c.

ĐKXĐ: \(\left[{}\begin{matrix}x>1\\x< -2\end{matrix}\right.\)

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

\(\Leftrightarrow x+4-2\sqrt[]{\dfrac{x+2}{x-1}}=0\)

\(\Leftrightarrow x+4=2\sqrt[]{\dfrac{x+2}{x-1}}\) (\(x\ge-4\))

\(\Leftrightarrow x^2+8x+16=\dfrac{4\left(x+2\right)}{x-1}\)

\(\Rightarrow x^3+7x^2+4x-24=0\)

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

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

a.

\(\Leftrightarrow2x^2-11x+21=3\sqrt[3]{4\left(x-1\right)}\)

Do \(2x^2-11x+21=2\left(x-\dfrac{11}{4}\right)^2+\dfrac{47}{8}>0\Rightarrow3\sqrt[3]{4\left(x-1\right)}>0\Rightarrow x-1>0\)

Ta có:

\(VT=2x^2-11x+21-3\sqrt[3]{4x-4}=2\left(x^2-6x+9\right)+x+3-3\sqrt[3]{4\left(x-1\right)}\)

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

\(\Rightarrow VT\ge x+3-3\sqrt[3]{4\left(x-1\right)}=\left(x-1\right)+2+2-3\sqrt[3]{4\left(x-1\right)}\)

\(\Rightarrow VT\ge3\sqrt[3]{\left(x-1\right).2.2}-3\sqrt[3]{4\left(x-1\right)}=0\)

Đẳng thức xảy ra khi và chỉ khi:

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

Vậy pt có nghiệm duy nhất \(x=3\)

b.

ĐKXD: \(x\ge-1\)

Phương trình: \(2\left(x+1\right)-\left(3x-2\right)\sqrt[]{x+1}+x^2-x=0\)

Đặt \(\sqrt[]{x+1}=t\ge0\)

\(\Rightarrow2t^2-\left(3x-2\right)t+x^2-x=0\)

\(\Delta=\left(3x-2\right)^2-8\left(x^2-x\right)=\left(x-2\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{3x-2+x-2}{4}=x-1\\t=\dfrac{3x-2-x+2}{4}=\dfrac{x}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt[]{x+1}=x-1\left(x\ge1\right)\\\sqrt[]{x+1}=\dfrac{x}{2}\left(x\ge0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=x^2-2x+1\left(x\ge1\right)\\x+1=\dfrac{x^2}{4}\left(x\ge0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2+2\sqrt[]{2}\end{matrix}\right.\)

a: \(2\cdot sin\left(x+\dfrac{\Omega}{5}\right)+\sqrt{3}=0\)

=>\(2\cdot sin\left(x+\dfrac{\Omega}{5}\right)=-\sqrt{3}\)

=>\(sin\left(x+\dfrac{\Omega}{5}\right)=-\dfrac{\sqrt{3}}{2}\)

=>\(\left[{}\begin{matrix}x+\dfrac{\Omega}{5}=-\dfrac{\Omega}{3}+k2\Omega\\x+\dfrac{\Omega}{5}=\dfrac{4}{3}\Omega+k2\Omega\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-\dfrac{8}{15}\Omega+k2\Omega\\x=\dfrac{4}{3}\Omega-\dfrac{\Omega}{5}+k2\Omega=\dfrac{17}{15}\Omega+k2\Omega\end{matrix}\right.\)

b: \(sin\left(2x-50^0\right)=\dfrac{\sqrt{3}}{2}\)

=>\(\left[{}\begin{matrix}2x-50^0=60^0+k\cdot360^0\\2x-50^0=300^0+k\cdot360^0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}2x=110^0+k\cdot360^0\\2x=350^0+k\cdot360^0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=55^0+k\cdot180^0\\x=175^0+k\cdot180^0\end{matrix}\right.\)

c: \(\sqrt{3}\cdot tan\left(2x-\dfrac{\Omega}{3}\right)-1=0\)

=>\(\sqrt{3}\cdot tan\left(2x-\dfrac{\Omega}{3}\right)=1\)

=>\(tan\left(2x-\dfrac{\Omega}{3}\right)=\dfrac{1}{\sqrt{3}}\)

=>\(2x-\dfrac{\Omega}{3}=\dfrac{\Omega}{6}+k2\Omega\)

=>\(2x=\dfrac{1}{2}\Omega+k2\Omega\)

=>\(x=\dfrac{1}{4}\Omega+k\Omega\)

\(a,\) Đặt \(x^2+2x=a\), pt trở thành:

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

\(\left[{}\begin{matrix}\Delta\left(1\right)=4+4=8\\\Delta\left(2\right)=4+8=12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{-2-\sqrt{8}}{2}\\x=\dfrac{-2+\sqrt{8}}{2}\end{matrix}\right.\\\left[{}\begin{matrix}x=\dfrac{-2-\sqrt{12}}{2}\\x=\dfrac{-2+\sqrt{12}}{2}\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1-\sqrt{2}\\x=-1+\sqrt{2}\\x=-1-\sqrt{3}\\x=-1+\sqrt{3}\end{matrix}\right.\)

\(b,\) Đặt \(x^2+x=b\), pt trở thành:

\(b\left(b+1\right)-6=0\\ \Leftrightarrow b^2+b-6=0\\ \Leftrightarrow\left[{}\begin{matrix}b=2\\b=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+x-2=0\\x^2+x+3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\\x\in\varnothing\left[x^2+x+3=\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}>0\right]\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

\(d,x^4-2x^3+x=2\\ \Leftrightarrow x^4-2x^3+x-2=0\\\Leftrightarrow\left(x^3+1\right)\left(x-2\right)=0 \\ \Leftrightarrow\left(x+1\right)\left(x-2\right)\left(x^2+x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\\x^2+x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\\x\in\varnothing\left[x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\right]\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)

Lời giải:

a. 

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

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

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

$\Leftrightarrow x^2+2x-1=0$ hoặc $x^2+2x-2=0$

$\Leftrightarrow x=-1\pm \sqrt{2}$ hoặc $x=-1\pm \sqrt{3}$

b.

PT $\Leftrightarrow (x^2+x)^2+(x^2+x)-6=0$

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

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

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

$\Leftrightarrow x^2+x-2=0$ (chọn) hoặc $x^2+x+3=0$ (loại do $x^2+x+3=(x+0,5)^2+2,75>0$)

$\Leftrightarrow x=-1\pm \sqrt{3}$

c. Nghiệm khá xấu. Bạn coi lại đề.

d.

PT $\Leftrightarrow x^3(x-2)+(x-2)=0$

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

$\Leftrightarrow x^3+1=0$ hoặc $x-2=0$

$\Leftrightarrow x=-1$ hoặc $x=2$

d: Ta có: \(x^4-2x^3+x=2\)

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

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

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

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

a.

$x^2-11=0$

$\Leftrightarrow x^2=11$

$\Leftrightarrow x=\pm \sqrt{11}$

b. $x^2-12x+52=0$

$\Leftrightarrow (x^2-12x+36)+16=0$

$\Leftrightarrow (x-6)^2=-16< 0$ (vô lý)

Vậy pt vô nghiệm.

c.

$x^2-3x-28=0$

$\Leftrightarrow x^2+4x-7x-28=0$

$\Leftrightarrow x(x+4)-7(x+4)=0$

$\Leftrightarrow (x+4)(x-7)=0$

$\Leftrightarrow x+4=0$ hoặc $x-7=0$

$\Leftrightarrow x=-4$ hoặc $x=7$

d.

$x^2-11x+38=0$

$\Leftrightarrow (x^2-11x+5,5^2)+7,75=0$

$\Leftrightarrow (x-5,5)^2=-7,75< 0$ (vô lý)

Vậy pt vô nghiệm

e.

$6x^2+71x+175=0$

$\Leftrightarrow 6x^2+21x+50x+175=0$

$\Leftrightarrow 3x(2x+7)+25(2x+7)=0$

$\Leftrightarrow (3x+25)(2x+7)=0$

$\Leftrightarrow 3x+25=0$ hoặc $2x+7=0$

$\Leftrightarrow x=-\frac{25}{3}$ hoặc $x=-\frac{7}{2}$

f.

$x^2-(\sqrt{2}+\sqrt{8})x+4=0$

$\Leftrightarrow x^2-\sqrt{2}x-2\sqrt{2}x+4=0$

$\Leftrightarrow x(x-\sqrt{2})-2\sqrt{2}(x-\sqrt{2})=0$

$\Leftrightarrow (x-\sqrt{2})(x-2\sqrt{2})=0$

$\Leftrightarrow x-\sqrt{2}=0$ hoặc $x-2\sqrt{2}=0$

$\Leftrightarrow x=\sqrt{2}$ hoặc $x=2\sqrt{2}$

g.

$(1+\sqrt{3})x^2-(2\sqrt{3}+1)x+\sqrt{3}=0$

$\Leftrightarrow (1+\sqrt{3})x^2-(1+\sqrt{3})x-(\sqrt{3}x-\sqrt{3})=0$

$\Leftrightarrow (1+\sqrt{3})x(x-1)-\sqrt{3}(x-1)=0$

$\Leftrightarrow (x-1)[(1+\sqrt{3})x-\sqrt{3}]=0$

$\Leftrightarrow x-1=0$ hoặc $(1+\sqrt{3})x-\sqrt{3}=0$

$\Leftrightarrow x=1$ hoặc $x=\frac{3-\sqrt{3}}{2}$

1.

\(\Leftrightarrow cos\left(2x+\dfrac{4\pi}{3}\right)=0\)

\(\Leftrightarrow2x+\dfrac{4\pi}{3}=\dfrac{\pi}{2}+k\pi\)

\(\Leftrightarrow2x=-\dfrac{5\pi}{6}+k\pi\)

\(\Leftrightarrow x=-\dfrac{5\pi}{12}+\dfrac{k\pi}{2}\)

b.

\(\Leftrightarrow2+2cos\left(2x+\dfrac{\pi}{3}\right)-3=0\)

\(\Leftrightarrow cos\left(2x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)

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

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

c.

\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{6}\right)=\dfrac{\sqrt{3}}{2}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k\pi\\x=k\pi\end{matrix}\right.\)

\(1.\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}.\Leftrightarrow\dfrac{x-1-3x}{3}=\dfrac{x-2}{2}.\Leftrightarrow\dfrac{-2x-1}{3}-\dfrac{x-2}{2}=0.\)

\(\Leftrightarrow\dfrac{-4x-2-3x+6}{6}=0.\Rightarrow-7x+4=0.\Leftrightarrow x=\dfrac{4}{7}.\)

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

\(\Leftrightarrow\left(x-2\right)\left(2x-1-x\right)=0.\Leftrightarrow\left(x-2\right)\left(x-1\right)=0.\)

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

\(3.3x^2-4x+1=0.\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{3}\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=\dfrac{1}{3}.\end{matrix}\right.\)

\(4.\left|2x-4\right|=0.\Leftrightarrow2x-4=0.\Leftrightarrow x=2.\)

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

\(1,\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\\ \Leftrightarrow\dfrac{x-1}{3}-x=\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{2\left(x-1\right)-6x}{6}=\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow2\left(x-1\right)-6x=3\left(x-2\right)\\ \Leftrightarrow2x-2-6x=3x-6\\ \Leftrightarrow-4x-2=3x-6\)

\(\Leftrightarrow3x-6+4x+2=0\\ \Leftrightarrow7x-4=0\\ \Leftrightarrow x=\dfrac{4}{7}\)

\(2,\left(x-2\right)\left(2x-1\right)=x^2-2x\\ \Leftrightarrow2x^2-4x-x+2=x^2-2x\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

\(3,3x^2-4x+1=0\\ \Leftrightarrow\left(3x^2-3x\right)-\left(x-1\right)=0\\ \Leftrightarrow3x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(4,\left|2x-4\right|=0\\ \Leftrightarrow2x-4=0\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\)

\(5,\left|3x+2\right|=4\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)

\(6,\left|2x-5\right|=\left|-x+2\right|\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=-x+2\\2x-5=x-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=7\\x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)

a:

ĐKXĐ: \(x\notin\left\{\dfrac{3}{2};1\right\}\)

 \(y=\dfrac{\left(x-2\right)^2}{\left(2x-3\right)\left(x-1\right)}=\dfrac{x^2-4x+4}{2x^2-2x-3x+3}\)

=>\(y=\dfrac{x^2-4x+4}{2x^2-5x+3}\)

=>\(y'=\dfrac{\left(x^2-4x+4\right)'\left(2x^2-5x+3\right)-\left(x^2-4x+4\right)\left(2x^2-5x+3\right)'}{\left(2x^2-5x+3\right)^2}\)

=>\(y'=\dfrac{\left(2x-4\right)\left(2x^2-5x+3\right)-\left(2x-5\right)\left(x^2-4x+4\right)}{\left(2x^2-5x+3\right)^2}\)

=>\(y'=\dfrac{4x^3-10x^2+6x-8x^2+20x-12-2x^3+8x^2-8x+5x^2-20x+20}{\left(2x^2-5x+3\right)^2}\)

=>\(y'=\dfrac{2x^3-5x^2-2x+8}{\left(2x^2-5x+3\right)^2}\)

b:

ĐKXĐ: x<>-3

 \(y=\left(x+3\right)+\dfrac{4}{x+3}\)

=>\(y'=\left(x+3+\dfrac{4}{x+3}\right)'=1+\left(\dfrac{4}{x+3}\right)'\)

\(=1+\dfrac{4'\left(x+3\right)-4\left(x+3\right)'}{\left(x+3\right)^2}\)

=>\(y'=1+\dfrac{-4}{\left(x+3\right)^2}=\dfrac{\left(x+3\right)^2-4}{\left(x+3\right)^2}\)

y'=0

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

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

=>(x+5)(x+1)=0

=>x=-5 hoặc x=-1

c:

ĐKXĐ: x<>-2

 \(y=\dfrac{\left(5x-1\right)\left(x+1\right)}{x+2}\)

=>\(y=\dfrac{5x^2+5x-x-1}{x+2}=\dfrac{5x^2+4x-1}{x+2}\)

=>\(y'=\dfrac{\left(5x^2+4x-1\right)'\left(x+2\right)-\left(5x^2+4x-1\right)\left(x+2\right)'}{\left(x+2\right)^2}\)

=>\(y'=\dfrac{\left(5x+4\right)\left(x+2\right)-\left(5x^2+4x-1\right)}{\left(x+2\right)^2}\)

=>\(y'=\dfrac{5x^2+10x+4x+8-5x^2-4x+1}{\left(x+2\right)^2}\)

=>\(y'=\dfrac{10x+9}{\left(x+2\right)^2}\)

\(y'\left(-1\right)=\dfrac{10\cdot\left(-1\right)+9}{\left(-1+2\right)^2}=\dfrac{-1}{1}=-1\)

d: 

ĐKXĐ: x<>2

\(y=x-2+\dfrac{9}{x-2}\)

=>\(y'=\left(x-2+\dfrac{9}{x-2}\right)'=1+\left(\dfrac{9}{x-2}\right)'\)

\(=1+\dfrac{9'\left(x-2\right)-9\left(x-2\right)'}{\left(x-2\right)^2}\)

=>\(y'=1+\dfrac{-9}{\left(x-2\right)^2}=\dfrac{\left(x-2\right)^2-9}{\left(x-2\right)^2}\)

y'=0

=>\(\dfrac{\left(x-2\right)^2-9}{\left(x-2\right)^2}=0\)

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

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

=>(x-5)(x+1)=0

=>x=5 hoặc x=-1