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Trả lời giúp bạn nè:
VT = S(S - 2b)(S -2c) + S(S-2c)(S - 2a) + S(S - 2a)(S - 2b)
= S((S - 2b)(S -2c) + (S-2c)(S - 2a) + (S - 2a)(S - 2b) )
= S ( S2 -2cS -2bS + 4bc + S2 - 2aS - 2cS +4ac + S2 -2bS -2aS +4ab )
= S ( 3S2 - 4cS -4bS - 4aS + 4bc + 4ac + 4ab)
= 3S3 - 4cS2 - 3bS2 - 4aS2 + 4bcS + 4acS + 4abS
= S3 + S3 + S3 - 4cS2 - 3bS2 - 4aS2 + 4bcS + 4acS + 4abS
= S2 (S -4c ) + S2 (S -4b ) + S2 (S -4a )
= S2 ( S -4c + S - 4b + S - 4a)
= S2 (3S - 4(c + b + a)
= S2 (3S - 4S)
= 3S3 - 4S3
= -S3 ( 1 )
VP = (S - 2a)(S - 2b)(S - 2c) + 8abc
= (S2 -2bS -2aS + 4ab)(S - 2c) + 8abc
= S3 - 2cS2 - 2bS2 + 4bcS - 2aS2 + 4acS + 4abS - 8abc + 8abc
= S3 - 2cS2 - 2bS2 - 2aS2 + 4bcS + 4acS + 4abS
= S2 (S -2c ) - S2 (2b + 2a )
= S2 ( S - 2c - 2b - 2a )
= S2 ( S - 2( c + b + a))
= S3 - 2S3
= -S3 ( 2 )
Từ (1) và (2) suy ra :
S(S - 2b)(S -2c) + S(S-2c)(S - 2a) + S(S - 2a)(S - 2b) = (S - 2a)(S - 2b)(S - 2c) + 8abc
Trả lời giúp bạn nè:
VT = S(S - 2b)(S -2c) + S(S-2c)(S - 2a) + S(S - 2a)(S - 2b)
= S((S - 2b)(S -2c) + (S-2c)(S - 2a) + (S - 2a)(S - 2b) )
= S ( S2 -2cS -2bS + 4bc + S2 - 2aS - 2cS +4ac + S2 -2bS -2aS +4ab )
= S ( 3S2 - 4cS -4bS - 4aS + 4bc + 4ac + 4ab)
= 3S3 - 4cS2 - 3bS2 - 4aS2 + 4bcS + 4acS + 4abS
= S3 + S3 + S3 - 4cS2 - 3bS2 - 4aS2 + 4bcS + 4acS + 4abS
= S2 (S -4c ) + S2 (S -4b ) + S2 (S -4a )
= S2 ( S -4c + S - 4b + S - 4a)
= S2 (3S - 4(c + b + a)
= S2 (3S - 4S)
= 3S3 - 4S3
= -S3 ( 1 )
VP = (S - 2a)(S - 2b)(S - 2c) + 8abc
= (S2 -2bS -2aS + 4ab)(S - 2c) + 8abc
= S3 - 2cS2 - 2bS2 + 4bcS - 2aS2 + 4acS + 4abS - 8abc + 8abc
= S3 - 2cS2 - 2bS2 - 2aS2 + 4bcS + 4acS + 4abS
= S2 (S -2c ) - S2 (2b + 2a )
= S2 ( S - 2c - 2b - 2a )
= S2 ( S - 2( c + b + a))
= S3 - 2S3
= -S3 ( 2 )
Từ (1) và (2) suy ra :
S(S - 2b)(S -2c) + S(S-2c)(S - 2a) + S(S - 2a)(S - 2b) = (S - 2a)(S - 2b)(S - 2c) + 8abc
\(x+y+z=1\\ \Rightarrow\left\{{}\begin{matrix}x=1-y-z\\y=1-x-z\\z=1-x-y\end{matrix}\right.\)
\(S=\dfrac{\left(xy+z\right)\left(yz+x\right)\left(zx+y\right)}{\left(1-x\right)^2\left(1-y\right)^2\left(1-z\right)^2}\)
\(\Rightarrow S=\dfrac{\left(xy+1-x-y\right)\left(yz+1-y-z\right)\left(zx+1-x-z\right)}{\left(x+y+z-x\right)^2\left(x+y+z-y\right)^2\left(x+y+z-z\right)^2}\)
\(\Rightarrow S=\dfrac{\left[\left(xy-x\right)-\left(y-1\right)\right]\left[\left(yz-y\right)-\left(z-1\right)\right]\left[\left(zx-x\right)-\left(z-1\right)\right]}{\left(y+z\right)^2\left(x+z\right)^2\left(x+y\right)^2}\)
\(\Rightarrow S=\dfrac{\left[x\left(y-1\right)-\left(y-1\right)\right]\left[y\left(z-1\right)-\left(z-1\right)\right]\left[x\left(z-1\right)-\left(z-1\right)\right]}{\left(y+z\right)^2\left(x+z\right)^2\left(x+y\right)^2}\)
\(\Rightarrow S=\dfrac{\left(x-1\right)\left(y-1\right)\left(y-1\right)\left(z-1\right)\left(x-1\right)\left(z-1\right)}{\left(y+z\right)^2\left(x+z\right)^2\left(x+y\right)^2}\)
\(\Rightarrow S=\dfrac{\left(x-1\right)^2\left(y-1\right)^2\left(z-1\right)^2}{\left(y+z\right)^2\left(x+z\right)^2\left(x+y\right)^2}\)
\(\Rightarrow S=\dfrac{\left(x-x-y-z\right)^2\left(y-x-y-z\right)^2\left(z-x-y-z\right)^2}{\left(y+z\right)^2\left(x+z\right)^2\left(x+y\right)^2}\)
\(\Rightarrow S=\dfrac{\left(-y-z\right)^2\left(-x-z\right)^2\left(-x-y\right)^2}{\left(y+z\right)^2\left(x+z\right)^2\left(x+y\right)^2}\)
\(\Rightarrow S=\dfrac{\left(y+z\right)^2\left(x+z\right)^2\left(x+y\right)^2}{\left(y+z\right)^2\left(x+z\right)^2\left(x+y\right)^2}\)
\(\Rightarrow S=1\)
Ta có: \(S=a+b+c\left(1\right)\)
Thay \(\left(1\right)\)vào ta được:
\(\left(S-2b\right).\left(S-2c\right)=\left(a+b+c-2b\right).\)\(\left(a+b+c-2c\right)\)
\(=\left(a-b+c\right).\left(a+b-c\right)\)
\(=a^2+ab-ac-ba-b^2+bc+ca+cb-c^2\)
\(=a^2-b^2-c^2+2.bc\left(2\right)\)
Tương tự, ta được:
\(\left(S-2c\right).\left(S-2a\right)=b^2-c^2-a^2+2.ca\left(3\right)\)
\(\left(S-2a\right).\left(S-2b\right)=c^2-a^2-b^2+2.ab\left(4\right)\)
Từ \(\left(2\right);\left(3\right);\left(4\right)\Rightarrow\)Tổng bằng:
\(a^2-b^2-c^2+2bc+b^2-c^2-a^2+2ca+c^2-a^2\)\(-b^2+2ab\)
\(=2ab+2bc+2ca-a^2-b^2-c^2\)
Vậy tổng trên \(=2ab+2bc+2ca-a^2-b^2-c^2.\)
\(yz\left(y+z\right)+zx\left(z-x\right)-xy\left(x+y\right)\)
\(=yz\left(y+z\right)+zx\left(z-x\right)-xy\left[\left(y+z\right)-\left(z-x\right)\right]\)
\(=yz\left(y+z\right)+zx\left(z-x\right)-xy\left(y+z\right)+xy\left(z-x\right)\)
\(=y\left(y+z\right)\left(z-x\right)+x\left(z-x\right)\left(z-y\right)\)
\(=\left(z-x\right)\left(yz-xy+xz-xy\right)\)
thi cấp tỉnh mà với có 1 số bài thi vào chuyên đại học với cấp 3 nữa
Bài 2: Ta có:
\(\left(2x+5y+1\right)\left(2020^{\left|x\right|}+y+x^2+x\right)=105\) là số lẻ
\(\Rightarrow\left\{{}\begin{matrix}2x+5y+1\\2020^{\left|x\right|}+y+x^2+x\end{matrix}\right.\) đều lẻ
\(\Rightarrow y⋮2\)\(\Rightarrow2020^{\left|x\right|}⋮̸2\Leftrightarrow\left|x\right|=0\Leftrightarrow x=0\).
Thay vào tìm được y...
Ta có \(x^2+1=x^2+xy+yz+xz=\left(x+y\right)\left(x+z\right)\)
\(y^2+1=\left(y+z\right)\left(y+x\right)\)
\(z^2+1=\left(z+x\right)\left(z+y\right)\)
Khi đó
\(S=x.\sqrt{\left(y+z\right)^2}+y.\sqrt{\left(x+z\right)^2}+z.\sqrt{\left(x+y\right)^2}=2\left(xy+yz+xz\right)=2\)
a)\(x^3+y^3+z^3-3xyz\\ \left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\\ =\left[\left(x+y\right)^3+z^3\right]-\left[3xyz+3xy\left(x+y\right)\right]\\=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right] \\ =\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)\\ =\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\\ =\left(x+y+z\right)\left(x^2+y^2+x^2-xy-xz-yz\right)\)