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\(a,\left(x-2\right)^2-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)
\(\Leftrightarrow24x=-10\)
\(\Leftrightarrow x=-\dfrac{5}{12}\)
Vậy:....
\(b,\left(5x+1\right)^2-\left(5x+3\right)\left(5x-3\right)=30\)
\(\Leftrightarrow25x^2+10x+1-25^2+9=30\)
\(\Leftrightarrow10x=20\)
\(\Rightarrow x=2\)
Vậy :....
\(c,\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)\(\Leftrightarrow x^3+27-x\left(x^2-4\right)=15\)
\(\Leftrightarrow x^3+27-x^3+4x=15\)
\(\Leftrightarrow4x=15-27=-12\)
\(\Leftrightarrow x=-3\)
vậy : .....
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(x^3+5x^2+3x-9=0\)
\(\Leftrightarrow x^3+3x^2+2x^2+6x-3x-9=0\)
\(\Leftrightarrow x^2\left(x+3\right)+2x\left(x+3\right)-3\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+2x-3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\left(x+3\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
Vậy PT có nghiệm là \(\left\{1;-3\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15=\left[\left(x+1\right)\left(x+7\right)\right]\left[\left(x+3\right)\left(x+5\right)\right]+15=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15=0\)\(Dat:x^2+8x+7=a\Rightarrow a\left(a+8\right)+15=0\Leftrightarrow a^2+8a+15=0\Leftrightarrow\left(a+3\right)\left(a+5\right)=0\Leftrightarrow\left[{}\begin{matrix}a=-3\\a=-5\end{matrix}\right.\)\(+,a=-5\Rightarrow x^2+8x+7=-5\Leftrightarrow x^2+8x+16=4\Leftrightarrow\left(x+4\right)^2=4\Rightarrow\left[{}\begin{matrix}x+4=-2\\x+4=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\left(thoaman\right)\\x=2\left(loai\right)\end{matrix}\right.\)\(+,a=-3\Rightarrow x^2+8x+7=-3\Leftrightarrow x^2+8x+16=6\Leftrightarrow\left(x+4\right)^2=6\Leftrightarrow\left[{}\begin{matrix}x+4=-\sqrt{6}\\x+4=\sqrt{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\left(\sqrt{6}+4\right)\left(thoaman\right)\\x=\sqrt{6}-4\left(thoaman\right)\end{matrix}\right.\) \(\Rightarrow x\in\left\{\sqrt{6}-4;-\sqrt{6}-4;-6\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)(3x-1)(4x-8)=0
⇔3x-1=0 hoặc 4x-8=0
1.3x-1=0⇔3x=1⇔x=1/3
2.4x-8=0⇔4x=8⇔x=2
phương trình có 2 nghiệm:x=1/3 và x=2
b)(x-2)(1-3x)=0
⇔x-2=0 hoặc 1-3x=0
1.x-2=0⇔x=2
2.1-3x=0⇔-3x=1⇔x=-1/3
phương trình có 2 nghiệm:x=2 và x=-1/3
c)(x-3)(x+4)-(x-3)(2x-1)=0
⇔(x+4)(2x-1)=0
⇔x+4=0 hoặc 2x-1=0
1.x+4=0⇔x=-4
2.2x-1=0⇔2x=1⇔x=1/2
phương trình có hai nghiệm:x=-4 và x=1/2
d)(x+1)(x+2)=2x(x+2)
⇔(x+1)(x+2)-2x(x+2)=0
⇔2x(x+1)=0
⇔2x=0 hoặc x+1=0
1.2x=0⇔x=0
2.x+1=0⇔x=-1
phương trình có 2 nghiệm:x=0 và x=-1
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{3x-102}{3x-24}\) \(ĐK:x\ne8\)
\(\Leftrightarrow\frac{3}{2\left(x-8\right)}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{3x-102}{3\left(x-8\right)}\)
\(\Leftrightarrow\frac{3.3}{6.\left(x-8\right)}+\frac{6.\left(3x-20\right)}{6\left(x-8\right)}-\frac{2\left(3x-102\right)}{6\left(x-8\right)}=\frac{-1}{8}\)
\(\Leftrightarrow\frac{9+18x-120-6x+204}{6\left(x-8\right)}=\frac{-1}{8}\)
\(\Leftrightarrow\frac{12x+93}{6\left(x-8\right)}=\frac{-1}{8}\)
\(\Leftrightarrow8\left(12x+93\right)=-6\left(x-8\right)\)
\(\Leftrightarrow96x+744=-6x+48\)
\(\Leftrightarrow102x=-696\)
\(\Leftrightarrow x=\frac{-116}{17}\) (nhận)
Vậy .....
b) \(\frac{1}{3-x}+\frac{14}{x^2-9}=\frac{x-4}{3+x}+\frac{7}{3+x}\) \(ĐK:x\ne\pm3\)
\(\Leftrightarrow\frac{1}{3-x}+\frac{14}{\left(x-3\right)\left(3+x\right)}=\frac{x-4}{3+x}+\frac{7}{3+x}\)
\(\Leftrightarrow-\frac{3+x}{\left(x-3\right)\left(3+x\right)}+\frac{14}{\left(x-3\right)\left(3+x\right)}=\frac{\left(x-4\right)\left(x-3\right)}{\left(3+x\right)\left(x-3\right)}+\frac{7\left(x-3\right)}{\left(3+x\right)\left(x-3\right)}\)
\(\Leftrightarrow\frac{-3-x+14}{\left(x-3\right)\left(x+3\right)}=\frac{\left(x-4\right)\left(x-3\right)}{\left(3+x\right)\left(x-3\right)}+\frac{7\left(x-3\right)}{\left(3+x\right)\left(x-3\right)}\)
\(\Leftrightarrow-3-x+14=x^2-3x-4x+12+7x-21\)
\(\Leftrightarrow x=-5\) (nhận)
Vậy ....
![](https://rs.olm.vn/images/avt/0.png?1311)
Với dạng bài này ta chỉ việc chia hoocne là ra nhé!
\(C1:x^4+x^3-8x^2-9x-9=0\\ \Leftrightarrow\left(x-3\right)\left(x^3+4x^2+4x+3\right)\\ \Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x^2+x+1\right)\\ \Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+3=0\\x^2+x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x^2+x+1=0\left(VN\right)\end{matrix}\right.\)
\(C2:x^4+2x^3-3x^2-8x-4=0\\ \Leftrightarrow\left(x+1\right)\left(x^3+x^2-4x-4\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x+1\right)\left(x^2-4\right)=0\\ \Leftrightarrow\left(x+1\right)^2\left(x^2-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x+1\right)^2=0\\x^2-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-2\end{matrix}\right.\)
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có: x(x+3)(x^2+3x+4)=-4
\(\Leftrightarrow\)(x^2+3x)(x^2+3x+4)+4=0
\(\Leftrightarrow\)(x^2+3x)\(^2\)+4(x^2+3x)+4=0
\(\Leftrightarrow\)(x^2+3x+2)\(^2\)=0
\(\Leftrightarrow\)x\(^2\)+3x+2=0
\(\Leftrightarrow\)(x+1)(x+2)=0
\(\Leftrightarrow\)x+1=0 hoặc x+2=0
*) Nếu x+2=0\(\Leftrightarrow\)x=-2
*) Nếu x+1=0\(\Leftrightarrow\)x=-1
Vậy S={ 2;-1}
chỗ y(x+1) sữa thành 9(x+1) nha
giúp mình vs nha