Rảnh rỗi thật sự .-.

a)\(3x^2-4x=0<=>x(3x-4)=0\)
TH1: x=0

TH2 3x-4=0 <=>x=4/3

KL:.....

b) .

<=> (x+3)(x-1+2x)=0

TH1: x+3=0 <=> x=-3

TH2  x-1=0  <=> x=1

KL:.....

c) \(9x^2+6x+1=0. <=>(3x+1)^2=0<=>3x+1=0<=>x=-1/3 ​\)

KL:......
d) \(x^2−4x=4.<=>(x-2)^2=0<=>x-2=0<=>x=2\)

KL:....

a) \(3x^2-4x=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{3}\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{3}\end{matrix}\right.\)

c) \(9x^2+6x+1=0\)

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

\(\Leftrightarrow3x+1=0\Leftrightarrow x=-\dfrac{1}{3}\)

d) \(x^2-4x=4\)

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

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

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

\(x^3+9x=0\)

<=> \(x\left(x^2+9\right)=0\)

<=> \(\orbr{\begin{cases}x=0\\x^2+9=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=0\\x\in\varnothing\end{cases}}\)

<=> \(x=0\)

\(9x^2-4-2\left(3x-2\right)^2=0\)

<=> \(\left(9x^2-4\right)-2\left(3x-2\right)^2=0\)

<=> \(\left[\left(3x\right)^2-2^2\right]-2\left(3x-2\right)^2=0\)

<=> \(\left(3x-2\right)\left(3x+2\right)-2\left(3x-2\right)^2=0\)

<=> \(\left(3x-2\right)\left[\left(3x+2\right)-2\left(3x-2\right)\right]=0\)

<=> \(\left(3x-2\right)\left(3x+2-6x+4\right)=0\)

<=> \(\left(3x-2\right)\left(-3x+6\right)=0\)

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

<=> \(3\left(3x-2\right)\left(2-x\right)=0\)

<=> \(\orbr{\begin{cases}3x-2=0\\2-x=0\end{cases}}\)

<=> \(\orbr{\begin{cases}3x=2\\x=2\end{cases}}\)

<=> \(\orbr{\begin{cases}x=\frac{2}{3}\\x=2\end{cases}}\)

\(\left(x^3-x^2\right)-4x+8x-4=0\)

<=> \(\left(x^3-x^2\right)+\left(4x-4\right)=0\)

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

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

<=> \(\orbr{\begin{cases}x-1=0\\x^2+4=0\end{cases}}\)

<=> \(x=1\)

\(\left(25x^2-10x\right):\left(-5x\right)-3\left(x-2\right)=4\)

<=> \(5x\left(5x-2\right)\left(-\frac{1}{5x}\right)-3\left(x-2\right)=4\)

<=> \(-\left(5x-2\right)-3\left(x-2\right)=4\)

<=> \(\left(5x-2\right)+3\left(x-2\right)=-4\)

<=> \(5x-2+3x-6=-4\)

<=> \(8x-8=-4\)

<=> \(8\left(x-1\right)=-4\)

<=> \(x-1=-\frac{1}{2}\)

<=> \(x=-\frac{3}{2}\)

a) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)

\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)

\(\Leftrightarrow24x+25=15\)

\(\Leftrightarrow24x=-10\)

hay \(x=-\dfrac{5}{12}\)

b) Ta có: \(2x^3-50x=0\)

\(\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)

c) Ta có: \(5x^2-4\left(x^2-2x+1\right)-5=0\)

\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)

\(\Leftrightarrow x^2+8x-9=0\)

\(\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=1\end{matrix}\right.\)

d) Ta có: \(x^3-x=0\)

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

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

e) Ta có: \(27x^3-27x^2+9x-1=1\)

\(\Leftrightarrow\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2-1^3=1\)

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

\(\Leftrightarrow3x-1=1\)

\(\Leftrightarrow3x=2\)

hay \(x=\dfrac{2}{3}\)

1: Ta có: \(4x^2-36=0\)

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

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

2: Ta có: \(\left(x-1\right)^2+x\left(4-x\right)=11\)

\(\Leftrightarrow x^2-2x+1+4x-x^2=11\)

\(\Leftrightarrow2x=10\)

hay x=5

a/ \(\Leftrightarrow9x^2=36\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=6\\3x=-6\end{matrix}\right.\)

\(\Leftrightarrow x=\pm2\)

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

\(\Leftrightarrow x=\pm1\)

c/ Có \(\left|x+4\right|\ge0\forall x\)

=> \(\left|x+4\right|+5\ge5>0\forall x\)

\(\Rightarrow\left|x+4\right|+5=0\left(vô-lí\right)\)

\(\Rightarrow x\in\varnothing\)

d/ \(\sqrt{2x}-3-1=0\)

\(\Leftrightarrow\sqrt{2x}=4\)

\(\Leftrightarrow2x=16\)

\(\Leftrightarrow x=8\)

a: \(\sqrt{\left(x+1\right)^2}=5\)(ĐKXĐ: \(x\in R\))

=>|x+1|=5

=>\(\left[{}\begin{matrix}x+1=5\\x+1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=-6\left(nhận\right)\end{matrix}\right.\)

b: Sửa đề: \(5\sqrt{9x-9}-\sqrt{4\left(x-1\right)}+\sqrt{36\left(x-1\right)}-18=0\)

ĐKXĐ: x>=1

\(PT\Leftrightarrow5\cdot3\sqrt{x-1}-2\sqrt{x-1}+6\sqrt{x-1}-18=0\)

=>\(15\sqrt{x-1}-2\sqrt{x-1}+6\sqrt{x-1}=18\)

=>\(19\sqrt{x-1}=18\)

=>\(\sqrt{x-1}=\dfrac{18}{19}\)

=>\(x-1=\left(\dfrac{18}{19}\right)^2=\dfrac{324}{361}\)

=>\(x=\dfrac{324}{361}+1=\dfrac{324+361}{361}=\dfrac{685}{361}\)

Lời giải:

a. PT $\Leftrightarrow |x+1|=5$

$\Leftrightarrow x+1=\pm 5\Leftrightarrow x=4$ hoặc $x=-6$

b. ** Sửa $x-9$ thành $x-1$

ĐKXĐ: $x\geq 1$

PT $\Leftrightarrow 5\sqrt{x-1}-2\sqrt{x-1}+6\sqrt{x-1}-18=0$

$\Leftrightarrow (5-2+6)\sqrt{x-1}=18$

$\Leftrightarrow 9\sqrt{x-1}=18$

$\Leftrightarrow \sqrt{x-1}=2$

$\Leftrightarrow x-1=4$

$\Leftrightarrow x=5$ (tm)