a: Ta có: \(\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=16\)

\(\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=16\)

\(\Leftrightarrow9x+7=16\)

\(\Leftrightarrow9x=9\)

hay x=1

\(5,\dfrac{4}{x-2}+\dfrac{x}{x+1}-\dfrac{x^2-2}{\left(x-2\right)\left(x+1\right)}=0\left(dkxd:x\ne2;-1\right)\)

\(\Rightarrow4\left(x+1\right)+x\left(x-2\right)-x^2-2=0\)

\(\Rightarrow4x+4+x^2-2x-x^2-2=0\)

\(\Rightarrow2x+2=0\)

\(\Rightarrow x=-1\left(loai\right)\)

Vậy \(S=\varnothing\)

\(4,\dfrac{x}{x-3}-\dfrac{1}{x+2}=0\left(dkxd:x\ne3;-2\right)\)

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

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

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

\(\Rightarrow S=\varnothing\)

giúp em tl những câu tl trên vs

Câu 1:

\(\left(x-2\right)\left(x^2+2x+4\right)+25x=x\left(x+5\right)\left(x-5\right)+8\)

\(\Leftrightarrow x^3-8+25x=x\left(x^2-25\right)+8\)

\(\Leftrightarrow x^3-8+25x=x^3-25x+8\)

\(\Leftrightarrow x^3-8+25x-x^3+25x-8=0\)

\(\Leftrightarrow50x-16=0\)

\(\Leftrightarrow50x=16\)

\(\Leftrightarrow x=\dfrac{8}{25}\)

Câu 2 :

\(\dfrac{x+5}{4}+\dfrac{3+2x}{3}=\dfrac{6x-1}{3}-\dfrac{1-2x}{12}\)

<=> \(\dfrac{3\left(x+5\right)}{12}+\dfrac{4\left(3+2x\right)}{12}=\dfrac{4\left(6x-1\right)}{12}-\dfrac{1-2x}{12}\)

<=>\(\dfrac{3x+15+12+8x}{12}=\dfrac{24x-4-1+2x}{12}\)

<=> 3x + 15 + 12 + 8x = 24x - 4 - 1 +2x

<=> 11x+27 = 26x -5

<=> ( 26x - 5 ) - ( 11x + 27 ) = 0

<=> 15x - 32 = 0

<=> 15x = 32

<=> x = \(\dfrac{32}{15}\)

Câu 3:

x - 4/3 - 3x - 1/12 = 3x + 1/4 + 9x - 2/8

<=> 4x - 16 - 3x + 1/12 = 6x + 2 + 9x - 2/8

<=> x - 15/12 = 15x/8

<=> 8x - 120 = 180x

<=> 120 = -172x <=> x = -172/120 = -43/30

=>8(x+1/x)^2+4[(x+1/x)^2-2]^2-4[(x+1/x)^2-2](x+1/x)^2=(x+4)^2

Đặt x+1/x=a(a>=2)

=>8a^2+4[a^2-2]^2-4[a^2-2]*a^2=(x+4)^2

=>8a^2+4a^4-16a^2+16-4a^4+8a^2=(x+4)^2

=>(x+4)^2=16

=>x+4=4 hoặc x+4=-4

=>x=-8;x=0

Điều kiện: \(x\ne0\)

\(\Leftrightarrow8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)\left[\left(x^2+\dfrac{1}{x^2}\right)-\left(x+\dfrac{1}{x}\right)^2\right]=\left(x+4\right)^2\)

\(\Leftrightarrow8\left(x+\dfrac{1}{x}\right)^2-8\left(x^2+\dfrac{1}{x^2}\right)=\left(x+4\right)^2\\ \Leftrightarrow\left(x+4\right)^2=16\\ \Rightarrow\left\{{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\)

Vì \(x\ne0\) nên \(S=\left\{-8\right\}\)

b) Đặt \(x^2+2x+3=a\)(a>0)

Ta có: \(\dfrac{x^2+2x+7}{\left(x+1\right)^2+2}=x^2+2x+4\)

\(\Leftrightarrow\dfrac{x^2+2x+7}{x^2+2x+1+2}=x^2+2x+4\)

\(\Leftrightarrow\dfrac{x^2+2x+7}{x^2+2x+3}=x^2+2x+4\)

\(\Leftrightarrow\dfrac{a+4}{a}=a+1\)

\(\Leftrightarrow a^2+a=a+4\)

\(\Leftrightarrow a^2=4\)

\(\Leftrightarrow\left[{}\begin{matrix}a=2\left(nhận\right)\\a=-2\left(loại\right)\end{matrix}\right.\)

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

\(\Leftrightarrow x^2+2x+1=0\)

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

\(\Leftrightarrow x+1=0\)

hay x=-1

Vậy: S={-1}

ĐKXĐ của cả 2 pt trên đều là `x in RR`

`a,1/(x^2-2x+2)+2/(x^2-2x+3)=6/(x^2-2x+4)`

Đặt `a=x^+2x+3(a>=2)` ta có:

`1/(a-1)+2/a=6/(a+1)`

`<=>a(a+1)+2(a-1)(a+1)=6a(a-1)`

`<=>a^2+a+2(a^2-1)=6a^2-6a`

`<=>a^2+a+2a^2-2=6a^2-6a`

`<=>3a^2-5a+2=0`

`<=>3a^2-3a-2a+2=0`

`<=>3a(a-1)-2(a-1)=0`

`<=>(a-1)(3a-2)=0`

`a>=2=>a-1>=1>0`

`a>=2=>3a-2>=4>0`

Vậy pt vô nghiệm

`(x^2+2x+7)/((x+1)^2+2)=x^2+2x+4`

`<=>(x^2+2x+7)=(x^2+2x+4)(x^2+2x+3)`

Đặt `a=x^2+2x+3(a>=2)`

`pt<=>a+4=a(a+1)`

`<=>a^2+a=a+4`

`<=>a^2=4`

`<=>a=2` do `a>=2`

`<=>(x+1)^2+2=2`

`<=>(x+1)^2=0`

`<=>x=-1`

Vậy `S={-1}`

1/ \(\dfrac{4x+7}{x-1}=\dfrac{12x+5}{3x+4}\) (1)

Điều kiện: \(\left\{{}\begin{matrix}x-1\ne0\\3x+4\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne-\dfrac{4}{3}\end{matrix}\right.\)

(1) \(\Leftrightarrow\left(4x+7\right)\left(3x+4\right)=\left(12x+5\right)\left(x-1\right)\\\Leftrightarrow12x^2+16x+21x+28=12x^2-12x+5x-5\\ \Leftrightarrow\left(16+21+12-5\right)x=-5-28\\ \Leftrightarrow44x=-33\\ \Leftrightarrow x=-\dfrac{3}{4}\) (Thỏa mãn)

Vậy \(x=-\dfrac{3}{4}\).

2/ \(\dfrac{x}{x-1}-\dfrac{2x}{x^2-1}=0\) (2)

Điều kiện: \(x\ne\pm1\)

(2)\(\Leftrightarrow\dfrac{x}{x-1}-\dfrac{2x}{\left(x-1\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{2x}{\left(x-1\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{x\left(x+1\right)-2x}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow x\left(x+1\right)-2x=0\\ \Leftrightarrow x^2+x-2x=0\\ \Leftrightarrow x^2-x=0\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

So sánh với điều kiện \(\Rightarrow x=0\) là nghiệm của PT.

3/ \(\dfrac{1}{3-x}-\dfrac{14}{x^2-9}=1\) (3)

Điều kiện: \(x\ne\pm3\)

(3)\(\Leftrightarrow\dfrac{1}{3-x}-\dfrac{14}{\left(x-3\right)\left(x+3\right)}=1\\ \Leftrightarrow-\dfrac{\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{14}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\\ \Leftrightarrow-\left(x+3\right)-14=\left(x-3\right)\left(x+3\right)\\ \Leftrightarrow-x-17=x^2-9\Leftrightarrow x^2+x+8=0\) (Vô nghiệm do \(x^2+x+8>0\qquad\forall x\)).

Vậy PT vô nghiệm.

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

Điều kiện: \(x\ne\pm1\)

(4)\(\Leftrightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\\ \Leftrightarrow\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\\ \Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2=4\\ \Leftrightarrow\left(x^2+2x+1\right)-\left(x^2-2x+1\right)=4\Leftrightarrow4x=4\Leftrightarrow x=1\) (loại)

Vậy PT vô nghiệm.

5/ \(x+\dfrac{1}{x}=x^2+\dfrac{1}{x^2}\) (5)

Điều kiện: \(x\ne0\)

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

Đặt \(t=x+\dfrac{1}{x}\), ta có: \(t=t^2-2\\ \Leftrightarrow t^2-t-2=0\Leftrightarrow\left(t-2\right)\left(t+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t=2\\t=-1\end{matrix}\right.\)

Với \(t=2\) ta có: \(x+\dfrac{1}{x}=2\Leftrightarrow x^2+1=2x\Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\) (thỏa mãn)

Với \(t=-1\) ta có: \(x+\dfrac{1}{x}=-1\Leftrightarrow x^2+1=-x\Leftrightarrow x^2+x+1=0\) (vô nghiệm).

Vậy \(x=1\) là nghiệm PT.

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

Điều kiện: \(x\ne-1\)

(6)\(\Leftrightarrow\dfrac{x-1}{x^2+4}-\dfrac{x-1}{x+1}=0\\ \Leftrightarrow\left(x-1\right)\left(\dfrac{1}{x^2+4}-\dfrac{1}{x+1}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\\dfrac{1}{x^2+4}-\dfrac{1}{x+1}=0\end{matrix}\right.\)

\(x-1=0\Leftrightarrow x=1\) (Thỏa mãn)

\(\dfrac{1}{x^2+4}-\dfrac{1}{x+1}=0\Leftrightarrow\dfrac{1}{x^2+4}=\dfrac{1}{x+1}\Leftrightarrow x^2+4=x+1\\ \Leftrightarrow x^2-x+3=0\) (vô nghiệm).

Vậy \(x=1\) là nghiệm PT.

1) ĐKXĐ: \(x\notin\left\{1;-\dfrac{4}{3}\right\}\)

Ta có: \(\dfrac{4x+7}{x-1}=\dfrac{12x+5}{3x+4}\)

\(\Leftrightarrow\left(4x+7\right)\left(3x+4\right)=\left(12x+5\right)\left(x-1\right)\)

\(\Leftrightarrow12x^2+16x+21x+28=12x^2+12x+5x-5\)

\(\Leftrightarrow12x^2+37x+28-12x^2-17x+5=0\)

\(\Leftrightarrow20x+33=0\)

\(\Leftrightarrow20x=-33\)

\(\Leftrightarrow x=-\dfrac{33}{20}\)(nhận)

Vậy: \(S=\left\{-\dfrac{33}{20}\right\}\)

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

Ta có: \(\dfrac{x}{x-1}-\dfrac{2x}{x^2-1}=0\)

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

Suy ra: \(x^2+x-2x=0\)

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

\(\Leftrightarrow x\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=1\left(loại\right)\end{matrix}\right.\)

Vậy: S={0}

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

Ta có: \(\dfrac{1}{3-x}-\dfrac{14}{x^2-9}=1\)

\(\Leftrightarrow\dfrac{-1}{x-3}-\dfrac{14}{\left(x-3\right)\left(x+3\right)}=1\)

\(\Leftrightarrow\dfrac{-\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{14}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)

Suy ra: \(-x-3-14=x^2-9\)

\(\Leftrightarrow x^2-9=-x-17\)

\(\Leftrightarrow x^2-9+x+17=0\)

\(\Leftrightarrow x^2+x+8=0\)

\(\Leftrightarrow x^2+2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{31}{4}=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{31}{4}=0\)(vô lý)

Vậy: \(S=\varnothing\)

4) 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: \(S=\varnothing\)

5) ĐKXĐ: \(x\ne0\)

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

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

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

\(\Leftrightarrow x^4+x^2=x^5+x\)

\(\Leftrightarrow x^5+x-x^4-x^2=0\)

\(\Leftrightarrow x\left(x^4-x^3-x+1\right)=0\)

\(\Leftrightarrow x\left[x^3\left(x-1\right)-\left(x-1\right)\right]=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x^3-1\right)=0\)

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

mà \(x^2+x+1>0\)

nên \(x\cdot\left(x-1\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x-1=0\end{matrix}\right.\Leftrightarrow x=1\)

Vậy: S={1}

6) ĐKXĐ: \(x\in R\)

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

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

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

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

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

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

mà \(x^2-x+3>0\)

nên x-1=0

hay x=1(nhận)

Vậy: S={1}

a) Ta có: \(\left(x-1\right)\left(x-2\right)\left(x^2+x+1\right)\left(x^2+2x+4\right)-x^6+9x^3\)

\(=\left(x-1\right)\left(x^2+x+1\right)\left(x-2\right)\left(x^2+2x+4\right)-x^6+9x^3\)

\(=\left(x^3-1\right)\left(x^3-8\right)-x^6+9x^3\)

\(=x^6-9x^3+8-x^6+9x^3=8\)

b) Ta có: \(\left(\dfrac{1}{3}+2x\right)\left(\dfrac{1}{9}-\dfrac{2}{3}x+4x^2\right)-\left(2x-\dfrac{1}{3}\right)\left(4x^2+\dfrac{2}{3}x+\dfrac{1}{4}\right)\)

\(=\dfrac{1}{27}+8x^3-8x^3+\dfrac{1}{27}\)

\(=\dfrac{2}{27}\)

c) Ta có: \(\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3x\left(1-x\right)\)

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

=0

d) Ta có: \(\left(x^2-y^2\right)\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)-x^6+y^6\)

\(=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)-x^6+y^6\)

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

\(=x^6-y^6-x^6+y^6=0\)

\(\Leftrightarrow\dfrac{2}{-x^2+6x-8}=\dfrac{x-1}{x-2}+\dfrac{x+3}{x-4}\\ \Leftrightarrow\left\{{}\begin{matrix}2=\left(-x^2+6x-8\right)\left(\dfrac{x-1}{x-2}+\dfrac{x+3}{x-4}\right)\\-x^2+6x-8\ne0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2=-2x^2+4x+2\\-x^2+6x-8\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\\-x^2+6x-8\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x=0\\-x^2+6x-8\ne0\end{matrix}\right.\\\left\{{}\begin{matrix}x=2\\-x^2+6x-8\ne\end{matrix}\right.\end{matrix}\right.\\\Rightarrow x=0\)