a^2+b^2+c^2-a-b-c+3/4. Tính a,b,c
K
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Những câu hỏi liên quan
7 tháng 6 2019
a) (x - 2)(x2 - 2x + 4)(x - 2)( x2 + 2x + 4)
= (x - 2)2(x - 2)2(x + 2)2
= (x - 2)4(x + 2)2
b) (a + b + c)3 - (b + c - a)3 - (a - b + c)3 - (a + b - c)3
Đặt a+b-c=x, c+a-b=y, b+c-a=z
=>x+y+z=a+b-c+c+a-b+b+c-a=a+b+c
Ta có hằng đẳng thức:
(x+y+z)^3-3x-3y-3z=3(x+y)(x+z)(y+z)
=>(a+b+c)^3-(b+c-a)^3-(a+c-b)^3-(a+b-c)^3=(x+y+z)^3-x^3-y^3-z^3
=3(x+y)(x+z)(y+z)
=3(a+b-c+c+a-b)(c+a-b+b+c-a)(b+c-a+a+b-c)
=3.2a.2b.2c
=24abc
c) (a + b)3 + (b + c)3 + (c + a)3 - 3(a + b)(b + c)(c + a)
Đặt x = a+b; y = b+c; z = c+a ta có:
x3+y3+z3−3xyz
= (x+y)3−3xy(x−y)+z3−3xyz
=[(x+y)3+z3]−3xy(x+y+z)
=(x+y+z)3−3z(x+y)(x+y+z)−3xy(x−y−z)
=(x+y+z)[(x+y+z)2−3z(x+y)−3xy]
=(x+y+z)(x2+y2+z2+2xy+2xz+2yz−3xz−3yz−3xy)
=(x+y+z)(x2+y2+z2−xy−yz−yx)
Thay vào ta có:
(a+b+b+c+c+a)[(a+b)2+(b+c)2+(c+a)2−(a+b)(b+c)−(b+c)(c+a)−(c+a)(a+b)]
=(2a+2b+2c)(a2−ab−ac+b2−bc+c2)
=2(a+b+c)(a2−ab−ac+b2−bc+c2)
KN
7 tháng 6 2019
a)\(\left(x-2\right)\left(x^2-2x+4\right)\left(x-2\right)\left(x^2+2x-4\right)\)
\(=\left(x-2\right)^2\left(x^2-2x+4\right)\left(x^2+2x-4\right)\)
\(=\left(x-2\right)^2\left(x^4+4x^2+16\right)\)
\(=x^6-4x^5+8x^4-16x^3+32x^2-64x+64\)
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
\(\begin{array}{l}A + B + C\\ = (3{x^4} - 2{x^3} - x + 1) + ( - 2{x^3} + 4{x^2} + 5x) + ( - 3{x^4} + 2{x^2} + 5)\\ = 3{x^4} - 2{x^3} - x + 1 - 2{x^3} + 4{x^2} + 5x - 3{x^4} + 2{x^2} + 5\\ = (3{x^4} - 3{x^4}) + ( - 2{x^3} - 2{x^3}) + (4{x^2} + 2{x^2}) + ( - x + 5x) + (1 + 5)\\ = 0 + ( - 4{x^3}) + 6{x^2} + 4x + 6\\ = - 4{x^3} + 6{x^2} + 4x + 6\\A - B + C\\ = (3{x^4} - 2{x^3} - x + 1) - ( - 2{x^3} + 4{x^2} + 5x) + ( - 3{x^4} + 2{x^2} + 5)\\ = 3{x^4} - 2{x^3} - x + 1 + 2{x^3} - 4{x^2} - 5x - 3{x^4} + 2{x^2} + 5\\ = (3{x^4} - 3{x^4}) + ( - 2{x^3} + 2{x^3}) + ( - 4{x^2} + 2{x^2}) + ( - x - 5x) + (1 + 5)\\ = 0 + 0 + ( - 2{x^2}) - 6x + 6\\ = - 2{x^2} - 6x + 6\\A - B - C\\ = (3{x^4} - 2{x^3} - x + 1) - ( - 2{x^3} + 4{x^2} + 5x) - ( - 3{x^4} + 2{x^2} + 5)\\ = 3{x^4} - 2{x^3} - x + 1 + 2{x^3} - 4{x^2} - 5x + 3{x^4} - 2{x^2} - 5\\ = (3{x^4} + 3{x^4}) + ( - 2{x^3} + 2{x^3}) + ( - 4{x^2} - 2{x^2}) + ( - x - 5x) + (1 - 5)\\ = 6{x^4} + 0 + ( - 6{x^2}) - 6x + ( - 4)\\ = 6{x^4} - 6{x^2} - 6x - 4\end{array}\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 1 2021
\(B=\dfrac{a^3+c^3+3ac\left(a+c\right)-b^3-3ac\left(a+c\right)+3abc}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2}\)
\(=\dfrac{\left(a+c\right)^3-b^3-3ac\left(a+c-b\right)}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2}\)
\(=\dfrac{\left(a+c-b\right)\left[\left(a+c\right)^2+b\left(a+c\right)+b^2\right]-3ac\left(a+c-b\right)}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2}\)
\(=\dfrac{\left(a+c-b\right)\left(a^2+b^2+c^2+ab+bc-ac\right)}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2}\)
\(=\dfrac{-2\left(2a^2+2b^2+2c^2+2ab+2bc-2ca\right)}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2}\)
\(=\dfrac{-2\left[\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2\right]}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2}=-2\)
NQ
1
ML
10 tháng 6 2016
\(a^2+b^2+c^2=1\Rightarrow\left|a\right|;\left|b\right|;\left|c\right|\le1\Rightarrow a;b;c\le1.\)
\(a^3+b^3+c^3=a^2+b^2+c^2\Rightarrow a^2\left(1-a\right)+b^2\left(1-b\right)+c^2\left(1-c\right)=0\)
Do \(a;b;c\le1\) nên \(a^2\left(1-a\right)+b^2\left(1-b\right)+c^2\left(1-c\right)\ge0\)
Dấu bằng xảy ra khi \(\hept{\begin{cases}a^2+b^2+c^2=1\\a;b;c\in\left\{0;1\right\}\end{cases}\Leftrightarrow\left(a;b;c\right)=\left(0;0;1\right);\left(0;1;0\right);\left(1;0;0\right)}\)
26 tháng 6 2015
Ta có
x + y = 2
=> (x+y)^2 = 4
=> x^2 + 2xy + y^2 = 4
=> 10 + 2xy= 4
=> 2xy = -6
=> xy= -3
x^3 + y^3 = ( x+Y) ( x^2 - xy + y^2) = 2 ( 10 -- 3) = 2( 10 + 3 ) = 2.13 = 26