\(\left|x-3\right|-x=7\)
Đạo hàm của hàm số \(y=\left(-x^2+3x+7\right)^7\) là:
A. \(y'=7\left(-2x+3\right)\left(-x^2+3x+7\right)^6\)
B. \(y'=7\left(-x^2+3x+7\right)^6\)
C. \(y'=\left(-2x+3\right)\left(-x^2+3x+7\right)^6\)
D. \(y'=7\left(-2x+3\right)\left(-x^2+3x+7\right)^6\)
\(y'=7\left(-x^2+3x+7\right)^6.\left(-x^2+3x+7\right)'\)
\(=7\left(-2x+3\right)\left(-x^2+3x+7\right)^6\)
Giải các PT sau
a)\(\left(3\left(x+1\right)-2\left(x+3\right)\right)^3+\left(2\left(x+3\right)-x+5\right)^3+\left(x-5-3\left(x+1\right)\right)^3=0\)
b)\(\left(x-2\right)^3+\left(x-4\right)^3+\left(x-7\right)^3+3\left(x-2\right)\left(x-4\right)\left(x-7\right)=0\)
1)\(4\left(x-5\right)-3\left(x+7\right)=-19\)
2)\(7\left(x-3\right)-5\left(3-x\right)=11x-5\)
3)\(4\left(2-x\right)+4\left(x-3\right)=14\)
4)\(-5\left(2-x\right)+4\left(x-3\right)=10x-15\)
5)\(7\left(x-9\right)-5\left(6-x\right)=-5+11x\)
6)\(-7\left(3x-5\right)+2\left(7x-14\right)=28\)
7)\(4\left(x-5\right)-3\left(x+7\right)=5.\left(-4\right)\)
a ) Ta có : 4(x - 5) - 3(x + 7) = -19
<=> 4x - 20 - 3x - 21 = -19
=> x - 41 = -19
=> x = -19 + 41
=> x = 22
b) Ta có " 7(x - 3) - 5(3 - x) = 11x - 5
<=> 7x - 21 - 15 + 5x = 11x - 5
<=> 12x - 36 = 11x - 5
=> 12x - 11x = -5 + 36
=> x = 31
\(\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}+\frac{1}{\left(x+7\right)\left(x+9\right)}+\frac{1}{\left(x+9\right)\left(x+11\right)}\)
1.Cho x+y+z=0. CMR:
a) \(5\left(x^3+y^3+z^3\right)\left(x^2+y^2+z^2\right)=6\left(x^5+y^5+z^5\right)\)
b) \(x^7+y^7+z^7=7xyz\left(x^2y^2+y^2z^2+z^2x^2\right)\)
c) \(10\left(x^7+y^7+z^7\right)=7\left(x^2+y^2+z^2\right)\left(x^5+y^5+z^5\right)\)
d) \(2\left(x^5+y^5+z^5\right)=5xyz\left(x^2+y^2+z^2\right)\)
2. Tìm n∈ N để biểu thức sau là số nguyên tố
a) \(A=n^3-4n^2-4n-1\)
b) \(B=n^3-6n^2+9n-2\)
c) \(C=n^{1975}+n^{1973}+1\)
Vì bài dài nên mình sẽ tách ra nhé.
1a. Ta có:
$x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)=-2(xy+yz+xz)$
$x^3+y^3+z^3=(x+y+z)^3-3(x+y)(y+z)(x+z)=-3(x+y)(y+z)(x+z)$
$=-3(-z)(-x)(-y)=3xyz$
$\Rightarrow \text{VT}=-30xyz(xy+yz+xz)(1)$
------------------------
$x^5+y^5=(x^2+y^2)(x^3+y^3)-x^2y^2(x+y)$
$=[(x+y)^2-2xy][(x+y)^3-3xy(x+y)]-x^2y^2(x+y)$
$=(z^2-2xy)(-z^3+3xyz)+x^2y^2z$
$=-z^5+3xyz^3+2xyz^3-6x^2y^2z+x^2y^2z$
$=-z^5+5xyz^3-5x^2y^2z$
$\Rightarrow 6(x^5+y^5+z^5)=6(5xyz^3-5x^2y^2z)$
$=30xyz(z^2-xy)=30xyz[z(-x-y)-xy]=-30xyz(xy+yz+xz)(2)$
Từ $(1);(2)$ ta có đpcm.
1b.
$x^4+y^4=(x^2+y^2)^2-2x^2y^2=[(x+y)^2-2xy]^2-2x^2y^2$
$=(z^2-2xy)^2-2x^2y^2=z^4+2x^2y^2-4xyz^2$
$x^3+y^3=(x+y)^3-3xy(x+y)=-z^3+3xyz$
Do đó:
$x^7+y^7=(x^4+y^4)(x^3+y^3)-x^3y^3(x+y)$
$=(z^4+2x^2y^2-4xyz^2)(-z^3+3xyz)+x^3y^3z$
$=7x^3y^3z-14x^2y^2z^3+7xyz^5-z^7$
$\Rightarrow \text{VT}=7x^3y^3z-14x^2y^2z^3+7xyz^5$
$=7xyz(x^2y^2-2xyz^2+z^4)$
$=7xyz(xy-z^2)$
$=7xyz[xy+z(x+y)]^2=7xyz(xy+yz+xz)^2$
$=7xyz[x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)]$
$=7xyz(x^2y^2+y^2z^2+z^2x^2)$ (đpcm)
1c. Sử dụng kq phần a,b:
\(10(x^7+y^7+z^7)=70xyz(xy+yz+xz)^2\)
\(=-35xyz(xy+yz+xz).-2(xy+yz+xz)=-35xyz(x+y+z)(x^2+y^2+z^2)\)
\(=\frac{7}{6}.-30xyz(xy+yz+xz)(x^2+y^2+z^2)=\frac{7}{6}.6(x^5+y^5+z^5).(x^2+y^2+z^2)\)
\(=7(x^5+y^5+z^5)(x^2+y^2+z^5)\)
(đpcm)
1d. Áp dụng kq phần a
$6(x^5+y^5+z^5)=-30xyz(xy+y+xz)=15xyz.-2(xy+yz+xz)=15xyz(x^2+y^2+z^2)$
$\Rightarrow 2(x^5+y^5+z^5)=5xyz(x^2+y^2+z^2)$ (đpcm)
Mình cần gấp bài này !!!
1/Tìm x, biết :
a/ \(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}+\frac{x}{\left(x+3\right)\left(x+34\right)}\)
b/ \(\frac{3}{\left(x-4\right)\left(x-7\right)}+\frac{6}{\left(x-7\right)\left(x-13\right)}+\frac{15}{\left(x-13\right)\left(x-28\right)}-\frac{1}{x-28}=\frac{-1}{20}\)
tìm x biết:
A) \(\frac{7}{\left(x+3\right)\left(x+10\right)}\)\(+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
B)\(\frac{3}{\left(x-4\right)\left(x-7\right)}+\frac{6}{\left(x-7\right)\left(x-13\right)}+\frac{15}{\left(x-13\right)\left(x-28\right)}-\frac{1}{x-28}=-\frac{5}{2}\)
A) \(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}\)
\(=\frac{\left(x+10\right)-\left(x+3\right)}{\left(x+3\right)\left(x+10\right)}+\frac{\left(x+21\right)-\left(x+10\right)}{\left(x+10\right)\left(x+21\right)}+\frac{\left(x+34\right)-\left(x+21\right)}{\left(x+21\right)\left(x+34\right)}\)
\(=\frac{1}{x+3}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+21}+\frac{1}{x+21}-\frac{1}{x+34}\)
\(=\frac{1}{x+3}-\frac{1}{x+34}\)
\(=\frac{\left(x+34\right)-\left(x+3\right)}{\left(x+3\right)\left(x+34\right)}\)\(=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
\(\Rightarrow\left(x+34\right)-\left(x+3\right)=x\)
\(\Rightarrow x=31\)
Vậy, x = 31
Bạn áp dụng: \(\frac{k}{x\cdot\left(x+k\right)}=\frac{1}{x}-\frac{1}{x+k}\) với \(x,k\inℝ;x\ne0;x\ne-k\)
Chứng minh: \(\frac{1}{x}-\frac{1}{x+k}=\frac{x+k}{x\left(x+k\right)}-\frac{x}{x\left(x+k\right)}=\frac{x+k-x}{x\left(x+k\right)}=\frac{k}{x\left(x+k\right)}\)
B) \(\frac{\left(x-4\right)-\left(x-7\right)}{\left(x-7\right)\left(x-4\right)}+\frac{\left(x-7\right)-\left(x-13\right)}{\left(x-13\right)\left(x-7\right)}+\frac{\left(x-13\right)-\left(x-28\right)}{\left(x-28\right)\left(x-13\right)}\)
\(=\frac{1}{x-7}-\frac{1}{x-4}+\frac{1}{x-13}-\frac{1}{x-7}+\frac{1}{x-28}-\frac{1}{x-13}\)
\(=\frac{1}{x-28}-\frac{1}{x-4}=-\frac{5}{2}+\frac{1}{x-28}\)
\(\Leftrightarrow\frac{1}{x-28}-\frac{1}{x-4}-\frac{1}{x-28}=-\frac{5}{2}\)
\(\Leftrightarrow\frac{1}{x-4}=\frac{5}{2}\)
=> 5x - 20 = 2
=> 5x = 22
\(\Rightarrow x=\frac{22}{5}=4,4\)
Vậy, x = 4,4
Tìm x:
\(\frac{3}{\left(x+1\right)\left(x+4\right)}+\frac{3}{\left(x+4\right)\left(x+7\right)}+\frac{3}{\left(x+7\right)\left(x+10\right)}=\frac{1}{2}\)
Điều kiện: \(x\ne-1;-4;-7;-10\)
Ta có:
\(\frac{3}{\left(x+1\right)\left(x+4\right)}=\frac{1}{x+1}-\frac{1}{x+4}\)
\(\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{x+4}-\frac{1}{x+7}\)
\(\frac{3}{\left(x+7\right)\left(x+10\right)}=\frac{1}{x+7}-\frac{1}{x+10}\)
Vậy:
\(\frac{1}{x+1}-\frac{1}{x+10}=\frac{1}{2}\)
Biến đổi tiếp để tìm x, sau đó đối chiếu với điều kiện khác -1; -4; -7; -11 để loại nghiệm
\(\frac{2x+9}{\left(x+1\right)\left(x+8\right)}-\frac{2x+15}{\left(x+8\right)\left(x+7\right)}+\frac{2x+10}{\left(x+7\right)\left(x+3\right)}=\frac{4}{3}\)
(2x+9)/(x+1)(x+8)-(2x+15)/(x+8)(x+7)+(2x+10)/(x+7)(x+3)=4/3
(x+1+x+8)/(x+1)(x+8)-(x+8+x+7)/(x+8)(x+7)+(x+7+x+3)/(x+7)(x+3)=4/3
1/(x+8)+1/(x+1)-1/(x+7)-1/(x+8)+1/(x+7)+1/(x+3)=4/3
1/(x+1)+1/(x+3)=4/3
(x+3+x+1)/(x+3)(x+1)=4/3
(2x+4)/(x+3)(x+1)=4/3
=>(2x+4).3=(x+3)(x+1).4
6(x+2)=4(x+3)(x+1)
3(x+2)=2(x+3)(x+1)
3x+6=2(x^2+4x+3)
3x+6=2x^2+8x+6
2x^2+8x+6-3x-6=0
2x^2+5x=0
x(2x+5)=0
=> x=0 hoac 2x+5=0
=> x=0 hoac x=-5/2
tính nhanh biểu thức sau:
A=\(\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}+\frac{1}{\left(x+7\right)\left(x+9\right)}+\frac{1}{\left(x+9\right)\left(x+11\right)}\)
Ta có: \(A=\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+.....+\frac{1}{\left(x+9\right)\left(x+11\right)}\)
\(\Rightarrow A=\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+5}+....+\frac{1}{x+9}-\frac{1}{x+11}\)
\(\Rightarrow A=\frac{1}{x+1}-\frac{1}{x+11}\)
\(\Rightarrow A=\frac{x+11-x+1}{\left(x+1\right)\left(x+11\right)}=\frac{12}{\left(x+1\right)\left(x+11\right)}\)