a, ĐKXĐ: ...

\(\sqrt{3x^2-2x+6}+3-2x=0\)

\(\Leftrightarrow\sqrt{3x^2-2x+6}=2x-3\)

\(\Leftrightarrow3x^2-2x+6=4x^2-12x+9\)

\(\Leftrightarrow4x^2-10x+3=0\)

.....

b, ĐKXĐ: ...

\(\sqrt{x+1}+\sqrt{x-1}=4\\ \Leftrightarrow x+1+x-1+2\sqrt{\left(x+1\right)\left(x-1\right)}=16\\ \Leftrightarrow2\sqrt{x^2-1}=16-2x\\ \Leftrightarrow\sqrt{x^2-1}=8-x\\ \Leftrightarrow x^2-1=64-16x+x^2\\ \Leftrightarrow65-16x=0\\ \Leftrightarrow x=\dfrac{65}{16}\)

ta có : x^5+2x^4+3x^3+3x^2+2x+1=0

\(\Leftrightarrow\)x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0

\(\Leftrightarrow\)(x^5+x^4)+(x^4+x^3)+(2x^3+2x^2)+(x^2+x)+(x+1)=0

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

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

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

\(\Leftrightarrow\)(x+1)[x^2(x^2+x+1)+(x^2+x+1)]=0

\(\Leftrightarrow\)(x+1)(x^2+x+1)(x^2+1)=0

VÌ x^2+x+1=(x+\(\dfrac{1}{2}\))^2+\(\dfrac{3}{4}\)\(\ne0\) và x^2+1\(\ne0\)

\(\Rightarrow\)x+1=0

\(\Rightarrow\)x=-1

CÒN CÂU B TỰ LÀM (02042006)

b: x^4+3x^3-2x^2+x-3=0

=>x^4-x^3+4x^3-4x^2+2x^2-2x+3x-3=0

=>(x-1)(x^3+4x^2+2x+3)=0

=>x-1=0

=>x=1

cái này dễ đợi tí mình giải cho, gõ đáp số mất khá nhiều thời gian

\(\frac{x}{2x-6}+\frac{x}{2x+2}-\frac{2x}{\left(x+1\right).\left(x-3\right)}=0\)

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

\(\Leftrightarrow\frac{\left(x+1\right).x}{2.\left(x-3\right).\left(x+1\right)}+\frac{x.\left(x-3\right)}{2.\left(x+1\right).\left(x-3\right)}-\frac{4x}{2.\left(x+1\right).\left(x-3\right)}=0\)

tự làm tiếp nha bạn :)))

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

\(x=3;x=2;x=-1;x=-\frac{1}{2}\)

\(\frac{x-2}{18}-\frac{2x+5}{12}>\frac{x+6}{9}-\frac{x-3}{6}\)

\(\Leftrightarrow\frac{2\left(x-2\right)}{36}-\frac{3\left(2x+5\right)}{36}>\frac{4\left(x+6\right)}{36}-\frac{6\left(x-3\right)}{36}\)

\(\Leftrightarrow2x-4-6x-15>4x+24-6x+18\)

\(\Leftrightarrow2x-6x-4x+6x>24+18+4+15\)

\(\Leftrightarrow-2x>61\)

\(\Leftrightarrow x< -\frac{61}{2}\)

Vậy nghiệm của bất phương trình là \(x< -\frac{61}{2}\)

Bài b và c làm cách mình thì dễ hiểu hơn nhiều :3

\(\left(2x-2\right)\left(2x+3\right)\le0\)

TH1 : \(\hept{\begin{cases}2x-3\le0\\2x+3\ge0\end{cases}< =>\hept{\begin{cases}2x\le3\\2x\ge-3\end{cases}}}\)

\(< =>\hept{\begin{cases}x\le\frac{3}{2}\\x\ge-\frac{3}{2}\end{cases}}\)

TH2 : \(\hept{\begin{cases}2x-3\ge0\\2x+3\le0\end{cases}< =>\hept{\begin{cases}2x\ge3\\2x\le-3\end{cases}}}\)

\(< =>\hept{\begin{cases}x\ge\frac{3}{2}\\x\le-\frac{3}{2}\end{cases}}\)

Vậy ...

\(\left(3-2x\right)\left(4x+8\right)\ge0\)

TH1 : \(\hept{\begin{cases}3-2x\ge0\\4x+8\ge0\end{cases}}\)

\(< =>\hept{\begin{cases}3\ge2x\\4x\ge-8\end{cases}< =>\hept{\begin{cases}\frac{3}{2}\ge x\\x\ge-\frac{8}{4}=-2\end{cases}}}\)

TH2 : \(\hept{\begin{cases}3-2x\le0\\4x+8\le0\end{cases}}\)

\(< =>\hept{\begin{cases}3\le2x\\4x\le-8\end{cases}}< =>\hept{\begin{cases}x\ge\frac{3}{2}\\x\ge-2\end{cases}}\)

Vậy ...

(x - 1)(2x² - 10) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x^2-10=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\2x^2=10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x^2=5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\sqrt{5}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là: \(S=\left\{1;\sqrt{5}\right\}\)
(2x - 7)² - 6(2x - 7)(x - 3) = 0

\(\Leftrightarrow\left(2x-7\right)\left(2x-7-6x+18\right)=0\)

\(\Leftrightarrow\left(2x-7\right)\left(11-4x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-7=0\\11-4x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=7\\4x=11\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=\frac{11}{4}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là: \(S=\left\{\frac{7}{2};\frac{11}{4}\right\}\)
(5x + 3)(x² + 4) = 0

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

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

\(\Leftrightarrow x=-\frac{3}{5}\)

Vậy phương trình có tập nghiệm là: \(S=\left\{-\frac{3}{5}\right\}\)

a)

\(\left(x-1\right)\cdot\left(2x^2-10\right)=0\\ \Leftrightarrow\left(x-1\right)\cdot2\cdot\left(x^2-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-1=0\\x^2-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=\pm\sqrt{5}\end{matrix}\right.\)

b)

\(\left(2x-7\right)^2-6\cdot\left(6x-7\right)\cdot\left(x-3\right)=0\\ \Leftrightarrow\left(2x-7\right)\cdot\left[\left(2x-7\right)-6\cdot\left(x-3\right)\right]=0\\ \Leftrightarrow\left(2x-7\right)\cdot\left(2x-7-6x+18\right)=0\\ \Leftrightarrow\left(2x-7\right)\cdot\left(11-4x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-7=0\\11-4x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=\frac{11}{4}\end{matrix}\right.\)

c)

\(\left(5x+3\right)\cdot\left(x^2+4\right)=0\)

Vì \(\left(x^2+4\right)>0\Rightarrow\left(loại\right)\)

\(\Rightarrow5x+3=0\\ \Rightarrow x=-\frac{3}{5}\)

a, \(cos^2x-cosx=0\)

\(\Leftrightarrow cosx\left(cosx-1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=0\end{matrix}\right.\)

b, \(2sin2x+\sqrt{2}sin4x=0\)

\(\Leftrightarrow2sin2x+2\sqrt{2}sin2x.cos2x=0\)

\(\Leftrightarrow sin2x\left(1+\sqrt{2}cos2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\1+\sqrt{2}cos2x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\\cos2x=-\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\\2x=\dfrac{3\pi}{4}+k2\pi\\2x=\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\\x=\dfrac{3\pi}{8}+k\pi\\x=\dfrac{\pi}{8}+k\pi\end{matrix}\right.\)

a, \(cos^2x-cosx=0\)

\(\Leftrightarrow cosx\left(cosx-1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=k2\pi\end{matrix}\right.\) (k ∈ Z)

Vậy...

b, \(2sin2x+\sqrt{2}sin4x=0\)

\(\Leftrightarrow2sin2x+2\sqrt{2}sin2x.cos2x=0\)

\(\Leftrightarrow2sin2x\left(1+\sqrt{2}cos2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\cos2x=\dfrac{-\sqrt{2}}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x=k\pi\\2x=\pm\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\\x=\pm\dfrac{3\pi}{8}+k\pi\end{matrix}\right.\)

Vậy...

c, \(8cos^2x+2sinx-7=0\)

\(\Leftrightarrow8\left(1-sin^2x\right)+2sinx-7=0\)

\(\Leftrightarrow8sin^2x-2sinx-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\sinx=-\dfrac{1}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\\x=arcsin\left(-\dfrac{1}{4}\right)+k2\pi\\x=\pi-arcsin\left(-\dfrac{1}{4}\right)+k2\pi\end{matrix}\right.\)

Vậy...

d, \(4cos^4x+cos^2x-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos^2x=\dfrac{3}{4}\\cos^2x=-1\left(loai\right)\end{matrix}\right.\) 

\(\Leftrightarrow\dfrac{cos2x+1}{2}=\dfrac{3}{4}\)

\(\Leftrightarrow cos2x=\dfrac{1}{2}\)

\(\Leftrightarrow2x=\pm\dfrac{\pi}{3}+k2\pi\)

\(\Leftrightarrow x=\pm\dfrac{\pi}{6}+k\pi\)

Vậy...

e, \(\sqrt{3}tanx-6cotx+\left(2\sqrt{3}-3\right)=0\) (ĐK: \(x\ne\dfrac{k\pi}{2}\))

\(\Leftrightarrow\sqrt{3}tanx-\dfrac{6}{tanx}+\left(2\sqrt{3}-3\right)=0\)

\(\Leftrightarrow\sqrt{3}tan^2x+\left(2\sqrt{3}-3\right)tanx-6=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=\sqrt{3}\\tanx=-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k\pi\left(tm\right)\\x=arctan\left(-2\right)+k\pi\end{matrix}\right.\)

Vậy...

c, \(8cos^2x+2sinx-7=0\)

\(\Leftrightarrow-8sin^2x+2sinx+1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\sinx=-\dfrac{1}{4}\end{matrix}\right.\)

Với \(sinx=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

Với \(sinx=-\dfrac{1}{4}\Leftrightarrow\left[{}\begin{matrix}x=arcsin\left(-\dfrac{1}{4}\right)+k2\pi\\x=\pi-arcsin\left(-\dfrac{1}{4}\right)+k2\pi\end{matrix}\right.\)

d, \(4cos^4x+cos^2x-3=0\)

\(\Leftrightarrow\left(4cos^2x-3\right)\left(cos^2x+1\right)=0\)

\(\Leftrightarrow4cos^2x-3=0\left(\text{Vì }cos^2x+1>0\right)\)

\(\Leftrightarrow cos^2x=\dfrac{3}{4}\)

\(\Leftrightarrow cosx=\pm\dfrac{\sqrt{3}}{2}\)

Với \(cosx=\dfrac{\sqrt{3}}{2}\Leftrightarrow x=\pm\dfrac{\pi}{3}+k2\pi\)

Với \(cosx=-\dfrac{\sqrt{3}}{2}\Leftrightarrow x=\pm\dfrac{5\pi}{6}+k2\pi\)