(a+b+c)2+(a-b-c)2+(b-c-a)2-+(c-a-b)2
bài rút gọn giúp minh với
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15 tháng 7 2021
B1
a, \(=>A=\left(x+y+x-y\right)\left(x+y-x+y\right)=2x.2y=4xy\)
b, \(=>B=\left[\left(x+y\right)-\left(x-y\right)\right]^2=\left[x+y-x+y\right]^2=\left[2y\right]^2=4y^2\)
c,\(\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^2-1\right)\)
\(=\)\(\left(x+1\right)\left(x^2-x+1\right)\left(x-1\right)\left(x^2+x+1\right)=\left(x^3+1^3\right)\left(x^3-1^3\right)=x^6-1\)
d, \(\left(a+b-c\right)^2+\left(a-b+c\right)^2-2\left(b-c\right)^2\)
\(=\left(a+b-c\right)^2-\left(b-c\right)^2+\left(a-b+c\right)^2-\left(b-c\right)^2\)
\(=\left(a+b-c+b-c\right)\left(a+b-c-b+c\right)\)
\(+\left(a-b+c+b-c\right)\left(a-b+c-b+c\right)\)
\(=a\left(a+2b-2c\right)+a\left(a-2b\right)\)
\(=a\left(a+2b-2c+a-2b\right)=a\left(2a-2c\right)=2a^2-2ac\)
B2:
\(\)\(x+y=3=>\left(x+y\right)^2=9=>x^2+2xy+y^2=9\)
\(=>xy=\dfrac{9-\left(x^2+y^2\right)}{2}=\dfrac{9-\left(17\right)}{2}=-4\)
\(=>x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=3\left(17+4\right)=63\)
15 tháng 7 2021
Bài 1:
a) Ta có: \(\left(x+y\right)^2-\left(x-y\right)^2\)
\(=x^2+2xy+y^2-x^2+2xy+y^2\)
=4xy
b) Ta có: \(\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\)
\(=\left(x+y-x+y\right)^2\)
\(=\left(2y\right)^2=4y^2\)
c) Ta có: \(\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^2-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)
\(=\left(x^3-1\right)\left(x^3+1\right)\)
\(=x^6-1\)
d) Ta có: \(\left(a+b-c\right)^2+\left(a+b+c\right)^2-2\left(b-c\right)^2\)
\(=\left(a+b-c\right)^2-\left(b-c\right)^2+\left(a+b+c\right)^2-\left(b-c\right)^2\)
\(=\left(a+b-c-b+c\right)\left(a+b-c+b-c\right)+\left(a+b+c-b+c\right)\left(a+b+c+b-c\right)\)
\(=a\cdot\left(a+2b-2c\right)+\left(a+2c\right)\left(a-2b\right)\)
\(=a^2+2ab-2ac+a^2-2ab+2ac-4bc\)
\(=2a^2-4bc\)
M
1
HK
18 tháng 9 2017
Ta có:\(\left(a-b+c\right)^2+\left(a-b+c\right)^2-2\left(b-c\right)^2\)
\(=2\left(a-b+c\right)^2-2\left(b-c\right)^2\\ =2\left(\left(a-b+c\right)^2-\left(b-c\right)^2\right)\)
\(=2\left(a-b+c-b+c\right)\left(a-b+c+b-c\right)\\ =2\left(a-2b+2c\right)a \)
\(=2a^2-4ab+4ac\)
13 tháng 6 2017
Ta có : \(\left(a+b\right)^2=a^2+2ab+b^2\)
Thay số từ đề bài vào rùi tính thui :
\(15^2=a^2+2\cdot7+b^2\)
\(\Leftrightarrow225=a^2+b^2+14\)
\(\Leftrightarrow a^2+b^2=225-14=211\)
TK NKA !!!
26 tháng 9 2017
Umk !!! giúp liền nàk
\(a+b+c=0\Rightarrow\hept{\begin{cases}a+b=-c\\a+c=-b\\b+c=-a\end{cases}}\)nên
\(\frac{a^2}{a^2-b^2-c^2}+\frac{b^2}{b^2-c^2-a^2}+\frac{c^2}{c^2-a^2-b^2}\)
\(=\frac{a^2}{\left(a-b\right)\left(a+b\right)-c^2}+\frac{b^2}{\left(b-c\right)\left(b+c\right)-a^2}+\frac{c^2}{\left(c-a\right)\left(c+a\right)-b^2}\)
\(=\frac{a^2}{-c\left(a-b\right)-c^2}+\frac{b^2}{-a\left(b-c\right)-a^2}+\frac{c^2}{-b\left(c-a\right)-b^2}\)
\(=\frac{a^2}{-ac+bc-c^2}+\frac{b^2}{-ab+ac-a^2}+\frac{c^2}{-bc+ab-b^2}\)
\(=\frac{a^2}{-c\left(a+c\right)+bc}+\frac{b^2}{-a\left(a+b\right)+ac}+\frac{c^2}{-b\left(b+c\right)+ab}\)
\(=\frac{a^2}{-c\left(-b\right)+bc}+\frac{b^2}{\left(-a\right)\left(-c\right)+ac}+\frac{c^2}{-b\left(-a\right)+ab}\)
\(=\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2ab}=\frac{a^3+b^3+c^3}{2abc}\)
Mà a + b +c = 0 nên \(a^3+b^3+c^3=3abc\) (tự chứng minh)
Do đó \(\frac{a^3+b^3+c^3}{2abc}=\frac{3abc}{2abc}=\frac{3}{2}\)
Vậy \(\frac{a^2}{a^2-b^2-c^2}+\frac{b^2}{b^2-c^2-a^2}+\frac{c^2}{c^2-a^2-b^2}=\frac{3}{2}\)
IL
26 tháng 9 2017
trả ơn này
Vì a + b + c = 0
\(\Rightarrow\)a2 = b2 + c2 + 2bc \(\Rightarrow\) a2 - b2 - c2 = 2bc
\(\Rightarrow\)b2 = a2 + c2 + 2bc\(\Rightarrow\) b2 - a2 - c2 = 2bc
\(\Rightarrow\) c2 = a2 + c2 +2ab\(\Rightarrow\)c2 - b2 - a2 = 2ab
còn lại tự làm nhé
19 tháng 9 2021
\(\left(a-b\right)^2=a^2+b^2-2ab\\ \Rightarrow49=a^2+b^2-120\Rightarrow a^2+b^2=169\)
\(\left(a+b\right)^2=a^2+b^2+2ab=169+120=289\\ \Rightarrow a+b=17\)
\(a^2-b^2=\left(a-b\right)\left(a+b\right)=7\cdot17=119\)
\(a^4+b^4=\left(a^2+b^2\right)^2-2a^2b^2=169^2-2\cdot60^2\\ =28561-7200=21361\)
19 tháng 9 2021
\(2\left(x^2+y^2\right)=\left(x-y\right)^2\\ \Rightarrow2x^2+2y^2=x^2-2xy+y^2\\ \Rightarrow x^2+2xy+y^2=0\\ \Rightarrow\left(x+y\right)^2=0\Rightarrow x+y=0\Rightarrow x=-y\)
17 tháng 10 2021
\(a,=x^3-16x-x^2-1-x^2+1=x^3-2x^2-16x\\ b,=y^4-81-y^4+4=-77\\ d,=a^2+b^2+c^2+2ab-2bc-2ac+a^2-2ac+c^2-2ab-2ac\\ =2a^2+b^2+2c^2-2bc-6ac\)
đặt a - b-c=x; b-c-a=y; c-a-b=z
=> a + b + c = ...
Thay vào ròi lm tiếp nha
Bài làm:
Đặt \(\hept{\begin{cases}a-b-c=x\\b-c-a=y\\c-a-b=z\end{cases}}\)=> \(a+b+c=-\left(x+y+z\right)\)
Thay vào:
Bt = \(x^2+y^2+z^2-\left(x+y+z\right)^2\)
\(=x^2+y^2+z^2-x^2-y^2-z^2-2\left(xy+yz+zx\right)\)
\(=-2\left(xy+yz+zx\right)\)
Xét: \(xy=\left(a-b-c\right)\left(b-c-a\right)=\left(b+c-a\right)\left(c+a-b\right)\)
\(=\left[c-\left(a-b\right)\right]\left[c+\left(a-b\right)\right]\)
\(=c^2-\left(a-b\right)^2\)
\(=c^2-a^2+2ab-b^2\)
Tương tự: \(yz=a^2-b^2+2bc-c^2\) ; \(zx=b^2-c^2+2ca-a^2\)
=> \(-2\left(xy+yz+zx\right)=2\left(a^2+b^2+c^2-2ab-2bc-2ca\right)\)