Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Câu 1 :
\(\text{ a) }12-2x-x^2=0\\ \Leftrightarrow2\left(6-x-x^2\right)=0\\ \Leftrightarrow6-x-x^2=0\\ \Leftrightarrow6-3x+2x-x^2=0\\ \Leftrightarrow\left(6-3x\right)+\left(2x-x^2\right)=0\\ \Leftrightarrow3\left(2-x\right)+x\left(2-x\right)=0\\ \Leftrightarrow\left(3+x\right)\left(2-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3+x=0\\2-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
Vậy \(x=-3\) hoặc \(x=2\)
\(\text{b) }\left(x^2-\dfrac{1}{2}x\right):2x-\left(3x-1\right):\left(3x-1\right)=0\\ \Leftrightarrow\dfrac{1}{2}x-\dfrac{1}{4}-1=0\\ \Leftrightarrow\dfrac{1}{2}x-\dfrac{5}{4}=0\\ \Leftrightarrow\dfrac{1}{2}x=\dfrac{5}{4}\\ \Leftrightarrow x=\dfrac{5}{2}\)
Vậy \(x=\dfrac{5}{2}\)
Câu 2:
\(N=x^2+5y^2+2xy-2y+2005\\ N=x^2+4y^2+y^2+2xy-2y+1+2004\\ N=\left(x^2+2xy+y^2\right)+\left(4y^2-2y+1\right)+2004\\ N=\left(x+y\right)^2+\left(2y-1\right)^2+2004\\ \text{Do }\left(x+y\right)^2\ge0\forall x;y\\ \left(2y-1\right)^2\ge0\forall y\\ \Rightarrow\left(x+y\right)^2+\left(2y-1\right)^2\ge0\forall x;y\\ \Rightarrow N=\left(x+y\right)^2+\left(2y-1\right)^2+2004\ge0\forall x;y\\ \text{Dấu "=" xảy ra khi : }\left\{{}\begin{matrix}\left(x+y\right)^2=0\\\left(2y-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\2y-1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-y\\y=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{2}\end{matrix}\right.\)
Vậy \(N_{\left(Min\right)}=2004\) khi \(x=-\dfrac{1}{2};y=\dfrac{1}{2}\)
bài 1 : a. x^3 +27 -54-x^3 =-27
b. 8x^3 +y^3 -8x^3 +y^3 =2y^3
c. (2x-1+2x+2)(2x-1-2x-2)=(4x+1).(-3)=-12x-3
d. a^3 +b^3 +3ab(a+b) -3ab(a+b)=a^3+b^3
Bài 1:
a) $9x^2-2x-1=(3x)^2-2.3x.\frac{1}{3}+(\frac{1}{3})^2-\frac{10}{9}$
$=(3x-\frac{1}{3})^2-\frac{10}{9}$
$\geq 0-\frac{10}{9}=\frac{-10}{9}$
Vậy GTNN của biểu thức là $\frac{-10}{9}$. Giá trị này đạt tại $3x-\frac{1}{3}=0\Leftrightarrow x=\frac{1}{9}$
b)
$(2x-5)(x-1)=2x^2-7x+5=2(x^2-\frac{7}{2}x)+5$
$=2[x^2-2.\frac{7}{4}x+(\frac{7}{4})^2]-\frac{9}{8}$
$=2(x-\frac{7}{4})^2-\frac{9}{8}$
$\geq 2.0-\frac{9}{8}=-\frac{9}{8}$
Vậy GTNN của biểu thức là $\frac{-9}{8}$ tại $x=\frac{7}{4}$
a) \(A=x^2+3x+4=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)
\(minA=\dfrac{7}{4}\Leftrightarrow x=-\dfrac{3}{2}\)
b) \(B=2x^2-x+1=2\left(x-\dfrac{1}{4}\right)^2+\dfrac{7}{8}\ge\dfrac{7}{8}\)
\(minB=\dfrac{7}{8}\Leftrightarrow x=\dfrac{1}{4}\)
c) \(C=5x^2+2x-3=5\left(x+\dfrac{1}{5}\right)^2-\dfrac{16}{5}\ge-\dfrac{16}{5}\)
\(minC=-\dfrac{16}{5}\Leftrightarrow x=-\dfrac{1}{5}\)
d) \(D=4x^2+4x-24=\left(2x+1\right)^2-25\ge-25\)
\(minD=-25\Leftrightarrow x=-\dfrac{1}{2}\)
e) \(E=x^2+6x-11=\left(x+3\right)^2-20\ge-20\)
\(minE=-20\Leftrightarrow x=-3\)
f) \(G=\dfrac{1}{4}x^2+x-\dfrac{1}{3}=\left(\dfrac{1}{2}x+1\right)^2-\dfrac{4}{3}\ge-\dfrac{4}{3}\)
\(minG=-\dfrac{4}{3}\Leftrightarrow x=-2\)
\(A=x^2+3x+4=\left(x^2+3x+\dfrac{9}{4}\right)+\dfrac{7}{4}=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\)
Do \(\left(x+\dfrac{3}{2}\right)^2\ge0\forall x\)
\(\Rightarrow A=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)
\(minA=\dfrac{7}{4}\Leftrightarrow x+\dfrac{3}{2}=0\Leftrightarrow x=-\dfrac{3}{2}\)
Mấy câu còn lại làm tương tự nhé em^^
\(M=\left(x^2-2xy+y^2\right)+\left(x^2+x+\dfrac{1}{4}\right)-\dfrac{1}{4}=\left(x-y\right)^2+\left(x+\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\\ M_{min}=-\dfrac{1}{4}\Leftrightarrow x=y=-\dfrac{1}{2}\)