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22 tháng 9 2020

đk: \(x\le-2;x\ge0\)

Ta có: \(\sqrt{x\left(x+1\right)}+\sqrt{x\left(x+2\right)}=2\sqrt{x^2}\)

\(\Leftrightarrow x\left(x+1\right)+2\sqrt{x\left(x+1\right)x\left(x+2\right)}+x\left(x+2\right)=4x^2\)

\(\Leftrightarrow x^2+x+2\sqrt{x^2\left(x^2+3x+2\right)}+x^2+2x=4x^2\)

\(\Leftrightarrow2x^2-3x=2\sqrt{x^4+3x^3+2x^2}\)

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

\(\Leftrightarrow4x^4-12x^3+9x^2=4x^4+12x^3+8x^2\)

\(\Leftrightarrow24x^3-x^2=0\)

\(\Leftrightarrow x^2\left(24x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\24x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\left(tm\right)\\x=\frac{1}{24}\left(ktm\right)\end{cases}}\)

Vậy x = 0

AH
Akai Haruma
Giáo viên
22 tháng 6 2023

Bạn nên viết lại đề bài cho sáng sủa, rõ ràng để người đọc dễ hiểu hơn.

f: =>4(x^2+4x-5)-x^2-7x-10=3(x^2+x-2)

=>4x^2+16x-20-x^2-7x-10-3x^2-3x+6=0

=>6x-24=0

=>x=4

e: =>8x+16-5x^2-10x+4(x^2-x-2)=4-x^2

=>-5x^2-2x+16+4x^2-4x-8=4-x^2

=>-6x+8=4

=>-6x=-4

=>x=2/3

d: =>2x^2+3x^2-3=5x^2+5x

=>5x=-3

=>x=-3/5

b: =>2x^2-8x+3x-12+x^2-7x+10=3x^2-12x-5x+20

=>-12x-2=-17x+20

=>5x=22

=>x=22/5

a,\(2x-5=3x+15\)

\(3x-2x=-5-15\)

\(x=-20\)

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

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

\(4x=8\)

\(x=2\)

b: =>2x^2-8x+3x-12+x^2-7x+10=3x^2-17x+20

=>-12x-2=-17x+20

=>5x=22

=>x=22/5

c: =>24x^2+16x-9x-6-4x^2-16x-7x-28=20x^2-4x+5x-1

=>-16x-34=x-1

=>-17x=33

=>x=-33/17

d: =>2x^2+3x^2-3=5x^2+5x

=>5x=-3

=>x=-3/5

e: =>8x+16-5x^2-10x+4x^2-4x-8=4-x^2

=>-6x+8=4

=>-6x=-4

=>x=2/3

f: =>4(x^2+4x-5)-x^2-7x-10=3x^2+3x-6

=>4x^2+16x-20-4x^2-10x+4=0

=>6x=16

=>x=8/3

1 tháng 9 2021

a. (x - 2)(x + 2) - (x - 3)2 = 9

<=> x2 - 22 - (x - 3)2 = 32

<=> x - 2 - (x - 3) = 3

<=> x - 2 - x + 3 = 3

<=> x - x = 3 - 3 + 2

<=> 0 = 2 (Vô lí)

Vậy nghiệm của PT là S = \(\varnothing\)

b: Ta có: \(\left(x-1\right)\left(x^2+1\right)-\left(x+1\right)\left(x^2-x+1\right)=x\left(2-x\right)\)

\(\Leftrightarrow x^3+x-x^2-1-x^3-1=2x-x^2\)

\(\Leftrightarrow-x^2+x-2-2x+x^2=0\)

\(\Leftrightarrow-x=2\)

hay x=-2

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

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

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

Suy ra: \(x^2+2x+1-\left(x^2-2x+1\right)=4\)

\(\Leftrightarrow x^2+2x+1-x^2+2x-1=4\)

\(\Leftrightarrow4x=4\)

hay x=1(loại)

Vậy: \(S=\varnothing\)

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

Ta có: \(\dfrac{x+2}{x-2}+\dfrac{x}{x+2}=2\)

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

Suy ra: \(x^2+4x+4+x^2-2x=2x^2-8\)

\(\Leftrightarrow2x^2+2x+4-2x^2-8=0\)

\(\Leftrightarrow2x-4=0\)

\(\Leftrightarrow2x=4\)

hay x=2(loại)

Vậy: \(S=\varnothing\)

21 tháng 11 2017

1)

\(\dfrac{7x-1}{2x^2+6x}=\dfrac{7x-12}{x\left(x+3\right)}\)

\(\dfrac{3-2x}{x^2-9}=\dfrac{3-2x}{\left(x-3\right)\left(x+3\right)}\)

MTC: \(x\left(x-3\right)\left(x+3\right)\)

\(\dfrac{7x-1}{2x^2+6x}=\dfrac{7x-12}{x\left(x+3\right)}=\dfrac{\left(x-3\right)\left(7x-12\right)}{x\left(x-3\right)\left(x+3\right)}=\dfrac{7x^2-12x-21x+36}{x\left(x-3\right)\left(x+3\right)}=\dfrac{7x^2-33x+36}{x\left(x-3\right)\left(x+3\right)}\)

\(\dfrac{3-2x}{x^2-9}=\dfrac{3-2x}{\left(x-3\right)\left(x+3\right)}=\dfrac{ x\left(3-2x\right)}{x\left(x-3\right)\left(x+3\right)}\dfrac{3x-2x^2}{x\left(x-3\right)\left(x+3\right)}\)

2)

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

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

MTC: \(2x\left(1-x\right)^2\)

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

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

21 tháng 11 2017

Phần còn lại nhé :v

3.

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

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

\(x+1=x+1\)

MTC: \(\left(x+1\right)\left(x^2-x+1\right)\)

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

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

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

4.

\(5x\)

\(x-2y=x-2y=-\left(2y-x\right)\)

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

MTC: \(-10x\left(2y-x\right)\left(2y+x\right)\)

\(\dfrac{7}{5x}=\dfrac{7\left(2y-x\right)\left(2y+x\right)-2}{5x\left(2y-x\right)\left(2y+x\right)-2}=\dfrac{-14\left(2y-x\right)\left(2y+x\right)}{-10x\left(2y-x\right)\left(2y+x\right)}\)

\(\dfrac{4}{x-2y}=\dfrac{4\left(2y-x\right)\left(2y+x\right)10x}{-\left(2y-x\right)\left(2y+x\right)10x}=\dfrac{40x\left(2y-x\right)\left(2y+x\right)}{-10x\left(2y-x\right)\left(2y+x\right)}\)

\(\dfrac{x-y}{8y^2-2x^2}=\dfrac{\left(x-y\right)-5x}{2\left(2y-x\right)\left(2y+x\right)-5x}=\dfrac{-5x\left(x-y\right)}{-10x\left(2y-x\right)\left(2y+x\right)}\)

5.

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

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

\(x^2+x+1\)

MTC: \(x\left(x-1\right)\left(x^2+x+1\right)\)

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

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

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

6.

\(x^2-2ax+a^2=\left(x-a\right)^2\)

\(x^2-ax=x\left(x-a\right)\)

MTC: \(x\left(x-a\right)^2\)

\(\dfrac{x}{x^2-2ax+a^2}=\dfrac{x.x}{\left(x-a\right)^2x}=\dfrac{x^2}{x\left(x-a\right)^2}\)

\(\dfrac{x+a}{x^2-ax}=\dfrac{\left(x+a\right)\left(x-a\right)}{x\left(x-a\right)\left(x-a\right)}=\dfrac{x^2-a^2}{x\left(x-a\right)^2}\)