Câu 1:

Theo bài ra ta có:

\(a^{12}+b^{12}=a^{12}+a^{11}b-a^{11}b-ab^{11}+ab^{11}+b^{12}\)

\(=a^{11}\left(a+b\right)-ab\left(a^{10}+b^{10}\right)+b^{11}\left(a+b\right)\)

\(=\left(a+b\right)\left(a^{11}+b^{11}\right)-ab\left(a^{10}+b^{10}\right)\)

\(=\left(a+b\right)\left(a^{12}+b^{12}\right)-ab\left(a^{12}+b^{12}\right)\)(gt cho rồi nhé)

\(=\left(a^{12}+b^{12}\right)\left(a+b-ab\right)\)

\(\Rightarrow a+b-ab=1\)

\(\Leftrightarrow a+b-ab-1=0\)

\(\Leftrightarrow a\left(1-b\right)-\left(1-b\right)=0\)

\(\Leftrightarrow\left(1-b\right)\left(a-1\right)=0\)

\(\)\(\Leftrightarrow\left[{}\begin{matrix}b=1\\a=1\end{matrix}\right.\)

=> a^20 + b^20 = 2

:)) đừng ném đá nhá

ket ban roblox voi minh di 

acc minh la duclong444 va viduclong4

KillerUnknow666

1/ Đặt \(\left\{{}\begin{matrix}2x^2+1=a\\2-5x=b\end{matrix}\right.\) \(\Rightarrow2x^2-5x+3=a+b\)

Ta được:

\(a^3+b^3=\left(a+b\right)^3\)

\(\Leftrightarrow a^3+b^3=a^3+b^3+3ab\left(a+b\right)\)

\(\Leftrightarrow ab\left(a+b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=0\\b=0\\a+b=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}2x^2+1=0\left(vn\right)\\2-5x=0\\2x^2-5x+3=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{2}{5}\\x=1\\x=\frac{3}{2}\end{matrix}\right.\)

2/

\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)

\(\Leftrightarrow\frac{a}{b-c}=-\frac{b}{c-a}-\frac{c}{a-b}=\frac{b}{a-c}+\frac{c}{b-a}\)

\(\Leftrightarrow\frac{a}{b-c}=\frac{b\left(b-a\right)+c\left(a-c\right)}{\left(a-c\right)\left(b-a\right)}=\frac{b^2-ab+ac-c^2}{\left(a-b\right)\left(c-a\right)}\)

\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}=\frac{b^2-ab+ac-c^2}{\left(a-b\right)\left(c-a\right)\left(b-c\right)}\)

Tương tự ta có: \(\frac{b}{\left(c-a\right)^2}=\frac{c^2+ab-bc-a^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\) ; \(\frac{c}{\left(a-b\right)^2}=\frac{a^2+bc-ac-b^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

Cộng vế với vế:

\(\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(c-a\right)^2}+\frac{c}{\left(a-b\right)^2}=\frac{b^2-ab+ac-c^2+c^2+ab-bc-a^2+a^2+bc-ca-b^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)

c/

\(M=x^2+5y^2-4xy+2x-8y+2018\)

\(M=\left(x^2+4y^2+1-4xy+2x-4y\right)+\left(y^2-4y+4\right)+2013\)

\(M=\left(x-2y+1\right)^2+\left(y-2\right)^2+2013\ge2013\)

\(M_{min}=2013\) khi \(\left\{{}\begin{matrix}x-2y+1=0\\y-2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)

\(a+b+c=0\)

⇔\(\left(a+b+c\right)^2=0\)

⇔\(a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)

⇔\(2018+2\left(ab+bc+ca\right)=0\)

⇔\(ab+bc+ca=-1009\)

⇔\(\left(ab+bc+ca\right)^2=\left(-1009\right)^2=1009^2\)

⇔\(a^2b^2+b^2c^2+c^2a^2+2\left(ab^2c+abc^2+a^2bc\right)=1009^2\)

⇔\(a^2b^2+b^2c^2+c^2a^2+2abc\left(b+c+a\right)=1009^2\)

⇔\(a^2b^2+b^2c^2+c^2a^2=1009^2\)

\(a^2+b^2+c^2=2018\)

⇔\(\left(a^2+b^2+c^2\right)^2=2018^2\)

⇔\(a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=2018^2\)

⇔\(a^4+b^4+c^4+2\cdot1009^2=2018^2\)

⇔\(a^4+b^4+c^4=2018^2-2\cdot1009^2=2036162\)

a) \(\dfrac{2a+3c}{2b+3d}\) = \(\dfrac{2a-3c}{2b-3d}\)

Từ \(\dfrac{a}{b}\) = \(\dfrac{c}{d}\) = k ( k \(\in\) Q, k \(\ne\) 0 )

=> \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

VP = \(\dfrac{2a+3c}{2b+3d}\) = \(\dfrac{2.b.k+3.d.k}{2b+3d}\) = \(\dfrac{k.\left(2b+3d\right)}{2b+3d}\) = k (1)

VT = \(\dfrac{2a-3c}{2b-3d}\) = \(\dfrac{2.b.k-3.d.k}{2b-3d}\) = \(\dfrac{k.\left(2b-3d\right)}{2b-3d}\) = k (2)

Từ (1) và (2) ta có: \(\dfrac{2a+3c}{2b+3d}\) = \(\dfrac{2a-3c}{2b-3d}\)

hay: (2a+3c).(3b-3d) = (2a-3c).(2b+3d)

b: Đặt a/b=c/d=k

=>a=bk; c=dk

\(\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}=\dfrac{\left(bk+dk\right)^2}{\left(b+d\right)^2}=k^2\)

\(\dfrac{\left(a-c\right)^2}{\left(b-d\right)^2}=\dfrac{\left(bk-dk\right)^2}{\left(b-d\right)^2}=k^2\)

Do đó: \(\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}=\dfrac{\left(a-c\right)^2}{\left(b-d\right)^2}\)

c: \(\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{b^3k^3+b^3}{d^3k^3+d^3}=\dfrac{b^3}{d^3}\)

\(\dfrac{a^3-b^3}{c^3-d^3}=\dfrac{b^3k^3-b^3}{d^3k^3-d^3}=\dfrac{b^3}{d^3}\)

Do đó: \(\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{a^3-b^3}{c^3-d^3}\)

d: \(\dfrac{a^{2018}-b^{2018}}{a^{2018}+b^{2018}}=\dfrac{b^{2018}k^{2018}-b^{2018}}{b^{2018}k^{2018}+b^{2018}}=\dfrac{k^{2018}-1}{k^{2018}+1}\)

\(\dfrac{c^{2018}-d^{2018}}{c^{2018}+d^{2018}}=\dfrac{k^{2018}-1}{k^{2018}+1}\)

Do đó: \(\dfrac{a^{2018}-b^{2018}}{a^{2018}+b^{2018}}=\dfrac{c^{2018}-d^{2018}}{c^{2018}+d^{2018}}\)