Khôi Bùi

\(B=10^2+8^2+...+2^2-\left(9^2+7^2+...+1^2\right)\)

\(=10^2+8^2+...+2^2-9^2-7^2-...-1^2\)

\(=\left(10^2-9^2\right)+\left(8^2-7^2\right)+...+\left(2^2-1\right)\)

\(=\left(10-9\right)\left(10+9\right)+\left(8-7\right)\left(8+7\right)+...+\left(2-1\right)\left(2+1\right)\)

\(=10+9+8+7+...+2+1\)

\(=\dfrac{10.11}{2}\)

\(=55\)

a ) \(A=2015.2017=\left(2016-1\right)\left(2016+1\right)=2016^2-1\)

Do \(2016^2>2016^2-1\)

\(\Rightarrow B>A\)

b ) \(C=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(=2^{32}-1< 2^{32}=D\)

Vậy \(C< D\)

Ta có : \(\left\{{}\begin{matrix}\left|5x+1\right|\ge0\forall x\\\left|2y-1\right|\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow3+\left|5x+1\right|+\left|2y-1\right|\ge3\forall x;y\)

\(\Rightarrow\dfrac{12}{3+\left|5x+1\right|+\left|2y-1\right|}\le\dfrac{12}{3}=4\forall x;y\)

Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left|5x+1\right|=0\\\left|2y-1\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x+1=0\\2y-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x=-1\\2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{5}\\y=\dfrac{1}{2}\end{matrix}\right.\)

Vậy Max của b/t trên là : \(4\Leftrightarrow x=-\dfrac{1}{5};y=\dfrac{1}{2}\)

\(\left(2-n\right)\left(n^2-3n+1\right)+n\left(n^2+12\right)+8\)

\(=2n^2-n^3-6n+3n^2+2-n+n^3+12n+8\)

\(=\left(2n^2+3n^2\right)+\left(n^3-n^3\right)+\left(12n-6n-n\right)+\left(8+2\right)\)

\(=5n^2+5n+10\)

\(=5\left(n^2+n+2\right)⋮5\forall n\in Z\left(đpcm\right)\)

\(x^4-1+x^2\)

\(=\left(x^4+x^2+\dfrac{1}{4}\right)-\dfrac{5}{4}\)

\(=\left(x^2+\dfrac{1}{2}\right)^2-\left(\dfrac{\sqrt{5}}{2}\right)^2\)

\(=\left(x^2+\dfrac{1}{2}-\dfrac{\sqrt{5}}{2}\right)\left(x^2+\dfrac{1}{2}+\dfrac{\sqrt{5}}{2}\right)\)

\(=\left(x^2-\dfrac{\sqrt{5}-1}{2}\right)+\left(x^2+\dfrac{\sqrt{5}+1}{2}\right)\)

b ) \(a+b+c+d=0\)

\(\Leftrightarrow a+b=-\left(c+d\right)\)

\(\Leftrightarrow\left(a+b\right)^3=-\left(c+d\right)^3\)

\(\Leftrightarrow\left(a+b\right)^3+\left(c+d\right)^3=0\)

\(\Leftrightarrow a^3+b^3+c^3+d^3+3a^2b+3b^2a+3c^2d+3d^2c=0\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=-3a^2b-3b^2a-3c^2d-3d^2c\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(-a^2b-b^2a-c^2d-d^2c\right)\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left[-ab\left(a+b\right)-cd\left(c+d\right)\right]\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left[ab\left(c+d\right)-cd\left(c+d\right)\right]\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(ab-cd\right)\left(c+d\right)\left(đpcm\right)\)

a ) Ta có : \(a+b+c=0\)

\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+ac+bc\right)=0\)

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

\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2=4\left(ab+ac+bc\right)^2\)

\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2=4\left(a^2b^2+b^2c^2+c^2a^2+2ab^2c+2a^2bc+2c^2ab\right)\)

\(\Leftrightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)+8abc\left(a+b+c\right)\)

\(\Leftrightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+a^2c^2\right)+8abc.0\)

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

Lại có : \(\dfrac{\left(a^2+b^2+c^2\right)^2}{2}=\dfrac{a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)}{2}\)

\(=\dfrac{a^4+b^4+c^4+a^4+b^4+c^4}{2}=\dfrac{2\left(a^4+b^4+c^4\right)}{2}\)

\(=a^4+b^4+c^4\left(đpcm\right)\)

Nếu \(x=9\Rightarrow10=x+1\)

Thay \(10=x+1\) vào A , ta được :

\(A=x^{14}-\left(x+1\right)x^{13}+\left(x+1\right)x^{12}-\left(x+1\right)x^{11}+...+\left(x+1\right)x^2-\left(x+1\right)x+x+1\)

\(=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-x^{12}-x^{11}+...+x^3+x^2-x^2-x+x+1\)

\(=1\)

Vậy \(A=1\) tại \(x=9\)

a ) \(\left(x+y\right)^3+\left(x-y\right)^3-2x^3\)

\(=x^3+3x^2y+3y^2x+y^3+x^3-3x^2y+3y^2x-y^3-2x^3\)

\(=\left(x^3+x^3-2x^3\right)+\left(y^3-y^3\right)+\left(3x^2y-3x^2y\right)+\left(3y^2x+3y^2x\right)\)

\(=6y^2x\)

b ) \(\left(x+y\right)^2-\left(x-y\right)^2+\left(x+y\right)\left(x-y\right)\)

\(=\left(x+y-x+y\right)\left(x+y+x-y\right)+x^2-y^2\)

\(=2y.2x+x^2-y^2\)

\(=x^2-y^2+4xy\)

c ) \(\left(3x+1\right)^2+2\left(9x^2-1\right)+\left(3x-1\right)^2\)

\(=\left(3x+1\right)^2+2\left(3x+1\right)\left(3x-1\right)+\left(3x-1\right)^2\)

\(=\left(3x+1+3x-1\right)^2\)

\(=\left(6x\right)^2=36x^2\)

d ) \(\left(a+b+c\right)^2-2\left(a+b+c\right)\left(b+c\right)+\left(b+c\right)^2\)

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

\(=a^2\)

Ta có hình vẽ :

Ta có : E là TĐ AD \(\Rightarrow2DE=AD\left(1\right)\)

F là TĐ BC \(\Rightarrow2EF=BC\left(2\right)\)

Hình thang ABCD có : \(\left\{{}\begin{matrix}AE=ÀD\left(GT\right)\\BF=FC\left(GT\right)\end{matrix}\right.\)

\(\Rightarrow EF\) là đường trung bình của hình thang ABCD

\(\Rightarrow EF=\dfrac{AB+CD}{2}\)

\(\Rightarrow2EF=AB+CD\left(3\right)\)

Từ ( 1 ) ; ( 2 ) ; ( 3 )

\(\Rightarrow AD+BC+AB+CD=2DE+2FC+2EF\)

\(\Rightarrow AD+BC+AB+CD=2\left(DE+EF+FC\right)\)

\(\Rightarrow AD+BC+AB+CD=2.5=10\left(cm\right)\)

hay Chu vi hình thang ABCD \(=10\left(cm\right)\)

Vậy chu vi hình thang ABCD là : \(10\left(cm\right)\)

Hình như sai đề :

Ta có : \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)

\(\Leftrightarrow\dfrac{bc}{abc}+\dfrac{ac}{abc}+\dfrac{ab}{abc}=0\)

\(\Leftrightarrow\dfrac{ab+ac+bc}{abc}=0\)

\(\Leftrightarrow ab+ac+bc=0\) ( do \(a;b;c\ne0\) ) ( 1 )

Từ ( 1 ) \(\Rightarrow ab+bc=-ac\)

\(\Rightarrow\left(ab+bc\right)^2=\left[-\left(ac\right)\right]^2\)

\(\Rightarrow a^2b^2+b^2c^2+2ab^2c=a^2c^2\) ( * )

CMTT , ta được : \(\left\{{}\begin{matrix}b^2c^2+c^2a^2+2bc^2a=a^2b^2\\c^2a^2+a^2b^2+2a^2cb=b^2c^2\end{matrix}\right.\) ( *' )

Thay ( * ) và ( * ') vào E , ta được :

\(E=\dfrac{a^2b^2c^2}{a^2b^2+b^2c^2-\left(a^2b^2+b^2c^2+2b^2ac\right)}+\dfrac{a^2b^2c^2}{b^2c^2+c^2a^2-\left(b^2c^2+c^2a^2+2bc^2a\right)}\)

\(+\dfrac{a^2b^2c^2}{c^2a^2+a^2b^2-\left(c^2a^2+a^2b^2+2a^2cb\right)}\)

\(=\dfrac{a^2b^2c^2}{-2b^2ac}+\dfrac{a^2b^2c^2}{-2c^2ab}+\dfrac{a^2b^2c^2}{-2a^2cb}\)

\(=\dfrac{-ac}{2}+\dfrac{-ab}{2}+\dfrac{-bc}{2}\)

\(=\dfrac{-\left(ac+ab+bc\right)}{2}\)

\(=\dfrac{0}{2}=0\)

Vậy \(E=0\)

Gọi 2 số lẻ liên tiếp là : \(2k+1;2k+3\left(k\in N\right)\)

Theo bài ra ta có :

\(\left(2k+3\right)^3-\left(2k+1\right)^3=6938\)

\(\Leftrightarrow\left(2k+3-2k-1\right)\left[\left(2k+3\right)^2+\left(2k+3\right)\left(2k+1\right)+\left(2k+1\right)^2\right]=6938\)

\(\Leftrightarrow2\left(4k^2+12k+9+4k^2+8k+3+4k^2+4k+1\right)=6938\)

\(\Leftrightarrow2\left(12k^2+24k+13\right)=6938\)

\(\Leftrightarrow12k^2+24k+13=3469\)

\(\Leftrightarrow12\left(k^2+2k+1\right)+1=3469\)

\(\Leftrightarrow12\left(k+1\right)^2=3468\)

\(\Leftrightarrow\left(k+1\right)^2=289\)

\(\Leftrightarrow\left[{}\begin{matrix}k+1=17\\k+1=-17\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}k=16\\k=-18\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2k+1=33\\2k+3=35\end{matrix}\right.\)

Vậy 2 số lẻ liên tiếp là : \(33;35\)

a ) \(\left(a-b+c\right)^2-\left(a+b+c\right)^2\)

\(=\left(a-b+c-a-b-c\right)\left(a-b+c+a+b+c\right)\)

\(=-2b\left(2a+2c\right)\)

\(=-4ab-4bc\left(đpcm\right)\)

b ) \(6,3-5x+x^2\)

\(=x^2-5x+\dfrac{63}{10}\)

\(=x^2-5x+\dfrac{25}{4}+\dfrac{1}{20}\)

\(=\left(x-\dfrac{5}{2}\right)^2+\dfrac{1}{20}\ge\dfrac{1}{20}>0\forall x\left(đpcm\right)\)

:D

a ) \(C=5x-3x^2+2\)

\(=-3\left(x^2-\dfrac{5}{3}x-\dfrac{2}{3}\right)\)

\(=-3\left(x^2-2x.\dfrac{5}{6}+\dfrac{25}{36}-\dfrac{49}{36}\right)\)

\(=-3\left[\left(x-\dfrac{5}{6}\right)^2-\dfrac{49}{36}\right]\)

\(=-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{49}{12}\le\dfrac{49}{12}\forall x\)

Dấu " = " xảy ra \(\Leftrightarrow x-\dfrac{5}{6}=0\Leftrightarrow x=\dfrac{5}{6}\)

Vậy GTLN của C là : \(\dfrac{49}{12}\Leftrightarrow x=\dfrac{5}{6}\)

b ) \(D=-8x^2+4xy-y^2+3\)

\(=-\left(4x^2-4xy+y^2\right)-4x^2+3\)

\(=-\left(2x-y\right)^2-4x^2+3\le3\forall x;y\)

Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}2x-y=0\\4x^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=y\\x^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x=y\\x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=0\end{matrix}\right.\)

Vậy GTLN của D là : \(3\Leftrightarrow x=y=0\)

a ) \(\left(3x-1\right)^2+\left(3x+1\right)^2+2\left(3x+1\right)\left(1-3x\right)\)

\(=\left(1-3x\right)^2+\left(3x+1\right)^2+2\left(3x+1\right)\left(1-3x\right)\)

\(=\left(1-3x+3x+1\right)^2\)

\(=2^2=4\)

b ) \(\left(2x-3y\right)^2+\left(2x+3y\right)^2+\left(2x-3y\right)\left(2x+3y\right)\)

\(=4x^2-12xy+9y^2+4x^2+12xy+9y^2+4x^2-9y^2\)

\(=\left(4x^2+4x^2+4x^2\right)+\left(9y^2+9y^2-9y^2\right)+\left(12xy-12xy\right)\)

\(=12x^2+9y^2\)

Nếu \(x=25\)

\(\Rightarrow\left\{{}\begin{matrix}26=x+1\\27=x+2\\47=2x-3\\77=3x+2;50=2x;24=x-1\end{matrix}\right.\) ( * )

Thay ( * ) vào C , ta được :

\(C=x^7-\left(x+1\right)x^6+\left(x+2\right)x^5-\left(2x-3\right)x^4-\left(3x+2\right)x^3+2x.x^2+x-\left(x-1\right)\)

\(=x^7-x^7-x^6+x^6+2x^5-2x^5+3x^4-3x^4-2x^3+2x^3+x-x+1\)

\(=1\)

Vậy \(C=1\) tại \(x=25\)

a ) \(\left(x+3\right)^2-x\left(x-5\right)=31\)

\(\Leftrightarrow x^2+6x+9-x^2+5x=31\)

\(\Leftrightarrow11x+9=31\)

\(\Leftrightarrow11x=22\)

\(\Leftrightarrow x=2\)

Vậy \(x=2\)

b ) \(\left(4x+3\right)\left(x-2\right)-\left(2x-1\right)^2=50\)

\(\Leftrightarrow4x^2+3x-8x-6-\left(4x^2-4x+1\right)=50\)

\(\Leftrightarrow4x^2-5x-6-4x^2+4x-1=50\)

\(\Leftrightarrow\left(4x^2-4x^2\right)-\left(5x-4x\right)-\left(6+1\right)=50\)

\(\Leftrightarrow-x-7=50\)

\(\Leftrightarrow-x=57\)

\(\Leftrightarrow x=-57\)

Vậy \(x=-57\)

c ) \(\left(3x+5\right)\left(3x-5\right)-9x\left(x+20\right)=18\)

\(\Leftrightarrow9x^2-25-9x^2-180x=18\)

\(\Leftrightarrow-25-180x=18\)

\(\Leftrightarrow180x=-43\)

\(\Leftrightarrow x=-\dfrac{43}{180}\)

Vậy \(x=-\dfrac{43}{180}\)

Sửa đề : CMR : \(2bc+b^2+c^2-a^2=4p\left(p-a\right)\)

Ta có : \(VT=2bc+b^2+c^2-a^2\)

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

\(=\left(b+c-a\right)\left(b+c+a\right)\)

\(=\left(b+c+a-2a\right).2p\)

\(=\left(2p-2a\right).2p\)

\(=4p^2-4ap\)

\(=4p\left(p-a\right)=VP\left(đpcm\right)\)

\(A=8x^2+3y^2-16x+6y+100\)

\(=\left(8x^2-16x+8\right)+\left(3y^2+6y+3\right)+89\)

\(=2\left(4x^2-8x+4\right)+3\left(y^2+2y+1\right)+89\)

\(=2\left[\left(2x\right)^2-2.2x.2+4\right]+3\left(y+1\right)^2+89\)

\(=2\left(2x-2\right)^2+3\left(y+1\right)^2+89\ge89\forall x;y\)

Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}2x-2=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=2\\y=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

Vậy Min A là : \(89\Leftrightarrow x=1;y=-1\)

1 ) \(x^2+xy+y^2=0\)

\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2\right)=0\)

\(\Leftrightarrow x^3-y^3=0\)

\(\Leftrightarrow x^3=y^3\)

\(\Leftrightarrow x=y\)

\(\Leftrightarrow x^2+xy+y^2=x^2+x^2+x^2\)

\(\Leftrightarrow0=3x^2\)

\(\Leftrightarrow x^2=0\)

\(\Leftrightarrow x=0\)

\(\Leftrightarrow y=0\)

Vậy \(x=y=0\)

2 ) \(2x^2+y^2+2xy-2x+1=0\)

\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow\left(x+y\right)^2+\left(x-1\right)^2=0\left(1\right)\)

Do \(\left\{{}\begin{matrix}\left(x+y\right)^2\ge0\forall x;y\\\left(x-1\right)^2\ge0\forall x\end{matrix}\right.\)

\(\Rightarrow\left(x+y\right)^2+\left(x-1\right)^2\ge0\forall x;y\left(2\right)\)

Từ ( 1 ) ; ( 2 )

\(\Rightarrow\left\{{}\begin{matrix}x+y=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-y\\x=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=-1\\x=1\end{matrix}\right.\)

Vậy \(x=1;y=-1\)

\(\left(3a-2x\right)^2-\left(a-2x\right)^2\)

\(=\left(3a-2x-a+2x\right)\left(3a-2x+a-2x\right)\)

\(=2a\left(4a-4x\right)\)

\(=8a^2-8ax\)

1 ) \(C=x^2+y^2-8x+4y+27\)

\(=\left(x^2-8x+16\right)+\left(y^2-4y+4\right)+7\)

\(=\left(x-4\right)^2+\left(y-2\right)^2+7\ge7\forall x;y\)

Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-4=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=2\end{matrix}\right.\)

Vậy GTNN của C là : \(7\Leftrightarrow x=4;y=2\)

2 ) a ) \(B=-3x^2+2x-1\)

\(=-3\left(x^2-\dfrac{2}{3}x+\dfrac{1}{3}\right)\)

\(=-3\left(x^2-2x.\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{2}{9}\right)\)

\(=-3\left[\left(x-\dfrac{1}{3}\right)^2+\dfrac{2}{9}\right]\)

\(=-3\left(x-\dfrac{1}{3}\right)^2-\dfrac{2}{3}\le-\dfrac{2}{3}\forall x\)

Dấu " = " xảy ra \(\Leftrightarrow x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\)

Vậy GTLN của B là : \(-\dfrac{2}{3}\Leftrightarrow x=\dfrac{1}{3}\)

b ) \(C=-5x^2+20x-49\)

\(=-5\left(x^2-4x+\dfrac{49}{5}\right)\)

\(=-5\left(x^2-4x+4+\dfrac{29}{5}\right)\)

\(=-5\left[\left(x-2\right)^2+\dfrac{29}{5}\right]\)

\(=-5\left(x-2\right)^2-29\le-29\forall x\)

Dấu " = " xảy ra \(\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Vậy GTLN của C là : \(-29\Leftrightarrow x=2\)

2 ) b )

\(a+b+c+d=0\)

\(\Leftrightarrow a+b=-\left(c+d\right)\)

\(\Leftrightarrow\left(a+b\right)^3=-\left(c+d\right)^3\)

\(\Leftrightarrow a^3+b^3+3a^2b+3b^2a=-c^3-3c^2d-3d^2c-d^3\)

\(\Leftrightarrow a^3+b^3+3a^2b+3b^2a+c^3+3c^2d+3d^2c+d^3=0\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=-3a^2b-3b^2a-3c^2d-3d^2c\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=-3ab\left(a+b\right)-3cd\left(c+d\right)\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=3ab\left(c+d\right)-3cd\left(c+d\right)\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(ab-cd\right)\left(c+d\right)\) \(\left(đpcm\right)\)

\(\left(x-1\right)^3-x^3+3x^2-3x-1\)

\(=x^3-3x^2+3x-1-x^3+3x^2-3x-1\)

\(=-2\left(đpcm\right)\)

Do điện tích hạt nhân của A là : 12+

\(\Rightarrow p=12\)

Mà số p = số e \(\Rightarrow e=12\)

\(\Rightarrow p+e=24\)

Nguyên tử A có tổng số hạt là : 36

\(\Rightarrow p+e+n=36\)

\(\Rightarrow24+n=36\)

\(\Rightarrow n=12\)

Vậy \(p=e=n=12\)

a ) \(VT=a\left(1-b\right)+a\left(a^2-1\right)\)

\(=a-ab+a^3-a\)

\(=a^3-ab\)

\(=a\left(a^2-b\right)=VP\left(đpcm\right)\)

b ) \(VP=a^3+b^3+c^3-3abc\)

\(=\left(a^3+b^3+3a^2b+3b^2a\right)+c^3-3abc-3a^2b-3b^2a\)

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

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2-3ab\right]\)

\(=\left(a+b+c\right)\left[a^2+2ab+b^2-ac-bc+c^2-3ab\right]\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=VP\left(đpcm\right)\)

Ta có hình vẽ :

Ta có : \(\left\{{}\begin{matrix}\widehat{B}-\widehat{D}=10^o\\\widehat{A}-\widehat{B}=21^o\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\widehat{B}=\widehat{D}+10^o\\\widehat{A}=21^o+\widehat{B}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\widehat{B}=\widehat{D}+10^o\\\widehat{A}=21^o+\widehat{D}+10^o=31^o+\widehat{D}\end{matrix}\right.\)

\(\Delta ADC\) cân tại D ( AD = DC ) \(\Rightarrow\widehat{DAC}=\widehat{DCA}\left(1\right)\)

\(\Delta ABC\) cân tại B ( AB = BC ) \(\Rightarrow\widehat{BAC}=\widehat{BCA}\left(2\right)\)

Từ ( 1 ) ; ( 2 ) \(\Rightarrow\widehat{DAC}+\widehat{BAC}=\widehat{DCA}+\widehat{BCA}\)

\(\Rightarrow\widehat{A}=\widehat{C}\)

Mà \(\widehat{A}=31^o+\widehat{D}\Rightarrow\widehat{C}=31^o+\widehat{D}\)

Xét tứ giác ABCD có :

\(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^o\)

\(\Rightarrow31^o+\widehat{D}+10^o+\widehat{D}+31^o+\widehat{D}+\widehat{D}=360^o\)

\(\Rightarrow4\widehat{D}+72^o=360^o\)

\(\Rightarrow4\widehat{D}=288^o\)

\(\Rightarrow\widehat{D}=72^o\)

\(\Rightarrow\left\{{}\begin{matrix}\widehat{B}=72^o+10^o=82^o\\\widehat{A}=\widehat{C}=72^o+31^o=103^o\end{matrix}\right.\)

Vậy \(\widehat{A}=\widehat{C}=103^o;\widehat{B}=82^o;\widehat{D}=72^o\)

a ) Nếu \(x=71\) \(\Rightarrow70=x-1\)

Thay \(70=x-1\) vào A , ta được :

\(A=x^5-\left(x-1\right)x^4-\left(x-1\right)x^3-\left(x-1\right)x^2-\left(x-1\right)x\)

\(=x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x\)

\(=x\)

\(=71\)

Vậy \(A=71\) tại \(x=71\)

b ) Ta có : \(x=35\)

\(\Rightarrow\left\{{}\begin{matrix}36=x+1\\37=x+2\\69=2x-1\\34=x-1\end{matrix}\right.\) ( * )

Thay ( * ) vào B , ta được :

\(B=x^5-\left(x+1\right)x^4+\left(x+2\right)x^3-\left(2x-1\right)x^2-\left(x-1\right)x+15\)

\(=x^5-x^5-x^4+x^4+2x^3-2x^3+x^2-x^2+x+15\)

\(=x+15\)

\(=35+15=50\)

Vậy \(B=50\) tại \(x=35\)

a ) \(4x\left(5x+2\right)-\left(10x-3\right)\left(2x+7\right)=133\)

\(\Leftrightarrow20x^2+8x-\left(20x^2-6x+70x-21\right)=133\)

\(\Leftrightarrow20x^2+8x-20x^2+6x-70x+21=133\)

\(\Leftrightarrow-56x+21=133\)

\(\Leftrightarrow-56x=112\)

\(\Leftrightarrow x=-2\)

Vậy \(x=-2\)

b ) \(3\left(6x-5\right)\left(4x+1\right)-\left(8x+3\right)\left(9x-2\right)=203\)

\(\Leftrightarrow\left(18x-15\right)\left(4x+1\right)-\left(72x^2+27x-16x-6\right)=203\)

\(\Leftrightarrow72x^2-60x+18x-15-72x^2-27x+16x+6=203\)

\(\Leftrightarrow\left(72x^2-72x^2\right)+\left(18x+16x-60x-27x\right)-\left(15-6\right)=203\)

\(\Leftrightarrow-53x-9=203\)

\(\Leftrightarrow-53x=212\)

\(\Leftrightarrow x=-4\)

Vậy \(x=-4\)

\(x^2\left(x+1\right)-x\left(x+1\right)\)

\(=x^3+x^2-x^2-x\)

\(=x^3-x\)

\(=x\left(x^2-1\right)\)

\(=x\left(x-1\right)\left(x+1\right)\)