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a) \(\frac{x+5}{4}-\frac{2x-3}{3}=\frac{6x-1}{8}+\frac{2x-1}{12}\)
<=> \(\frac{x}{4}+\frac{5}{4}-\frac{2x}{3}+1=\frac{6x}{8}-\frac{1}{8}+\frac{2x}{12}-\frac{1}{12}\)
<=> \(-\frac{4}{3}x=-\frac{59}{24}\)
<=> \(x=\frac{59}{32}\)
Vậy S = { 59/32}
b) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)
<=> \(\frac{x^2+14x+40}{12}-\frac{-x^2-2x+8}{4}=\frac{x^2+8x-20}{3}\)
<=> \(\left(\frac{x^2}{12}+\frac{x^2}{4}-\frac{x^2}{3}\right)+\left(\frac{14}{12}x+\frac{2}{4}x-\frac{8}{3}x\right)=-\frac{20}{8}+\frac{8}{4}-\frac{40}{12}\)
<=> \(-x=-8\)
<=> x = 8
Vậy S = { 8 }
\(\Rightarrow2x\left(x+2\right)\left(x+2\right)-8x^2=2\left(x^3-8\right).\)
\(\Rightarrow\left(2x^2+4\right)\left(x+2\right)-8x^2=2x^3-8\)
\(\Rightarrow2x^3+4x^2+4x+8=2x^2-8\)
\(\Rightarrow2x^3+4x^2-2x^2+4x=-8-8\)
\(\Rightarrow2x^3+2x^2+4x=-16\)
\(\Rightarrow2x\left(x^2+x+2\right)=-16\)tự giải tiếp nhé!
Giải:
\(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-3\right)\left(x-2\right)}=\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-1\right)\left(x-4\right)}\)
ĐKXĐ: \(x\ne\left\{1;2;3;4\right\}\)
\(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-3\right)\left(x-2\right)}=\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-1\right)\left(x-4\right)}\)
\(\Rightarrow\left(x-3\right)\left(x-4\right)+\left(x-1\right)\left(x-4\right)=\left(x-1\right)\left(x-2\right)+\left(x-2\right)\left(x-3\right)\)
\(\Leftrightarrow\left(x-4\right)\left[\left(x-3\right)+\left(x-1\right)\right]=\left(x-2\right)\left[\left(x-1\right)+\left(x-3\right)\right]\)
\(\Leftrightarrow x-4=x-2\)
\(\Leftrightarrow0x=2\)
Vậy ...
~~~~~e)~~~~~
\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
Đặt \(x^2+x+1=v\)
Ta có: \(v.\left(v+1\right)-12\)
\(=v^2+v-12\)
\(=v^2-3v+4v-12\)
\(=v\left(v-3\right)+4\left(v-3\right)\)
\(=\left(v-3\right)\left(v+4\right)\)
\(=\left(x^2+x+1-3\right)\left(x^2+x+1+4\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x+5\right)\)
~~~~~g)~~~~~
\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)(nhân cái đầu vs cái cuối, hai cái giữa nhân vs nhau)
Đặt \(x^2+5x+5=t\)
Ta có: \(\left(t-1\right)\left(t+1\right)-24\)
\(=t^2-1-24\)
\(=t^2-25\)
\(=\left(t-5\right)\left(t+5\right)\)
\(=\left(x^2+5x+5-5\right)\left(x^2+5x+5+5\right)\)
\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)
~~~~~h)~~~~~
\(\left(x^2+x+1\right)\left(x^2+3x+1\right)+x^2\)
Đặt \(x^2+2x+1=n\)
Ta có: \(\left(n-x\right)\left(n+x\right)+x^2\)
\(=n^2-x^2+x^2\)
\(=n^2\)
\(=\left(x^2+2x+1\right)^2\)
\(=\left(\left(x+1\right)^2\right)^2\)
\(=\left(x+1\right)^4\)
~~~~~~~~~~~~~~~~~~~~
(Mong là mình làm đúng, chúc you học tốt nha, tíck cho mìk với nhé!)
\(2x\left(x+2\right)^2-8x^2=2\left(x-2\right)\left(x^2+2x+4\right)\)
\(\Leftrightarrow2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\)
\(\Leftrightarrow2x^3+8x^2+8x-8x^2=2x^3-16\)
\(\Leftrightarrow8x=-16\)
\(\Leftrightarrow x=-2\)
ĐK: \(x\in R\backslash\left\{-4,-3,-2,-1\right\}\)
PT ban đầu
\(\Leftrightarrow\frac{x+2-x-1}{\left(x+1\right)\left(x+2\right)}+\frac{x+3-x-2}{\left(x+2\right)\left(x+3\right)}+\frac{x+4-x-3}{\left(x+3\right)\left(x+4\right)}+\frac{x+5-x-4}{\left(x+4\right)\left(x+5\right)}=\frac{1}{x+1}-403\\ \Leftrightarrow\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}=\frac{1}{x+1}-403\\ \Leftrightarrow\frac{1}{x+5}=403\\ \Leftrightarrow x+5=\frac{1}{403}\Leftrightarrow x=\frac{-2014}{403}\)
Chúc bạn học tốt nha
.
Sr bạn nha, nhưng điều kiện là \(x\in R\backslash\left\{-5,-4,-3,-2,-1\right\}\). (Xét thiếu :>)
Chúc bạn học tốt nha.
\(48x\left(x+1\right)\left(x^3-4\right)=\left(x^4+8x+12\right)^2\)
\(\Leftrightarrow4\left(12x+12\right)\left(x^4-4x\right)=\left(x^4+8x+12\right)^2\)
Đặt \(\left\{{}\begin{matrix}x^4-4x=a\\12x+12=b\end{matrix}\right.\)
\(\Rightarrow4ab=\left(a+b\right)^2\)
\(\Leftrightarrow4ab=a^2+a^2+2ab\)
\(\Leftrightarrow\left(a-b\right)^2=0\)
\(\Leftrightarrow a-b=0\)
\(\Leftrightarrow x^4-16x-12=0\)
\(\Leftrightarrow\left(x^2-2x-2\right)\left(x^2+2x+6\right)=0\)
\(\Leftrightarrow x^2-2x-2=0\)
\(\Rightarrow x=1\pm\sqrt{3}\)
cho e hỏi vs ạ. sao từ \(x^4-16x-12=0\) lại ra \(\left(x^2-2x-2\right)\left(x^2+2x+6\right)=0\) ạ?