Rút gọn:
a. (2a+b)^2-(2a+b)(2a-b)-2a(b-a).
b. (a+b-c)^2-(a+b)^2+2c(a+b).
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bài 1 : a +b , rút gọn và tính
(-a+b-c)-(a-b-c)= -a+b -c-a+b+c= -2a+2b= -2.1+2.-1=-2+-2 = -4
a)A=(-2a + 3b - 4c) - (-2a - 3b -4c)
=-2a+3b-4c+2a+3b+4c
=(-2a+2a)+(3b+3b)+(-4c+4c)
=2.3b
=6b (1)
b) thay b=-1 vào -6b ta được:
6.(-1)=-6
Vậy A=-6
a/ \(\frac{\left(x+2\right)\left(x+6\right)}{x+2}\) = X + 6
b/\(\frac{3a+3b+3c}{a+b+c}\) = \(\frac{3\left(a+b+c\right)}{a+b+c}\)= 3
c/\(\frac{\left(2a+b\right)c+c\left(b-a\right)}{\left(a+b\right)c}\)= \(\frac{c\left(2a+b+b-a\right)}{\left(a+b\right)c}\)= \(\frac{c\left(a+2b\right)}{\left(a+b\right)c}\)= \(\frac{a+2b}{a+b}\)
Cám ơn bạn nhìu nha, mik ngồi đợi mãi câu trả lời đây, thanks you bạn nhìu
1.
\(2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4>0\\ \Leftrightarrow a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2< 0\\ \Leftrightarrow\left(a^4+b^4+c^4+2a^2b^2-2b^2c^2-2c^2a^2\right)-4a^2b^2< 0\\ \Leftrightarrow\left(a^2+b^2-c^2\right)^2-4a^2b^2< 0\\ \Leftrightarrow\left(a^2+b^2-c^2-2ab\right)\left(a^2+b^2-c^2+2ab\right)< 0\\ \Leftrightarrow\left[\left(a-b\right)^2-c^2\right]\left[\left(a+b\right)^2-c^2\right]< 0\\ \Leftrightarrow\left(a-b+c\right)\left(a-b-c\right)\left(a+b-c\right)\left(a+b+c\right)< 0\left(1\right)\)
Vì a,b,c là độ dài 3 cạnh của 1 tg nên \(\left\{{}\begin{matrix}a+c>b\\a-b< c\\a+b>c\\a+b+c>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a-b+c>0\\a-b-c< 0\\a+b-c>0\\a+b+c>0\end{matrix}\right.\)
Do đó \(\left(1\right)\) luôn đúng (do 3 dương nhân 1 âm ra âm)
Từ đó ta được đpcm
1. Phuc will look through a new English book tomorrow.
2. Which ethnic group has the largest population in Vietnam?
a + b = c => (a + b)² = c² <=> a²+ b² + 2ab = c²
=> c^4 = (a² + b² + 2ab)²
=> c^4 = a^4 + b^4 + 6a²b² + 4a^3.b + 4a.b^3
vậy: a^4 + b^4 + c^4 = 2a^4 + 2b^4 + 6a²b² + 4a^3.b + 4a.b^3
= 2a^4 + 2a²b² + 4a^3.b + 2b^4 + 2a²b² + 4a.b^3 + 2a²b²
= 2a²(a² + b² + 2ab) + 2b²(b² + a² + 2ab) + 2a²b²
= 2a²(a + b)² + 2b²(a + b)² + 2a²b²
= 2a²b² + 2(a + b)²(a² + b²)
= 2a²b² + 2c²(a² +b²)
= 2a²b² + 2b²c² + 2c²a² (đpcm)
a ) \(\left(2a+b\right)^2-\left(2a+b\right)\left(2a-b\right)-2a\left(b-a\right)\)
\(=4a^2+4ab+b^2-\left(4a^2-b^2\right)-2ab+2a^2\)
\(=4a^2+4ab+b^2-4a^2+b^2-2ab+2a^2\)
\(=2a^2+2ab+2b^2\)
\(=\left(a^2+2ab+b^2\right)+a^2+b^2\)
\(=\left(a+b\right)^2+a^2+b^2\)
b ) \(\left(a+b-c\right)^2-\left(a+b\right)^2+2c\left(a+b\right)\)
\(=\left(a+b\right)^2-2\left(a+b\right)c+c^2-\left(a+b\right)^2+2c\left(a+b\right)\)
\(=\left[\left(a+b\right)^2-\left(a+b\right)^2\right]+\left[2c\left(a+b\right)-2\left(a+b\right)c\right]+c^2\)
\(=c^2\)
@Khôi Bùi
\(\left(2a+b\right)^2-\left(2a+b\right)\left(2a-b\right)-2a\left(b-a\right)\)
\(=\left(2a+b\right)\left[\left(2a+b\right)-\left(2a-b\right)\right]-2a\left(b-a\right)\)
\(=2b\left(2a+b\right)-2a\left(b-a\right)\)
\(=4ab+2b^2-2ab+2a^2=2\left(a^2+ab+b^2\right)\)
\(\left(a+b-c\right)^2-\left(a+b\right)^2+2c\left(a+b\right)\)
\(=\left(a+b-c+a+b\right)\left(a+b-c-a-b\right)+2c\left(a+b\right)\)
\(=-c\left(2a+2b-c\right)+2c\left(a+b\right)=\)
\(-2c\left(a+b\right)+c^2+2c\left(a+b\right)=c^2\)