Giải phương trình sau
\(\left(x-2\right)\left(x+2\right)-\left(2x+1\right)^2=x\cdot\left(2-3x\right)\)
Những câu hỏi liên quan
Giải phương trìnha. frac{1}{27}cdotleft(x-3right)^3-frac{1}{125}cdotleft(x-5right)^30b.125x^3-left(2x+1right)^3-left(3x-1right)^30c.left(x-3right)^3+left(x+1right)^38cdotleft(x-1right)^3d.left(x^2-3x+2right)cdotleft(x^2+15x+56right)+80e.left(2x^2-3x+1right)cdotleft(2x^2+5x+1right)-9x^20f.left(x+6right)^4+left(x+8right)^4272
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Giải phương trình
a. \(\frac{1}{27}\cdot\left(x-3\right)^3-\frac{1}{125}\cdot\left(x-5\right)^3=0\)
b.\(125x^3-\left(2x+1\right)^3-\left(3x-1\right)^3=0\)
c.\(\left(x-3\right)^3+\left(x+1\right)^3=8\cdot\left(x-1\right)^3\)
d.\(\left(x^2-3x+2\right)\cdot\left(x^2+15x+56\right)+8=0\)
e.\(\left(2x^2-3x+1\right)\cdot\left(2x^2+5x+1\right)-9x^2=0\)
f.\(\left(x+6\right)^4+\left(x+8\right)^4=272\)
Tìm x :
3xcdotleft(x-2right)-2xcdotleft(2x-1right)left(1-xright)cdotleft(1+xright)
left(5x+3right)cdotleft(3x-5right)-left(x-2right)cdotleft(2x+1right)6xcdotleft(3x+1right)-x^2
left(2x-1right)cdotleft(2x+1right)-3cdotleft(x-1right)left(1-4xright)cdotleft(1-xright)
left(2x^2+1right)cdotleft(3x^2-1right)-left(4x^2-3right)cdotleft(x^2+1right)xcdotleft(2x^3+1right)
GIÚP MK ĐI MAI MK PHẢI NỘP RÙI !
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Tìm x :
\(3x\cdot\left(x-2\right)-2x\cdot\left(2x-1\right)=\left(1-x\right)\cdot\left(1+x\right)\)
\(\left(5x+3\right)\cdot\left(3x-5\right)-\left(x-2\right)\cdot\left(2x+1\right)=6x\cdot\left(3x+1\right)-x^2\)
\(\left(2x-1\right)\cdot\left(2x+1\right)-3\cdot\left(x-1\right)=\left(1-4x\right)\cdot\left(1-x\right)\)
\(\left(2x^2+1\right)\cdot\left(3x^2-1\right)-\left(4x^2-3\right)\cdot\left(x^2+1\right)=x\cdot\left(2x^3+1\right)\)
GIÚP MK ĐI MAI MK PHẢI NỘP RÙI !
1> 3x(x-2)-2x(2x-1)=(1-x)(1+x)
⇔\(3x^2\)-6x-\(4x^2\)+2x=1-\(x^2\)
⇔-1\(x^2\) - 4x= 1- \(x^2\)
⇔ -1\(x^2\) -4x+ \(x^2\) = 1
⇔-4x=1
⇔ x = \(\dfrac{-1}{4}\)
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giải các phương trình sau
a. \(\left(x-3\right)\cdot\left(x-5\right)\cdot\left(x-6\right)\cdot\left(x-10\right)=24x^2\)
b. \(\left(x-6\right)^4+\left(x-8\right)^4=272\)
c. \(x^4-3x^3+2x^2-9x+9=0\)
cho hàm số \(f\left(x\right)=x^3-3x^2+2\)
a, giải bất phương trình \(f'\left(x\right)\le0\)
b, giải phương trình \(f'=\left(x^2-3x+2\right)=0\)
c, đặt \(g\left(x\right)=f\left(1-2x\right)+x^2-x+2022\) giải bất phương trình\(g'\left(x\right)\ge0\)
\(a,f'\left(x\right)=3x^2-6x\\ f'\left(x\right)\le0\Leftrightarrow3x^2-6x\le0\\ \Leftrightarrow3x\left(x-2\right)\le0\Leftrightarrow0\le x\le2\)
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Lời giải:
a. $f'(x)\leq 0$
$\Leftrightarrow 3x^2-6x\leq 0$
$\Leftrightarrow x(x-2)\leq 0$
$\Leftrightarrow 0\leq x\leq 2$
b.
$f'(x)=x^2-3x+2=0$
$\Leftrightarrow 3x^2-6x=x^2-3x+2=0$
$\Leftrightarrow 3x(x-2)=(x-1)(x-2)=0$
$\Leftrightarrow x-2=0$
$\Leftrightarrow x=2$
c.
$g(x)=f(1-2x)+x^2-x+2022$
$g'(x)=(1-2x)'f(1-2x)'_{1-2x}+2x-1$
$=-2[3(1-2x)^2-6(1-2x)]+2x-1$
$=-24x^2+2x+5$
$g'(x)\geq 0$
$\Leftrightarrow -24x^2+2x+5\geq 0$
$\Leftrightarrow (5-12x)(2x-1)\geq 0$
$\Leftrightarrow \frac{-5}{12}\leq x\leq \frac{1}{2}$
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Chứng minh biểu thức không phụ thuộc x :
1, \(\left(3x-1\right)^2-2\cdot\left(2x-3\right)\cdot\left(2x+3\right)-\left(x-3\right)^2\)
2, \(\left(3x+2\right)^3-\left(3x-2\right)^3-3\cdot\left(6x-1\right)\cdot\left(6x+1\right)\)
3, \(\left(3x-5\right)^2+3\cdot\left(x+1\right)\cdot\left(x-1\right)-\left(4x-3\right)^2+\left(2x+2\right)\cdot\left(2x+1\right)\)
Giải các bất phương trình sau :
\(a.4\left(x-3\right)^2-\left(2x-1\right)^2\ge12\)
\(b.\left(x-4\right)\left(x+4\right)\ge\left(x+3\right)^2+5\)
c. \(\left(3x-1\right)^2-9\left(x+2\right)\left(x-2\right)< 5x\)
\(a,4\left(x-3\right)^2-\left(2x-1\right)^2\ge12\)
\(\Leftrightarrow4x^2-24x+36-4x^2-4x+1\ge12\)
\(\Leftrightarrow-28x+37\ge12\)
\(\Leftrightarrow-28x\ge12-37\)
\(\Leftrightarrow-28x\ge-25\)
\(\Leftrightarrow x\le\dfrac{25}{28}\)
Vậy \(S=\left\{x\left|x\le\dfrac{25}{28}\right|\right\}\)
b, \(\left(x-4\right)\left(x+4\right)\ge\left(x+3\right)^2+5\)
\(\Leftrightarrow x^2-16\ge x^2+6x+9+5\)
\(\Leftrightarrow x^2-x^2-6x\ge9+5+16\)
\(\Leftrightarrow-6x\ge30\)
\(\Leftrightarrow x\le-5\)
Vậy \(S=\left\{x\left|x\le-5\right|\right\}\)
\(c,\left(3x-1\right)^2-9\left(x+2\right)\left(x-2\right)< 5x\)
\(\Leftrightarrow9x^2-6x-1-9x^2+36< 5x\)
\(\Leftrightarrow9x^2-9x^2-6x-5x+36+1< 0\)
\(\Leftrightarrow-11x+37< 0\)
\(\Leftrightarrow-11x< -37\)
\(\Leftrightarrow x>\dfrac{37}{11}\)
vậy \(S=\left\{x\left|x>\dfrac{37}{11}\right|\right\}\)
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Giải các phương trình sau:1, dfrac{x-1}{3}-xdfrac{2x-4}{4}2, left(x-2right)left(2x-1right)x^2-2x3, 3x^2-4x+104, left|2x-4right|05, left|3x+2right|46, left|2x-5right|left|-x+2right|*Giúp mình với mình đg cần gấp ạ T_T
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Giải các phương trình sau:
1, \(\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\)
2, \(\left(x-2\right)\left(2x-1\right)=x^2-2x\)
3, \(3x^2-4x+1=0\)
4, \(\left|2x-4\right|=0\)
5, \(\left|3x+2\right|=4\)
6, \(\left|2x-5\right|=\left|-x+2\right|\)
*Giúp mình với mình đg cần gấp ạ T_T
\(1.\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}.\Leftrightarrow\dfrac{x-1-3x}{3}=\dfrac{x-2}{2}.\Leftrightarrow\dfrac{-2x-1}{3}-\dfrac{x-2}{2}=0.\)
\(\Leftrightarrow\dfrac{-4x-2-3x+6}{6}=0.\Rightarrow-7x+4=0.\Leftrightarrow x=\dfrac{4}{7}.\)
\(2.\left(x-2\right)\left(2x-1\right)=x^2-2x.\Leftrightarrow\left(x-2\right)\left(2x-1\right)-x\left(x-2\right)=0.\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1-x\right)=0.\Leftrightarrow\left(x-2\right)\left(x-1\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=1.\end{matrix}\right.\)
\(3.3x^2-4x+1=0.\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{3}\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=\dfrac{1}{3}.\end{matrix}\right.\)
\(4.\left|2x-4\right|=0.\Leftrightarrow2x-4=0.\Leftrightarrow x=2.\)
\(5.\left|3x+2\right|=4.\Leftrightarrow\left[{}\begin{matrix}3x+2=4.\\3x+2=-4.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}.\\x=-2.\end{matrix}\right.\)
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\(1,\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\\ \Leftrightarrow\dfrac{x-1}{3}-x=\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{2\left(x-1\right)-6x}{6}=\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow2\left(x-1\right)-6x=3\left(x-2\right)\\ \Leftrightarrow2x-2-6x=3x-6\\ \Leftrightarrow-4x-2=3x-6\)
\(\Leftrightarrow3x-6+4x+2=0\\ \Leftrightarrow7x-4=0\\ \Leftrightarrow x=\dfrac{4}{7}\)
\(2,\left(x-2\right)\left(2x-1\right)=x^2-2x\\ \Leftrightarrow2x^2-4x-x+2=x^2-2x\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(3,3x^2-4x+1=0\\ \Leftrightarrow\left(3x^2-3x\right)-\left(x-1\right)=0\\ \Leftrightarrow3x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(4,\left|2x-4\right|=0\\ \Leftrightarrow2x-4=0\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\)
\(5,\left|3x+2\right|=4\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)
\(6,\left|2x-5\right|=\left|-x+2\right|\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=-x+2\\2x-5=x-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=7\\x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)
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giải phương trình\(\left(x-1\right)\cdot\left(x^2+3x-2\right)-\left(x^3-1\right)=0\)
(x-1)(x^2+3x-2-x^2-x-1)=(x-1)(2x-3)=0=> x=1 hoăc x=3/2
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(x-1)(x2+3x-2)-(x3-1)=0
<=>(x-1)(x2+3x-2)-(x-1)(x2+x+1)=0
<=>(x-1)(x2+3x-2-(x2+x+1))=0
<=>(x-1)(x2+3x-2-x2-x-1)=0
<=>(x-1)(2x-3)=0
<=>x-1=0 hay 2x-3=0
<=>x=1 hay x=\(\frac{3}{2}\)
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giải phương trình\(\left(x-1\right)\cdot\left(x^2+3x-2\right)-\left(x^3-1\right)=0\)
<=>(x-1)(x2+3x-2) - (x-1)(x2+x+1)=0<=>(x-1)(x2+3x-2-x2-x-1)=0<=>(x-1)(2x-3)=0<=>x-1=0 hoặc 2x-3=0<=>x=1 hoặc x=3/2
VẬY S=1;3/2 :)))))))))))))))))))))))))
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