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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
cái trên thì bn dùng BĐT Bunhiakovshi nha
cái dưới hơi rườm tí mik ko bt lm đúng ko
\(f\left(x\right)=x\left(x+1\right)\left(x+2\right)\left(ax+b\right)\)
\(f\left(x-1\right)=\left(x-1\right)x\left(x+1\right)\left(ax-a+b\right)\)
\(\Rightarrow f\left(x\right)-f\left(x-1\right)=x\left(x+1\right)\left(x+2\right)\left(ax+b\right)-\)
\(\left(x-1\right)x\left(x+1\right)\left(ax-a+b\right)\)
\(=x\left(x+1\right)\left[\left(x+2\right)\left(ax+b\right)-\left(x-1\right)\left(ax-a+b\right)\right]\)
\(=x\left(x+1\right)[x\left(ax+b\right)+2\left(ax+b\right)-x\left(ax-a+b\right)\)
\(+\left(ax-a+b\right)]\)
\(=x\left(x+1\right)(ax^2+bx+2ax+2b-ax^2+ax\)
\(-bx+ax-a+b)\)
\(=x\left(x+1\right)\left(4ax-a+3b\right)\)
Mà theo đề \(f\left(x\right)-f\left(x-1\right)=x\left(x+1\right)\left(2x+1\right)\)
Đồng nhất hệ số là ra
Đặt biểu thức trên là A
Đặt \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\ne0\)
\(\Rightarrow x=ak,y=bk,z=ck\)
Nên \(A=\frac{\text{[}\left(ak\right)^2+\left(bk\right)^2+\left(ck\right)^2\text{]}.\left(a^2+b^2+c^2\right)}{\left(a.ak+b.bk+c.bk\right)^2}\)
\(=\frac{\left(a^2k^2+b^2k^2+c^2k^2\right).\left(a^2+b^2+c^2\right)}{\left(a^2k+b^2k+c^2k\right)^2}\)
\(=\frac{k^2\left(a^2+b^2+c^2\right).\left(a^2+b^2+c^2\right)}{\text{[}k\left(a^2+b^2+c^2\right)\text{]}^2}\)
\(=\frac{k^2.\left(a^2+b^2+c^2\right)^2}{k^2.\left(a^2+b^2+c^2\right)}\)
\(=1\)
Vậy A=1
à quên sửa dòng trên chỗ A=1 cái chỗ mẫu là \(k^2.\left(a^2+b^2+c^2\right)^2\)nhen :v
Đặt \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\Rightarrow x=ak,y=bk,z=ck\)
Ta có: \(\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)=\left(a^2k^2+b^2k^2+c^2k^2\right)\left(a^2+b^2+c^2\right)=k^2\left(a^2+b^2+c^2\right)^2\) (1)
\(\left(ax+by+cz\right)^2=\left(a.ak+b.bk+c.ck\right)^2=\left(a^2k+b^2k+c^2k\right)^2=\left[k\left(a^2+b^2+c^2\right)\right]^2=k^2\left(a^2+b^2+c^2\right)^2\)(2)
Từ (1),(2) => đpcm
Đặt \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\Rightarrow x=ka,y=kb,z=kc\)
Ta có VT=\(\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)=\left(k^2a^2+k^2b^2+k^2c^2\right)\left(a^2+b^2+c^2\right)\)=
=\(k^2\left(a^2+b^2+c^2\right)^2\)
Mà \(\left(ax+by+cz\right)^2=\left(a^2k+b^2k+c^2k\right)^2=k^2\left(a^2+b^2+c^2\right)^2\)
=> VT=VP
=> ĐPCM
Đặt \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\ne̸0\) thì \(x=ak;y=bk;z=ck.\)
Do đó : \(\frac{\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)}{\left(ax+by+cz\right)^2}\)
\(=\frac{\left(a^2k^2+b^2k^2+c^2k^2\right)\left(a^2+b^2+c^2\right)}{\left(a^2k+b^2k+c^2k\right)^2}=\frac{k^2\left(a^2+b^2+c^2\right)^2}{k^2\left(a^2+b^2+c^2\right)^2}=1.\)
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) \(\left(a+b\right)^2=2\left(a^2+b^2\right)\)
\(\Leftrightarrow a^2+b^2+2ab=2a^2+2b^2\)
\(\Leftrightarrow a^2-2ab+b^2=0\)
\(\Leftrightarrow\left(a-b\right)^2=0\Leftrightarrow a=b\)