Phân tích:
a) -6x^3 + 18x^2 + 60x
b) x^7 + x^2 + 1
c) x^10 + x^5 + 1
Mình cần gấp, khoảng 5h chiều nay. Các bạn giúp mk nhé!
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AH
Akai Haruma
Giáo viên
31 tháng 7 2021
Lời giải:
a. ĐKXĐ: $x\geq -9$
PT $\Leftrightarrow x+9=7^2=49$
$\Leftrightarrow x=40$ (tm)
b. ĐKXĐ: $x\geq \frac{-3}{2}$
PT $\Leftrightarrow 4\sqrt{2x+3}-\sqrt{4(2x+3)}+\frac{1}{3}\sqrt{9(2x+3)}=15$
$\Leftrightarrow 4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15$
$\Leftrgihtarrow 3\sqrt{2x+3}=15$
$\Leftrightarrow \sqrt{2x+3}=5$
$\Leftrightarrow 2x+3=25$
$\Leftrightarrow x=11$ (tm)
AH
Akai Haruma
Giáo viên
31 tháng 7 2021
c.
PT \(\Leftrightarrow \left\{\begin{matrix} 2x+1\geq 0\\ x^2-6x+9=(2x+1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ 3x^2+10x-8=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ (3x-2)(x+4)=0\end{matrix}\right.\)
\(\Leftrightarrow x=\frac{2}{3}\)
d. ĐKXĐ: $x\geq 1$
PT \(\Leftrightarrow \sqrt{(x-1)+4\sqrt{x-1}+4}-\sqrt{(x-1)+6\sqrt{x-1}+9}=9\)
\(\Leftrightarrow \sqrt{(\sqrt{x-1}+2)^2}-\sqrt{(\sqrt{x-1}+3)^2}=9\)
\(\Leftrightarrow \sqrt{x-1}+2-(\sqrt{x-1}+3)=9\)
\(\Leftrightarrow -1=9\) (vô lý)
Vậy pt vô nghiệm.
20 tháng 7 2021
a) \(A=x^2-6x+10=\left(x^2-6x+9\right)+1=\left(x-3\right)^2+1\ge1\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=3\). \(min_A=1\)
b) \(B=3x^2+x-2=3\left(x^2+\dfrac{1}{3}x-\dfrac{2}{3}\right)=3\left(x^2+\dfrac{1}{3}x+\dfrac{1}{36}-\dfrac{25}{36}\right)=3\left(x+\dfrac{1}{6}\right)^2-\dfrac{25}{12}\ge\dfrac{-25}{12}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=-\dfrac{1}{6}\). \(min_B=\dfrac{-25}{12}\)
c) \(C=\dfrac{4}{x^2}-\dfrac{3}{x}-1=\left(\dfrac{4}{x^2}-\dfrac{3}{x}+\dfrac{9}{16}\right)-\dfrac{25}{16}=\left(\dfrac{2}{x}+\dfrac{2}{3}\right)^2-\dfrac{25}{16}\ge\dfrac{-25}{16}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=-3\). \(min_C=\dfrac{-25}{16}\)
d) \(D=x^2+y^2-x+3y+7=\left(x^2-x+\dfrac{1}{4}\right)+\left(y^2+3y+\dfrac{9}{4}\right)+\dfrac{9}{2}=\left(x-\dfrac{1}{2}\right)^2+\left(y+\dfrac{3}{2}\right)^2+\dfrac{9}{2}\ge\dfrac{9}{2}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-3}{2}\end{matrix}\right.\). \(min_D=\dfrac{9}{2}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
11 tháng 5 2023
\(\dfrac{1}{2022}\) \(\times\) \(\dfrac{2}{5}\) + \(\dfrac{1}{2022}\) \(\times\) \(\dfrac{7}{5}\) - \(\dfrac{1}{2022}\) \(\times\) \(\dfrac{8}{10}\)
= \(\dfrac{1}{2022}\) \(\times\) ( \(\dfrac{2}{5}\) + \(\dfrac{7}{5}\) - \(\dfrac{8}{10}\))
= \(\dfrac{1}{2022}\) \(\times\) ( \(\dfrac{9}{5}\) - \(\dfrac{4}{5}\))
= \(\dfrac{1}{2022}\) \(\times\) \(\dfrac{5}{5}\)
= \(\dfrac{1}{2022}\times1\)
= \(\dfrac{1}{2022}\)
D
1
25 tháng 7 2018
1, = 6(x^2 .y -z^2 +10x+25 )
2, = x^2 -2x-9x+18
= x(x-2)-9(x-2)
= (x-9)(x-2)
16 tháng 8 2016
\(xy\left(x-y\right)+yz\left(y-z\right)+zx\left(z-x\right)=x^2y-xy^2+y^2z-yz^2+z^2z-zx^2=x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(z-y\right)\)
\(x^2\left(y-z\right)-y^2\left(x-z\right)-z^2\left(y-z\right)=\left(y-z\right)\left(x-z\right)\left(x+z\right)-y^2\left(x-z\right)=\left(x-z\right)\left(xy-yz-zx-z^2-y^2\right)\)
t cx k bt có đúng hay k đâu nha, nhớ xem kĩ lại
8 tháng 6 2020
a, \(A=2\left(x-1,5\right)-5=0\)
\(2x-3-5=0\Leftrightarrow2x-8=0\Leftrightarrow2x=8\Leftrightarrow x=4\)
b, \(B=-3x+8+6x-9=0\)
\(3x-1=0\Leftrightarrow3x=1\Leftrightarrow x=\frac{1}{3}\)
c, \(C=6x-18x^3=0\)
\(6x\left(1-3x^2\right)=0\Leftrightarrow\orbr{\begin{cases}6x=0\\1-3x^2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\3x^2=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x^2=\frac{1}{3}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=\pm\frac{1}{\sqrt{3}}\end{cases}}}\)
a)\(-6x^3+18^2+60x\)
\(=6x.-x^2+6x.3x+6x.10\)
=\(6x\left(-x^2+3x+10\right)\)
b)\(x^7+x^2+1\)
=\(\left(x^7-x\right)+\left(x^2+x+1\right)\)
=\(x\left(x^6-1\right)+\left(x^2+x+1\right)\)
\(=x\left(x^3-1\right).\left(x^3+1\right)+\left(x^2+x+1\right)\)
=\(x\left(x-1\right).\left(x^2+x+1\right).\left(x^3+1\right)+\left(x^2+x+1\right)\)
=\(\left(x^2+x+1\right).\left[x.\left(x-1\right).\left(x^3+1\right)+1\right]\)
\(=\left(x^2+x+1\right).\left[\left(x^2-x\right).\left(x^3+1\right)+1\right]\)
\(=\left(x^2+x+1\right).\left(x^5+x^4+x^2-x+1\right)\)
Bạn học tốt nhé!
c)\(x^{10}+x^5+1\)
\(=x^{10}+x^9-x^9+x^8-x^8+x^7-x^7+x^6-x^6+\)\(x^5+x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x-x+1\)
=\(\left(x^{10}+x^9+x^8\right)-\left(x^9+x^8+x^7\right)+\left(x^7+x^6+x^5\right)\)
\(-\left(x^6+x^5+x^4\right)+\left(x^5+x^4+x^3\right)-\left(x^3+x^2+x\right)\)
\(+\left(x^2+x+1\right)\)
=\(x^8\left(x^2+x+1\right)-x^7\left(x^2+x+1\right)+x^5\left(x^2+x+1\right)\)\(-x^4\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)\)
\(+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right).\left(x^8-x^7+x^5-x^4+x^3-x+1\right)\)
Lộn xộn quá ha, chúc bạn học thật giỏi!