\(D=\dfrac{x^2}{x^2-1}+\dfrac{1}{x^2-x^4}=\dfrac{x^4}{x^2\left(x^2-1\right)}-\dfrac{1}{x^2\left(x^2-1\right)}=\dfrac{x^4-1}{x^2\left(x^2-1\right)}=\dfrac{\left(x^2-1\right)\left(x^2+1\right)}{x^2\left(x^2-1\right)}=\dfrac{x^2+1}{x^2}=1+\dfrac{1}{x^2}\)
do \(x\ne0,\pm1\Rightarrow\dfrac{1}{x^2}>0\Rightarrow1+\dfrac{1}{x^2}>1\Rightarrow D>1\left(đpcm\right)\)

\(D=\dfrac{x^2}{x^2-1}+\dfrac{1}{x^2-x^4}\\ =\dfrac{x^4\left(1-x\right)}{\left(x-1\right)\left(x+1\right)\left(1-x\right)x^2}+\dfrac{x-1}{x^2\left(1-x\right)\left(1+x\right)\left(x-1\right)}\\ =\dfrac{x^4-x^5+x-1}{x^2\left(1-x\right)\left(1+x\right)\left(x-1\right)}\\ =\dfrac{-\left(x-1\right)^2\left(x^2+1\right)\left(x+1\right)}{-x^2\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{x^2+1}{x^2}>1\left(đpcm\right)\)

(x² + 1 luôn lớn hơn x²)

Bài 1. (a) Điều kiện: \(x\ne\pm1\).

Ta có: \(A=\left(\dfrac{x-2}{x-1}-\dfrac{x+3}{x+1}+\dfrac{3}{x-1}\right):\left(1-\dfrac{x+3}{x+1}\right)\)

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

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

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

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

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

Vậy: \(A=\dfrac{2}{1-x}\)

(b) \(A=3\Leftrightarrow\dfrac{2}{1-x}=3\)

\(\Rightarrow1-x=\dfrac{2}{3}\Leftrightarrow x=\dfrac{1}{3}\left(TM\right)\)

Vậy: \(x=\dfrac{1}{3}\)

Bài 2. (a) Phương trình tương đương với:

\(\dfrac{3\left(3x-2\right)}{12}+\dfrac{6\left(x+3\right)}{12}=\dfrac{4\left(x-1\right)}{12}+\dfrac{x+1}{12}\)

\(\Rightarrow3\left(3x-2\right)+6\left(x+3\right)=4\left(x-1\right)+x+1\)

\(\Leftrightarrow9x-6+6x+18=4x-4+x+1\)

\(\Leftrightarrow10x=-15\Leftrightarrow x=-\dfrac{3}{2}\)

Vậy: Phương trình có tập nghiệm \(S=\left\{-\dfrac{3}{2}\right\}\).

(b) Điều kiện: \(x\ne\pm1\). Phương trình tương đương với:

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

\(\Rightarrow2\left(x+1\right)+2\left(x-1\right)=2x^2+2\)

\(\Leftrightarrow2x+2+2x-2=2x^2+2\)

\(\Leftrightarrow2x^2-4x+2=0\Leftrightarrow2\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow2\left(x-1\right)^2=0\Rightarrow x-1=0\Leftrightarrow x=1\left(KTM\right)\)

Vậy: Phương trình có tập nghiệm \(S=\varnothing\)

3:

1: =>15x-9x+6=45-10x+25

=>6x+6=-10x+70

=>16x=64

=>x=4

2: =>x^2+4x-16-16=0

=>x^2+4x-32=0

=>(x+8)(x-4)=0

=>x=4 hoặc x=-8

3: ĐKXĐ: x<>4; x<>-4

\(PT\Leftrightarrow\dfrac{x+4+\left(x+2\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{5x-4}{\left(x-4\right)\left(x+4\right)}\)

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

=>x^2-x-4=5x-4

=>x^2-6x=0

=>x(x-6)=0

=>x=0 hoặc x=6

4: \(\Leftrightarrow5\left(4x+1\right)-x+2>=3\left(2x-3\right)\)

=>20x+5-x+2>=6x-9

=>19x+7>=6x-9

=>13x>=-16

=>x>=-16/13

bài ý ở đâu ?

bam vao "cau hoi tuong tu"

2:

1: =7x(x-y)-5(x-y)

=(x-y)(7x-5)

2: =(x^2-y^2)-(4x-4y)

=(x-y)(x+y)-4(x-y)

=(x-y)(x+y-4)

3: =(x^2+2xy+y^2)-(2x+2y)+1

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

=(x+y-1)^2

1) ( a + b ) ( a + b )

  = a² + ab + ab + b²

  = a² + 2ab + b²

2) ( a+ b ) ( a - b )

  = a² - ab + ab -b²

  = a² - b²

3) ( a - b ) ( a + b )

  = a² + ab - ab - b²

  = a² - b²

Hk tốt

a) \(5x\left(x-3\right)+x-3=0\)

\(\left(5x+1\right)\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}5x+1=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-1}{5}\\x=3\end{cases}}\)

vậy \(\orbr{\begin{cases}x=\frac{-1}{5}\\x=3\end{cases}}\)

câu b) bạn xem kĩ lại đề đi

sorry nha mik viết nhầm

b)x⁴-4x³+4x²-9=0

Ta có \(\frac{x}{y}< \frac{x+m}{y+m}\)khi 0<x<y,m>0

Áp dụng ta được

\(\frac{a+b}{a+b+c}< \frac{a+b+d}{a+b+c+d}\)

\(\frac{b+c}{b+c+d}< \frac{a+b+c}{a+b+c+d}\)

....................................................

Khi đó

\(VT< \frac{a+b+d+a+b+c+c+d+b+d+a+c}{a+b+c+d}=3\)

Vậy VT<3

=>x=8/3 bạn thay vào rồi giải là ra mà

ban co the noi ro ra dc khong ak