Bài 1 : Tính
a) ( x + 1 ) \(^2\)+ ( x - 2 ) ( x + 3 ) - 4 x
b) ( 6x\(^5\)y\(^2\)-9x\(^4\)y\(3\)+ 12x\(^3\)y\(^4\)) : 3x\(^3\)y\(2\)
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19 tháng 10 2021
\(a,=3x\left(y-z\right)-y\left(y-z\right)=\left(3x-y\right)\left(y-z\right)\\ b,=x^3\left(x-1\right)+x\left(x-1\right)=x\left(x^2+1\right)\left(x-1\right)\\ c,=x\left(y+z\right)+y\left(y+z\right)=\left(x+y\right)\left(y+z\right)\\ d,=\left(x-3\right)^2\\ e,=\left(x+2\right)^3\\ f,=\left(2x-x+y\right)\left(2x+x-y\right)=\left(x+y\right)\left(3x-y\right)\\ g,=\left(y+1\right)\left(5x-2\right)\\ h,=\left(x+2\right)^2\\ i,=x^2\left(x^2-2\right)\\ k,=3x\left(x-4y\right)\)
YN
5 tháng 4 2022
`Answer:`
1) \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)
\(=[x\left(x+3\right)][\left(x+1\right)\left(x+2\right)]+1\)
\(=\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)
\(=\left(x^2+3x\right)^2+2.\left(x^2+3x\right)+1\)
\(=\left(x^2+3x+1\right)^2\)
2) \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)
\(=[\left(4x+1\right)\left(3x+2\right)][\left(12x-1\right)\left(x+1\right)]-4\)
\(=\left(12x^2+8x+3x+2\right)\left(12x^2+12x-x-1\right)-4\)
\(=[\left(12x^2+11x+0,5\right)+1,5][\left(12x^2+11x+0,5\right)-1,5]-4\)
\(=\left(12x^2+11x+0,5\right)^2-\left(1,5\right)^2-4\)
\(=\left(12x^2+11x+0,5\right)^2-\left(2,5\right)^2\)
\(=\left(12x^2+11x+0,5-2,5\right)\left(12x^2+11x+0,5+2,5\right)\)
\(=\left(12x^2+11x-2\right)\left(12x^2+11x+3\right)\)
3) \(\left(x^2+6x+5\right)\left(x^2+10x+21\right)+15\)
\(=\left(x^2+x+5x+5\right)\left(x^2+3x+7x+21\right)+15\)
\(=\left(x+1\right)\left(x+5\right)\left(x+3\right)\left(x+7\right)+15\)
\(=[\left(x+1\right)\left(x+7\right)][\left(x+5\right)\left(x+3\right)]+15\)
\(=\left(x^2+x+7x+7\right)\left(x^2+3x+5x+15\right)+15\)
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
Đặt \(v=x^2+=8x+11\)
Đa thức có dạng sau: \(\left(v-4\right)\left(v+4\right)+15\)
\(=v^2-4^2+15\)
\(=v^2-1\)
\(=\left(v+1\right)\left(v-1\right)\)
\(=\left(x^2+8x+11+1\right)\left(x^2+8x+11-1\right)\)
\(=\left(x^2+8x+12\right)\left(x^2+8x+10\right)\)
4) \(\left(x^2-a\right)^2-6x^2+4x+2a\)
\(=\left(x^2-a\right)\left(x^2-a\right)-6x^2+4x+2a\)
\(=\left(x^2-a\right).x^2-a\left(x^2-a\right)-6x^2+4x+2a\)
\(=x^4-ax^2-a.\left(x^2-a\right)-6x^2+4x+2a\)
\(=x^4-ax^2-\left(ax^2-aa\right)-6x^2+4x+2a\)
\(=x^4-2ax^2+a^2-6x^2+2a+4x\)
6) \(a^2-b^2-c^2+2bc-2a+1\)
\(=\left(a^2-2a+1\right)-\left(b^2-2bc+c^2\right)\)
\(=\left(a-1\right)^2-\left(b-c\right)^2\)
\(=\left(a-b+c-1\right)\left(a+b-c-1\right)\)
7) \(4a^2-4b^2+16bc-16c^2\)
\(=4a^2-\left(4b^2-16bc+16c^2\right)\)
\(=\left(2a\right)^2-\left(2b-4c\right)^2\)
\(=\left(2a-2b+4c\right)\left(2a+2b-4c\right)\)
\(=2.\left(a-b-2c\right).2\left(a+b-2c\right)\)
\(=4\left(a-b-2c\right)\left(a+b-2c\right)\)
14 tháng 9 2017
d)5.(x-y)-y(x-y)
=(x-y)(5-y)
e) y.(x-z)+7(z-x)
=y.(x-z)-7(x-z)
=(x-z)(y-7)
T
24 tháng 10 2023
Bài 1.
a)
\((x-2)(2x-1)-(2x-3)(x-1)-2\\=2x^2-x-4x+2-(2x^2-2x-3x+3)-2\\=2x^2-5x+2-(2x^2-5x+3)-2\\=2x^2-5x+2-2x^2+5x-3-2\\=(2x^2-2x^2)+(-5x+5x)+(2-3-2)\\=-3\)
b)
\(x(x+3y+1)-2y(x-1)-(y+x+1)x\\=x^2+3xy+x-2xy+2y-xy-x^2-x\\=(x^2-x^2)+(3xy-2xy-xy)+(x-x)+2y\\=2y\)
Bài 2.
a)
\((14x^3+12x^2-14x):2x=(x+2)(3x-4)\\\Leftrightarrow 14x^3:2x+12x^2:2x-14x:2x=3x^2-4x+6x-8\\ \Leftrightarrow 7x^2+6x-7=3x^2+2x-8\\\Leftrightarrow (7x^2-3x^2)+(6x-2x)+(-7+8)=0\\\Leftrightarrow 4x^2+4x+1=0\\\Leftrightarrow (2x)^2+2\cdot 2x\cdot 1+1^2=0\\\Leftrightarrow (2x+1)^2=0\\\Leftrightarrow 2x+1=0\\\Leftrightarrow 2x=-1\\\Leftrightarrow x=\frac{-1}2\)
b)
\((4x-5)(6x+1)-(8x+3)(3x-4)=15\\\Leftrightarrow 24x^2+4x-30x-5-(24x^2-32x+9x-12)=15\\\Leftrightarrow 24x^2-26x-5-(24x^2-23x-12)=15\\\Leftrightarrow 24x^2-26x-5-24x^2+23x+12=15\\\Leftrightarrow -3x+7=15\\\Leftrightarrow -3x=8\\\Leftrightarrow x=\frac{-8}3\\Toru\)
a: \(\left(x+1\right)^2+\left(x+3\right)\left(x-2\right)-4x\)
\(=x^2+2x+1+x^2+x-6-4x\)
\(=2x^2-x-6\)