Ta thấy nếu một trong hai số \(x,y\) bằng 0 thì số kia cũng bằng 0. Do đó \(x=y=0\) là một nghiệm của pt đã cho.

Xét \(x,y\ne0\) . Gọi \(\operatorname{gcd}\left(x,y\right)=d\), khi đó \(\begin{cases}x=da\\ y=db\end{cases}\) với \(\operatorname{gcd}\left(a,b\right)=1\) và \(d,a,b\ne0\). Khi đó pt đã cho thành:

\(\left(da\right)^2\left(da+db\right)=\left(db\right)^2\left(da-db\right)^2\)

\(\lrArr a^2\left(a+b\right)=db^2\left(a-b\right)^2\) (1)

Vì \(\operatorname{gcd}\left(a,b\right)=1\) nên \(\operatorname{gcd}\left(b,a+b\right)=\operatorname{gcd}\left(a,a-b\right)=1\) (thuật toán Euclid).

Từ (1) suy ra \(a^2\vert db^2\left(a-b\right)^2\), nhưng vì \(\operatorname{gcd}\left(a,b\right)=\operatorname{gcd}\left(a,a-b\right)=1\) nên \(a^2\vert d\). Đặt \(d=ka^2\) thì (1) thành

\(a+b=kb^2\left(a-b\right)^2\) (2)

Từ (2) suy ra \(b^2\left(a-b\right)^2\vert a+b\), suy ra \(\begin{cases}b^2\vert a+b\\ \left(a-b\right)^2\vert a+b\end{cases}\)

Ta có \(b^2\vert a+b\) thì \(b\vert a+b\) thì \(b\vert a\), nhưng do \(\operatorname{gcd}\left(a,b\right)=1\) nên \(b=\pm1\)

Tương tự, suy ra \(a-b=\pm1\)

Ta lập bảng sau:

Nếu \(\left(a,b\right)=\left(2,1\right)\) thì \(k=3\), suy ra \(d=12\), dẫn đến \(\left(x,y\right)=\left(24,12\right)\), thử lại thỏa mãn.

Nếu \(\left(a,b\right)=\left(-2,-1\right)\) thì \(k=-3\), suy ra \(d=-12\), cũng dẫn đến \(\left(x,y\right)=\left(24,12\right)\).

Vậy có hai cặp số \(\left(a,b\right)\) thỏa mãn yêu cầu bài toán là \(\left(0,0\right)\) và \(\left(24,12\right)\).

\(\frac{2}{x-3}\) ≤ \(\frac23\)

\(\frac{1}{x-3}\) ≤ \(\frac13\)

\(\frac{1}{x-3}-\frac13\) ≤ 0

\(\frac{3-x+3}{3\left(x-3\right)}\) ≤ 0

\(\frac{\left(3+3\right)-x}{3\left(x-3\right)}\) ≤ 0

\(\frac{6-x}{3\left(x-3\right)}\) ≤ 0

6 - \(x\) = 0 ⇒ \(x=6\); \(x-3=0\) ⇒ \(x=3\)

Lập bảng xét dấu ta có:

Theo bảng trên ta có: \(x\) ≥ 6 hoặc \(x\) < 3

\(x^2\) - 25 - ( 5 - 7 + \(x-5\)).(7 - 4\(x\)) = 0

(\(x^2\) - 25) - (\(x-5\)).(7 - 4\(x\)) = 0

\(\left(x-5\right)\left(x+5\right)\) - (\(x\) - 5).(7 - 4\(x\)) = 0

(\(x\) - 5)(\(x\) + 5 - 7 + 4\(x\)) = 0

(\(x\) - 5){(\(x+4x\)) + (5 - 7)} = 0

(\(x\) - 5).{5\(x\) - 2} = 0

\(\left[\begin{array}{l}x-5=0\\ 5x-2=0\end{array}\right.\)

\(\left[\begin{array}{l}x=5\\ x=\frac25\end{array}\right.\)

Vậy \(x\) ∈ {\(\frac25\); 5}

\(a.\left(5x-4\right)\left(4x+^{}6\right)=0\)

\(\left[\begin{array}{l}5x-4=0\Rightarrow x=\frac45\\ 4x+6=0\Rightarrow x=-\frac32\end{array}\right.\)

vậy x = \(\frac45\) hoặc \(x=-\frac32\)

\(b.3x^2+6x=x+2\)

\(3x\cdot\left(x+2\right)=x+2\)

\(3x\cdot\left(x+2\right)-\left(x+2\right)=0\)

\(\left(3x-1\right)\left(x+2\right)=0\)

\(\left[\begin{array}{l}3x-1=0\Rightarrow x=\frac13\\ x+2=0\Rightarrow x=-2\end{array}\right.\)

vậy x \(=\frac13\) hoặc x=-2

\(c.x^2\left(2x+1\right)+4x+2=0\)

\(x^2\left(2x+1\right)+2\cdot\left(2x+1\right)=0\)

\(\left(2x+1\right)\left(x^2+2\right)=0\)

\(\left[\begin{array}{l}2x+1=0\Rightarrow x=-\frac12\\ x^2+2=0\Rightarrow x\notin O\end{array}\right.\)

vậy \(x=-\frac12\)

\(d.x^3-5x^2-4x+20=0\)

\(x^2\cdot\left(x-5\right)-4\cdot\left(x-5\right)=0\)

\(\left(x^2-4\right)\left(x-5\right)=0\)

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

\(\left[\begin{array}{l}x-2=0\Rightarrow x=2\\ x+2=0\Rightarrow x=-2\\ x-5=0\Rightarrow x=5\end{array}\right.\)

vậy x = 2 hoặc x = -2 hoặc x = 5

\(e.\left(2x+5\right)^2=16=4^2=\left(-4\right)^2\)

\(\left[\begin{array}{l}2x+5=4\Rightarrow x=-\frac12\\ 2x+5=-4\Rightarrow x=-\frac92\end{array}\right.\)

vậy \(x=-\frac12\) hoặc \(x=-\frac92\)

a: ĐKXĐ: x∉{4;-5}

ta có: \(\frac{2x+3}{x-4}=\frac{2x-1}{x+5}\)

=>(2x+3)(x+5)=(2x-1)(x-4)

=>\(2x^2+10x+3x+15=2x^2-8x-x+4\)

=>13x+15=-9x+4

=>22x=4-15=-11

=>\(x=-\frac{11}{22}=-\frac12\) (nhận)

b: ĐKXĐ: x∉{5;-1}

\(2-\frac{x+3}{x-5}+\frac{1-x}{x+1}=0\)

=>\(\frac{2\left(x-5\right)\left(x+1\right)}{\left(x-5\right)\left(x+1\right)}-\frac{\left(x+3\right)\left(x+1\right)}{\left(x-5\left)\left(x+1\right)\right.\right.}-\frac{\left(x-1\right)\left(x-5\right)}{\left(x-5\right)\left(x+1\right)}=0\)

=>2(x-5)(x+1)-(x+3)(x+1)-(x-1)(x-5)=0

=>\(2\left(x^2+x-5x-5\right)-\left(x^2+4x+3\right)-\left(x^2-6x+5\right)=0\)

=>\(2x^2-8x-10-x^2-4x-3-x^2+6x-5=0\)

=>-6x-18=0

=>-6x=18

=>x=-3(nhận)

c: ĐKXĐ: x∉{2;-2}

\(\frac{x-2}{x+2}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{x^2-4}\)

=>\(\frac{x-2}{x+2}-\frac{3}{x-2}=\frac{2x-22}{\left(x-2\right)\left(x+2\right)}\)

=>\(\frac{\left(x-2\right)^2-3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{2x-22}{\left(x-2\right)\left(x+2\right)}\)

=>\(\left(x-2\right)^2-3\left(x+2\right)=2x-22\)

=>\(x^2-4x+4-3x-6-2x+22=0\)

=>\(x^2-9x+20=0\)

=>(x-4)(x-5)=0

=>\(\left[\begin{array}{l}x-4=0\\ x-5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4\left(nhận\right)\\ x=5\left(nhận\right)\end{array}\right.\)

d: ĐKXĐ: x∉{2;-2}

Ta có: \(\frac{12}{x^2-4}-\frac{x+1}{x-2}+\frac{x+7}{x+2}=0\)

=>\(\frac{12}{\left(x-2\right)\left(x+2\right)}-\frac{x+1}{x-2}+\frac{x+7}{x+2}=0\)

=>\(\frac{12-\left(x+1\right)\left(x+2\right)+\left(x+7\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=0\)

=>12-(x+1)(x+2)+(x+7)(x-2)=0

=>\(12-\left(x^2+3x+2\right)+\left(x^2-2x+7x-14\right)=0\)

=>\(12-x^2-3x-2+x^2+5x-14=0\)

=>2x-4=0

=>2x=4

=>x=2(loại)

e: ĐKXĐ: x∉{2;4}

\(\frac{x-1}{x-2}+\frac{2}{\left(x-2\right)\left(x-4\right)}=\frac{x+3}{x-4}\)

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

=>(x-1)(x-4)+2=(x+3)(x-2)

=>\(x^2-5x+4+2=x^2-2x+3x-6\)

=>-5x+6=x-6

=>-6x=-12

=>x=2(loại)

bạn ơi chứng minh j vậy

Qua D, kẻ đường thẳng DM⊥ID tại D và cắt BC tại M

Ta có: \(\hat{ADI}+\hat{IDC}=\hat{ADC}=90^0\)

\(\hat{IDC}+\hat{CDM}=\hat{IDM}=90^0\)

Do đó: \(\hat{ADI}=\hat{CDM}\)

Xét ΔADI vuông tại A và ΔCDM vuông tại C có

AD=CD

\(\hat{ADI}=\hat{CDM}\)

Do đó: ΔADI=ΔCDM

=>DI=DM

Xét ΔDME vuông tại D có DC là đường cao

nên \(\frac{1}{DM^2}+\frac{1}{DE^2}=\frac{1}{DC^2}\)

=>\(\frac{1}{DI^2}+\frac{1}{DE^2}=\frac{1}{DC^2}\) không đổi

a: \(\sin\alpha=cos\alpha\)

=>\(\sin\alpha=\sin\left(90^0-\alpha\right)\)

=>\(\alpha=90^0-\alpha\)

=>\(2\cdot\alpha=90^0\)

=>\(\alpha=\frac{90^0}{2}=45^0\)

b: \(\tan\alpha=\cot\alpha\)

=>\(\tan\alpha=\frac{1}{tan\alpha}\)

=>\(\tan^2\alpha=1\)

=>\(\tan\alpha=1\)

=>\(\alpha=45^0\)

`\sqrt{x^2-x+1}=1` (ĐK: `x\inR)`

`<=>x^2-x+1=1^2`

`<=>x^2-x+1=1`

`<=>x^2-x=1-1`

`<=>x^2-x=0`

`<=>x(x-1)=0`

`TH1:x=0`

`TH2;x-1=0`

`<=>x=1`

Vậy: `S={0;1}`

1: ĐKXĐ: x∉{0;-1}

Ta có: \(\frac{x-1}{x}+\frac{1-2x}{x\left(x+1\right)}=\frac{1}{x+1}\)

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

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

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

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

=>x(x-3)=0

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

2: ĐKXĐ: x∉{0;4}

ta có: \(\frac{5}{x}+\frac{x-3}{x-4}=\frac{x^2-10}{x\left(x-4\right)}\)

=>\(\frac{5\left(x-4\right)+x\left(x-3\right)}{x\left(x-4\right)}=\frac{x^2-10}{x\left(x-4\right)}\)

=>\(5\left(x-4\right)+x\left(x-3\right)=x^2-10\)

=>\(5x-20+x^2-3x=x^2-10\)

=>2x-20=-10

=>2x=10

=>x=5(nhận)

3: ĐKXĐ: x∉{0;3}

Ta có: \(\frac{x+3}{x-3}=\frac{3}{x^2-3x}+\frac{1}{x}\)

=>\(\frac{x+3}{x-3}=\frac{3}{x\left(x-3\right)}+\frac{1}{x}\)

=>\(\frac{x\left(x+3\right)}{x\left(x-3\right)}=\frac{3}{x\left(x-3\right)}+\frac{x-3}{x\left(x-3\right)}\)

=>\(x\left(x+3\right)=3+x-3=x\)

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

=>\(x^2+2x=0\)

=>x(x+2)=0

=>\(\left[\begin{array}{l}x=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\left(loại\right)\\ x=-2\left(nhận\right)\end{array}\right.\)

4: ĐKXĐ: x∉{0;3}

Ta có: \(\frac{3}{x^2-3x}+\frac{1}{x}=\frac{x+4}{x-3}\)

=>\(\frac{3}{x\left(x-3\right)}+\frac{1}{x}=\frac{x+4}{x-3}\)

=>\(\frac{3+x-3}{x\left(x-3\right)}=\frac{x\left(x+4\right)}{x\left(x-3\right)}\)

=>\(x=x\left(x+4\right)\)

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

=>x(x+3)=0

=>\(\left[\begin{array}{l}x=0\left(loại\right)\\ x=-3\left(nhận\right)\end{array}\right.\)

5: ĐKXĐ: x∉{0;4}

ta có: \(\frac{x+4}{x-4}-\frac{1}{x}=\frac{4}{x^2-4x}\)

=>\(\frac{x+4}{x-4}-\frac{1}{x}=\frac{4}{x\left(x-4\right)}\)

=>\(\frac{x\left(x+4\right)-\left(x-4\right)}{x\left(x-4\right)}=\frac{4}{x\left(x-4\right)}\)

=>\(x\left(x+4\right)-x+4=4\)

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

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

=>x(x+3)=0

=>\(\left[\begin{array}{l}x=0\\ x+3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\left(loại\right)\\ x=-3\left(nhận\right)\end{array}\right.\)

6: ĐKXĐ: x∉{3;-1}

Ta có: \(\frac{x}{x-3}+\frac{x}{x+1}=\frac{2x^2-4}{\left(x-3\right)\left(x+1\right)}\)

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

=>\(x\left(x+1\right)+x\left(x-3\right)=2x^2-4\)

=>\(x^2+x+x^2-3x=2x^2-4\)

=>-2x=-4

=>x=2(nhận)

7: ĐKXĐ: x∉{0;2}

ta có: \(\frac{x+2}{x-2}-\frac{6}{x}=\frac{9}{x^2-2x}\)

=>\(\frac{x+2}{x-2}-\frac{6}{x}=\frac{9}{x\left(x-2\right)}\)

=>\(\frac{x\left(x+2\right)-6\left(x-2\right)}{x\left(x-2\right)}=\frac{9}{x\left(x-2\right)}\)

=>x(x+2)-6(x-2)=9

=>\(x^2+2x-6x+12-9=0\)

=>\(x^2-4x+3=0\)

=>(x-1)(x-3)=0

=>\(\left[\begin{array}{l}x-1=0\\ x-3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\left(nhận\right)\\ x=3\left(nhận\right)\end{array}\right.\)

8: ĐKXĐ: x∉{0;2}

ta có: \(\frac{2}{x^2-2x}+\frac{1}{x}=\frac{x+2}{x-2}\)

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

=>\(\frac{2+x-2}{x\left(x-2\right)}=\frac{x\left(x+2\right)}{x\left(x-2\right)}\)

=>x(x+2)=x

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

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

=>x(x+1)=0

=>\(\left[\begin{array}{l}x=0\left(loại\right)\\ x=-1\left(nhận\right)\end{array}\right.\)

9: ĐKXĐ: x∉{0;-5}

\(\frac{x-5}{x}+\frac{x-3}{x+5}=\frac{x-25}{x^2+5x}\)

=>\(\frac{x-5}{x}+\frac{x-3}{x+5}=\frac{x-25}{x\left(x+5\right)}\)

=>\(\frac{\left(x-5\right)\left(x+5\right)+x\left(x-3\right)}{x\left(x+5\right)}=\frac{x-25}{x\left(x+5\right)}\)

=>\(\left(x-5\right)\left(x+5\right)+x\left(x-3\right)=x-25\)

=>\(x^2-25+x^2-3x-x+25=0\)

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

=>2x(x-2)=0

=>x(x-2)=0

=>\(\left[\begin{array}{l}x=0\left(loại\right)\\ x=2\left(nhận\right)\end{array}\right.\)

10:

ĐKXĐ: x∉{0;6}

\(\frac{x+6}{x-6}-\frac{6}{x^2-6x}=\frac{1}{x}\)

=>\(\frac{x+6}{x-6}-\frac{6}{x\left(x-6\right)}=\frac{1}{x}\)

=>\(\frac{x\left(x+6\right)}{x\left(x-6\right)}-\frac{6}{x\left(x-6\right)}=\frac{x-6}{x\left(x-6\right)}\)

=>\(x^2+6x-6=x-6\)

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

=>x(x+5)=0

=>\(\left[\begin{array}{l}x=0\\ x+5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\left(loại\right)\\ x=-5\left(nhận\right)\end{array}\right.\)

11: ĐKXĐ: x∉{0;7}

Ta có: \(\frac{x+7}{x-7}-\frac{7}{x^2-7x}=\frac{1}{x}\)

=>\(\frac{x+7}{x-7}-\frac{7}{x\left(x-7\right)}=\frac{1}{x}\)

=>\(\frac{x\left(x+7\right)-7}{x\left(x-7\right)}=\frac{x-7}{x\left(x-7\right)}\)

=>x(x+7)-7=x-7

=>x(x+7)=x

=>x(x+7)-x=0

=>x(x+6)=0

=>\(\left[\begin{array}{l}x=0\\ x+6=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\left(loại\right)\\ x=-6\left(nhận\right)\end{array}\right.\)

12: ĐKXĐ: x∉{0;-4}

ta có: \(\frac{x+5}{x}-\frac{x-7}{x+4}=\frac{x^2+35}{x^2+4x}\)

=>\(\frac{x+5}{x}-\frac{x-7}{x+4}=\frac{x^2+35}{x\left(x+4\right)}\)

=>\(\frac{\left(x+5\right)\left(x+4\right)-x\left(x-7\right)}{x\left(x+4\right)}=\frac{x^2+35}{x\left(x+4\right)}\)

=>\(\left(x+5\right)\left(x+4\right)-x\left(x-7\right)=x^2+35\)

=>\(x^2+9x+20-x^2+7x=x^2+35\)

=>\(x^2+35=16x+20\)

=>\(x^2-16x+15=0\)

=>(x-1)(x-15)=0

=>\(\left[\begin{array}{l}x-1=0\\ x-15=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x...