^2=0\)
\(\hept{\begin{cases}bx=ay\\bz=cy\\az=cx\end{cases}\Rightarrow\hept{\begin{cases}\frac{a}{x}=\frac{b}{y}\\\frac{b}{y}=\frac{c}{z}\\\frac{a}{x}=\frac{c}{z}\end{cases}}\Rightarrow\left(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\right)}\)
c) \(\left(a+b\right)^2=2\left(a^2+b^2\right)\)
\(\Leftrightarrow a^2+b^2+2ab=2a^2+2b^2\)
\(\Leftrightarrow a^2-2ab+b^2=0\)
\(\Leftrightarrow\left(a-b\right)^2=0\Leftrightarrow a=b\)
CMR: Nếu:
a) \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)\(\forall x,y\ne0\) thì \(\frac{a}{x}=\frac{b}{y}\)
b) \(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)=\left(ax+by+cz\right)^2\forall x,y,z\ne0\) thì\(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)
c)\(\left(a+b\right)^2=2\left(a^2+b^2\right)\) thì \(a=b\)
a, Tương đương : \(a^2x^2+a^2y^2+b^2x^2+b^2y^2\) = \(a^2x^2+2axby+b^2y^2\)
\(a^2y^2-2axby+b^2x^2=0\)
\(\left(ay-bx\right)^2\) = 0
\(ay-bx=0\)
\(ay=bx\)
\(\frac{a}{x}=\frac{b}{y}\) dpcm
Câu b, c làm tương tự câu a
Cho a,b,c,d là các số thực bất kỳ thỏa mãn \(\left(a^2+b^2+c^2\right)\cdot\left(x^2+y^2+z^2\right)=\left(ax+by+cz\right)^2\)
CMR:\(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\left(a,b,c\ne0\right)\)
bài này là bđt bunhia copxi khi xảy ra dấu =
\(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\ge\left(ax+by+cz\right)^2\)
c/m nhân tung ra thôi bạn
!@@@
Cho \(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)và \(x;y;z\ne0\)
CMR \(\frac{x^2+y^2+z^2}{\left(ax+by+cz\right)^2}=\frac{1}{a^2+b^2+c^2}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có: \(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}=\frac{a+b+c}{x+y+z}=k\)
\(\Rightarrow\hept{\begin{cases}a=kx;b=ky;c=kz\Rightarrow a^2=k^2x^2;b^2=k^2y^2;c^2=k^2z^2\\a+b+c=k\left(x+y+z\right)\end{cases}}\)
Có: \(\frac{x^2+y^2+z^2}{\left(ax+by+cz\right)^2}=\frac{x^2+y^2+z^2}{\left(kx^2+ky^2+kz^2\right)^2}=\frac{x^2+y^2+z^2}{k^2\left(x^2+y^2+z^2\right)^2}=\frac{1}{k^2\left(x^2+y^2+z^2\right)}\)
\(=\frac{1}{k^2x^2+k^2y^2+k^2z^2}=\frac{1}{a^2+b^2+c^2}\)(đpcm)
rút gọn các phân thức
a) \(\frac{x^2-16}{4x-x^2}\left(x\ne0,x\ne4\right)\) d) \(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ab}\)
b) \(\frac{5\left(x-y\right)-3\left(y-x\right)}{10\left(x-y\right)}\left(x\ne y\right)\) c) \(\frac{\left(x+y\right)^2-z^2}{x+y+z}\left(x+y+z\ne0\right)\)
e)\(\frac{a^3+b^3+c^3}{a^2+b^2+c^2-ab-bc-ac}\)
Lời giải:
a) \(\frac{x^2-16}{4x-x^2}=\frac{(x-4)(x+4)}{x(4-x)}=\frac{x+4}{-x}\)
b) \(\frac{5(x-y)-3(y-x)}{10(x-y)}=\frac{5(x-y)+3(x-y)}{10(x-y)}=\frac{8(x-y)}{10(x-y)}=\frac{8}{10}=\frac{4}{5}\)
c)
\(\frac{(x+y)^2-z^2}{x+y+z}=\frac{(x+y-z)(x+y+z)}{x+y+z}=x+y-z\)
d)
Biểu thức không rút gọn được
e)
\(\frac{a^3+b^3+c^3}{a^2+b^2+c^2-ab-bc-ac}=\frac{(a+b)^3-3ab(a+b)+c^3}{a^2+b^2+c^2-ab-bc-ac}=\frac{(a+b+c)[(a+b)^2-c(a+b)+c^2]-3ab(a+b)}{a^2+b^2+c^2-ab-bc-ac}\)
\(=\frac{(a+b+c)(a^2+b^2+c^2-ac-bc+2ab)-3ab(a+b+c)+3abc}{a^2+b^2+c^2-ab-bc-ac}\)
\(=\frac{(a+b+c)(a^2+b^2+c^2-ab-bc-ac)+3abc}{a^2+b^2+c^2-ab-bc-ac}=a+b+c+\frac{3abc}{a^2+b^2+c^2-ab-bc-ac}\)
Cho \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\ne0\)
Tính \(\frac{\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)}{\left(ax^2+by^2+cz^2\right)}\)