Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,\Leftrightarrow-\dfrac{1}{2}x=\dfrac{1}{4}\Leftrightarrow x=-\dfrac{1}{2}\\ b,\Leftrightarrow\dfrac{1}{6}:x=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\Leftrightarrow x=\dfrac{1}{6}:\dfrac{5}{6}=\dfrac{1}{5}\\ c,\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=3\\x+\dfrac{1}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}\\x=-\dfrac{16}{5}\end{matrix}\right.\)
\(d,\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{22}{9}-\dfrac{7}{3}=\dfrac{1}{9}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{3}\\x+\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{6}\\x=-\dfrac{5}{6}\end{matrix}\right.\\ e,\Leftrightarrow2\left|x\right|=2-\dfrac{1}{2}=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{3}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
\(f,\Leftrightarrow\left|x+\dfrac{1}{2}\right|=1+\dfrac{1}{6}=\dfrac{7}{6}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{7}{6}\\x+\dfrac{1}{2}=-\dfrac{7}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
e: ta có: \(2\left|x\right|+\dfrac{1}{2}=2\)
\(\Leftrightarrow2\left|x\right|=\dfrac{3}{2}\)
\(\Leftrightarrow\left|x\right|=\dfrac{3}{4}\)
hay \(x\in\left\{\dfrac{3}{4};-\dfrac{3}{4}\right\}\)
\(A=2x^3+6x^2-3x+\dfrac{1}{2}=2\cdot\dfrac{1}{3}^3+6\cdot\dfrac{1}{3}^2-3\cdot\dfrac{1}{3}+\dfrac{1}{2}\)
=13/54
\(-3x\left(x+2\right)^2+\left(x+3\right)\left(x-1\right)\left(x+1\right)-\left(2x-3\right)^2\)
\(=-3x\left(x^2+4x+4\right)+\left(x+3\right)\left(x^2-1\right)-\left(4x^2-12x+9\right)\)
\(=-3x^3-12x^2-12x+x^3-x+3x^2-3-4x^2+12x-9\)
\(=-2x^3-13x^2-x-12\)
( x - 3 )3 = ( x - 2 )2
=> x3 - 3.x2.3 + 3. x . 32 - 33 = x2 - 2. x.2 + 22
=> x3 - 9x2 + 27x - 27 = x2 - 4x + 4
=> x3 - 10x2 + 31x - 31 = 0
Phân tích đa thức thành nhân tử.
Chờ tí để mình đi hỏi.
\(\left|x\right|=2\Leftrightarrow x=\pm2\)
- Với \(x=2\Rightarrow y=1-2=-1\)
\(M=x\left(y+7\right)-19=2\left(-1+7\right)-19=-7\)
- Với \(x=-2\Rightarrow y=1-\left(-2\right)=3\)
\(M=x\left(y+7\right)-19=-2\left(3+7\right)-19=-39\)
P(x)=2x^4-x-2x^3+1
Q(x)=5x^2-x^3+4x-8
tính h(x)=P+Q
h(x)=P+Q=2x^4-x-2x^3+1+5x^2-x^3+4x-8
\(=2x^4-3x^3+5x^2+3x-7\)
k(x)=Q-P=2x^4-x-2x^3+1-(5x^2-x^3+4x-8)
\(=2x^4-2x^3-x+1-5x^2+x^3-4x+8\)
\(=2x^4-x^3-5x^2-5x+9\)
=>3^x(1+3^2)=810
<=>3^x.10=810
<=>3^x=81
<=>x=4