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\(x^2+5x=0\Leftrightarrow x\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
\(x^2-2x-xy+2y=\left(x^2-xy\right)-2\left(x-y\right)=x\left(x-y\right)-2\left(x-y\right)=\left(x-y\right)\left(x-2\right)\)
\(x^2-4x-4=25\)
\(\Leftrightarrow x^2-4x+4=33\)
\(\Leftrightarrow\left(x-2\right)^2=33\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=\sqrt{33}\\x-2=-\sqrt{33}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{33}+2\\x=2-\sqrt{33}\end{cases}}\)
\(x^2-4x-4=25\)
\(\Rightarrow x^2-4x-4-25=0\)
\(\Rightarrow x^2-4x-29=0\)
\(\Rightarrow x^2-4x+4-33=0\)
\(\Rightarrow\left(x^2-4x+4\right)-33=0\)
\(\Rightarrow\left(x-2\right)^2=33\)
\(\Rightarrow\left(x-2\right)^2=\left(\sqrt{33}\right)^2=\left(-\sqrt{33}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}x-2=\sqrt{33}\\x-2=-\sqrt{33}\end{cases}\Rightarrow\orbr{\begin{cases}x=2+\sqrt{33}\\x=2-\sqrt{33}\end{cases}}}\)
Vậy \(x\in\left\{2+\sqrt{33};2-\sqrt{33}\right\}\)
\(x^2-4x+4=25\)
\(\Leftrightarrow\left(x-2\right)^2=25\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=5\\x-2=-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=7\\x=-3\end{cases}}\)
a)
⇔ \(x^2-16=9\)
⇔ \(x^2=25\)
⇔ \(x=\pm5\)
b)
⇔ \(x^2-4x+4-25x^2+20x-4=0\)
⇔ \(16x-24x^2=0\)
⇔ \(8x\left(2-3x\right)=0\)
⇒ \(\left[{}\begin{matrix}x=0\\2-3x=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(x=0\) hoặc \(x=\dfrac{2}{3}\)
c)
⇔ \(3x^2-10x-20=0\)
⇔ \(x^2-2.x.\dfrac{5}{3}+\dfrac{25}{9}-\dfrac{205}{9}=0\)
⇔ \(\left(x-\dfrac{5}{3}\right)^2=\dfrac{205}{9}\)
⇒ \(\left[{}\begin{matrix}x-\dfrac{5}{3}=\sqrt{\dfrac{205}{9}}\\x-\dfrac{5}{3}=-\sqrt{\dfrac{205}{9}}\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{\sqrt{\text{205}}}{\text{3}}+\dfrac{5}{3}\\x=-\dfrac{\sqrt{\text{205}}}{\text{3}}+\dfrac{5}{3}\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{15+\text{9}\sqrt{\text{205}}}{\text{9}}\\\text{x}=-\dfrac{15+\text{9}\sqrt{\text{205}}}{\text{9}}\end{matrix}\right.\)
Vậy...
d)
⇔ \(\left(x^2+x\right)^2-49=\left(x^2+x\right)^2-7x\)
⇔ 7x = 49
⇔ x=7
Vậy...
Q=3x+9y+15z+x+x4+y+y9+z+z25
\ge 108+2.2+2.3+2.5=128≥108+2.2+2.3+2.5=128
Dấu "=" xảy ra khi x+3y+5z=36, x=\dfrac{4}x, y=\dfrac{9}y, z=\dfrac{25}z\Rightarrow x=2,y=3,z=5x+3y+5z=36,x=x4,y=y9,z=z25⇒x=2,y=3,z=5
bạn tham khảo nhé
\(a,=x^2-4x+4-\dfrac{15}{4}=\left(x-2\right)^2-\dfrac{15}{4}=\left(x-2-\dfrac{\sqrt{15}}{2}\right)\left(x-2+\dfrac{\sqrt{15}}{2}\right)\\ b,=?\\ c,\Rightarrow x^2+7x-8=0\\ \Rightarrow\left(x+8\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-8\\x=1\end{matrix}\right.\\ d,Sửa:x^3-3x^2=-27+9x\\ \Rightarrow x^3-3x^2+9x-27=0\\ \Rightarrow x^2\left(x-3\right)+9\left(x-3\right)=0\\ \Rightarrow\left(x^2+9\right)\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-9\left(vô.lí\right)\\x=3\end{matrix}\right.\\ \Rightarrow x=3\\ e,\Rightarrow x\left(x-3\right)-7x+21=0\\ \Rightarrow x\left(x-3\right)-7\left(x-3\right)=0\\ \Rightarrow\left(x-7\right)\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\\ f,\Rightarrow x^2\left(x-2\right)+\left(x-2\right)=0\\ \Rightarrow\left(x^2+1\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=2\end{matrix}\right.\\ \Rightarrow x=2\)
\(g,\Rightarrow x^2-4x+4=0\\ \Rightarrow\left(x-2\right)^2=0\\ \Rightarrow x=2\\ h,Sửa:x^3-x^2+x=1\\ \Rightarrow x^2\left(x-1\right)+\left(x-1\right)=0\\ \Rightarrow\left(x^2+1\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=1\end{matrix}\right.\\ \Rightarrow x=1\)
B = 2\(x^2\) - 4\(x\) - 8
B = 2(\(x^2\) - 2\(x\) + 4) - 16
B = 2(\(x-2\))2 - 16
Vì (\(x-2\))2 ≥ 0 ∀ \(x\) ⇒ 2(\(x-2\))2 ≥ 0 ∀ \(x\)
⇒ 2(\(x-2\))2 - 16 ≥ -16 ∀ \(x\)
Dấu bằng xảy ra khi (\(x-2\))2 = 0 ⇒ \(x-2=0\) ⇒ \(x=2\)
Vậy Bmin = -16 khi \(x=2\)
Tìm min của C biết:
C = \(x^2\) - 2\(xy\) + 2y2 + 2\(x\) - 10y + 17
C = (\(x^2\) - 2\(xy\) + y2) + 2(\(x\) - y) + y2 - 8y + 16 + 1
C = (\(x\) - y)2 + 2(\(x\) - y) + 1 + (y2 - 8y + 16)
C = (\(x-y+1\))2 + (y - 4)2
Vì (\(x\) - y + 1)2 ≥ 0 ∀ \(x;y\); (y - 4)2 ≥ 0 ∀ y
Dấu bằng xảy ra khi: \(\left\{{}\begin{matrix}x-y+1=0\\y-4=0\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x-y+1=0\\y=4\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}x-4+1=0\\y=4\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=-1+4\\y=4\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\)
Vậy Cmin = 0 khi (\(x;y\)) = (3; 4)
\(a,8x-3=5x+12\\ \Leftrightarrow8x-5x=12+3\\ \Leftrightarrow3x=15\\ \Leftrightarrow x=\dfrac{15}{3}=5\)
\(b,x-12+4x=25+2x-1\\ \Leftrightarrow x+4x-2x=25-1+12\\ \Leftrightarrow3x=36\\ \Leftrightarrow x=\dfrac{36}{3}=12\)
\(c,7-\left(2x+4\right)=-\left(x+4\right)\\ \Leftrightarrow7-2x-4=-x-4\\ \Leftrightarrow-2x+x=-4+4-7\\ \Leftrightarrow-x=-7\\ \Leftrightarrow x=7\)
\(d,3-4x\left(45-2x\right)=8x^2+x-300\\ \Leftrightarrow3-100x+8x^2=8x^2+x-300\\ \Leftrightarrow8x^2-8x^2-100x-x=-300-3\\ \Leftrightarrow-101x=-303\\ \Leftrightarrow x=\dfrac{-303}{-101}=3\)
Đề câu d của bạn hình như sai dấu ý
x2 - 4x + 4 = 25
x2 - 4x = 25 - 4
x2 - 4x = 21
x2 = 21 + 4x
x.x = 21 + 4x
\(\Rightarrow\) 3x = 21
x = 21 : 3
x = 7
pt<=>(x+2)2=25=>x+2=5 hoac x+2=-5=>x=3 hoac x=-7