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2) b)
Do \(a+b+c=9\Rightarrow\left(a+b+c\right)^2=81\)
\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=81\)
\(\Rightarrow2\left(ab+bc+ac\right)=81-141=-60\)
\(ab+bc+ac=-60:2=-30\)
a, B=x^3 + 3xy +y^3 = x^3 +3xy(x+y)+y^3 (vì x+y=1)
= (x+y)^3
= 1^3 =1
b, (a+b+c)^2 =a^2 +b^2 +c^2 +2ab +2bc +2ac
9^2 = 141 +2(ab+bc+ac)
-60 = 2(ab+bc+ac)
ab+ac+bc=-30
Vậy M=-30
c, N =(x+y)^3 -3(x+y)(x^2+y^2) +2(x^3+y^3)
= x^3 + 3x^2 .y + 3xy^2 + -3(x^3+xy^2 +x^2 .y+y^3)+ 2x^3 +2y^3
= x^3 +3x^2 .y + 3xy^2 - 3x^3 -3xy^2 -3x^2 .y -3y^3 +2x^3 +2y^3
= 0
Vậy N=0 .Chúc bạn học tốt.
\(A=x^2+2xy+y^2-4x-4y+q\)
\(=\left(x+y\right)^2-4\left(x+y\right)+q\)
\(=3^2-4.3+q\)
\(=q-3\)
\(A=3\left(x^2+y^2\right)-2\left(x^3-y^3\right)\)
\(A=3x^2+3y^2-2\left(x-y\right)\left(x^2+xy+y^2\right)\)
Mà x- y = 1
\(\Rightarrow A=3x^2+3y^2-2\left(x^2+xy+y^2\right)\)
\(\Leftrightarrow A=3x^2+3y^2-2x^2-2xy-2y^2\)
\(\Leftrightarrow A=x^2-2xy+y^2\)
\(\Leftrightarrow A=\left(x-y\right)^2=1^2=1\)
A = 3( x2 + y2 ) - 2( x3 - y3 )
= 3x2 + 3y2 - 2( x - y )( x2 + xy + y2 )
= 3x2 + 3y2 - 2( x2 + xy + y2 )
= 3x2 + 3y2 - 2x2 - 2xy - 2y2
= x2 - 2xy + y2
= ( x - y )2
= 12 = 1
\(a)\)\(M=x^3-3xy\left(x-y\right)-y^3-x^2+2xy-y^2\) ( đề nhầm đúng ko bn )
\(M=\left(x^3-3x^2y+3xy^2-y^3\right)-\left(x^2-2xy+y^2\right)\)
\(M=\left(x-y\right)^3-\left(x-y\right)^2\)
\(M=7^3-7^2\)
\(M=294\)
Chúc bạn học tốt ~
bài 1
bài 2
ta có: \(\left(x+y\right)^3=x^3+y^3+3x^2y+3xy^2\)
\(\Leftrightarrow\)\(\left(x+y\right)^3=x^3+y^3+3xy\left(x+y\right)\)
mà x+y=1 nên
1=\(x^3+y^3+3xy.1\)
Vậy =1
\(2;x^3+y^3+3xy\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+3xy\)
\(=x^2-xy+y^3+3xy\)
\(=\left(x+y\right)^2=1\)
\(1;\left(a+b+c\right)^3=0\)
\(\Rightarrow\left[\left(a+b\right)+c\right]^3=0\)
\(\Rightarrow\left(a+b\right)^3+3.\left(a+b\right)^2.c+3\left(a+b\right).c^2+c^3=0\)
\(\Rightarrow a^3+3a^2b+3ab^2+b^3+3\left(a^2+2ab+b^2\right)c+3ac^2+3bc^2+c^3=0\)
\(\Rightarrow\left(a^3+b^3+c^3\right)+3a^2b+3ab^2+3a^2c+6abc+3b^2c+3ac^2+3bc^2=0\)
\(.\)M= bn ghi lại đề nha ^.^
\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\left[\left(a^2+2ab+b^2\right)-2ab\right]+6a^2b^2\left(a+b\right)\)
\(=1^3-3ab.1+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2.1\)
\(=1-3ab+3ab\left(1-2ab\right)+6a^2b^2\)
\(M=1-3ab+3ab-6a^2b^2+6a^2b^2\)\(=1\)
k cho mình nha bn thanks nhìu <3 <3 (^3^)
2. \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)(1)
Đặt \(x^2+5x+4=t\)
(1) = \(t.\left(t+2\right)-24\)
\(=t^2+2t+1-25\)
\(=\left(t+1\right)^2-25\)
\(=\left(t+1-5\right)\left(t+1+5\right)\)
\(=\left(t-4\right)\left(t+6\right)\)(2)
Thay \(t=x^2+5x+4\)vào (2) ta có:
(2) = \(\left(x^2+5x+4-4\right)\left(x^2+5x+4+6\right)\)
\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)\(=x\left(x+5\right)\left(x^2+5x+10\right)\)
k mình nha bn <3 thanks
Viết lại :
a) \(M=\left(x+y\right)^3+2\left(x+y\right)^2\)
b) \(N=\left(x-y\right)^3-\left(x-y\right)^2\)
a) M=(x+y)3+2x2+4xy+2y2
M=73+(2x+2y)2=4(x+y)2=73+4.72=343+196=539
b)N=(x-y)3-x2+2xy-y2
N=-53-(x2-2xy+y2)=-125-(x-y)2=-125-(-5)2=-150
\(A=3\left(x+y\right)^2-6xy-\left(x+y\right)^3+3xy\left(x+y\right)\)
Mà x+y=2
\(\Rightarrow A=3.2^2-6xy-2^3+6xy\)
\(=12-8\)
\(=4\)