1) Đơn giản biểu thức
a) 4a(b-c+2a)
b)-(m-n)-(2m+n-p)
c)-(x-y)+(-3x-2y+z)
d)-(2a-2b)+(2a-3b+c)
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a) (2a - b)(b + 4a) + 2a(b - 3a)
= 2a(b + 4a) - b(b + 4a) + 2ab - 6a^2
= 2ab + 8a^2 - b^2 - 4ab + 2ab - 6a^2
= (8a^2 - 6a^2) + (2ab + 2ab - 4ab) - b^2
= 2a^2 - b^2
b) .(3a - 2b)(2a - 3b) - 6a(a - b)
= 3a(2a - 3b) - 2b(2a - 3b) - (6a^2 - 6ab)
= 6a^2 - 9ab - (4ab - 6b^2) - (6a^2 - 6ab)
= 6a^2 - 9ab - 4ab + 6b^2 - 6a^2 + 6ab
= 6b^2 + (6a^2 - 6a^2) + (6ab - 4ab - 9ab)
= 6b^2 - 7ab
c. 5b(2x - b) - (8b - x)(2x - b)
= 10bx - 5b^2 - 8b(2x - b) + x(2x - b)
= 10bx - 5b^2 - 16bx + 8b^2 + 2x^2 - bx
= (10bx - 16bx - bx) + 2x^2 + (8b^2 - 5b^2)
= -7bx + 2x^2 + 3b^2
d. 2x(a + 15x) + (x - 6a)(5a + 2x)
= 2ax + 30x^2 + x(5a + 2x) - 6a(5a + 2x)
= 2ax + 30x^2 + 5ax + 2x^2 - 30a^2 - 12ax
= (30x^2 + 2x^2) + (2ax + 5ax - 12ax) - 30a^2
= 32x^2 - 5ax - 30a^2
Chúc bạn hok tốt !!!
Sửa đề c/m : \(\frac{a}{x+2y+z}=\frac{b}{2x+y-z}=\frac{c}{4x-4y+z}\)
Ta có \(\frac{x}{a+2b+c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}\)
=> \(\frac{a+2b+c}{x}=\frac{2a+b-c}{y}=\frac{4a-4b+c}{z}\)
Từ (1) => \(\frac{a+2b+c}{x}=\frac{4a+2b-2c}{2y}=\frac{4a-4b+c}{z}=\frac{a+2b+c+4a+2b-2c+4a-4b+c}{x+2y+z}\)
\(=\frac{9a}{x+2y+z}\)(2)
Từ (1) => \(\frac{2a+4b+2c}{2x}=\frac{2a+b-c}{y}=\frac{4a-4b+c}{z}=\frac{2a+4b+2c+2a+b-c-4a+4b-c}{2x+y-z}\)
\(=\frac{9b}{2x+y-z}\)(3)
Từ (1) => \(\frac{4a+8b+4c}{4x}=\frac{8a+4b-4c}{4y}=\frac{4a-4b+c}{z}\)
\(=\frac{4a+8a+4c-8a-4b+4c+4a-4b+c}{4x-4y+z}=\frac{9c}{4x-4y+z}\)(4)
Từ (2)(3)(4) => \(\frac{9a}{x+2y+z}=\frac{9b}{2x+y-z}=\frac{9c}{4x-4y+z}\)
=> \(\frac{a}{x+2y+z}=\frac{b}{2x+y-z}=\frac{c}{4x-4y+z}\)(đpcm)
\(a)4a\left(b-c+2a\right)\)
\(=4ab-4ac+8a^2\)
\(b)-\left(m-n\right)-\left(2m+n-p\right)\)
\(=-m+n-2m-n+p\)
\(=\left(-m-2m\right)+\left(n-n\right)+p\)
\(=p-3m\)
\(c)-\left(x-y\right)+\left(-3x-2y+z\right)\)
\(=-x+y-3x-2y+z\)
\(=\left(-x-3x\right)+\left(y-2y\right)+z\)
\(=z-4x-y\)
\(d)-\left(2a-2b\right)+\left(2a-3b+c\right)\)
\(=-2a+2b+2a-3b+c\)
\(=\left(-2a+2a\right)+\left(2b-3b\right)+c\)
\(=c-b\)