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) \(x^2-4x=4\)

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

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

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

a: =>8x^2-20x+20x-50+4x(4x^2-12x+9)=0

=>8x^2-50+8x^3-48x^2+36x=0

=>8x^3-40x^2+36x-50=0

=>\(x\simeq4,29\)

b: =>(2x-3-3x-1)(2x-3+3x+1)=0

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

=>x=2/5 hoặc x=-4

a) Ta có: \(\left(x^2-16\right)\left(\dfrac{x}{4}-\dfrac{4x+5}{3}\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+4\right)\left(\dfrac{3x-16x-20}{12}\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+4\right)\cdot\left(-13x-20\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x+4=0\\-13x-20=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\\-13x=20\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\\x=\dfrac{-20}{13}\end{matrix}\right.\)

Vậy: \(x\in\left\{4;-4;\dfrac{-20}{13}\right\}\)

b) Ta có: \(\left(4x-1\right)\left(x+5\right)=x^2-25\)

\(\Leftrightarrow\left(4x-1\right)\left(x+5\right)-\left(x^2-25\right)=0\)

\(\Leftrightarrow\left(4x-1\right)\left(x+5\right)-\left(x+5\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(4x-1-x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(3x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{4}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{-5;\dfrac{-4}{3}\right\}\)

c) Ta có: \(x\left(x+3\right)^3-\dfrac{x}{4}\cdot\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\cdot\left[x\left(x+3\right)^2-\dfrac{1}{4}x\right]=0\)

\(\Leftrightarrow\left(x+3\right)\left[x\left(x^2+6x+9\right)-\dfrac{1}{4}x\right]=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^3+6x^2+9x-\dfrac{1}{4}x\right)=0\)

\(\Leftrightarrow\left(x+3\right)\cdot x\cdot\left(x^2+6x+\dfrac{35}{4}\right)=0\)

\(\Leftrightarrow x\left(x+3\right)\left(x^2+6x+9-\dfrac{1}{4}\right)=0\)

\(\Leftrightarrow x\left(x+3\right)\left[\left(x+3\right)^2-\dfrac{1}{4}\right]=0\)

\(\Leftrightarrow x\left(x+3\right)\left(x+3-\dfrac{1}{2}\right)\left(x+3+\dfrac{1}{2}\right)=0\)

\(\Leftrightarrow x\left(x+3\right)\left(x+\dfrac{5}{2}\right)\left(x+\dfrac{7}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\x+\dfrac{5}{2}=0\\x+\dfrac{7}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=-\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{0;-3;-\dfrac{5}{2};-\dfrac{7}{2}\right\}\)

|5x-3| - 3x = 7

*Nếu \(x\ge\frac{3}{5}\)

5x - 3 - 3x = 7

2x = 10

x = 5 ( tm)

*Nếu \(x< \frac{3}{5}\)

3 - 5x - 3x = 7

-8x = 4 

x = \(-\frac{1}{2}\)( tm )

Làm hơi khó nhìn , thông cảm. Mệt rùi :)

|x - 3| + |x - 5| - 4x = -28

*Nếu x < 3

3 - x + 5 - x - 4x = -28

-6x = -36

x = 6 ( loại do ko tm khoảng đang xét )

* nếu 3 < x < 5

x - 3 + 5 - x - 4x = -28

-4x = -30

x= \(\frac{15}{2}\) ( loại do ko tm khaongr đang xét )

*Nếu x > 5

x - 3 + x - 5 - 4x = -28

-2x = -20

x = 10 ( tm)

Vậy x =10

|x + 2| + |x + 3/5| + |x+1/2| = 4x

Câu này cũng xét khoảng x < -2     

                                         -3/5 < x < -1/2

                                           x > -1/2

|2x-1| + ( 4x² - 1)² = 0

Vì |2x - 1| > 0 với mọi x

   ( 4x² - 1)² > 0 với mọi x

=> |2x-1| + (4x² - 1)² > 0 với mọi x

Dấu "=" xảy ra <=> 2x - 1= 4x² - 1 = 0

                       <=> x = 1/2

Đây gọi là phương pháp dùng bất đẳng thức

\(\left(2x-3\right)^2+2\left(4x^2-9\right)+\left(2x+3\right)^2=0\\ \Leftrightarrow\left(2x-3\right)^2+2\left(2x-3\right)\left(2x+3\right)+\left(2x+3\right)^2=0\\ \Leftrightarrow\left[\left(2x-3\right)+\left(2x+3\right)\right]^2=0\\ \Leftrightarrow\left(4x\right)^2=0\\ \Leftrightarrow16x^2=0\Leftrightarrow x=0\)

Đặt \(\left|3-4x\right|=t\)

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

\(\Leftrightarrow t^2-3t=0\)

\(\Leftrightarrow t\left(t-3\right)=0\)

\(\Rightarrow t=0;t-3=0\)

\(\Rightarrow y=0;t=3\)

Với \(t=0\) thì \(\left|3-4x\right|=0\Rightarrow x=\frac{3}{4}\)

Với \(t=3\) thì \(\left|3-4x\right|=3\Rightarrow x=\frac{3}{2};0\)

Vậy \(x=0;\frac{3}{4};\frac{3}{2}\)

Đặt  \(\left|4x-3\right|=t\)ta có: \(t^2-3t=0\Leftrightarrow t\left(t-3\right)=0\)

<=> t=0 hoặc t-3=0 <=> |4x-3|=0 hoặc |4x-3|-3=0 

TH1: \(\left|4x-3\right|=0\Leftrightarrow4x-3=0\Leftrightarrow4x=3\Leftrightarrow x=\frac{3}{4}\)

TH2: \(\left|4x-3\right|-3=0\Leftrightarrow\left|4x-3\right|=3\) <=> 4x-3=-3 hoặc 4x-3=3

<=>4x=0 hoặc 4x=6 <=> x=0 hoặc x=3/2

Vậy \(x\in\){0;3/4;3/2}

a) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=46\)

\(\Leftrightarrow16x^2-9-16x^2+40x-25=46\)

\(\Leftrightarrow40x=46+9+25=80\)

\(\Leftrightarrow x=2\)

b) \(\left(x+1\right)^3+2x-\left(x-1\right)^3-3\left[\left(x+1\right)^2+\left(x-1\right)^2\right]+5=0\)

\(=x^3+3x^2+3x+1+2x-x^3+3x^2-3x+1-3\left(x^2+2x+1+x^2-2x+1\right)+5=0\)

\(=6x^2+2x+2-3\left(2x^2+2\right)+5=0\)

\(\Leftrightarrow6x^2+2x+2-6x^2-6+5=0\)

\(\Leftrightarrow2x=-2+6-5=-1\)

\(\Leftrightarrow x=\frac{1}{2}\)

\(B1\\ a,2x+10y=2\left(x+5y\right)\\ b,x^2+4x+4=x^2+2.2x+2^2=\left(x+2\right)^2\\ c,x^2-y^2+10y-25\\ =\left(x^2-y^2\right)+5\left(2y-5\right)\\ =\left(x-y\right)\left(x+y\right)+5\left(2y-5\right)\\ B2\)

\(a,x^2-3x+x-3=0\\ =>x\left(x-3\right)+\left(x-3\right)=0\\ =>\left(x+1\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+1=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\\ b,2x\left(x-3\right)-\dfrac{1}{2}\left(4x^2-3\right)=0\\ =>2x^2-6x-2x^2+\dfrac{3}{2}=0\\ =>-6x=-\dfrac{3}{2}\\ =>x=\left(-\dfrac{3}{2}\right):\left(-6\right)\\ =>x=\dfrac{1}{4}\\ c,x^2-\left(x-3\right)\left(2x-5\right)=9\\ =>x^2-2x^2+6x+5x-15=9\\ =>-x^2+11-15-9=0\\ =>-x^2+11x-24=0\\ =>-x^2+8x+3x-24=0\\ =>-x\left(x-8\right)+3\left(x-8\right)=0\\ =>\left(3-x\right)\left(x-8\right)=0\\ =>\left[{}\begin{matrix}3-x=0\\x-8=0\end{matrix}\right.=>\left[{}\begin{matrix}x=3\\x=8\end{matrix}\right.\)

(\(x-3\))² + (2y - 1)² = 0

          (\(x\) - 3)² ≥ 0 ∀ \(x\)

        (2y - 1)² ≥ 0 ∀ y

⇔ (\(x\) - 3)² + (2y - 1)²= 0

⇔ \(\left\{{}\begin{matrix}x-3=0\\3y-1=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{3}\end{matrix}\right.\)

(4\(x-3\))⁴ + (y + 2)² ≤ 0

(4\(x\) - 3)⁴ ≥ 0 ∀ \(x\)

(y + 2)² ≥ 0 ∀ y

⇔(4\(x\) - 3)⁴   + (y+2)² ≥ 0

⇔ (4\(x\) - 3)⁴ + (y + 2)² ≤ 0 ⇔

⇔\(\left\{{}\begin{matrix}4x-3=0\\y+2=0\end{matrix}\right.\)

⇔ \(\left\{{}\begin{matrix}x=\dfrac{3}{4}\\y=-2\end{matrix}\right.\)