Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Lời giải:
$\frac{x^3+8}{x^2-2x+1}.\frac{x^2+3x+2}{1-x^2}=\frac{(x^3+8)(x^2+3x+2)}{(x^2-2x+1)(1-x^2)}$
$=\frac{(x+2)(x^2-2x+4)(x+1)(x+2)}{(x-1)^2(1-x)(x+1)}$
$=\frac{(x+2)^2(x^2-2x+4)}{-(x-1)^3}$
Ta có :
\(4x\left(x-1\right)-3\left(x^2-5\right)-x^2=\left(x-3\right)-\left(x+4\right)\)
\(\Leftrightarrow\)\(4x^2-4x-3x^2+15=x-3-x-4\)
\(\Leftrightarrow\)\(x^2-4x+15=-7\)
\(\Leftrightarrow\)\(\left(x^2-2.x.2+2^2\right)+11=-7\)
\(\Leftrightarrow\)\(\left(x-2\right)^2=-18\)
Mà \(\left(x-2\right)^2\ge0\) \(\left(\forall x\inℝ\right)\)
\(\Rightarrow\)\(x\in\left\{\varnothing\right\}\)
Vậy không có giá trị nào của x thoã mãn đề bài
Chúc bạn học tốt ~
\(x-\dfrac{1}{3}=\dfrac{2}{3}.\dfrac{9}{14}+\dfrac{3}{7}\)
\(x-\dfrac{1}{3}=\dfrac{1}{7}+\dfrac{3}{7}\)
\(x-\dfrac{1}{3}=\dfrac{4}{7}\)
\(x=\dfrac{19}{21}\)
a: =(x-y)^2+2(x-y)
=(x-y)(x-y+2)
c: =(x-3)(x+3)+(x-3)^2
=(x-3)(x+3+x-3)
=2x(x-3)
d: =(x+3)(x^2-3x+9)-4x(x+3)
=(x+3)(x^2-7x+9)
e: =(x^2-8x+7)(x^2-8x+15)-20
=(x^2-8x)^2+22(x^2-8x)+85
=(x^2-8x+17)(x^2-8x+5)
a.
\(\left(x^2-x+1\right)\left(x^2-x+2\right)=12\)
Đặt \(x^2-x+1=y\) ta được:
\(y\left(y+1\right)=12\)
\(\Leftrightarrow y^2+y-12=0\)
\(\Leftrightarrow y^2+4y-3y-12=0\)
\(\Leftrightarrow\left(y-3\right)\left(y+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=3\\y=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+1=3\\x^2-x+1=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x-2=0\\x^2-x+5=0\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
b.
\(3y^3-7y^2-7y+3=0\)
\(\Leftrightarrow3\left(y^3+1\right)-7y\left(y+1\right)=0\)
\(\Leftrightarrow3\left(y+1\right)\left(y^2-y+1\right)-7y\left(y+1\right)=0\)
\(\Leftrightarrow\left(y+1\right)\left(3y^2-3y+3-7y\right)=0\)
\(\Leftrightarrow\left(y+1\right)\left(3y^2-10y+3\right)=0\)
\(\Leftrightarrow\left(y+1\right)\left(3y-1\right)\left(y-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=-1\\y=\dfrac{1}{3}\\y=3\end{matrix}\right.\)
a: \(x+\dfrac{3}{9}=\dfrac{7}{6}\cdot\dfrac{2}{3}\)
=>\(x+\dfrac{1}{3}=\dfrac{14}{18}=\dfrac{7}{9}\)
=>\(x=\dfrac{7}{9}-\dfrac{1}{3}=\dfrac{7}{9}-\dfrac{3}{9}=\dfrac{4}{9}\)
b: \(x-\dfrac{2}{3}=\dfrac{1}{8}:\dfrac{5}{4}\)
=>\(x-\dfrac{2}{3}=\dfrac{1}{8}\cdot\dfrac{4}{5}=\dfrac{1}{10}\)
=>\(x=\dfrac{1}{10}+\dfrac{2}{3}=\dfrac{3+20}{30}=\dfrac{23}{30}\)
\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}+\dfrac{x+y-3}{z}\\ =\dfrac{y+z+1+x+z+2+x+y-3}{x+y+z}=\dfrac{2\left(z+y+x\right)}{x+y+z}=2\\ \to\left\{{}\begin{matrix}y+z+1=2x\\x+z+2=2y\\x+y-3=2z\end{matrix}\right.\to\left\{{}\begin{matrix}x+y+z=3x-1\\x+y+z=3y-2\\x+y+z=3z+3\end{matrix}\right.\)
Mặt khác \(\dfrac{1}{x+y+z}=2\to x+y+z=\dfrac{1}{2}\)
\(\to\left\{{}\begin{matrix}3x-1=\dfrac{1}{2}\\3y-2=\dfrac{1}{2}\\3z+3=\dfrac{1}{2}\end{matrix}\right.\to\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{5}{6}\\z=-\dfrac{5}{6}\end{matrix}\right.\)
\(\dfrac{1}{2}-\dfrac{5}{12}x=\dfrac{2}{3}\)
\(\dfrac{5}{12}x=\dfrac{1}{2}-\dfrac{2}{3}=\dfrac{3}{6}-\dfrac{4}{6}\)
\(\dfrac{5}{12}x=\dfrac{-1}{6}\)
\(x=\dfrac{-1}{6}:\dfrac{5}{12}=\dfrac{-1}{6}.\dfrac{12}{5}\)
\(x=\dfrac{-2}{5}\)
=1
=4
ở dưới là 4 nhân 3 nhân 3 à nếu thế thì =36
còn ko thì 3 nhân 3 = 9
ok bn dễ thg kb face nha