\(x\left(x-5\right)+3\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

Vậy tập nghiệm pt là: \(S=\left\{5;-3\right\}\)

x(x-5)+3(x-5)=0

=>(x-5)(x+3)=0

=>\(\left[{}\begin{matrix}x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

\(2x^3+10x^2=0\)

=>\(2x^2\left(x+5\right)=0\)

=>\(x^2\left(x+5\right)=0\)(Vì 2>0)

=>\(\left[{}\begin{matrix}x^2=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

\(2x^3+10x^2=0\Leftrightarrow x^2\left(2x+10\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

`x^2-6x=0`

`<=>x(x-6)=0`

TH1: `x =0 `

TH2: `x - 6=0<=>x=6`

Vậy: ... 

\(x^2-6x=0\Leftrightarrow x\left(x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

c:

Bài 1:

Thay $3=x^2+xy+y^2$ vào PT(2) thì:

$2x^3=(x+y)(x^2+xy+y^2-2xy)$

$\Leftrightarrow 2x^3=(x+y)(x^2-xy+y^2)=x^3+y^3$

$\Leftrightarrow x^3=y^3\Leftrightarrow x=y$. 

Thay vào PT(1) thì: $3x^2=3\Leftrightarrow x^2=1\Leftrightarrow x=\pm 1$

$\Rightarrow y=\pm 1$ (tương ứng)

Vậy HPT có nghiệm $(x,y)=(\pm 1, \pm 1)$

Bài 2:

Thay $2=xy(x+y)$ vào PT(2) thì:

$x^3+y^3+3xy(x+y)=8y^3$

$\Leftrightarrow (x+y)^3=(2y)^3$

$\Leftrightarrow x+y=2y\Leftrightarrow x=y$. 

Thay vào PT(1): $x^2.2x=2$

$\Leftrightarrow 2x^3=2\Leftrightarrow x^3=1\Leftrightarrow x=1$.

$\Rightarrow y=x=1$

Vậy HPT có nghiệm $(x,y)=(1,1)$

Bài 1:

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

\(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{16}{x^2-1}\)

=>\(\dfrac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{16}{\left(x-1\right)\left(x+1\right)}\)

=>\(\left(x+1\right)^2-\left(x-1\right)^2=16\)

=>\(\left(x+1+x-1\right)\left(x+1-x+1\right)=16\)

=>4x=16

=>x=4(nhận)

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

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

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

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

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

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

=>\(x^2+x-2=x^2+1\)

=>x-2=1

=>x=3(nhận)

a: ĐKXĐ: \(x\notin\left\{0;-1;4\right\}\)

\(\dfrac{11}{x}=\dfrac{9}{x+1}+\dfrac{2}{x-4}\)

=>\(\dfrac{11}{x}=\dfrac{9\left(x-4\right)+2\left(x+1\right)}{\left(x+1\right)\left(x-4\right)}\)

=>\(\dfrac{11}{x}=\dfrac{11x-34}{x^2-3x-4}\)

=>\(11\left(x^2-3x-4\right)=x\left(11x-34\right)\)

=>\(11x^2-33x-44=11x^2-34x\)

=>-33x-44=-34x

=>-33x+34x=44

=>x=44(nhận)

b: ĐKXĐ: \(x\ne4\)

\(\dfrac{14}{3x-12}-\dfrac{2+x}{x-4}=\dfrac{3}{8-2x}-\dfrac{5}{6}\)

=>\(\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)

=>\(\dfrac{28}{6\left(x-4\right)}-\dfrac{6\left(x+2\right)}{6\left(x-4\right)}=\dfrac{-9}{6\left(x-4\right)}-\dfrac{5\left(x-4\right)}{6\left(x-4\right)}\)

=>28-6(x+2)=-9-5(x-4)

=>28-6x-12=-9-5x+20

=>-6x+16=-5x+11

=>-6x+5x=11-16

=>-x=-5

=>x=5(nhận)

c: ĐKXĐ: \(x\notin\left\{\dfrac{1}{3};-\dfrac{1}{3}\right\}\)

\(\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\)

=>\(\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}=\dfrac{\left(1-3x\right)^2-\left(1+3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}\)

=>\(\left(1-3x\right)^2-\left(1+3x\right)^2=12\)

=>\(9x^2-6x+1-9x^2-6x-1=12\)

=>-12x=12

=>x=-1(nhận)

d: ĐKXĐ: \(x\notin\left\{0;5;-5\right\}\)

\(\dfrac{x+5}{x^2-5x}-\dfrac{x+25}{2x^2-50}=\dfrac{x-5}{2x^2+10x}\)

=>\(\dfrac{x+5}{x\left(x-5\right)}-\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}=\dfrac{x-5}{2x\left(x+5\right)}\)

=>\(\dfrac{2\left(x+5\right)^2}{2x\left(x-5\right)\left(x+5\right)}-\dfrac{x\left(x+25\right)}{2x\left(x-5\right)\left(x+5\right)}=\dfrac{\left(x-5\right)^2}{2x\left(x+5\right)\left(x-5\right)}\)

=>\(2\left(x+5\right)^2-x\left(x+25\right)=\left(x-5\right)^2\)

=>\(2x^2+20x+50-x^2-25x=x^2-10x+25\)

=>-5x+50=-10x+25

=>5x=-25

=>x=-5(loại)

Bài 2:

a: ĐKXĐ: \(x\notin\left\{2;5\right\}\)

\(\dfrac{6x+1}{x^2-7x+10}+\dfrac{5}{x-2}=\dfrac{3}{x-5}\)

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

=>6x+1+5x-25=3x-6

=>11x-24=3x-6

=>8x=18

=>x=9/4(nhận)

b: ĐKXĐ: \(x\notin\left\{0;2;-2\right\}\)

\(\dfrac{2}{x^2-4}-\dfrac{x-1}{x\left(x-2\right)}+\dfrac{x-4}{x\left(x+2\right)}=0\)

=>\(\dfrac{2x}{x\left(x-2\right)\left(x+2\right)}-\dfrac{\left(x-1\right)\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-4\right)\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}=0\)

=>2x-(x-1)(x+2)+(x-4)(x-2)=0

=>\(2x-\left(x^2+x-2\right)+x^2-6x+8=0\)

=>\(x^2-4x+8-x^2-x+2=0\)

=>-5x+10=0

=>x=2(loại)

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

\(\dfrac{1}{3-x}-\dfrac{1}{x+1}=\dfrac{x}{x-3}-\dfrac{\left(x-1\right)^2}{x^2-2x-3}\)

=>\(\dfrac{-1}{x-3}-\dfrac{1}{x+1}-\dfrac{x}{x-3}+\dfrac{\left(x-1\right)^2}{\left(x-3\right)\left(x+1\right)}=0\)

=>\(\dfrac{\left(-1-x\right)\left(x+1\right)-x+3}{\left(x-3\right)\left(x+1\right)}+\dfrac{\left(x-1\right)^2}{\left(x-3\right)\left(x+1\right)}=0\)

=>-(x+1)^2-x+3+(x-1)²=0

=>\(-x^2-2x-1-x+3+x^2-2x+1=0\)

=>-5x+3=0

=>\(x=\dfrac{3}{5}\left(nhận\right)\)

d: ĐKXĐ: \(x\notin\left\{2;-3\right\}\)

\(\dfrac{1}{x-2}-\dfrac{6}{x+3}=\dfrac{5}{6-x^2-x}\)

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

=>x+3-6(x-2)=-5

=>x+3-6x+12+5=0

=>-5x+20=0

=>x=4(nhận)

e: ĐKXĐ: x<>-2

\(\dfrac{2}{x+2}-\dfrac{2x^2+16}{x^3+8}=\dfrac{5}{x^2-2x+4}\)

=>\(\dfrac{2}{x+2}-\dfrac{2x^2+16}{\left(x+2\right)\left(x^2-2x+4\right)}-\dfrac{5}{x^2-2x+4}=0\)

=>\(\dfrac{2\left(x^2-2x+4\right)-2x^2-16-5x-10}{\left(x+2\right)\left(x^2-2x+4\right)}=0\)

=>\(2x^2-4x+8-2x^2-5x-26=0\)

=>-9x-18=0

=>x=-2(loại)

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

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

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

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

=>2(x^2-1)=2(x+2)^2

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

=>\(x^2+4x+4-x^2+1=0\)

=>4x+5=0

=>\(x=-\dfrac{5}{4}\left(nhận\right)\)

Bài 3:

c:

=>\(\dfrac{x}{x-1}+\dfrac{x}{x-2}+\dfrac{x}{x-3}=\dfrac{3x-12}{x-6}\)

=>

ĐKXĐ: \(x\notin\left\{1;2;\dfrac{3\pm\sqrt{7}}{2}\right\}\)

 \(\dfrac{4}{x^2-3x+2}-\dfrac{3}{2x^2-6x+1}+1=0\)

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

=>\(8x^2-24x+4-3x^2+9x-6=-\left(x^2-3x+2\right)\left[2\cdot\left(x^2-3x\right)+1\right]\)

=>\(5x^2-15x-2=-\left[2\left(x^2-3x\right)^2+5\left(x^2-3x\right)+2\right]\)

=>\(5\left(x^2-3x\right)-2+2\left(x^2-3x\right)^2+5\left(x^2-3x\right)+2=0\)

=>\(2\left(x^2-3x\right)^2+10\left(x^2-3x\right)=0\)

=>\(\left(x^2-3x\right)^2+5\left(x^2-3x\right)=0\)

=>\(\left(x^2-3x\right)\left(x^2-3x+5\right)=0\)

mà \(x^2-3x+5=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}>=\dfrac{11}{4}>0\forall x\)

nên x(x-3)=0

=>\(\left[{}\begin{matrix}x=0\left(nhận\right)\\x=3\left(nhận\right)\end{matrix}\right.\)

a:

ĐKXĐ: \(x\notin\left\{8;9;10;11\right\}\)

 \(\dfrac{8}{x-8}+\dfrac{11}{x-11}=\dfrac{9}{x-9}+\dfrac{10}{x-10}\)

=>\(\left(\dfrac{8}{x-8}+1\right)+\left(\dfrac{11}{x-11}+1\right)=\left(\dfrac{9}{x-9}+1\right)+\left(\dfrac{10}{x-10}+1\right)\)

=>\(\dfrac{x}{x-8}+\dfrac{x}{x-11}-\dfrac{x}{x-9}-\dfrac{x}{x-10}=0\)

=>\(x\left(\dfrac{1}{x-8}+\dfrac{1}{x-11}-\dfrac{1}{x-9}-\dfrac{1}{x-10}\right)=0\)

=>x=0(nhận)

b:

ĐKXĐ: \(x\notin\left\{3;4;5;6\right\}\)

 \(\dfrac{x}{x-3}-\dfrac{x}{x-5}=\dfrac{x}{x-4}-\dfrac{x}{x-6}\)

=>\(\dfrac{x\left(x-5\right)-x\left(x-3\right)}{\left(x-3\right)\left(x-5\right)}=\dfrac{x\left(x-6\right)-x\left(x-4\right)}{\left(x-4\right)\left(x-6\right)}\)

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

=>\(x\left(\dfrac{1}{\left(x-3\right)\left(x-5\right)}-\dfrac{1}{\left(x-4\right)\left(x-6\right)}\right)=0\)

=>\(x\cdot\dfrac{\left(x-4\right)\left(x-6\right)-\left(x-3\right)\left(x-5\right)}{\left(x-3\right)\left(x-5\right)\left(x-4\right)\left(x-6\right)}=0\)

=>\(x\left(x^2-10x+24-x^2+8x-15\right)=0\)

=>x(-2x+9)=0

=>\(\left[{}\begin{matrix}x=0\left(nhận\right)\\x=\dfrac{9}{2}\left(nhận\right)\end{matrix}\right.\)

Sai đề! Làm sao số dư lớn hơn số chia được?

\(-5x^3+xy^2z^3\)có bậc 6 vì  6 > 3 

Giải:

+ Xét hạng  tử thứ nhất là: 5\(x^3\) vậy hạng tử này có bậc là 3

+ Xét hạng tử thứ hai là: \(xy^2z^3\)

     \(x\)  có bậc là 1

     y² có bậc là 2

     z³ có bậc là 3 

Vậy hạng tử \(xy^2z^3\) có bậc là: 1 + 2 + 3 = 6

+ Bậc của hạng tử \(xy^2z^3\) lớn hơn bậc của hạng tử - 5\(x^3\) nên đó là bậc của đa thức vì vậy bậc của đa thức là 6

2:

a: Xét ΔBHA vuông tại H và ΔBKC vuông tại K có

\(\widehat{HBA}\) chung

Do đó: ΔBHA~ΔBKC

=>\(\dfrac{BH}{BK}=\dfrac{BA}{BC}\)(2)

=>\(\dfrac{BH}{BA}=\dfrac{BK}{BC}\)

=>\(BH\cdot BC=BK\cdot BA\)

b: Xét ΔBHK và ΔBAC có

\(\dfrac{BH}{BA}=\dfrac{BK}{BC}\)

\(\widehat{HBK}\) chung

Do đó: ΔBHK~ΔBAC
=>\(\widehat{BHK}=\widehat{BAC}=70^0\)

c: Xét ΔBKH có BI là phân giác

nên \(\dfrac{IH}{IK}=\dfrac{BH}{BK}\left(1\right)\)

Xét ΔBAC có BD là phân giác

nên \(\dfrac{DA}{DC}=\dfrac{BA}{BC}\left(3\right)\)

Từ (1),(2),(3) suy ra \(\dfrac{IH}{IK}=\dfrac{DA}{DC}\)

=>\(IH\cdot DC=DA\cdot IK\)