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a) ( a - b + c ) - ( -b - a + c ) - [ - ( -a ) ]
= a - b + c + b + a - c - a
= 0
chắc là z ~~
a: (2x+1)(y-3)=1
\(\Leftrightarrow\left(2x+1;y-3\right)\in\left\{\left(1;1\right);\left(-1;-1\right)\right\}\)
hay \(\left(x,y\right)\in\left\{\left(0;4\right);\left(-1;2\right)\right\}\)
b: (x+1)(xy-1)=3
\(\Leftrightarrow\left(x+1;xy-1\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(-1;-3\right);\left(-3;-1\right)\right\}\)
\(\Leftrightarrow\left(x,xy-1\right)\in\left\{\left(0;3\right);\left(2;1\right);\left(-1;-2\right);\left(-4;-1\right)\right\}\)
\(\Leftrightarrow\left(x,y\right)\in\left\{\left(2;1\right);\left(-1;1\right);\left(-4;0\right)\right\}\)
a,(a+b)-(-c+a+b)
= a+b-c-a-b
= -c
b,-(x+y)+(-z+x+y)
= -x-y+(-z)+x+y
= -x+(-y)+x+y+(-z)
=-z
c,(m-n+p)+(-m+n+p)
= m-n+p+(-m)+n+p
=2p
cho biểu thức
a. A = 3/n+2 (n thuộc z, n khác 2). Tìm n sao cho n thuộc A.
b. B= -5/n-1n(n thuộc z, n khác 1). Tìm n sao cho n thuộc B
Đề sai rồi! Sửa đề: Cho \(S_1=\dfrac{b}{a}x+\dfrac{c}{a}z...\)
Giải:
Ta có:
\(S_1+S_2+S_3=\left(\dfrac{b}{a}x+\dfrac{c}{a}z\right)+\left(\dfrac{a}{b}x+\dfrac{c}{b}y\right)\)\(+\left(\dfrac{a}{c}z+\dfrac{b}{c}y\right)\)
\(=\left(\dfrac{b}{a}x+\dfrac{a}{b}x\right)+\left(\dfrac{c}{b}y+\dfrac{b}{c}y\right)+\left(\dfrac{c}{a}z+\dfrac{a}{c}z\right)\)
\(=\left(\dfrac{b}{a}+\dfrac{a}{b}\right)x+\left(\dfrac{c}{b}+\dfrac{b}{c}\right)y+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)z\)
Dễ thấy: \(\left\{{}\begin{matrix}\dfrac{b}{a}+\dfrac{a}{b}\ge2\\\dfrac{c}{b}+\dfrac{b}{c}\ge2\\\dfrac{c}{a}+\dfrac{a}{c}\ge2\end{matrix}\right.\)
\(\Rightarrow S_1+S_2+S_3\ge2x+2y+2z\)
\(=2\left(x+y+z\right)=2.1008=2016\)
Vậy \(S_1+S_2+S_3\ge2016\) (Đpcm)
\(\left(a+b\right)-\left(-c+a+b\right)\)
\(=a+b+c-a-b\)
\(=\left(a-a\right)+\left(b-b\right)+c\)
\(=c\)
\(-\left(x+y\right)+\left(-z+x+y\right)\)
\(=-x+-y+-z+x+y\)
\(=\left[\left(-x\right)+x\right]+\left[\left(-y\right)+y\right]+-z\)
\(=-z\)
\(\left(m-n+p\right)+\left(-m+n+p\right)\)
\(=m-n+p-m+n+p\)
\(=\left(m-m\right)+\left(n-n\right)+\left(p+p\right)\)
\(=2p\)
2p là gì