1:rút gọn các phân thức
a,(3x+2)2-(x+2)2/x3-x2
b,x4-1/x3+2x2-x-2
c,x2+7x+12/x2+5x+6
d,x10-x8+x6-x4+x2-1/x4-1
rút gọn phân thức
1 . 8x3-125 / 3(x-3)-(x-3)(8-4x)
2 . x4-y4 / y3-x3
3 . x10-x8-x7-x6-x5-x4-x3-x2+1 / x30+x24+x18+x12+x6+1
2: \(=\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}{-\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{-\left(x+y\right)\left(x^2+y^2\right)}{x^2+xy+y^2}\)
Rút gọn phân thức: Q = x 10 - x 8 - x 7 + x 6 + x 5 + x 4 - x 3 - x 2 + 1 x 30 + x 24 + x 18 + x 12 + x 6 + 1
Cho các phản ứng sau:
(1) FeS + X1 → X2↑ + X3
(2) X2 + CuSO4 → X4 ↓ (đen) + X5
(3) X2 + X6 → X7↓ (vàng) + X8
(4) X3 + X9 → X10
(5) X10 + HI → X3 + X1 + X11
(6) X1 + X12 → X9 + X8 + MnCl2
Các chất X4, X7, X10 và X12 lần lượt là
A. CuO, CdS, FeCl2, MnO2
B. CuS, S, FeCl2, KMnO4
C. CuS, CdS, FeCl3, MnO2
D. CuS, S, FeCl3, MnO2
X1: HCl X2: H2S X3: FeCl2
X4: CuS X5: H2SO4 X6: O2
X7: S X8: H2O X9: Cl2
X10: FeCl3 X11:I2 X12: MnO2
Đáp án D
Cho các phản ứng sau:
(1) FeS + X1 → X2↑ + X3
(2) X2 + CuSO4 → X4 ↓ (đen) + X5
(3) X2 + X6 → X7↓ (vàng) + X8
(4) X3 + X9 → X10
(5) X10 + HI → X3 + X1 + X11
(6) X1 + X12 → X9 + X8 + MnCl2
Các chất X4, X7, X10 và X12 lần lượt là
A. CuO, CdS, FeCl2, MnO2
B. CuS, S, FeCl2, KMnO4
C. CuS, CdS, FeCl3, MnO2
D. CuS, S, FeCl3, MnO2
Chọn D
X1: HCl X2: H2S
X3: FeCl2 X4: CuS
X5: H2SO4 X6: O2
X7: S X8: H2O
X9: Cl2 X10: FeCl3
X11:I2 X12: MnO2
Cho các phản ứng sau:
(1) FeS + X1 → X2↑ + X3
(2) X2 + CuSO4 → X4 ↓ (đen) + X5
(3) X2 + X6 → X7↓ (vàng) + X8
(4) X3 + X9 → X10
(5) X10 + HI → X3 + X1 + X11
(6) X1 + X12 → X9 + X8 + MnCl2
Các chất X4, X7, X10 và X12 lần lượt là
A. CuO, CdS, FeCl2, MnO2
B. CuS, S, FeCl2, KMnO4
C. CuS, CdS, FeCl3, MnO2
D. CuS, S, FeCl3, MnO2
Đáp án D
X1: HCl
X2: H2S
X3: FeCl2
X4: CuS
X5: H2SO4
X6: O2
X7: S
X8: H2O
X9: Cl2
X10: FeCl3
X11:I2
X12: MnO2
Bài 1: Phân tích các đa thức sau thành nhân tử
a)x2-y2-2x+2y e)x4+4y4
b)x2(x-1)+16(1-x) f)x4-13x2+36
c)x2+4x-y2+4 g) (x2+x)2+4x2+4x-12
d)x3-3x2-3x+1 h)x6+2x5+x4-2x3-2x2+1
a.
$x^2-y^2-2x+2y=(x^2-y^2)-(2x-2y)=(x-y)(x+y)-2(x-y)=(x-y)(x+y-2)$
b.
$x^2(x-1)+16(1-x)=x^2(x-1)-16(x-1)=(x-1)(x^2-16)=(x-1)(x-4)(x+4)$
c.
$x^2+4x-y^2+4=(x^2+4x+4)-y^2=(x+2)^2-y^2=(x+2-y)(x+2+y)$
d.
$x^3-3x^2-3x+1=(x^3+1)-(3x^2+3x)=(x+1)(x^2-x+1)-3x(x+1)$
$=(x+1)(x^2-4x+1)$
e.
$x^4+4y^4=(x^2)^2+(2y^2)^2+2.x^2.2y^2-4x^2y^2$
$=(x^2+2y^2)^2-(2xy)^2=(x^2+2y^2-2xy)(x^2+2y^2+2xy)$
f.
$x^4-13x^2+36=(x^4-4x^2)-(9x^2-36)$
$=x^2(x^2-4)-9(x^2-4)=(x^2-9)(x^2-4)=(x-3)(x+3)(x-2)(x+2)$
g.
$(x^2+x)^2+4x^2+4x-12=(x^2+x)^2+4(x^2+x)-12$
$=(x^2+x)^2-2(x^2+x)+6(x^2+x)-12$
$=(x^2+x)(x^2+x-2)+6(x^2+x-2)=(x^2+x-2)(x^2+x+6)$
$=[x(x-1)+2(x-1)](x^2+x+6)=(x-1)(x+2)(x^2+x+6)$
h.
$x^6+2x^5+x^4-2x^3-2x^2+1$
$=(x^6+2x^5+x^4)-(2x^3+2x^2)+1$
$=(x^3+x^2)^2-2(x^3+x^2)+1=(x^3+x^2-1)^2$
Rút gọn các phân thức sau
1) 9 - ( x + 5)2 / x2 + 4x + 4
2) 32x - 8x2 + 2x3 / x3 + 64
3) 5x3 + 5x / x4 -1
4) 3x2 - 12x + 12 / x4 - 8x
5) 2a2 - 2ab / ac + ad - bc -bd
6) x2 - xy / y2 - x2
7) 2 - 2a / a3 - 1
8) x7 - x4 / x6 - 1
9) ( x + 2 )2 - ( x - 2)2 / 16x
10) 24,5x2 - 0,5y2 / 3,5x2 - 0,5xy
11) a3 - 3a2 + 2a - 6 / a2 +2
12) ( a - b) ( c - d) / (b2- a2) ( d2 - c2)
Giúp mình với ạ, mình cảm ơn !
1: \(=\dfrac{-\left[\left(x+5\right)^2-9\right]}{\left(x+2\right)^2}=\dfrac{-\left(x+5-3\right)\left(x+5+3\right)}{\left(x+2\right)^2}\)
\(=\dfrac{-\left(x+2\right)\left(x+8\right)}{\left(x+2\right)^2}=\dfrac{-\left(x+8\right)}{x+2}\)
2: \(=\dfrac{2x\left(x^2-4x+16\right)}{\left(x+4\right)\left(x^2-4x+16\right)}=\dfrac{2x}{x+4}\)
3: \(=\dfrac{5x\left(x^2+1\right)}{\left(x^2-1\right)\left(x^2+1\right)}=\dfrac{5x}{x^2-1}\)
4: \(=\dfrac{3\left(x^2-4x+4\right)}{x\left(x^3-8\right)}=\dfrac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+4\right)}\)
\(=\dfrac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}\)
5: \(=\dfrac{2a\left(a-b\right)}{a\left(c+d\right)-b\left(c+d\right)}=\dfrac{2a\left(a-b\right)}{\left(c+d\right)\left(a-b\right)}=\dfrac{2a}{c+d}\)
6: \(=\dfrac{x\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}\cdot\left(-1\right)=\dfrac{-x}{x+y}\)
7: \(=\dfrac{2\left(1-a\right)}{-\left(1-a^3\right)}=\dfrac{-2\left(1-a\right)}{\left(1-a\right)\left(1+a+a^2\right)}=-\dfrac{2}{1+a+a^2}\)
8: \(=\dfrac{x^4\left(x^3-1\right)}{\left(x^3-1\right)\left(x^3+1\right)}=\dfrac{x^4}{x^3+1}\)
9: \(=\dfrac{\left(x+2-x+2\right)\left(x+2+x-2\right)}{16x}=\dfrac{4\cdot2x}{16x}=\dfrac{1}{2}\)
10: \(=\dfrac{0.5\left(49x^2-y^2\right)}{0.5x\left(7x-y\right)}=\dfrac{1}{x}\cdot\dfrac{\left(7x-y\right)\left(7x+y\right)}{7x-y}\)
\(=\dfrac{7x+y}{x}\)
Thực hiện phép chia:
a) ( x 3 - 3x - 2) : (x - 2);
b) ( x 3 + 6 x 2 + 8x - 3): ( x 2 + 3x -1);
c) (2 x 4 – 7 x 3 + 9 x 2 - 7x + 2): (2 x 2 - 5x + 2).
a) x 2 + 2x + 1. b) x + 3. c) x 2 – x + 1.
Bài 1: Giải phương trình:
a) ( x+1)2 (x+2) + ( x – 1)2 ( x- 2) = 12
b) x4 + 3x3 + 4x2 + 3x + 1 = 0
c) x5 – x4 + 3x3 + 3x2 –x + 1 = 0
Bài 2: Chứng minh rằng các phương trình sau vô nghiệm
a) x4 – x3 + 2x2 – x + 1 = 0
b) x4 + x3 + x2 + x + 1 = 0
c) x4 – 2x3 +4x2 – 3x +2 = 0
d) x6+ x5+ x4 + x3 + x2 + x + 1 = 0
1.
a/ \(\Leftrightarrow\left(x+1\right)\left(x^2+3x+2\right)+\left(x-1\right)\left(x^2-3x+2\right)-12=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+2\right)+3x\left(x+1\right)-3x\left(x-1\right)+\left(x-1\right)\left(x^2+2\right)-12=0\)
\(\Leftrightarrow2x\left(x^2+2\right)+6x^2-12=0\)
\(\Leftrightarrow x^3+3x^2+2x-6=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+4x+6\right)=0\Rightarrow x=1\)
b/ Nhận thấy \(x=0\) ko phải nghiệm, chia 2 vế cho \(x^2\)
\(x^2+\frac{1}{x^2}+3\left(x+\frac{1}{x}\right)+4=0\)
Đặt \(x+\frac{1}{x}=t\Rightarrow x^2+\frac{1}{x^2}=t^2-2\)
\(t^2-2+3t+4=0\Rightarrow t^2+3t+2=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{x}=-1\\x+\frac{1}{x}=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2+x+1=0\left(vn\right)\\x^2+2x+1=0\end{matrix}\right.\) \(\Rightarrow x=-1\)
1c/
\(\Leftrightarrow x^5+x^4-2x^4-2x^3+5x^3+5x^2-2x^2-2x+x+1=0\)
\(\Leftrightarrow x^4\left(x+1\right)-2x^3\left(x+1\right)+5x^2\left(x+1\right)-2x\left(x+1\right)+x+1=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^4-2x^3+5x^2-2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x^4-2x^3+5x^2-2x+1=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x^4-2x^3+x^2+x^2-2x+1+3x^2=0\)
\(\Leftrightarrow\left(x^2-x\right)^2+\left(x-1\right)^2+3x^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-x=0\\x-1=0\\x=0\end{matrix}\right.\) \(\Rightarrow\) không tồn tại x thỏa mãn
Vậy pt có nghiệm duy nhất \(x=-1\)
2.
a. \(x^4-x^3+x^2+x^2-x+1=0\)
\(\Leftrightarrow x^2\left(x^2-x+1\right)+x^2-x+1=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=0\left(vn\right)\\x^2-x+1=0\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}=0\left(vn\right)\end{matrix}\right.\)
Vậy pt vô nghiệm
b.
\(x^4+x^3+x^2+x+1=0\)
\(\Leftrightarrow x\left(x^3+1\right)+x^3+1+x^2=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^3+1\right)+x^2=0\)
\(\Leftrightarrow\left(x+1\right)^2\left(x^2-x+1\right)+x^2=0\)
Mà \(\left\{{}\begin{matrix}\left(x+1\right)^2\left(x^2-x+1\right)\ge0\\x^2\ge0\end{matrix}\right.\)
Nên dấu "=" xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}x+1=0\\x=0\end{matrix}\right.\) ko tồn tại x thỏa mãn
Giải các phương trình sau:
a, (9x2 - 4)(x + 1) = (3x +2)(x2 - 1)
b, (x - 1)2 - 1 + x2 = (1 - x)(x + 3)
c, (x2 - 1)(x + 2)(x - 3) = (x - 1)(x2 - 4)(x + 5)
d, x4 + x3 + x + 1 = 0
e, x3 - 7x + 6 = 0
f, x4 - 4x3 + 12x - 9 = 0
g, x5- 5x3 + 4x = 0
h, x4 - 4x3 + 3x2 + 4x - 4 = 0
a, \(\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)-\left(3x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(\left(9x^2-4\right)-\left(\left(3x+2\right)\left(x-1\right)\right)\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-\left(3x^2-x-2\right)\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-3x^2+x+2\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x+1\right)=0;3x^2+x-2=0\)
=> x=-1
với \(3x^2+x-2=0\)
ta sử dụng công thức bậc 2 suy ra : \(x=\dfrac{2}{3};x=-1\)
Vậy ghiệm của pt trên \(S\in\left\{-1;\dfrac{2}{3}\right\}\)
b: \(\Leftrightarrow x^2-2x+1-1+x^2=x+3-x^2-3x\)
\(\Leftrightarrow2x^2-2x=-x^2-2x+3\)
\(\Leftrightarrow3x^2=3\)
hay \(x\in\left\{1;-1\right\}\)
c: \(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x-3\right)-\left(x-1\right)\left(x-2\right)\left(x+2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+1\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-2x-3-x^2-3x+10\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(-5x+7\right)=0\)
hay \(x\in\left\{1;-2;\dfrac{7}{5}\right\}\)