2 nha bn

lớp 1 căng đét

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: \(S=\varnothing\)

a: Ta có: \(\left(x+1\right)^3-\left(x+2\right)\left(x-1\right)^2-3\left(x-3\right)\left(x+3\right)=5\)

\(\Leftrightarrow x^3+3x^2+3x+1-\left(x+2\right)\left(x^2-2x+1\right)-3\left(x^2-9\right)=5\)

\(\Leftrightarrow x^3+3x^2+3x+1-\left(x^3-2x^2+x+2x^2-4x+2\right)-3\left(x^2-9\right)=5\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x-2-3x^2+9=5\)

\(\Leftrightarrow6x=-3\)

hay \(x=-\dfrac{1}{2}\)

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

\(\Leftrightarrow x^3+3x^2+3x+1+x^3-3x^2+3x-1=x^3+6x^2+12x+8+x^3-6x^2+12x-8\)

\(\Leftrightarrow2x^3+6x=2x^3+24x\)

\(\Leftrightarrow x=0\)

c: Ta có: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-1=-10\)

\(\Leftrightarrow12x=-11\)

hay \(x=-\dfrac{11}{12}\)

Lời giải:

a. $x(3x+1)+(x-1)^2-(2x+1)(2x-1)=0$

$\Leftrightarrow (3x^2+x)+(x^2-2x+1)-(4x^2-1)=0$

$\Leftrightarrow 3x^2+x+x^2-2x+1-4x^2+1=0$

$\Leftrightarrow (3x^2+x^2-4x^2)+(x-2x)+(1+1)=0$

$\Leftrightarrow -x+2=0$

$\Leftrightarrow x=2$

b.

$(x+1)^3+(2-x)^3-9(x-3)(x+3)=0$

$\Leftrightarrow [(x+1)+(2-x)][(x+1)^2-(x+1)(2-x)+(2-x)^2]-9(x-3)(x+3)=0$

$\Leftrightarrow 3[x^2+2x+1-(x-x^2+2)+(x^2-4x+4)]-9(x-3)(x+3)=0$

$\Leftrightarrow 3(3x^2-3x+3)-9(x^2-9)=0$

$\Leftrightarrow 9(x^2-x+1)-9(x^2-9)=0$

$\Leftrightarrow 9(x^2-x+1-x^2+9)=0$
$\Leftrightarrow 9(-x+10)=0$

$\Leftrightarrow -x+10=0\Leftrightarrow x=10$

c.

$(x-1)^3-(x+3)(x^2-3x+9)+3x^2=25$

$\Leftrightarrow (x^3-3x^2+3x-1)-(x^3+3^3)+3x^2=25$

$\Leftrightarrow x^3-3x^2+3x-1-x^3-27+3x^2=25$
$\Leftrightarrow (x^3-x^3)+(-3x^2+3x^2)+3x-28=25$

$\Leftrightarrow 3x-28=25$

$\Leftrightarrow x=\frac{53}{3}$

d.

$(x+2)^3-(x+1)(x^2-x+1)-6(x-1)^2=23$
$\Leftrightarrow (x^3+6x^2+12x+8)-(x^3+1)-6(x^2-2x+1)=23$

$\Leftrightarrow x^3+6x^2+12x+8-x^3-1-6x^2+12x-6=23$

$\Leftrightarrow (x^3-x^3)+(6x^2-6x^2)+(12x+12x)+(8-1-6)=23$
$\Leftrightarrow 24x+1=23$

$\Leftrgihtarrow 24x=22$

$\Leftrightarrow x=\frac{11}{12}$

e.

$(x+3)(x^2-3x+9)-x(x-2)(x+2)+11=0$

$\Leftrightarrow x^3+3^3-x(x^2-4)+11=0$

$\Leftrightarrow x^3+27-x^3+4x+11=0$

$\Leftrightarrow (x^3-x^3)+4x+(27+11)=0$

$\Leftrightarrow 4x+38=0$

$\Leftrightarrow x=\frac{-19}{2}$

f.

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

$\Leftrightarrow x(x-3)-(x-3)=0$

$\Leftrightarrow (x-3)(x-1)=0$

$\Leftrightarrow x-3=0$ hoặc $x-1=0$

$\Leftrightarrow x=3$ hoặc $x=1$

a: \(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{6}{\sqrt{x}-1}-\dfrac{2\sqrt{3}}{\sqrt{x}-1}\)

\(=\dfrac{\sqrt{x}-6-2\sqrt{3}}{\sqrt{x}-1}\)

b: \(=\dfrac{3-\sqrt{x}-1+\sqrt{x}+5\sqrt{x}}{\sqrt{x}-2}=\dfrac{5\sqrt{x}+2}{\sqrt{x}-2}\)

c: \(=\dfrac{2-6\sqrt{x}-1+\sqrt{x}-3+\sqrt{x}}{\sqrt{x}-4}\)

\(=\dfrac{-4\sqrt{x}-4}{x-4}\)

a: =>9x^2+12x+4-9x^2+12x-4=5x+38

=>24x=5x+38

=>19x=38

=>x=2

e: =>x^3+1-2x=x^3-x

=>-2x+1=-x

=>-x=-1

=>x=1

f: =>x^3-6x^2+12x-8+9x^2-1=x^3+3x^2+3x+1

=>12x-9=3x+1

=>9x=10

=>x=10/9

b: \(\Leftrightarrow3x^2-12x+12+9x-9=3x^2+3x-9\)

=>-3x+3=3x-9

=>-6x=-12

=>x=2

\(A=2x^3+3x^2-3-5x^2-5x=2x^3-2x^2-5x-3\\ B=125-150x+60x^2-8x^3-25+9x^2=-8x^3+69x^2-150x+100\\ C=\left(3x+1-2x+1\right)\left(3x+1+2x-1\right)=5x\left(x+2\right)=5x^2+10x\\ D=\left(2x+1-3+x\right)^2=\left(3x-2\right)^2=9x^2-12x+4\\ E=x^3-6x^2+12x-8-x^3+x+6x^2-18x=-5x-8\\ F=x^3-3x^2+3x-1-3+3x^2-x^3+1-3x=-3\)

Bài dài quá, lần sau chia nhỏ câu hỏi nhé!!!!!

đúng vậy

\(\frac{1-x}{1+x}+3=\frac{2x+3}{x+1}\left(ĐKXĐ:x\ne-1\right)\)

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

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

\(\Rightarrow1-x+3\left(x+1\right)=2x+3\)

\(\Leftrightarrow1-x+3x+3=2x+3\)

\(\Leftrightarrow2x+4=2x+3\)

\(\Leftrightarrow0x=-1\)(vô nghiệm)

Vậy phương trình vô nghiệm.

\(\frac{\left(x+2\right)^2}{2x-3}-1=\frac{x^2-10}{2x-3}\left(ĐKXĐ:x\ne\frac{3}{2}\right)\)

\(\Leftrightarrow\frac{x^2+4x+4}{2x-3}-\frac{2x-3}{2x-3}=\frac{x^2-10}{2x-3}\)

\(\Leftrightarrow\frac{x^2+4x+4-2x+3}{2x-3}=\frac{x^2-10}{2x-3}\)

\(\Rightarrow x^2+4x+4-2x+3=x^2-10\)

\(\Leftrightarrow2x+7=-10\)

\(\Leftrightarrow2x=-17\)

\(\Leftrightarrow x=\frac{-17}{2}\)(thỏa mãn ĐKXĐ)

Vậy phương trình có nghiệm duy nhất : \(x=\frac{-17}{2}\)

Trả lời:

a, \(\frac{1-x}{x+1}+3=\frac{2x+3}{x+1}\)\(\left(đkxđ:x\ne-1\right)\)

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

\(\Rightarrow1-x+3x+3=2x+3\)

\(\Leftrightarrow4+2x=2x+3\)

\(\Leftrightarrow2x-2x=3-4\)

\(\Leftrightarrow0x=-1\)(không thỏa mãn)

Vậy \(S=\varnothing\)

b, \(\frac{\left(x+2\right)^2}{2x-3}-1=\frac{x^2-10}{2x-3}\)\(\left(đkxđ:x\ne\frac{3}{2}\right)\)

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

\(\Rightarrow x^2+4x+4-2x+3=x^2-10\)

\(\Leftrightarrow x^2+2x+7=x^2-10\)

\(\Leftrightarrow x^2+2x-x^2=-10-7\)

\(\Leftrightarrow2x=-17\)

\(\Leftrightarrow x=\frac{-17}{2}\)(tm)

Vậy \(S=\left\{\frac{-17}{2}\right\}\)

c, \(\frac{5x-2}{2-2x}+\frac{2x-1}{2}=1+\frac{x^2+x-3}{x-1}\)\(\left(đkxđ:x\ne1\right)\)

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

\(\Leftrightarrow\frac{2-5x}{2\left(x-1\right)}+\frac{2x-1}{2}=1+\frac{x^2+x-3}{x-1}\)

\(\Leftrightarrow\frac{2-5x}{2\left(x-1\right)}+\frac{\left(2x-1\right)\left(x-1\right)}{2\left(x-1\right)}=\frac{2\left(x-1\right)}{2\left(x-1\right)}+\frac{2\left(x^2+x-3\right)}{2\left(x-1\right)}\)

\(\Rightarrow2-5x+2x^2-3x+1=2x-2+2x^2+2x-6\)

\(\Leftrightarrow2x^2-8x+3=2x^2+4x-8\)

\(\Leftrightarrow2x^2-8x-2x^2-4x=-8-3\)

\(\Leftrightarrow-12x=-13\)

\(\Leftrightarrow x=\frac{13}{12}\)(tm)

Vậy \(S=\left\{\frac{13}{12}\right\}\)