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TT
1
PT
16 tháng 10 2017
\(x^2+y^2=1\) hay sao.?
\( 2(x^6 + y^6) - 3(x^4 + y^4) \)
\(= 2x^4(x^2 - 1) + 2y^4(y^2 - 1) - (x^4 + y^4) \)
\(= - 2x^4 .y^2 - 2y^4 .x^2 - [(x^2 +y^2)^2 - 2x^2.y^2] \)
\(= - 2x^2y^2.(x^2 + y^2) - 1 + 2x^2.y^2 \)
\(= - 2x^2y^2 - 1 + 2x^2y^2 \)
\(=-1\)
17 tháng 8 2020
a)Ta có:
\(\left(x-2\right)^2-\left(x-3\right)\left(x-1\right)\\ =x^2-4x+4-x^2+4x-3\\ =1\)
Vậy biểu thức \(\left(x-2\right)^2-\left(x-3\right)\left(x-1\right)\)không phụ thuộc vào biến
b) Ta có:
\(\left(x-1\right)^3-\left(x+1\right)^3+6\left(x+1\right)\left(x-1\right)\\ =x^3-3x^2+3x-1-x^3-3x^2-3x-1+6x^2-6\\ =-8\)
Vậy.....
c) Ta có:
\(\left(x-3\right)\left(x+3\right)\left(x^2+9\right)-\left(x^2-2\right)\left(x^2+2\right)\\ =\left(x^2-9\right)\left(x^2+9\right)-x^4+4\\ =x^4-81-x^4+4=-77\)
Vậy....
d) Ta có: \(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x-5\right)+\left(3x-5\right)^2\\ =\left(3x+1-3x+5\right)^2\\ =6^2=36\)
Vậy....
TC
11 tháng 10 2019
Bài 1:Phân tích các đa thức sau:
\(a,4x^2-6x\\ =2x(2x-3)\\ b,x^3y-2x^2y+5xy\\ = xy(x^2-2x+5)\\ c,2x^2(x+1) +4x(x+1)\\ =2x(x+1)(x+2)\\ d,\frac{2}{5}x.(y-1) -\frac{2}{5}x.(1-y)\\ =\frac{2}{5}x.(y-1)+\frac{2}{5}x.(y-1)\\ =2.\bigg[\frac{2}{5}x.(y-1)\bigg]\)
Bài 2 tính bằng cách hợp lý
\(a, 8,4.84, 5+840.0, 155\\ =8,4.(84,5+100.0,155)\\ =8,4.100\\ =840\\ b, 0,78.1300+50.6, 5-39\\ =(0,78.1300-39)+50. 6,5\\ =0,78.(1300-50)+50. 6,5\\ =0,78.1250+50. 6,5\\ =50.(0,78.25+6,5)\\ =1300\\ c,0, 12.90-110.0, 6+36-25.6\\ =6.(15.0,12-110.0,1+6-25)\\ =6.-28,2\\ =-169.2\)
Bài 3 Phân tích các đa thức sau\(a, (3x+1) ^2-(3x-1) ^2\\ =(3x+1-3x+1)(3x+1+3x-1)\\ =2.6x\\ b, (x+y) ^2-(x-y) ^2\\ =(x+y-x+y)(x+y+x-y)\\ =2y.2x\\ =2.(x-y)\\ c,(x+y)^3-(x-y) ^3\\ =(x+y-x+y)\bigg[(x+y)^2+(x+y)(x-y)+(x-y)^2\bigg]\\ =2y(x^2+2xy+y^2+x^2-xy+xy-y^2+x^2-2xy+y^2)\\ =2y(3x^2+y^2)\)
11 tháng 10 2019
Bài 1:
\(a,4x^2-6x=2x\left(2x-3\right)\\ b,x^3y-2x^2y+5xy=xy\left(x^2-2x+5\right)\\ c,2x^2\left(x+1\right)+4x\left(x+1\right)=2x\left(x+1\right)\left(x+2\right)\\ d,\frac{2}{5}x\left(y-1\right)-\frac{2}{5}y\left(1-y\right)\\ =\frac{2}{5}\left(y-1\right)\left(x+y\right)\)
Bài 2:
\(a,8,4\cdot84,5+840\cdot0,155\\ =840\left(0,845+0,155\right)\\ =840\cdot1=840\\ b,0,78\cdot1300+50\cdot6,5-39\\ =39\cdot2\cdot13-39+25\cdot2\cdot6,5\\ =39\left(26-1\right)+25\cdot13\\ =39\cdot25+25\cdot13\\ =25\left(39+13\right)\\ =25\cdot52\\ =1300\\ c,0,12\cdot90-110\cdot0,6+36-25\cdot6\\ =6\cdot2\cdot0,9-6\cdot11+6\cdot6-25\cdot6\\ =6\left(1,8-11+6-25\right)\\ =-28,2\cdot6\\ =-169,2\)
Bài 3:
\(a,\left(3x+1\right)^2-\left(3x-1\right)^2\\ =\left(3x+1-3x+1\right)\left(3x+1+3x-1\right)\\ =2\cdot6x\\ =12x\\ b,\left(x+y\right)^2-\left(x-y\right)^2\\ =\left(x+y-x+y\right)\left(x+y+x-y\right)\\ =2y\cdot2x\\ =4xy\\ c,\left(x+y\right)^3-\left(x-y\right)^3\\ =\left(x+y-x+y\right)\left[\left(x+y\right)^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\\ =2y\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)\\ =2y\left(3x^2+y^2\right)\)
13 tháng 6 2018
\(6\left(x+1\right)^2-2\left(x+1\right)^3+2\left(x-1\right)\left(x^2+x+1\right)=1\)
⇔ \(6\left(x^2+2x+1\right)-2\left(x^3+3x^2+3x+1\right)+2\left(x^3-1\right)=1\)
⇔ \(6x^2+12x+6-2x^3-6x^2-6x-2+2x^3-2=1\)
⇔ 6x + 1 = 0
⇔ x = \(\dfrac{-1}{6}\)
KL.........
14 tháng 6 2022
a: \(B=\left(\dfrac{x+1}{2\left(x-1\right)}+\dfrac{3}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+3}{2\left(x+1\right)}\right)\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=\dfrac{x^2+2x+1+6-x^2-2x+3}{2\left(x+1\right)\left(x-1\right)}\cdot\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=\dfrac{10}{1}\cdot\dfrac{2}{5}=10\cdot\dfrac{2}{5}=4\)
b: \(\dfrac{x^2-36}{2x+10}\cdot\dfrac{3}{6-x}\)
\(=\dfrac{\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)}\cdot\dfrac{-3}{x-6}\)
\(=\dfrac{-3\left(x+6\right)}{2\left(x+5\right)}\)
c: \(\dfrac{5x+10}{4x-8}\cdot\dfrac{4-2x}{x+2}\)
\(=\dfrac{5\left(x+2\right)}{4\left(x-2\right)}\cdot\dfrac{-2\left(x-2\right)}{x+2}=\dfrac{-10}{4}=\dfrac{-5}{2}\)
d: \(\dfrac{1-4x^2}{x^2+4x}:\dfrac{2-4x}{3x}\)
\(=\dfrac{1-4x^2}{x\left(x+4\right)}\cdot\dfrac{3x}{2\left(1-2x\right)}\)
\(=\dfrac{\left(1-2x\right)\left(1+2x\right)}{x+4}\cdot\dfrac{3}{2\left(1-2x\right)}=\dfrac{3\left(2x+1\right)}{x+4}\)
(-x+6)(-x-2)
= -(x-6) . [- (x+2)]
= (x-6)(x+2)
= x(x+2) - 6(x+2)
= x2 + 2x - 6x - 12
= x2 - 4x -12