\(5X\left(X-2020\right)+X=2020\)

\(\Leftrightarrow5X^2-10100X+X=2020\)

\(\Leftrightarrow5X^2-10099X=2020\)

\(\Leftrightarrow5X^2-10099X-2020=0\)

\(\Leftrightarrow5X^2-10100X+x-2020=0\)

\(\Leftrightarrow5X\left(X-2020\right)+X-2020=0\)

\(\Leftrightarrow\left(X-2020\right)\left(5X+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2020\\x=-\frac{1}{5}\end{cases}}\)

\(4\left(x-5\right)^2-\left(2x+1\right)^2=0\)

\(\Leftrightarrow\left[2\left(x-5\right)\right]^2-\left(2x+1\right)^2=0\)

\(\Leftrightarrow\left[2\left(x-5\right)-2x-1\right]\left[2\left(x-5\right)+2x+1\right]=0\)

\(\Leftrightarrow\left(2x-10-2x-1\right)\left(2x-10+2x+1\right)=0\)

\(\Leftrightarrow-11\left(4x-9\right)=0\)

\(\Leftrightarrow x=\frac{9}{4}\)

\(a,5x\left(x-2020\right)+x=2020\)

\(< =>5x\left(x-2020\right)+x-2020=0\)

\(< =>\left(5x+1\right)\left(x-2020\right)=0\)

\(< =>\orbr{\begin{cases}5x+1=0\\x-2020=0\end{cases}}\)

\(< =>\orbr{\begin{cases}5x=-1\\x=2020\end{cases}< =>\orbr{\begin{cases}x=-\frac{1}{5}\\x=2020\end{cases}}}\)

\(b,4\left(x-5\right)^2-\left(2x+1\right)^2=0\)

\(< =>4\left(x^2-20x+25\right)-\left(4x^2+4x+1\right)=0\)

\(< =>4x^2-80x+100-4x^2-4x-1=0\)

\(< =>-84x+99=0< =>84x=99< =>x=\frac{99}{84}\)

a. g(x) - h(x) = -5x⁴ - x⁵ + 2x² - 2x³ - 1/4 + x⁵ + 2x³ + 3x⁴ - 7/4x + 2x⁴

g(x) - h(x) = (-5x⁴ + 3x⁴ + 2x⁴) + (- x⁵ + x⁵) + 2x² + (- 2x³ + 2x³) - 1/4 - 7/4x

g(x) - h(x) = 2x² - 1/4 - 7/4x

b.Ta có g(x) - h(x) = 0 khi 2x² - 1/4 - 7/4x = 0

⇒ 2x² - 7/4x = 1/4

⇒ x(2x - 7/4) = 1/4

⇒ x = 1/4

hoặc 2x - 7/4 = 0 ⇒ 2x = 7/4 ⇒ x = 7/8

Vậy nghiệm của đa thức g(x) - h(x) = 1/4 hoặc 7/8

a, 2(4x - 7 ) = 3(x + 1) + 18

⇌ 8x -14 = 3x + 3 + 18

⇌ 5x = 35 ⇌ x = 7

→ S = \(\left\{7\right\}\)

b, ( 2x - 1 )² - 4x ( x - 3 ) = -11

⇌ 4x² - 2x + 1 - 4x² + 12 = -11

⇌ 10x = -12

⇌ x = \(-\frac{12}{10}\)

→ S = \(\left\{-\frac{12}{10}\right\}\)

c, ( 2x - 5 )² - ( x + 2 )² = 0

⇌ ( 2x - 5 -x + 2 )² = 0

⇌ ( x - 3 )² = 0

⇌ x - 3 = 0 ⇌ x = 3

→ S = \(\left\{3\right\}\)

d, ( x - 6 ) ( x + 1 ) = 2(x + 1)

⇌ ( x - 6 - 2 ) ( x+ 1) = 0

⇌ x² - 7x - 8 =0

⇌ ( x - 8 ) ( x + 1 ) = 0

⇒\(\left\{{}\begin{matrix}x-8=0\\x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\x=-1\end{matrix}\right.\)

→ S = \(\left\{8;-1\right\}\)

e, \(\frac{x-3}{2}=2-\frac{1-2x}{5}\)

⇌ 5( x - 3) = 20 - 2(1 - 2x)

⇌ 5x - 4x = 15 + 20 + 2

⇌ x = 37

→ S = \(\left\{37\right\}\)

g, \(\frac{3x+2}{2}+\frac{5-2x}{3}=\frac{11}{6}\)

⇌ 3(3x + 2) + 2(5 - 2x) = 11

⇌ 6x + 6 + 10 - 4x = 11

⇌ 2x = -5

⇌ x = \(-\frac{5}{2}\)

→ S = \(\left\{-\frac{5}{2}\right\}\)

h, \(\frac{x-2}{x+2}-\frac{3}{x-2}=\frac{9x-66}{x^2-4}\)

⇌ (x - 2)² - 3(x - 2) = 9x - 66

⇌ x² - 4x + 4 - 3x - 6 = 9x - 66

⇌ x² -16 + 64 = 0

⇌ (x - 8)² = 0

⇌ x - 8 = 0

⇌ x = 8

→ S = \(\left\{8\right\}\)

a)2x-5/x+5=3=>2x-5=3(x+5)=3x+15

=>2x=3x+20=>x=-20

b)(x^2-6)/x=x+3/2

=>(x^2-6)/x - x=3/2

=>-6/x[quy đồng]=3/2

=>x=-4

c)Để (x^2+2x)−(3x+6)/x−3=0

thì  (x^2+2x)−(3x+6)=0

=x(x+2)-3(x+2)=(x-3)(x+2)=0

=>x=3 hoặc x=-2

Mà ở mẫu có x-3 nếu x=3 thì mẫu =0=>loại

Vậy x=2

d)5/3x+2=2x−1

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

Tìm ước của 5 rùi thay vào 3x+2 và 2x-1 rùi tìm x,cái đó dễ nên bn tự lm nhé

e)

(2x−1/x−1)+1=1/x−1

=>1/x-1-2x-1/x-1=1

=>-2x/x-1=1

=>-2x=x-1

=>x=1/3

g)(x+3/x+1)+(x−2/x)=2

=>quy đồng rùi tính và tìm x nhé bn,mk mỏi tay rùi

nhớ tick cho mk nha,mk siêng lắm ms ghi cho bn nhiều thế này nè,nhớ tick nha,thanks

a)  \(\frac{2x-5}{x+5}=3\)

  \(\Leftrightarrow2x-5=3\left(x+5\right)\)

  \(\Leftrightarrow2x-5=3x+15\)

  \(\Leftrightarrow2x-3x=15+5\)

  \(\Leftrightarrow-x=20\\ \)

   \(\Leftrightarrow x=-20\)

b) \(\frac{x^2-6}{x}=x+\frac{3}{2}\)

  \(\Leftrightarrow\frac{x^2-6}{x}=\frac{2x+3}{2}\)

  \(\Leftrightarrow2\left(x^2-6\right)=x\left(2x+3\right)\)

  \(\Leftrightarrow2x^2-12=2x^2+3x\)

  \(\Leftrightarrow3x=-12\)

  \(\Leftrightarrow x=-4\) 

c) \(\frac{\left(x^2+2x\right)-\left(3x+6\right)}{x-3}=0\)

  \(\Leftrightarrow\frac{x\left(x+2\right)-3\left(x+2\right)}{x-3}=0\)

  \(\Leftrightarrow\frac{\left(x+2\right)\left(x-3\right)}{x-3}=0\)

  \(\Leftrightarrow x+2=0\)

  \(\Leftrightarrow x=-2\)

d)  \(\frac{5}{3x+2}=2x-1\)

 \(\Leftrightarrow5=\left(2x-1\right)\left(3x+2\right)\)

 \(\Leftrightarrow5=6x^2+x-2\)

 \(\Leftrightarrow6x^2+x-7=0\)

 \(\Leftrightarrow\left[\begin{array}{nghiempt}1\\\frac{-7}{6}\end{array}\right.\)

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

   \(\Leftrightarrow2x-1+x-1=1\)

   \(\Leftrightarrow3x=3\)

   \(\Leftrightarrow x=1\)

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

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

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

  \(\Leftrightarrow x^2+3x+x^2-x-2=2x^2+2x\)

  \(\Leftrightarrow2x-2x-2=0\)

  \(\Leftrightarrow-2=0\)    \(\Rightarrow\)Phương trình vô nghiệm 

a/ ĐKXĐ: ...

\(\Leftrightarrow3\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)-7\)

Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow a^2=x+\frac{1}{4x}+1\)

\(\Rightarrow x+\frac{1}{4x}=a^2-1\)

Pt trở thành:

\(3a=2\left(a^2-1\right)-7\)

\(\Leftrightarrow2a^2-3a-9=9\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=3\)

\(\Leftrightarrow2x-6\sqrt{x}+1=0\)

\(\Rightarrow\sqrt{x}=\frac{3+\sqrt{7}}{2}\Rightarrow x=\frac{8+3\sqrt{7}}{2}\)

b/ ĐKXĐ:

\(\Leftrightarrow5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)

Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow x+\frac{1}{4x}=a^2-1\)

\(\Rightarrow5a=2\left(a^2-1\right)+4\Leftrightarrow2a^2-5a+2=0\)

\(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+\frac{1}{2\sqrt{x}}=2\\\sqrt{x}+\frac{1}{2\sqrt{x}}=\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x-4\sqrt{x}+1=0\\2x-\sqrt{x}+1=0\left(vn\right)\end{matrix}\right.\)

c/ ĐKXĐ: ...

\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)

\(\Leftrightarrow\frac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)

\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\frac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)

\(\Leftrightarrow2x^2-8x+5=0\)

d/ ĐKXĐ: ...

\(\Leftrightarrow x+1-\frac{15}{6}\sqrt{x}+\sqrt{x^2-4x+1}-\frac{1}{2}\sqrt{x}=0\)

\(\Leftrightarrow\frac{x^2-\frac{17}{4}x+1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{x^2-\frac{17}{4}x+1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}=0\)

\(\Leftrightarrow\left(x^2-\frac{17}{4}x+1\right)\left(\frac{1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}\right)=0\)

\(\Leftrightarrow x^2-\frac{17}{4}x+1=0\)

\(\Leftrightarrow4x^2-17x+4=0\)

e/ ĐKXĐ: ...

\(\Leftrightarrow x^2-1+2x\sqrt{\frac{x^2-1}{x}}=3x\)

Nhận thấy \(x=0\) không phải nghiệm, pt tương đương:

\(\frac{x^2-1}{x}+2\sqrt{\frac{x^2-1}{x}}=3\)

Đặt \(\sqrt{\frac{x^2-1}{x}}=a\ge0\)

\(a^2+2a=3\Leftrightarrow a^2+2a-3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-3\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow\sqrt{\frac{x^2-1}{x}}=1\Leftrightarrow x^2-1=x\Leftrightarrow x^2-x-1=0\)

f/ ĐKXĐ: ...

\(\Leftrightarrow x^2-6+x\sqrt{\frac{x^2-6}{x}}-6x=0\)

Nhận thấy \(x=0\) ko phải nghiệm, pt tương đương:

\(\frac{x^2-6}{x}+\sqrt{\frac{x^2-6}{x}}-6=0\)

Đặt \(\sqrt{\frac{x^2-6}{x}}=a\ge0\)

\(a^2+a-6=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-3\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{\frac{x^2-6}{x}}=2\Leftrightarrow x^2-4x-6=0\)

a, (x-5).(x-1) >0
<=> x-5>0 và x-1>0
<=> x-5>0
<=> x>5
x-1>0
<=> x>1
Vậy x>5
b, (2x-3).(x+1) <0
<=> 2x-3<0 và x+1<0
2x-3<0 <=> 2x<3 <=> x<2/3
x+1<0 <=> x<-1
Vậy x<2/3
c, 2x² - 3x +1>0
<=> 2x² - 2x- x +1>0
<=>(x-1). (2x-1) >0
<=> x-1>0 và 2x-1>0
x-1>0 <=> x>1
2x-1>0 <=> 2x>1 <=> x>1/2
Vậy x>1/2

f: =>35x-5=96-6x

=>41x=101

hay x=101/41

g: =>3(x-3)=90-5(1-2x)

=>3x-9=90-5+10x

=>3x-9=10x+85

=>-7x=94

hay x=-94/7

h.3x - 2/6 - 5 = 3 - 2(x + 7)/4

<=> 3x - 2 - 30/6 = 3 - 2(x + 7)/4

<=> 3x - 32/6 = 3 - 2x - 14/4

<=> 3x - 32/6 = -2x - 11/4

<=> 6x - 64/12 = -6x - 33/12

<=> 6x - 64 = -6x - 33 <=> 12x = 31 <=> x = 31/12