Lời giải:
PT $\Leftrightarrow \frac{x-342}{15}-1+\frac{x-323}{17}-2+\frac{x-300}{19}-3+\frac{x-273}{21}-4=0$

$\Leftrightarrow \frac{x-357}{15}+\frac{x-357}{17}+\frac{x-357}{19}+\frac{x-357}{21}=0$

$(x-357)\left(\frac{1}{15}+\frac{1}{17}+\frac{1}{19}+\frac{1}{21}\right)=0$

Dễ thấy: $\frac{1}{15}+\frac{1}{17}+\frac{1}{19}+\frac{1}{21}\neq 0$

$\Rightarrow x-357=0$

$\Rightarrow x=357$

\(\dfrac{x-342}{15}+\dfrac{x-323}{17}+\dfrac{x-300}{19}+\dfrac{x-273}{21}=10\)

<=>\(\dfrac{x-342}{15}-1+\dfrac{x-323}{17}-2+\dfrac{x-300}{19}-3+\dfrac{x-273}{21}-4=0\)

<=>\(\dfrac{x-357}{15}+\dfrac{x-357}{17}+\dfrac{x-357}{19}+\dfrac{x-357}{21}=0\)

<=>\(\left(x-357\right)\left(\dfrac{1}{15}+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{21}\right)=0\)

vì 1/15+1/17+1/19+1/21 khác 0=>x-357=0<=>x=357

vậy..................

chúc bạn học tốt ^^

\(2x^4+3x^3+8x^2+6x+5=0\)

\(\Leftrightarrow2x^4+2x^3+2x^2+x^3+x^2+x+5x^2+5x+5=0\)

\(\Leftrightarrow2x^2\left(x^2+x+1\right)+x\left(x^2+x+1\right)+5\left(x^2+x+1\right)=0\)

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

Mà \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)

\(2x^2+x+5=2\left[\left(x+\frac{1}{4}\right)^2+\frac{39}{16}\right]>0\forall x\)

Vậy tập nghiệm của pt là \(S=\varnothing\)

b, \(\frac{x-342}{15}+\frac{x-323}{17}+\frac{x-300}{19}+\frac{x-273}{21}=10\)

\(\Leftrightarrow\left(\frac{x-342}{15}-1\right)+\left(\frac{x-323}{17}-2\right)+\left(\frac{x-300}{19}-3\right)+\left(\frac{x-273}{21}-4\right)=0\)

\(\Leftrightarrow\frac{x-357}{15}+\frac{x-357}{17}+\frac{x-357}{19}+\frac{x-357}{21}=0\)

\(\Leftrightarrow\left(x-357\right)\left(\frac{1}{15}+\frac{1}{17}+\frac{1}{19}+\frac{1}{21}\right)=0\)

\(\Leftrightarrow x-357=0\Leftrightarrow x=357\) 

Vậy tập nghiệm của pt: \(S=\left\{357\right\}\)

Câu a)

\(2x^4+3x^3+8x^2+6x+5=0\)

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

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

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

\(\Rightarrow \left[\begin{matrix} x^2+x+1=0\\ 2x^2+x+5=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} (x+\frac{1}{2})^2+\frac{3}{4}=0\\ 2(x+\frac{1}{4})^2+\frac{39}{8}=0\end{matrix}\right.\) (vô lý)

Vậy pt vô nghiệm.

Cách khác:

PT \(\Leftrightarrow 4x^4+6x^3+16x^2+12x+10=0\)

\(\Leftrightarrow 3x^4+(x^4+6x^3+9x^2)+7x^2+12x+10=0\)

\(\Leftrightarrow 3x^4+(x^2+3x)^2+(4x^2+12x+9)+3x^2+1=0\)

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

(vô lý vì vế phải âm còn vế trái không âm)

Vậy pt vô nghiệm.

Câu b:

\(\frac{x-342}{15}+\frac{x-323}{17}+\frac{x-300}{19}+\frac{x-273}{21}=10\)

\(\Leftrightarrow \frac{x-342}{15}+\frac{x-323}{17}+\frac{x-300}{19}+\frac{x-273}{21}-10=0\)

\(\Leftrightarrow \frac{x-342}{15}-1+\frac{x-323}{17}-2+\frac{x-300}{19}-3+\frac{x-273}{21}-4=0\)

\(\Leftrightarrow \frac{x-357}{15}+\frac{x-357}{17}+\frac{x-357}{19}+\frac{x-357}{21}=0\)

\(\Leftrightarrow (x-357)\left(\frac{1}{15}+\frac{1}{17}+\frac{1}{19}+\frac{1}{21}\right)=0\)

Dễ thấy \(\frac{1}{15}+\frac{1}{17}+\frac{1}{19}+\frac{1}{21}\neq 0\), do đó $x-357=0$ hay $x=357$ là nghiệm duy nhất của pt.

a) Ta có: \(2x^4+3x^3+8x^2+6x+5=0\)

\(\Leftrightarrow 4x^4+6x^3+16x^2+12x+10=0\)

\(\Leftrightarrow (x^4+9x^2+4+6x^3+4x^2+12x)+(3x^4+3x^2+6)=0\)

\(\Leftrightarrow (x^2+3x+2)^2+3(x^4+x^2+\frac{1}{4})+\frac{21}{4}=0\)

\(\Leftrightarrow (x^2+3x+2)^2+3(x^2+\frac{1}{2})^2+\frac{21}{4}=0(*)\)

Thấy rằng \((x^2+3x+2)^2\geq 0; (x^2+\frac{1}{2})^2\geq 0\forall x\in\mathbb{R}\)

Do đó \((x^2+3x+2)^2+3(x^2+\frac{1}{2})^2+\frac{21}{4}\geq \frac{21}{4}>0\)

Suy ra \((*)\) vô nghiệm dẫn đến PT đầu tiên vô nghiệm (đpcm)

Ta có : \(\frac{x-342}{15}+\frac{x-323}{17}+\frac{x-300}{19}+\frac{x-273}{21}=10\)

=> \(\left(\frac{x-342}{15}-1\right)+\left(\frac{x-323}{17}-2\right)+\left(\frac{x-300}{19}-3\right)+\left(\frac{x-273}{21}-4\right)=0\)

=> \(\frac{x-357}{15}+\frac{x-357}{17}+\frac{x-357}{19}+\frac{x-357}{21}=0\)

=> \(\left(x-357\right)\left(\frac{1}{15}+\frac{1}{17}+\frac{1}{19}+\frac{1}{21}\right)=0\)

Vì \(\frac{1}{15}+\frac{1}{17}+\frac{1}{19}+\frac{1}{21}\ne0\)

=> x - 357 = 0

=> x = 357

Vậy x = 357

bn xem lại đề đc k ?

theo mk thì là = 10 chứ k phải =0

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