giải giúp mk vs ạ

bn ơi mk rất mún giải hộ pạn nhưng mk k bít để giải xin lỗi pạn nhìu

k mk nhé

ahihu tk k ai pít giải à 

mọi người ơi giúp với tui đang cần gấp .Ai làm nhanh nhất thì tui sẽ h cho

1: =>x^2+4x-21=0

=>(x+7)(x-3)=0

=>x=3 hoặc x=-7

2: =>(2x-5-4)(2x-5+4)=0

=>(2x-9)(2x-1)=0

=>x=9/2 hoặc x=1/2

3: =>x^3-9x^2+27x-27-x^3+27+9(x^2+2x+1)=15

=>-9x^2+27x+9x^2+18x+9=15

=>18x=15-9-27=-21

=>x=-7/6

6: =>4x^2+4x+1-4x^2-16x-16=9

=>-12x-15=9

=>-12x=24

=>x=-2

7: =>x^2+6x+9-x^2-4x+32=1

=>2x+41=1

=>2x=-40

=>x=-20

a: Ta có: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2+3x^2=-33\)

\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+6x^2+12x+1+3x^2=-33\)

\(\Leftrightarrow39x=-34\)

hay \(x=-\dfrac{34}{39}\)

b: Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x-2\right)\left(x+2\right)=1\)

\(\Leftrightarrow x^3-27-x^3+4x=1\)

\(\Leftrightarrow4x=28\)

hay x=7

c: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x-3\right)\left(x+3\right)=26\)

\(\Leftrightarrow x^3+8-x^3+9x=26\)

\(\Leftrightarrow x=2\)

1: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\notin\left\{4;9\right\}\end{matrix}\right.\)

Ta có: \(A=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

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

\(1,A=\dfrac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ A=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ A=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\left(x\ge0;x\ne4;x\ne9\right)\\ 2,A< 1\Leftrightarrow\dfrac{\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}-3}< 0\\ \Leftrightarrow\dfrac{4}{\sqrt{x}-3}< 0\Leftrightarrow\sqrt{x}-3< 0\Leftrightarrow0\le x< 9\)

\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2-9\right)=27\\ x.x^2-x.3x+x.9-x.x^2+x.9=27\\ x^3-3x^2+9x-x^3+9x=27\\ 3x^2+18x=27\\ 21x^2=27\\ x^2=\dfrac{9}{7}\\ \Rightarrow x=\sqrt{\dfrac{9}{7}}\)

\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2-9\right)=27\\ x.x^2-x.3x+x.9+3.x^2-3.3x+3.9-x.x^2+x.9=27\\ x^3-3x^2+9x+3x^2-9x+27-x^3+9x=27\\ 9x+27=27\\ 9x=0\\ x=0\)

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 $x=1$

\(a,\left(x-3\right)\left(x-1\right)=\left(x-3\right)^2\\ \Leftrightarrow\left(x-3\right)\left(x-1-x+3\right)=0\\ \Leftrightarrow2\left(x-3\right)=0\\ \Leftrightarrow x=3\)

\(b,4x^2-9=0\\ \Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

\(c,x^2+6x+9=0\\ \Leftrightarrow\left(x+3\right)^2=0\\ \Leftrightarrow x+3=0\\ \Leftrightarrow x=-3\)

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

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

\(\Leftrightarrow2\left(x-3\right)=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

b: Ta có: \(4x^2-9=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

c: Ta có: \(x^2+6x+9=0\)

\(\Leftrightarrow x+3=0\)

hay x=-3

`Answer:`

a. \(x^3+6x^2+12=19\)

\(\Leftrightarrow x^3+6x^2+12x-19=0\)

\(\Leftrightarrow x^3-x^2+7x^2-7x+19x-19=0\)

\(\Leftrightarrow x^2.\left(x-1\right)+7x\left(x-1\right)+19\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+7x+19\right)=0\)

Ta có \(x^2+7x+19=x^2+2x.3,5+12,25+6,75=\left(x+3,5\right)^2+6,75>0\)

\(\Rightarrow x-1=0\Leftrightarrow x=1\)

b. \(5\left(x+9\right)^2.\left(x-4\right)^3-10\left(x+9\right)^3.\left(x-4\right)^2=0\)

\(\Leftrightarrow5\left(x+9\right)^2.\left(x-4\right)^2.[x-4-2\left(x+9\right)]=0\)

\(\Leftrightarrow\left(x+9\right)^2.\left(x-4\right)^2.\left(x-4-2x-18\right)=0\)

\(\Leftrightarrow\left(x+9\right)^2.\left(x-4\right)^2.\left(-x-22\right)=0\)

\(\Leftrightarrow\left(x+9\right)^2=0\) hoặc \(\left(x-4\right)^2=0\) hoặc \(-x-22=0\)

\(\Leftrightarrow x+9=0\) hoặc \(x-4=0\) hoặc \(-x=22\)

\(\Leftrightarrow x=-9\) hoặc \(x=4\) hoặc \(x=-22\)

c. \(\left(2x+3\right)^2+\left(x-2\right)^2-2\left(2x+3\right)\left(x-2\right)\)

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

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

\(=\left(x+5\right)^2\)