X x 2 + X x3 =24
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,x^3+x^2+x+1=0\\ \Rightarrow x^2\left(x+1\right)+\left(x+1\right)=0\\ \Rightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{-1\right\}\)
\(b,x^3+x^2-x-1=0\\ \Rightarrow x^2\left(x+1\right)-\left(x+1\right)=0\\ \Rightarrow\left(x^2-1\right)\left(x+1\right)=0\\ \Rightarrow\left(x-1\right)\left(x+1\right)^2=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{-1;1\right\}\)
\(c,\left(x+1\right)^2\left(x+2\right)+\left(x+1\right)^2\left(x-2\right)=-24\\ \Rightarrow\left(x+1\right)^2\left(x+2+x-2\right)=-24\\ \Rightarrow2x\left(x^2+2x+1\right)=-24\\ \Rightarrow x^3+2x^2+x+12=0\\ \Rightarrow\left(x^3+3x^2\right)-\left(x^2+3x\right)+\left(4x+12\right)=0\\ \Rightarrow x^2\left(x+3\right)-x\left(x+3\right)+4\left(x+3\right)=0\\ \Rightarrow\left(x^2-x+4\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2+\dfrac{15}{4}=0\left(vô.lí\right)\\x=-3\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{-3\right\}\)
mình viết lại rồi làm bài mới nhất í ai nhanh nhất mình k luôn
Bài 1
a/ \(x\left(x^2+1\right)+2\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+1\right)=0\Rightarrow x=-2\)
b/
\(\Leftrightarrow x^3-6x^2+9x+5x^2-30x+45=0\)
\(\Leftrightarrow x\left(x-3\right)^2+5\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-3\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=3\end{matrix}\right.\)
1.
c/ \(\Leftrightarrow x^3+2x^2+2x+x^2+2x+2=0\)
\(\Leftrightarrow x\left(x^2+2x+2\right)+x^2+2x+2=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+2x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x^2+2x+2=0\left(vn\right)\end{matrix}\right.\)
d/
\(\Leftrightarrow x^4+x^3-2x^2-x^3-x^2+2x+4x^2+4x-8=0\)
\(\Leftrightarrow x^2\left(x^2+x-2\right)-x\left(x^2+x-2\right)+4\left(x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x^2-x+4\right)\left(x^2+x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+4=0\left(vn\right)\\x^2+x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
a: Ta có: \(\left(3x-2\right)\left(2x-1\right)-\left(6x^2-3x\right)=0\)
\(\Leftrightarrow2x-1=0\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x^3-\left(x+1\right)\left(x^2-x+1\right)=x\)
\(\Leftrightarrow x^3-x^3-1=x\)
hay x=-1
c: Ta có: \(56x^4+7x=0\)
\(\Leftrightarrow7x\left(8x^3+1\right)=0\)
\(\Leftrightarrow x\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)
d: Ta có: \(x^2-5x-24=0\)
\(\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)
\(g,x^4-16=\left(x^2-4\right)\left(x^2+4\right)=\left(x-2\right)\left(x+2\right)\left(x^2+4\right)\\ i,-x^2+10x-25=-\left(x-5\right)^2\\ k,x^3+3x^2+3x+1-27z^3\\ =\left(x+1\right)^3-27z^3\\ =\left(x+1-3z\right)\left[\left(x+1\right)^2+3z\left(x+1\right)+9z^2\right]\\ =\left(x-3z+1\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\\ m,\left(x+y\right)^2-25\left(x+y\right)+24=\left(x+y-5\right)^2-1=\left(x+y-4\right)\left(x+y-6\right)\)
g. x4 - 16
<=> x4 - 42
<=> (x2)2 - 42
<=> (x2 - 4)(x2 + 4)
i. -x2 + 10x - 25
<=> -(x2 - 10x + 25)
<=> -(x2 -10x + 52)
<=> -(x - 5)2
Bài làm :
\(x\times2+x\times3=24\)
\(x\times\left(2+3\right)=24\)
\(x\times5=24\)
\(x=\frac{24}{5}\)
Học tốt nhé
Xx(2+3)=24
Xx5=24
X=5/24