Hỏi đáp Toán lớp 8

a) \(\dfrac{x^2-x}{x+3}-\dfrac{x^2}{x-3}=\dfrac{7x^2-3x}{9-x^2}\)ĐKXĐ : \(x\ne\pm3\)

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

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

\(\Leftrightarrow-7x^2+3x-3x+7x^2=0\)

\(\Leftrightarrow0x=0\)( luôn đúng )

Vậy \(x\in R\left(x\ne\pm3\right)\)

b) \(\dfrac{2x-1}{x^3+1}=\dfrac{2}{x^2-x+1}-\dfrac{1}{x+1}\)ĐKXĐ : \(x\ne-1\)

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

\(\Rightarrow2x-1=2x+2-x^2+x-1\)

\(\Leftrightarrow2x+2-x^2+x-1-2x+1=0\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(\text{t/m ĐKXĐ}\right)\\x=-1\left(\text{không t/m ĐKXĐ}\right)\end{matrix}\right.\)

Vậy....

a) \(\dfrac{3}{x+1}-\dfrac{1}{x-2}=\dfrac{9}{\left(x+1\right)\left(2-x\right)}\)ĐKXĐ : \(x\ne-1;2\)

\(\Leftrightarrow\dfrac{3\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\dfrac{x+1}{\left(x+1\right)\left(x-2\right)}=\dfrac{-9}{\left(x+1\right)\left(x-2\right)}\)

\(\Rightarrow3x-6-x-1=-9\)

\(\Leftrightarrow2x=-2\)

\(\Leftrightarrow x=-1\)(không t/m ĐKXĐ )

Vậy....

b) \(\dfrac{x-4}{x-1}+\dfrac{x+4}{x+1}=2\)ĐKXĐ : \(x\ne\pm1\)

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

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

\(\Leftrightarrow\)\(6=0\)( vô lí )

Vậy pt vô nghiệm

Gọi số đó là \(\overline{ab}\Rightarrow a=2b\)

\(\overline{ab}=10a+b=10.2b+b=21b\)

Khi xen số 1 giữa 2 số thì ta được số mới là

\(\overline{a1b}=100a+10+b=100.2b+10+b=201b+10\)

Theo bài ra ta có pt:

\(201b+10-21b=550\Leftrightarrow180b=540\)

\(\Rightarrow b=3\Rightarrow a=2b=6\)

\(\Rightarrow\) số ban đầu là \(63\)

Áp dụng bđt Cauchy-Schawarz cũng đc bạn nhé :")))

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1^2^{ }}{a}+\dfrac{1^2}{b}+\dfrac{1^2}{c}\ge\dfrac{\left(1+1+1\right)^2}{a+b+c}=\dfrac{3^2}{1}=9\)Dấu bằng xảy ra khi \(a=b=c=\dfrac{1}{3}\)

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.

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:

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

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

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

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

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

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

Vậy pt có nghiệm $x=\pm 2$

Câu a:

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

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

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

Ta thấy: \((x^2+1)^2>0; (x-1)^2\geq 0, \forall x\in\mathbb{R}\)

Do đó \((x^2+1)^2+(x-1)^2+1>0\). Suy ra pt $(*)$ vô nghiệm.

Vậy pt đã cho vô nghiệm.