b: 4x^2-20x+25=(x-3)^2

=>(2x-5)^2=(x-3)^2

=>(2x-5)^2-(x-3)^2=0

=>(2x-5-x+3)(2x-5+x-3)=0

=>(3x-8)(x-2)=0

=>x=8/3 hoặc x=2

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

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

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

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

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

=>\(x\in\left\{0;1;-1\right\}\)

d: 2x^3+3x^2+2x+3=0

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

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

=>2x+3=0

=>x=-3/2

a: =>x^2(5x-7)-3(5x-7)=0

=>(5x-7)(x^2-3)=0

=>\(x\in\left\{\dfrac{7}{5};\sqrt{3};-\sqrt{3}\right\}\)

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

hay \(x\in\left\{-\dfrac{2}{3};\dfrac{-1+\sqrt{17}}{4};\dfrac{-1-\sqrt{17}}{4}\right\}\)

c: \(\Leftrightarrow\left[{}\begin{matrix}3x^2+10x+21=x^2-20x-9\\3x^2+10x+21=-x^2+20x+9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x^2+30x+30=0\\4x^2-10x+12=0\end{matrix}\right.\Leftrightarrow x\in\left\{\dfrac{-15+\sqrt{165}}{2};\dfrac{-15-\sqrt{165}}{2}\right\}\)

a)

\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+21}=5-2x-x^2\)

\(\Leftrightarrow\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+16}=6-\left(x+1\right)^2\)

\(VT\ge6;VP\le6\Rightarrow VT=VP=6\)

Vậy pt có một nghiệm duy nhất là \(x=-1\)

b)

\(\sqrt{4x^2+20x+25}+\sqrt{x^2-8x+16}=\sqrt{x^2+18x+81}\)

\(\Leftrightarrow\sqrt{\left(2x+5\right)^2}+\sqrt{\left(x-4\right)^2}=\sqrt{\left(x+9\right)^2}\)

\(\Leftrightarrow\left|2x+5\right|+\left|x-4\right|=\left|x+9\right|\)

Lập bảng xét dấu ra nhé ~^o^~

\(PT\Leftrightarrow\left(\sqrt{4x^2-20x+28}-2\right)=3x^2-15x+18\\ \Leftrightarrow\dfrac{4x^2-20x+24}{\sqrt{4x^2-20x+28}+2}=3\left(x-2\right)\left(x-3\right)\\ \Leftrightarrow\dfrac{4\left(x-2\right)\left(x-3\right)}{\sqrt{4x^2-20x+28}+2}-3\left(x-2\right)\left(x-3\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-3\right)\left(\dfrac{4}{\sqrt{4x^2-20x+28}+2}-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\\\dfrac{4}{\sqrt{4x^2-20x+28}+2}-3=0\left(1\right)\end{matrix}\right.\)

Vì \(\dfrac{4}{\sqrt{4x^2-20x+28}+2}\le2\Leftrightarrow\dfrac{4}{\sqrt{4x^2-20x+28}+2}-3\le-1< 0\)

Do đó \(\left(1\right)\) vô nghiệm

Vậy PT có nghiệm \(x=2;x=3\)

\(\sqrt{4x^2-20x+25}+2x=5\\ < =>\sqrt{\left(2x-5\right)^2}+2x=5\\ < =>\left|2x-5\right|+2x=5 \\ < =>\left[{}\begin{matrix}2x-5+2x=5\left(x\ge\dfrac{5}{2}\right)\\2x-5+2x=-5\left(x< \dfrac{5}{3}\right)\end{matrix}\right.< =>\left[{}\begin{matrix}4x=10< =>x=\dfrac{5}{2}\left(tmdk\right)\\4x=0< =>x=0\left(ktmdk\right)\end{matrix}\right.\\ =>x=\dfrac{5}{2}\)

\(\sqrt{\left(5-2x\right)^2}=5-2x\)

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

\(\Leftrightarrow5-2x\ge0\) (tính chất: \(\left|A\right|=A\Leftrightarrow A\ge0\))

\(\Leftrightarrow x\le\dfrac{5}{2}\)

Vậy nghiệm của pt là \(x\le\dfrac{5}{2}\)

\(\sqrt{4x^2-20x+25}+2x=5\left(đk:x\le\dfrac{5}{2}\right)\)

\(\Leftrightarrow\sqrt{\left(2x-5\right)^2}=5-2x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{5}{2}\\2x-5=2x-5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\in R\\x\le\dfrac{5}{2}\end{matrix}\right.\)

refer

https://lazi.vn/edu/exercise/1000869/giai-phuong-trinh-x4-x2-6-0

ta cho x⁴ là x² ta có pt:

x²-x-6=0

\(\Rightarrow\left\{{}\begin{matrix}x_1=3\\x_2=-2\end{matrix}\right.\)

\(x^4+9x^2=0\left(1\right)\\ < =>x^2\left(x^2+9\right)=0\\ < =>\left[{}\begin{matrix}x^2=0\\x^2+9=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x^2+9=0\left(2\right)\end{matrix}\right.\)

có

\(x^2\ge0\forall x\\ =>x^2+9>0\)

mâu thuẫn với (2)

=> (2) vô nghiệm

vậy ...

Ta có : x⁴+3x³+4x²+3x+1=0
⇔ ( x⁴ + x³ ) + ( 2x³ + 2x² ) + ( 2x² + 2x ) + ( x + 1 ) = 0

⇔ x³ ( x + 1 ) + 2x² ( x + 1 ) + 2x ( x+1 ) + ( x + 1 ) =0

⇔  ( x + 1 ) ( x³ + 2x² + 2x + 1 ) = 0

⇔ ( x + 1 ) [ ( x³ + 1 ) + ( 2x² + 2x ) ] = 0

⇔ ( x + 1 ) [ (x + 1 ) ( x² - x +1 ) + 2x ( x + 1 ) ] =0

⇔ ( x +1 ) ( x + 1 ) ( x² + x +1 ) =0
⇒ \(\left[{}\begin{matrix}x+1=0\\x^{2^{ }}+x+1=0\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}x=-1\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(VoLy\right)\end{matrix}\right.\)

Vậy x = -1

x⁴+3x³+4x²+3x+1=0

⇔(x⁴+2x³+x²)+(x³+2x²+1)+(x²+2x+1)=0

⇔x²(x²+2x+1)+x(x²​+2x+1)+(x²​+2x+1)=0

⇔x²(x+1)²+x(x+1)²+(x+1)²=0

⇔(x+1)²(x²+x+1)=0

Vì x²+x+1=x²+x+\(\dfrac{1}{4}\)+\(\dfrac{3}{4}\)=(x+\(\dfrac{1}{2}\))²+\(\dfrac{3}{4}\)>0 nên phương trình đã cho tương đương:

(x+1)²=0 ⇔(x+1)(x+1)=0 ⇔x=-1.