a/ ĐKXĐ: \(x\ne\frac{3\pi}{4}+k\pi\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\tan\left(x-\frac{\pi}{4}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\\x-\frac{\pi}{4}=k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=\frac{\pi}{4}+k\pi\end{matrix}\right.\)

b/

\(\Leftrightarrow sin3x=-cos\left(4x+50^0\right)\)

\(\Leftrightarrow sin3x=sin\left(4x-40^0\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-40^0=3x+k360^0\\4x-40^0=180^0-3x+k360^0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=40^0+k360^0\\x=\frac{220^0}{7}+\frac{k360^0}{7}\end{matrix}\right.\)

a) <=> 4x^3 - 12x^2 - x^2 + 3x + 6x - 18 = 0

<=> 4x^2 (x - 3) - x(x - 3) + 6(x - 3) = 0

<=> (x - 3)(4x^2 - x + 6) = 0

xét 2 th

. x - 3 = 0 <=> x = 3

. 4x^2 - x + 6 = 0

<=> 4x^2 + 2.(1/2)x + 1/4 + 23/4 = 0

<=> (4x + 1/2)^2 = -23/4

.... phần sau bạn tự làm nhé 

vậy pt trên có nghiệm là ...

. mik bận nên chỉ làm như vậy thôi.. những ý sau thì tách tương tự

c) => x³ + 2x² - 6x²  - 12x + 4x + 8 = 0

=> (x³ + 2x²)  -  (6x²  + 12x)  + (4x + 8) = 0

=> x². (x +2) - 6x. (x + 2) + 4.(x + 2) =0

=> (x +2).(x²  - 6x + 4) = 0

=> x+ 2 = 0 hoặc x²  - 6x + 4 = 0

+) x+ 2 =0 => x = -2

+) x²  - 6x + 4 = 0 => x²  - 2.x.3  + 9  - 5 = 0 => (x -3)²  = 5

=> x - 3 = \(\sqrt{5}\) hoặc x - 3 = - \(\sqrt{5}\)

=> x = 3 + \(\sqrt{5}\) hoặc x = 3 - \(\sqrt{5}\)

vậy...

Ta có: \(\sqrt{9x+9}+\sqrt{4x+4}-2\sqrt{16x+16}=\sqrt{x+1}-8\)

\(\Leftrightarrow3\sqrt{x+1}+2\sqrt{x+1}-8\sqrt{x+1}-\sqrt{x+1}=-8\)

\(\Leftrightarrow\sqrt{x+1}=2\)

\(\Leftrightarrow x+1=4\)

hay x=3

\(4x^2-9x+23=0\)

\(\Leftrightarrow\left(2x\right)^2-2\cdot2x\cdot\frac{9}{4}+\frac{81}{16}+\frac{287}{16}=0\)

\(\Leftrightarrow\left(2x-\frac{9}{4}\right)^2=\frac{-287}{16}\)( vô lý )

Vậy pt vô nghiệm

\(\sqrt{4x+4}+\sqrt{9x+9}=5\sqrt{2}\)

\(\sqrt{4\left(x+1\right)}+\sqrt{9\left(x+1\right)}=5\sqrt{2}\)

\(2\sqrt{x+1}+3\sqrt{x+1}=5\sqrt{2}\)

\(5\sqrt{x+1}=5\sqrt{2}\)

\(\sqrt{x+1}=\sqrt{2}\)

\(\left(\sqrt{x+1}\right)^2=\left(\sqrt{2}\right)^2\)

\(x+1=2\)

\(x=1\)

vậy \(x=1\)

\(\sqrt{4x+4}+\sqrt{9x+9}=5\sqrt{2}\)

\(\sqrt{4\left(x+1\right)}+\sqrt{9\left(x+1\right)}=5\sqrt{2}\)

\(\sqrt{2^2\left(x+1\right)}+\sqrt{3^2\left(x+1\right)}=5\sqrt{2}\)

\(2\sqrt{x+1}+3\sqrt{x+1}=5\sqrt{2}\)

\(5\sqrt{x+1}=5\sqrt{2}\)

\(\sqrt{x+1}=\sqrt{2}\)

\(\left(\sqrt{x+1}\right)^2=\left(\sqrt{2}\right)^2\)

\(\Rightarrow x+1=2\)

\(\Rightarrow x=1\)

vậy \(x=1\)

tất cả đều làm đúng hết 

a. Dat \(x^2=t\left(t\ge0\right)\)

Suy ra PT:\(\orbr{\begin{cases}t^2=-4t+1\left(1\right)\left(x< 0\right)\\t^2=4t+1\left(2\right)\left(x\ge0\right)\end{cases}}\)

(1)\(\Leftrightarrow t^2+4t-1=0\)

\(\Leftrightarrow\left(t+2\right)^2-5=0\)

\(\Leftrightarrow\left(t+2+\sqrt{5}\right)\left(t+2-\sqrt{5}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t=-2-\sqrt{5}\left(l\right)\\t=\sqrt{5}-2\left(n\right)\end{cases}}\)

Nghiem cua PT(1) la \(t=\sqrt{5}-2\)

(2)\(\Leftrightarrow t^2-4t-1=0\)

\(\Leftrightarrow\left(t-2\right)^2-5=0\)

\(\Leftrightarrow\left(t-2+\sqrt{5}\right)\left(t-2-\sqrt{5}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t=2-\sqrt{5}\left(l\right)\\t=2+\sqrt{5}\left(n\right)\end{cases}}\)

Nghiem cua PT(2) la \(t=2+\sqrt{5}\)

Suy ra:\(\orbr{\begin{cases}x=\sqrt{\sqrt{5}-2}\\x=\sqrt{\sqrt{5}+2}\end{cases}}\)

b.\(x^3-3x^2+9x-9=0\)

\(\Leftrightarrow\left(x-3\right)^3=-18\)

\(\Leftrightarrow x-3=-\sqrt[3]{18}\)

\(\Leftrightarrow x=3-\sqrt[3]{18}\)

\(b,x^3-3x^2+9x-9=0\)

\(\Rightarrow x^2\left(x-3\right)+9\left(x-3\right)+18=0\)

\(\Rightarrow\left(x^2+9\right)\left(x-3\right)=-18\)

từ đây bạn xét các TH nhá ! 

 Chú ý : Vì \(x^2+9\ge9\forall\) để xét ít Th hơn

a, x^4 - 4x - 1 = 0
<=> x^4 +2x^2 + 1 -2x^2 - 4x - 2 = 0 
<=> (x^2 + 1)^2 - 2(x + 1)^2 = 0 
<=> (x^2 + x*can(2) + 1 + can2))*(x^2 - x*can2) + 1 - can2)) = 0 
<=> x^2 + x*can(2) + 1 + can(2) = 0 (1) 
hoac x^2 - x*can(2) + 1 - can(2) = 0 (2) 
PT (1) vo nghiem 
PT (2) co 2 nghiem phan biet: 
x1 = (can2) + can(-2 + 4can(2)))/2 
x2 = (can(2) + can(-2 + 4can(2)))/2 

a) \(\sqrt[]{x^2-4x+4}=x+3\)

\(\Leftrightarrow\sqrt[]{\left(x-2\right)^2}=x+3\)

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

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

\(\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)

b) \(2x^2-\sqrt[]{9x^2-6x+1}=5\)

\(\Leftrightarrow2x^2-\sqrt[]{\left(3x-1\right)^2}=5\)

\(\Leftrightarrow2x^2-\left|3x-1\right|=5\)

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

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=2x^2-5\\3x-1=-2x^2+5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x-4=0\left(1\right)\\2x^2+3x-6=0\left(2\right)\end{matrix}\right.\)

Giải pt (1)

\(\Delta=9+32=41>0\)

Pt \(\left(1\right)\) \(\Leftrightarrow x=\dfrac{3\pm\sqrt[]{41}}{4}\)

Giải pt (2)

\(\Delta=9+48=57>0\)

Pt \(\left(2\right)\) \(\Leftrightarrow x=\dfrac{-3\pm\sqrt[]{57}}{4}\)

Vậy nghiệm pt là \(\left[{}\begin{matrix}x=\dfrac{3\pm\sqrt[]{41}}{4}\\x=\dfrac{-3\pm\sqrt[]{57}}{4}\end{matrix}\right.\)

hết cứu đi mà làm