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a) (x-y)(x4+x3y+x2y2+xy3+y4) = x(x4+x3y+x2y2+xy3+y4)-y(x4+x3y+x2y2+xy3+y4) =(x5+x4y+x3y2+x2y2+xy4)-(x4y+x3y2+x2y2+xy4+y5) = x5+x4y+x3y2+x2y2+xy4-x4y-x3y2-x2y2-xy4-y5 =x5-y5⇒Điều cần chứng minh
Các câu b d tương tự
\(\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)=x^5-y^5\)
Ta có VT:
\(\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)\)
\(=x.x^4+x.x^3y+x.x^2y^2+x.xy^3+x.y^4-y.x^4-y.x^3y-y.x^2y^2-y.xy^3-y.y^4\)
\(=x^5+x^4y+x^3y^2+x^2y^3+xy^4-x^4y-x^3y^2-x^2y^3-xy^4-y^5\)
\(=x^5-y^5\)
VT=VP
Vậy:...
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\) (*)
a) Từ (*) ta có:
\(\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\) (1)
\(\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\) (2)
Từ (1) và (2) suy ra \(\dfrac{a}{a-b}=\dfrac{c}{c-d}\)
b) Từ (*) ta có:
\(\dfrac{a}{b}=\dfrac{bk}{b}=k\) (3)
\(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=\dfrac{k\left(b+d\right)}{b+d}=k\) (4)
Từ (3) và (4) suy ra \(\dfrac{a}{b}=\dfrac{a+c}{b+d}\)
c) Từ (*) ta có:
\(\dfrac{a}{3a+b}=\dfrac{bk}{3bk+b}=\dfrac{bk}{b\left(3k+1\right)}=\dfrac{k}{3k+1}\) (5)
\(\dfrac{c}{3c+d}=\dfrac{dk}{3dk+d}=\dfrac{dk}{d\left(3k+1\right)}=\dfrac{k}{3k+1}\) (6)
Từ (5) và (6) suy ra \(\dfrac{a}{3a+b}=\dfrac{c}{3c+d}\)
d) Từ (*) ta có:
\(\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\) (7)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2.k^2+d^2.k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\) (8)
Từ (7) và (8) suy ra \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
e) Từ (*) ta có:
\(\dfrac{ab}{cd}=\dfrac{bk.b}{dk.d}=\dfrac{b^2}{d^2}=\dfrac{b}{d}\) (9)
\(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{b^2.k^2-b^2}{d^2.k^2-d^2}=\dfrac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\dfrac{b}{d}\) (10)
Từ (9) và (10) suy ra \(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\)
f) Từ (*) ta có:
\(\dfrac{ab}{cd}=\dfrac{bk.b}{dk.d}=\dfrac{b^2}{d^2}=\dfrac{b}{d}\) (11)
\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\dfrac{\left[b\left(k-1\right)\right]^2}{\left[d\left(k-1\right)\right]^2}=\dfrac{b}{d}\) (12)
Từ (11) và (12) suy ra \(\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
\(\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{a^2+ac}{ac+c^2}=\dfrac{a\left(a+c\right)}{c\left(a+c\right)}=\dfrac{a}{c}\left(đpcm\right)\)
1
\(M=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1\)
\(M=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}< \frac{a+c}{a+b+c}+\frac{b+a}{b+a+c}+\frac{c+b}{a+b+c}=2\)
=> M ko là số tự nhiên
2
\(a+b+c=0\)
\(\Rightarrow\left(a+b+c\right)^2=0\)
\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
Do \(a^2+b^2+c^2\ge0\Rightarrow ab+bc+ca\le0\)
3
\(\left(x+y\right)\cdot35=\left(x-y\right)\cdot2010=xy\cdot12\)
\(\Rightarrow35x+35y=2010x-2010y\)
\(\Rightarrow35-2010x=2010y-35y\)
\(\Rightarrow-175x=-245y\)
\(\Rightarrow\frac{x}{y}=\frac{245}{175}=\frac{7}{5}\)
\(\Rightarrow\frac{x}{7}=\frac{y}{5}\)
Đặt \(\frac{x}{7}=\frac{y}{5}=k\)
\(\Rightarrow x=7k;y=5k\)
\(\Rightarrow\left(5k+7k\right)\cdot35=35k^2\cdot12\)
\(\Rightarrow k=k^2\Rightarrow k=1\left(k\ne0\right)\)
Vậy \(x=7;y=5\)
bài 2 chưa thuyết phục lắm, nếu \(a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\) thì \(ab+bc+ca\ge0\) vẫn đúng, lẽ ra phải là \(ab+bc+ca=-\frac{\left(a^2+b^2+c^2\right)}{2}\le0\) *3*
Ta có: \(A\cdot C+B^2-2x^4y^4=x^3y\cdot xy^3+\left(x^2y^2\right)^2-2x^4y^4\)
\(\Leftrightarrow A\cdot C+B^2-2x^4y^4=x^4y^4+x^4y^4-2xy^4\)
\(\Leftrightarrow A\cdot C+B^2-2x^4y^4=0\)(đpcm)
A.C + B^2 - 2x^4.y^4
=(x^3.y)(x.y^3) + x^4.y^4 - 2x^4.y^4
=(x^4.y^4 + x^4.y^4) - 2x^4.y^4
=2x^4.y^4 - 2x^4.y^4
=0