a) Ta có: \(x^2-9x+20=0\)

\(\Leftrightarrow x^2-5x-4x+20=0\)

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

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

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

Vậy: x∈{4;5}

b) Ta có: \(x^3-4x^2+5x=0\)

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

Ta có: \(x^2-4x+5\)

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

Ta có: \(\left(x-2\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-2\right)^2+1\ge1>0\forall x\)

hay \(x^2-4x+5>0\forall x\)(2)

Từ (1) và (2) suy ra x=0

Vậy: x=0

c) Sửa đề: \(x^2-2x-15=0\)

Ta có: \(x^2-2x-15=0\)

\(\Leftrightarrow x^2+3x-5x-15=0\)

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

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

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

Vậy: x∈{-3;5}

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

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

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

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

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

\(\Leftrightarrow x\cdot\left[x\left(x^2+2x+2\right)-2\left(x^2+2x+2\right)\right]=0\)

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

Ta có: \(x^2+2x+2\)

\(=x^2+2x+1+1=\left(x+1\right)^2+1\)

Ta có: \(\left(x+1\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+1\right)^2+1\ge1>0\forall x\)

hay \(x^2+2x+2>0\forall x\)(4)

Từ (3) và (4) suy ra

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

Vậy: x∈{0;2}

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

\(\Leftrightarrow x^3-9x^2+27x-27+x^3+9x^2+27x+27=0\)\(\Leftrightarrow2x^3+54x^2=0\)

\(\Leftrightarrow x^2\left(2x+54\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\2x+54=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-27\end{matrix}\right.\)

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

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1=0\)\(\Leftrightarrow6x^2+2=0\)

\(\Leftrightarrow6x^2=-2\)

\(\Leftrightarrow x^2=-3\) ( vô lí)

Vậy pt vô nghiệm

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

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

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

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

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

\(d,4x^2+4x+1=0\)

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

\(\Rightarrow2x+1=0\)

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

\(e,\left(x+2\right)^2-\left(x+3\right)^2=0\)

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

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

\(\Leftrightarrow-2x-5=0\)

\(\Leftrightarrow-2x=5\Rightarrow x=-\dfrac{5}{2}\)

Học tốt nha you <3

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

\(\Leftrightarrow x^3-9x^2+27x-27+x^3+9x^2+27x+27=0\)\(\Leftrightarrow2x^3+54x^2=0\)

\(\Leftrightarrow x^2\left(2x+54\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\2x+54=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-27\end{matrix}\right.\)

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

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1=0\)\(\Leftrightarrow6x^2+2=0\)

\(\Leftrightarrow6x^2=-2\)

\(\Leftrightarrow x^2=-3\) ( vô lí)

Vậy pt vô nghiệm

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

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

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

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

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

\(d,4x^2+4x+1=0\)

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

\(\Rightarrow2x+1=0\)

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

\(e,\left(x+2\right)^2-\left(x+3\right)^2=0\)

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

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

\(\Leftrightarrow-2x-5=0\)

\(\Leftrightarrow-2x=5\Rightarrow x=-\dfrac{5}{2}\)

Học tốt nha you <3

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

\(\Leftrightarrow x^3-9x^2+27x-27+x^3+9x^2+27x+27=0\)\(\Leftrightarrow2x^3+54x^2=0\)

\(\Leftrightarrow x^2\left(2x+54\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\2x+54=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-27\end{matrix}\right.\)

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

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1=0\)\(\Leftrightarrow6x^2+2=0\)

\(\Leftrightarrow6x^2=-2\)

\(\Leftrightarrow x^2=-3\) ( vô lí)

Vậy pt vô nghiệm

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

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

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

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

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

\(d,4x^2+4x+1=0\)

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

\(\Rightarrow2x+1=0\)

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

\(e,\left(x+2\right)^2-\left(x+3\right)^2=0\)

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

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

\(\Leftrightarrow-2x-5=0\)

\(\Leftrightarrow-2x=5\Rightarrow x=-\dfrac{5}{2}\)

Học tốt nha you <3

a: \(x^3-4x^2-x+4=0\)

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

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

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

=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)

b: Sửa đề: \(x^3+3x^2+3x+1=0\)

=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)

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

=>x+1=0

=>x=-1

c: \(x^3+3x^2-4x-12=0\)

=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)

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

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

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

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

d: \(\left(x-2\right)^2-4x+8=0\)

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

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

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

=>(x-2)(x-6)=0

=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

1. \(\sqrt{x^2-4x+3}=x-2\)

<=> x² - 4x + 3 = (x - 2)²

<=> x² - 4x + 3 = x² - 4x + 4

<=> x² - x² - 4x + 4x = 1

<=> 0 = 1 (Vô lí)

vậy PT có nghiệm là S = \(\varnothing\)

2. \(\sqrt{4x^2-4x+1}=x-1\)

<=> \(\sqrt{\left(2x-1\right)^2}=x-1\)

<=> 2x - 1 = x - 1

<=> 2x - x = -1 + 1

<=> x = 0

1: ta có: \(\sqrt{x^2-4x+3}=x-2\)

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

2: Ta có: \(\sqrt{4x^2-4x+1}=x-1\)

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

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

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

1.

\(x^4-6x^2-12x-8=0\)

\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)

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

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

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

\(\Leftrightarrow x=1\pm\sqrt{5}\)

3.

ĐK: \(x\ge-9\)

\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)

\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)

Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)

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

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

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

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

\(\Leftrightarrow...\)

2.

ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)

\(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{1}{2x+\dfrac{1}{x}+4}+\dfrac{1}{2x+\dfrac{1}{x}-4}=\dfrac{3}{5}\)

Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)

\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)

Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:

\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)

\(\Leftrightarrow10b+40=3\left(b+8\right)b\)

\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)

TH1: \(b=2\Leftrightarrow...\)

TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)

a) Ta có: \(x^2+3x-10=0\)

\(\Leftrightarrow x^2+5x-2x-10=0\)

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

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

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

Vậy: S={-5;2}

b) Ta có: \(3x^2-7x+1=0\)

\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{1}{3}\right)=0\)

mà 3>0

nên \(x^2-\dfrac{7}{3}x+\dfrac{1}{3}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}-\dfrac{37}{36}=0\)

\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=\dfrac{37}{36}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{7}{6}=\dfrac{\sqrt{37}}{6}\\x-\dfrac{7}{6}=-\dfrac{\sqrt{37}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{37}+7}{6}\\x=\dfrac{-\sqrt{37}+7}{6}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{\sqrt{37}+7}{6};\dfrac{-\sqrt{37}+7}{6}\right\}\)

c) Ta có: \(3x^2-7x+8=0\)

\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{8}{3}\right)=0\)

mà 3>0

nên \(x^2-\dfrac{7}{3}x+\dfrac{8}{3}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}+\dfrac{47}{36}=0\)

\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=-\dfrac{47}{36}\)(vô lý)

Vậy: \(x\in\varnothing\)

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