Bài 4:

d: 

ĐKXĐ: \(x\notin\left\{1;-1;2;-2\right\}\)

\(\dfrac{x+4}{x-1}+\dfrac{x-4}{x+1}=\dfrac{x+8}{x-2}+\dfrac{x-8}{x+2}+6\)

=>\(\dfrac{\left(x+4\right)\left(x+1\right)+\left(x-4\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x+8\right)\left(x+2\right)+\left(x-8\right)\left(x-2\right)+6\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

=>\(\dfrac{2x^2+8}{\left(x-1\right)\left(x+1\right)}=\dfrac{2x^2+32+6x^2-24}{\left(x-2\right)\left(x+2\right)}\)

=>\(\dfrac{2x^2+8}{x^2-1}=\dfrac{8x^2+8}{x^2-4}\)

=>\(\left(2x^2+8\right)\left(x^2-4\right)=\left(8x^2+8\right)\left(x^2-1\right)\)

=>\(2x^4-32=8x^4-8\)

=>\(-6x^4=24\)

=>\(x^4=-4\left(loại\right)\)

Vậy: Phương trình vô nghiệm

c:

ĐKXĐ: \(x\notin\left\{-1;-3;-8;-10\right\}\)

 \(\dfrac{2}{x^2+4x+3}+\dfrac{5}{x^2+11x+24}+\dfrac{2}{x^2+18x+80}=\dfrac{9}{52}\)

=>\(\dfrac{2}{\left(x+1\right)\left(x+3\right)}+\dfrac{5}{\left(x+3\right)\left(x+8\right)}+\dfrac{2}{\left(x+8\right)\left(x+10\right)}=\dfrac{9}{52}\)

=>\(\dfrac{1}{x+1}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+8}+\dfrac{1}{x+8}-\dfrac{1}{x+10}=\dfrac{9}{52}\)

=>\(\dfrac{1}{x+1}-\dfrac{1}{x+10}=\dfrac{9}{52}\)

=>\(\dfrac{9}{\left(x+1\right)\left(x+10\right)}=\dfrac{9}{52}\)

=>(x+1)(x+10)=52

=>\(x^2+11x-42=0\)

=>(x+14)(x-3)=0

=>\(\left[{}\begin{matrix}x=-14\left(nhận\right)\\x=3\left(nhận\right)\end{matrix}\right.\)

b: 

ĐXKĐ: \(x\notin\left\{-2;-3;-4;-5;-6\right\}\)\(\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}=\dfrac{1}{8}\)

=>\(\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)

=>\(\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\)

=>\(\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)

=>\(\dfrac{x+6-x-2}{\left(x+2\right)\left(x+6\right)}=\dfrac{1}{8}\)

=>(x+2)(x+6)=32

=>\(x^2+8x-20=0\)

=>(x+10)(x-2)=0

=>\(\left[{}\begin{matrix}x=-10\left(nhận\right)\\x=2\left(nhận\right)\end{matrix}\right.\)

a: \(\dfrac{x^2}{x^2+2x+2}+\dfrac{x^2}{x^2-2x+2}-\dfrac{4x^2-20}{x^4+4}=\dfrac{322}{65}\)

=>\(\dfrac{x^2\left(x^2-2x+2\right)+x^2\left(x^2+2x+2\right)-4x^2+20}{\left(x^2+2x+2\right)\left(x^2-2x+2\right)}=\dfrac{322}{65}\)

=>\(\dfrac{x^4-2x^3+2x^2+x^4+2x^3+2x^2-4x^2+20}{x^4+4}=\dfrac{322}{65}\)

=>\(\dfrac{2x^4+20}{x^4+4}=\dfrac{322}{65}\)

=>\(322\left(x^4+4\right)=65\left(2x^4+20\right)\)

=>\(322x^4+1288-130x^4-1300=0\)

=>\(192x^4=12\)

=>\(x^4=\dfrac{1}{16}\)

=>\(\left[{}\begin{matrix}x=\dfrac{1}{2}\left(nhận\right)\\x=-\dfrac{1}{2}\left(nhận\right)\end{matrix}\right.\)

Bạn bấm vào biểu tượng  để nhập các công thức toán học cho rõ ràng nhé!

Vd:\(3^{10}\) 

\(\dfrac{3^{10}\cdot15^5}{25^3\cdot9^7}=\dfrac{3^{10}\cdot\left(3\cdot5\right)^5}{\left(5^2\right)^3\cdot\left(3^2\right)^7}=\dfrac{3^{10}\cdot3^5\cdot5^5}{5^6\cdot3^{14}}\)

\(=\dfrac{3^{15}}{5\cdot3^{14}}=\dfrac{3}{5}\)

watermelon

orange

mangosteen

jackfruit

fruit custard

watermelon

orange

mangosteen

jackfruit

fruit custard

Before my sister went on holiday, she had packed some boots and plenty of warm clothes.

chắc sắp sếp hoặc điền thêm từ

khong bt

Lấy điểm A bất kì nằm trên đường tròn đáy.

Khi đó góc tạo bởi đường sinh và mặt phẳng đáy chính là \(\widehat{SAO}=45^o\)

Do đó \(h=r=\dfrac{a}{\sqrt{2}}\)

\(\Rightarrow S_{xq}=\pi rl=\pi.\dfrac{a}{\sqrt{2}}.a=\dfrac{\pi a^2}{\sqrt{2}}\)

\(S_{tp}=S_{xq}+\pi r^2=\dfrac{\pi a^2}{\sqrt{2}}+\pi\left(\dfrac{a}{\sqrt{2}}\right)^2=\dfrac{\pi a^2\sqrt{2}+\pi a^2}{2}\) 

Thay n = 100 vào biểu thức , ta được;

\(N=156-224:100\)

\(=156-22,4\)

\(=133,6\)

Thay \(N=100\) vào \(N=156-224:n\), ta được:

\(100=156-224:n\)

\(224:n=156-100\)

\(224:n=56\)

\(n=224:56=4\)

111

222 + 333 - 111 - 444 + 111

= 555 - 111 - 444 + 111

= 444 - 444 + 111

= 0 + 111

= 111

a/

$\frac{1}{x}+\frac{1}{y}=\frac{1}{5}$

$\Rightarrow \frac{x+y}{xy}=\frac{1}{5}$

$\Rightarrow 5(x+y)=xy$

$\Rightarrow 5x+5y-xy=0$

$\Rightarrow x(5-y)+5y=0$

$\Rightarrow x(5-y)-5(5-y)=-25$

$\Rightarrow (x-5)(5-y)=-25$

$\Rightarrow (x-5)(y-5)=25$

Do $x,y$ nguyên nên $x-5,y-5$ nguyên. Mà tích $(x-5)(y-5)=25$ nên xảy ra các TH sau đây:

TH1: $x-5=1, y-5=25\Rightarrow x=6; y=30$

TH2: $x-5=-1, y-5=-25\Rightarrow x=4; y=-20$

TH3: $x-5=25, y-5=1\Rightarrow x=30; y=6$

TH4: $x-5=-25, y-5=-1\Rightarrow x=-20; y=4$

TH5: $x-5=5, y-5=5\Rightarrow x=10; y=10$

TH6: $x-5=-5, y-5=-5\Rightarrow x=0; y=0$

b/

$\frac{2}{x}+\frac{1}{y}=3$

$\Rightarrow \frac{x+2y}{xy}=3$

$\Rightarrow x+2y=3xy$

$\Rightarrow 3xy-x-2y=0$

$\Rightarrow x(3y-1)-2y=0$

$\Rightarrow 3x(3y-1)-6y=0$

$\Rightarrow 3x(3y-1)-2(3y-1)=2$

$\Rightarrow (3x-2)(3y-1)=2$

Do $x,y$ nguyên nên $3x-2, 3y-1$ cũng là số nguyên. Mà tích của chúng bằng 2 nên ta xét các TH sau:

TH1: $3x-2=1, 3y-1=2\Rightarrow x=y=1$

TH2: $3x-2=2, 3y-1=1\Rightarrow x=\frac{4}{3}$ (loại) 

TH3: $3x-2=-1, 3y-1=-2\Rightarrow x=\frac{1}{3}$ (loại)

TH4: $3x-2=-2, 3y-1=-1\Rightarrow x=y=0$ (loại do $x,y\neq 0$)

Vậy $x=y=1$

52+2=54 =))

54