Cho x = b^2 + c^2 - a^2/2ab; y = a^2 - (b - c)^2/(b + c)^2 - a^2. Tính giá trị M=x+y/1-xy
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
b: Ta có: \(x\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x^3-4x-x^4+1\)
\(=-x^4+x^3-4x+1\)
c: Ta có: \(\left(a+b-c\right)^2-\left(a-c\right)^2-2ab+2ab\)
\(=\left(a+b-c-a+c\right)\left(a+b-c+a-c\right)\)
\(=b\left(2a+b-2c\right)\)
\(=2ab+b^2-2bc\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A/x(x2-16)- (x4-1)=x3-16x - x4+1
B/ (y-3)(y+3)(y^2+9)-(y^2+2)(y^2-2)=
=(y2-9)(y2+9)- (y4-4)
=(y4-81)-y4+4= -81+4= -77
C/(a+b+c)2-(a-c.)2-2ab+2ab
=( a2 + b2 + c2 +2ab+2ac+2bc)- ( a2-2ac. + c2)- 2ab+2ab
=a2 + b2 + c2 +2ab+2ac+2bc - a2 + 2ac - c2 -2ab+2ab
=b2+2ab+4ac+2bc
=2a(b+2c)+b(b+2c)
=(2a+b)(b+2c)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(a+b+c=2x\)
\(\Rightarrow\left(a+b+c\right)^2=4x^2\)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac=4x^2\)
\(VP=4x\left(x-a\right)\)
\(=4x^2-4xa\)
\(=a^2+b^2+c^2+2ab+2bc+2ac-\left(2a+2b+2c\right)a\)
\(=a^2+b^2+c^2+2ab+2bc+2ac-2a^2-2ab-2ac\)
\(=2bc+b^2+c^2-a^2=VT\)
\(\Rightarrowđpcm\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đề có sai không cậu ? Tớ nghĩ x=.../2bc chứ
Cho \(x=\frac{b^2+c^2-a^2}{2ab},y=\frac{a^2-\left(b-c\right)^2}{\left(b+c\right)^2-a^2}\)
Tính giá trị của:
\(M=\frac{x+y}{1-xy}\)