( x2 + 3 )2
6). – x2 y(xy2 – 1/2 xy + 3/4 x2 y2 )
7). (3xy – x2 + y). 2/3 x2 y
8). (4x3 – 5xy + 2x)( – 1/2 xy)
9). 2x2 (x2 + 3x + 1/2 )
10). – 3/2 x4 y2 (6x4 − 10/9 x2 y3 – y5 )
11). 2 3 x3 (x + x2 – 3/4 x5 )
12). 2xy2 (xy + 3x2 y – 2/3 xy3 )
13). 3x(2x3 – 1/3 x2 – 4x)
14). 3/5 x3 y5 (7x4 + 5x2 y − 10/21 x4 y3 –y4 )
6: \(-x^2y\left(xy^2-\dfrac{1}{2}xy+\dfrac{3}{4}x^2y^2\right)\)
\(=-x^3y^3+\dfrac{1}{2}x^3y^2-\dfrac{3}{4}x^4y^3\)
7: \(\dfrac{2}{3}x^2y\cdot\left(3xy-x^2+y\right)\)
\(=2x^3y^2-\dfrac{2}{3}x^4y+\dfrac{2}{3}x^2y^2\)
8: \(-\dfrac{1}{2}xy\left(4x^3-5xy+2x\right)\)
\(=-2x^4y+\dfrac{5}{2}x^2y^2-x^2y\)
9: \(2x^2\left(x^2+3x+\dfrac{1}{2}\right)=2x^4+6x^3+x^2\)
10: \(-\dfrac{3}{2}x^4y^2\left(6x^4-\dfrac{10}{9}x^2y^3-y^5\right)\)
\(=-9x^8y^2+\dfrac{5}{3}x^6y^5+\dfrac{3}{2}x^4y^7\)
11: \(\dfrac{2}{3}x^3\left(x+x^2-\dfrac{3}{4}x^5\right)=\dfrac{2}{3}x^3+\dfrac{2}{3}x^5-\dfrac{1}{2}x^8\)
12: \(2xy^2\left(xy+3x^2y-\dfrac{2}{3}xy^3\right)=2x^2y^3+6x^3y^3-\dfrac{4}{3}x^2y^5\)
13: \(3x\left(2x^3-\dfrac{1}{3}x^2-4x\right)=6x^4-x^3-12x^2\)
1/(x2+5)(x2+4)+1/(x2+4)(x2+3)+1/((x2+3)(x2+2)+1/(x2+2)(x2+1)=-1
Ta có: \(\dfrac{1}{\left(x^2+5\right)\left(x^2+4\right)}+\dfrac{1}{\left(x^2+4\right)\left(x^2+3\right)}+\dfrac{1}{\left(x^2+3\right)\left(x^2+2\right)}+\dfrac{1}{\left(x^2+2\right)\left(x^2+1\right)}=-1\)
\(\Leftrightarrow\dfrac{1}{x^2+4}-\dfrac{1}{x^2+5}+\dfrac{1}{x^2+3}-\dfrac{1}{x^2+4}+\dfrac{1}{x^2+2}-\dfrac{1}{x^2+3}-\dfrac{1}{x^2+2}+\dfrac{1}{x^2+1}=-1\)
\(\Leftrightarrow\dfrac{1}{x^2+1}-\dfrac{1}{x^2+5}=-1\)
\(\Leftrightarrow\dfrac{\left(x^2+5\right)-\left(x^2+1\right)}{\left(x^2+1\right)\left(x^2+5\right)}=\dfrac{-1\left(x^2+1\right)\left(x^2+5\right)}{\left(x^2+1\right)\left(x^2+5\right)}\)
Suy ra: \(x^2+5-x^2-1=-\left(x^4+6x^2+5\right)\)
\(\Leftrightarrow4+x^4+6x^2+5=0\)
\(\Leftrightarrow x^4+6x^2+9=0\)
\(\Leftrightarrow\left(x^2+3\right)^2=0\)(Vô lý)
Vậy: \(S=\varnothing\)
\(\left(x^2+5\right)\left(x^2+4\right)+\dfrac{1}{\left(x^2+4\right)\left(x^2+3\right)}+\dfrac{1}{\left(x^2+3\right)\left(x^2+2\right)}+\dfrac{1}{\left(x^2+2\right)\left(x^2+1\right)}=-1\)
\(\Leftrightarrow\)\(\dfrac{x^4+9x^2+20}{\left(x^2+4\right)\left(x^2+3\right)\left(x^2+2\right)\left(x^2+1\right)}+\dfrac{1\left(x^2+2\right)\left(x^2+1\right)}{\left(x^2+4\right)\left(x^2+3\right)\left(x^2+2\right)\left(x^2+1\right)}+\dfrac{1\left(x^2+4\right)\left(x^2+1\right)}{\left(x^2+3\right)\left(x^2+2\right)\left(x^2+1\right)\left(x^2+4\right)}+\dfrac{1\left(x^2+4\right)\left(x^2+3\right)}{\left(x^2+2\right)\left(x^2+1\right)}=-\dfrac{\left(x^2+4\right)\left(x^2+3\right)\left(x^2+2\right)\left(x^2+1\right)}{\left(x^2+4\right)\left(x^2+3\right)\left(x^2+2\right)\left(x^2+1\right)}\)
\(\left(x^2+5\right)\left(x^2+4\right)+\left(x^2+2\right)\left(x^2+1\right)+\left(x^2+4\right)\left(x^2+1\right)+\left(x^2+4\right)\left(x^2+3\right)=\left(x^2+4\right)\left(x^2+3\right)\left(x^2+2\right)\left(x^2+1\right)\)
\(\left(x^2+4\right)\left(x^2+5+x^2+1+x^2+3\right)+\left(x^2+2\right)\left(x^2+1\right)\left(1-\left(x^2+4\right)\left(x^2+3\right)\right)=0\)
Nếu phương trình sau:x^2-2x-1=0 có 2 nghiệm x1,x2(x1<x2) thì hãy tính giá trị các đại lượng sau mà ko giải PT(bài này làm theo định lí Vi-et)
1.((x1^2+2)/x1)+((x2^2+2)/x2)
2.(x2/(x2^2-3))+(x1/(x1^2-3))
3.(x1^2/(x1.x2^2-1))+(x2^2/(x1^2.x2-1))
4.(x1/(3.x1.x2^2-1)+(x2/3.x1^2.x2-1)
5.(1/x1)-(1/x2)
6.(x1/(x2-1))+(x2/(x1-1))
7.((3x1-7)/x2)-((3x2-7)/x1)
Mọi người giúp mình với
Nếu phương trình sau:x^2-2x-1=0 có 2 nghiệm x1,x2(x1<x2) thì hãy tính giá trị các đại lượng sau mà ko giải PT(bài này làm theo định lí Vi-et)
1.((x1^2+2)/x1)+((x2^2+2)/x2)
2.(x2/(x2^2-3))+(x1/(x1^2-3))
3.(x1^2/(x1.x2^2-1))+(x2^2/(x1^2.x2-1))
4.(x1/(3.x1.x2^2-1)+(x2/3.x1^2.x2-1)
5.(1/x1)-(1/x2)
6.(x1/(x2-1))+(x2/(x1-1))
7.((3x1-7)/x2)-((3x2-7)/x1)
Mọi người giúp mình với
Cho (x2)^2=x1.x3;(x3)^2=x2.x4.Chứng minh rằng: (x1+x2+x3)^2/(x2+x3+x4)^2=x1^2+x2^2+x3^3/x2^2+x3^3+x4^4
a. x2(x – 2x3) b. (x2 + 1)(5 – x) c. (x – 2)(x2 + 3x – 4) d. (x – 2)(x – x2 + 4) e. (x2 – 1)(x2 + 2x) f. (2x – 1)(3x + 2)(3 – x) g. (x + 3)(x2 + 3x – 5) h. (xy – 2).(x3 – 2x – i. (5x3 – x2 + 2x – 3).(4x2 – x + 2
a: \(=x^3-2x^5\)
e: \(=x^4+2x^3-x^2-2x\)
Bài 1: Rút gọn
C) (x2 - 3) (x2 +3) - 5x2 (x + 1)2 - (x2 - 3x) ( x2 - 2x) + 4x (x + 2)2
D) -6x2 (x + 5)2 - ( x - 3)2 + (x2 - 2) (2x2 + 1) - 4x2 ( 3x - 4)2
A) -2x(3x+2)(3x-2)+5(x+2)2 - (x-1)(2x+1)(2x+1)
= -2x(9x2-4)+5(x2+4x+4) - (x-1)(4x2-1)
= -18x3+8x+5x2+20x+20-(4x3-x-4x2+1)
= -18x3+5x2+28x+20-4x3+x+4x2+1
= -22x3+9x2+29x+21
B) (7x-8)(7x+8)-10(2x+3)2+5x(3x-2)2-4x(x-5)2
= 49x2 - 64 -10(4x2+ 12x + 3) + 5x(9x2 - 12x +4) - 4x(x2 - 10x +25)
= 49x2 - 64 -40x2 - 120x - 30 + 45x3 - 60x2 - 20x - 4x3 + 40x2 -100x
= 41x3 -11x2 -240x -94
C) \(\left(x^2-3\right)\left(x^2+3\right)-5x^2\left(x+1\right)^2-\left(x^2-3x\right)\left(x^2-2x\right)+4x\left(x+2\right)^2\)
\(\left(x^4-9\right)-5x^2\left(x^2+2x+1\right)-\left(x^4-2x^3-3x^3+6x^2\right)+4x\left(x^2+4x+4\right)\)
\(x^4-9-5x^4-10x^3-5x^2-x^4+5x^3-6x^2+4x^3+16x^2+16x\)
\(-5x^4-x^3+5x^2+20x-9\)
D) \(-6x^2\left(x+5\right)^2-\left(x-3\right)^2+\left(x^2-2\right)\left(2x^2+1\right)-4x^2\left(3x-4\right)^2\)
\(-6x^2\left(x^2+10x+25\right)-\left(x^2-6x+9\right)+2x^4-3x^2-2-4x^2\left(9x^2-24x+16\right)\)
\(-6x^4-60x^3+150x^2-x^2+6x-9+2x^4-3x^2-2-36x^4+96x^3-64x^2\)
\(-40x^4+36x^3+82x^2+6x-11\)
cong phan thuc
a)x2+2/x2+4+5/x+2
b)x+y/2+x+2/2x2+4
c)8/(x2+3)(x2-1)+2/x2+3+1/X+1
c: \(=\dfrac{8}{\left(x^2+3\right)\left(x-1\right)\left(x+1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x-1\right)\left(x+1\right)}+\dfrac{\left(x^2+3\right)\left(x-1\right)}{\left(x^2+3\right)\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{1}{x-1}\)
Bài 1: Tính:
a) x2(x-2x3); b) (x2+1)(5-x); c) (x-2)(x2+3x-4); d) (x-2)(x-x2+4); e) (x2-1)(x2+2x); f) (2x-1)(3x+2)(3-x)
Bài 2: Tính:
a) (x-2y)2; b) (2x2+3)3; c) (x-2)(x2+2x+4); d) (2x-1)3
Bài 3: Rút gọn biểu thức:
a) (6x+1)2+(6x-1)2-2(1+6x)(6x-1); b) 3(22+1)(24+1)(28+1)(216+1); c) x(2x2-3)-x2(5x+1)+x2; d) 3x(x-2)-5x(1-x)-8(x2-3)
Bài 4: Tính nhanh:
a) 1012; b) 97.103; c) 772+232+77.46; d) 1052-52; e) A= (x-y)(x2+xy+y2)+2y3 tại x= \(\dfrac{2}{3}\) và y= \(\dfrac{1}{3}\)
Bạn chú ý đăng lẻ câu hỏi! 1/
a/ \(=x^3-2x^5\)
b/\(=5x^2+5-x^3-x\)
c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)
d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)
e/ \(=x^4-x^2+2x^3-2x\)
f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)
tìm x:
a)3(2x-3)+2(2-x)=-3
b)2x(x2-2)+x2(1-2x)-x2=-12
c)3x(2x+3)-(2x+5)(3x-2)=8
d)4x(x - 1) - 3(x2-5)-x2=(x-3)-(x+4)
e)2(3x-1)(2x+5)-6(2x-1)(x+2)=-6
a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)