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\(a,VT=\left(a+b+c\right)\left(a-b+c\right)\)
\(=\left(a+c+b\right)\left(a+c-b\right)\)
\(=\left(a+c\right)^2-b^2\)
\(=a^2+2ac+c^2-b^2=VP\)
\(b,VT=\left(3x+2y\right)\left(3x-2y\right)-\left(4x-2y\right)\left(4x+2y\right)\)
\(=9x^2-4y^2-16x^2+4y^2=-7x^2=VP\)
\(c,VT=x^3-1-x^3-1=-2=VP\)
\(d,VT=8x^3+1-8x^3+1=2=VP\)
\(e,VT=\left(x^2+2xy+4y^2\right)\left(x-2y-2x+1\right)\)
\(=\left(x^2+2xy+4y^2\right)\left(-x-2y+1\right)\)
\(=-x^3-2x^2y+x^2-2x^2y-4xy^2+2xy-4xy^2-8y^3+4y^2\)
( bn kiểm tra lại đề nhé)
\(1)\)
\(a)\)\(A=100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(A=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(A=100+99+98+97+...+2+1\)
\(A=\frac{100\left(100+1\right)}{2}\)
\(A=5050\)
\(b)\)\(B=3\left(2^2+1\right)\left(2^4+1\right).....\left(2^{64}+1\right)+1\)
\(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right).....\left(2^{64}+1\right)+1\)
\(B=\left(2^4-1\right)\left(2^4+1\right).....\left(2^{64}+1\right)+1\)
\(B=\left(2^8+1\right).....\left(2^{64}+1\right)+1\)
\(............\)
\(B=\left(2^{64}-1\right)\left(2^{64}+1\right)+1\)
\(B=2^{128}-1+1\)
\(B=2^{128}\)
Chúc bạn học tốt ~
\(1)\)
\(c)\)\(C=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)
\(C=\left(a+b\right)^2+2\left(a+b\right)c+c^2+\left(a+b\right)^2-2\left(a+b\right)c+c^2-2\left(a+b\right)^2\)
\(C=2\left(a+b\right)^2+2c^2-2\left(a+b\right)^2\)
\(C=2c^2\)
\(2)\)
\(a)\)\(VP=\left(a+b\right)^3-3ab\left(a+b\right)\)
\(VP=a^3+3a^2b+3ab^2+b^3-3ab\left(a+b\right)\)
\(VP=a^3+3ab\left(a+b\right)+b^3-3ab\left(a+b\right)\)
\(VP=a^3+b^3=VT\) ( đpcm )
\(b)\)\(VT=a^3+b^3+c^3-3abc\)
\(VT=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(VT=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(VT=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(VT=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=VP\) ( đpcm )
Từ đó suy ra :
\(i)\)\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow\)\(a^3+b^3+c^3-3abc=0\)\(\Rightarrow\)\(a+b+c=0\)
Hoặc \(a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow\)\(2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\)\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\)\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\)\(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow}a=b=c}\)
Chúc bạn học tốt ~
Ta có:\(x+y=a\)
=>\(x^2+2xy+y^2=a^2\)
=>\(x^2+y^2=a^2-2xy=a^2-2b\left(đpcm\right)\)
Ta lại có:\(x^3+3x^2y+3xy^2+y^3=a^3\)
=>\(x^3+y^3+3xy\left(x+y\right)=a^3\)
=>\(x^3+y^3=a^3-3xy\left(x+y\right)=a^3-3ab\left(đpcm\right)\)
b)\(a+b+c=0\) =>\(a^3+b^3+c^3+3a^2b+3ab^2+3b^2c+3bc^2+3c^2a+3a^2c+6abc=0\) =>\(a^3+b^3+c^3+3\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\) =>\(a^3+b^3+c^3+3\left(-a\right)\left(-b\right)\left(-c\right)=0\) =>\(a^3+b^3+c^3=3abc\left(đpcm\right)\)
1,2x2+2y2+z2+2xy+2xz+2yz+10x+6y+34=0
<=>(x2+y2+z2+2xy+2xz+2yz)+(x2+10x+25)+(y2+6y+9)=0
<=>(x+y+z)2+(x+5)2+(y+3)2=0
Mà \(\hept{\begin{cases}\left(x+y+z\right)^2\ge0\\\left(x+5\right)^2\ge0\\\left(y+3\right)^2\ge0\end{cases}\Rightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2\ge0}\)
\(\Rightarrow\hept{\begin{cases}\left(x+y+z\right)^2=0\\\left(x+5\right)^2=0\\\left(y+3\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x+y+z=0\\x=-5\\y=-3\end{cases}\Rightarrow}\hept{\begin{cases}z=8\\x=-5\\y=-3\end{cases}}}\)
2, A=2x2+4y2+4xy+2x+4y+9
=(x2+4xy+4y2)+(2x+4y)+x2+9
=[(x+2y)2+2(x+2y)+1]+x2+8
=(x+2y+1)2+x2+8
Vì \(\hept{\begin{cases}\left(x+2y+1\right)^2\ge0\\x^2\ge0\end{cases}}\Rightarrow\left(x+2y+1\right)^2+x^2\ge0\)
\(\Rightarrow\left(x+2y+1\right)^2+x^2+8\ge8\)
Dấu "=" xảy ra khi x=0,y=-1/2
Vậy Amin = 8 khi x=0,y=-1/2
Bài 1:
Ta có:\(2x^2+2y^2+z^2+2xy+2xz+2yz+10x+6y+34=0\)
\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2xz+2yz\right)+\left(x^2+10x+25\right)+\left(y^2+6y+9\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)^2+\left(x+5\right)^2+\left(y+3\right)^2=0\)
Vì 3 vế trên đều dương ,nên ta có
\(\hept{\begin{cases}x+y+z=0\\x+5=0\\y+3=0\end{cases}\Leftrightarrow\hept{\begin{cases}z=0-y-x\\x=-5\\y=-3\end{cases}}\Leftrightarrow\hept{\begin{cases}z=0+3+5=8\\x=-5\\y-3\end{cases}}}\)
Vậy ...........................................................................................................................
Có (a+b+c)2 = 3(ab+bc+ac)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac=3ab+3bc+3ac\)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac-3ab-3bc-3ac\)\(=0\)
\(\Rightarrow a^2+b^2+c^2-ab-bc-ac\)\(=0\)
\(\Rightarrow a^2+b^2+c^2=ab+bc+ac\)
\(\Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ac\)
\(\Rightarrow a^2-2ab+b^2+b^2-2bc+c^2+a^2-2ac+c^2\)\(=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
\(\Rightarrow a=b=c\)
Câu 1:
a: \(\left(y-1\right)^3+3\left(y+1\right)^2=\left(y+2\right)\left(y^2-2y+4\right)\)
\(\Leftrightarrow y^3-3y^2+3y-1+3y^2+6y+3=y^3+8\)
\(\Leftrightarrow y^3+9y+2=y^3+8\)
=>9y=6
hay y=2/3
b: \(x^2-2x+y^2+4y+5+\left(2z-3\right)^2=0\)
\(\Leftrightarrow x^2-2x+1+y^2+4y+4+\left(2z-3\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y+2\right)^2+\left(2z-3\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\\z=\dfrac{3}{2}\end{matrix}\right.\)
c: \(4x^2-12x+9=\left(5-x\right)^2\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(x-5\right)^2=0\)
\(\Leftrightarrow\left(2x-3-x+5\right)\left(2x-3+x-5\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x-8\right)=0\)
hay \(x\in\left\{-2;\dfrac{8}{3}\right\}\)
13.
M \(=\)\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)\)\(+16\)
\(=\)\(\left(x+2\right)\left(x+8\right)\left(x+4\right)\left(x+6\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)
\(=\left(x^2+10x+20-4\right)\left(x^2+10x+20+4\right)\) \(+16\)
\(=\left(x^2+10x+20\right)^2-16+16\)
\(=\left(x^2+10x+20\right)^2\) là một số chính phương
Nhiều quá, nhìn đã thấy ớn lạnh :(
Bạn nên chia nhỏ ra , post 1 hoặc 2 bài 1 lần thôi, đăng 1 lần 1 nùi thế này không ai dám làm đâu, bội thực chữ viết.