x2-3
Những câu hỏi liên quan
Rút gọn
a) (x2−1)3−(x4+x2+1)(x2−1)(x2−1)3−(x4+x2+1)(x2−1)
b) (x4−3x2+9)(−x2+3)−(3+x2)3(x4−3x2+9)(−x2+3)−(3+x2)3
c) (x+y)3−(x−y)3−6x2y
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: \(S=\varnothing\)
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\(\left(x^2+5\right)\left(x^2+4\right)+\dfrac{1}{\left(x^2+4\right)\left(x^2+3\right)}+\dfrac{1}{\left(x^2+3\right)\left(x^2+2\right)}+\dfrac{1}{\left(x^2+2\right)\left(x^2+1\right)}=-1\)
\(\Leftrightarrow\)\(\dfrac{x^4+9x^2+20}{\left(x^2+4\right)\left(x^2+3\right)\left(x^2+2\right)\left(x^2+1\right)}+\dfrac{1\left(x^2+2\right)\left(x^2+1\right)}{\left(x^2+4\right)\left(x^2+3\right)\left(x^2+2\right)\left(x^2+1\right)}+\dfrac{1\left(x^2+4\right)\left(x^2+1\right)}{\left(x^2+3\right)\left(x^2+2\right)\left(x^2+1\right)\left(x^2+4\right)}+\dfrac{1\left(x^2+4\right)\left(x^2+3\right)}{\left(x^2+2\right)\left(x^2+1\right)}=-\dfrac{\left(x^2+4\right)\left(x^2+3\right)\left(x^2+2\right)\left(x^2+1\right)}{\left(x^2+4\right)\left(x^2+3\right)\left(x^2+2\right)\left(x^2+1\right)}\)
\(\left(x^2+5\right)\left(x^2+4\right)+\left(x^2+2\right)\left(x^2+1\right)+\left(x^2+4\right)\left(x^2+1\right)+\left(x^2+4\right)\left(x^2+3\right)=\left(x^2+4\right)\left(x^2+3\right)\left(x^2+2\right)\left(x^2+1\right)\)
\(\left(x^2+4\right)\left(x^2+5+x^2+1+x^2+3\right)+\left(x^2+2\right)\left(x^2+1\right)\left(1-\left(x^2+4\right)\left(x^2+3\right)\right)=0\)
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6). – x2 y(xy2 – 1/2 xy + 3/4 x2 y2 )7). (3xy – x2 + y). 2/3 x2 y8). (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 )
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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: \(-\dfrac{1}{2}xy\left(4x^3-5xy+2x\right)\)
\(=-2x^4y+\dfrac{5}{2}x^2y^2-x^2y\)
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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: \(3x\left(2x^3-\dfrac{1}{3}x^2-4x\right)=6x^4-x^3-12x^2\)
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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: \(=x^4+2x^3-x^2-2x\)
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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
Tìm x, biết:a)
x
2
(x - 5) + 5 - x 0; b) 3
x
4
- 9
x
3
-9
x
2
+ 27x;c)
x
2
(x + 8) +
x
2
-8x; d) (x + 3)(
x
2
-3x + 5)
x
2...
Đọc tiếp
Tìm x, biết:
a) x 2 (x - 5) + 5 - x = 0; b) 3 x 4 - 9 x 3 = -9 x 2 + 27x;
c) x 2 (x + 8) + x 2 = -8x; d) (x + 3)( x 2 -3x + 5) = x 2 + 3x.
17 tháng 5 2022 lúc 18:54
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
(x-2)(x+2)(x2+4)-(x2-3)(x2+3)
\(=\left(x^2-4\right)\left(x^2+4\right)-\left(x^4-9\right)\\ =x^4-16-x^4+9=-7\)
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=\(\left(x^2-4\right)\left(x^2+4\right)-\left(x^4-9\right)\)
=\(x^4-16-x^4+9=7\)
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Phân tích đa thức thành nhân tử:a)
x
2
-3x + 2; b) 4
x
2
- 36x + 56;c) 2
x
2
+ 5x + 2; d)2
x
2
-9x + 7;e) 4
x
2
- 4x - 9
y
2
+ 12y - 3; g)
x...
Đọc tiếp
Phân tích đa thức thành nhân tử:
a) x 2 -3x + 2; b) 4 x 2 - 36x + 56;
c) 2 x 2 + 5x + 2; d)2 x 2 -9x + 7;
e) 4 x 2 - 4x - 9 y 2 + 12y - 3; g) x 4 - 2 x 3 -4 x 2 + 4x-3;
h) x 3 -x +3 x 2 y + 3x y 2 + y 3 -y.
a) (x - 1)(x - 2). b) 4(x - 2)(x - 7).
c) (x + 2)(2x +1). d) (x - l)(2x - 7).
e) (2x + 3y - 3)(2x - 3y +1). g) (x - 3)( x 3 + x 2 - x +1).
h) (x + y)(x + y-l)(x + y + l).
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