Tìm x biết :
a) \(x\left(x^2+x+1\right)-x^2\left(x-1\right)=2x+5\)
b)(x-3)(x-2)-(x+1)(x-5)=0
c)\(x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+4,5\right)=3,5\)
Tìm x biết:
\(a.3x^2-3x\left(x-2\right)=36\)
\(b.x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+4,5\right)=3,5\)
\(c.\left(3x^2-x+1\right)\left(x-1\right)+x^2\left(4-3x\right)=\frac{5}{2}\)
Giúp mk vs ạ <3 <3
a)\(\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow6x=36\Leftrightarrow x=6\)
Tìm x biết :
a) \(\left(x-2\right)^3+6\left(x+1\right)^2-x^3+12=0\)
b) \(\left(x-5\right)\left(x+5\right)-\left(x+3\right)^3+3\left(x-2\right)^2=\left(x+1\right)^2-\left(x+4\right)\left(x-4\right)+3x^2\)
c) \(\left(2x+3\right)^2+\left(x-1\right)\left(x+1\right)=5\left(x+2\right)^2-\left(x-5\right)\left(x+1\right)+\left(x+4\right)^2\)
d) \(\left(1-3x\right)^2-\left(x-2\right)\left(9x+1\right)=\left(3x-4\right)\left(3x+4\right)-9\left(x+3\right)^2\)
a/ \(x=\dfrac{-5}{12}\)
b/ \(x\approx-1,9526\)
c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)
d/ \(x=\dfrac{-20}{13}\)
a) (x-2)3+6(x+1)2-x3+12=0
⇒ x3-6x2+12x-8+6(x2+2x+1)-x3+12=0
⇒ x3-6x2+12x-8+6x2+12x+6-x3+12=0
⇒ 24x+10=0
⇒ 24x=-10
⇒ x=-5/12
a.
PT \(\Leftrightarrow x^3-6x^2+12x-8+6(x^2+2x+1)-x^3+12=0\)
\(\Leftrightarrow x^3-6x^2+12x-8+6x^2+12x+6-x^3+12=0\)
\(\Leftrightarrow 24x+10=0\Leftrightarrow x=\frac{-5}{12}\)
b. Bạn xem lại đề, nghiệm khá xấu không phù hợp với mức độ tổng thể của bài.
c.
PT $\Leftrightarrow (4x^2+12x+9)+(x^2-1)=5(x^2+4x+4)+(x^2-4x-5)+9(x^2+6x+9)$
$\Leftrightarrow 10x^2+42x+64=0$
$\Leftrightarrow x^2+(3x+7)^2=-15< 0$ (vô lý)
Do đó pt vô nghiệm.
d.
PT $\Leftrightarrow (1-6x+9x^2)-(9x^2-17x-2)=(9x^2-16)-9(x^2+6x+9)$
$\Leftrightarrow 11x+3=-54x-97$
$\Leftrightarrow 65x=-100$
$\Leftrightarrow x=\frac{-20}{13}$
Tìm x:
a, \(x-2=\left(x-2\right)^2\)
b,\(x+5=2\left(x+5\right)^2\)
c,\(\left(x^2+1\right)\left(2x-1\right)+2x=1\)
d,\(\left(x^2+3\right)\left(x+1\right)+x=1\)
a) \(x-2=\left(x-2\right)^2\)
\(\left(x-2\right)^2-\left(x-2\right)=0\)
\(\left(x-2\right)\left(x-2-1\right)=0\)
\(\left(x-2\right)\left(x-3\right)=0\)
\(\Rightarrow x-2=0\) hoặc \(x-3=0\)
*) \(x-2=0\)
\(x=2\)
*) \(x-3=0\)
\(x=3\)
Vậy \(x=2;x=3\)
b) \(x+5=2\left(x+5\right)^2\)
\(2\left(x+5\right)^2-\left(x+5\right)=0\)
\(\left(x+5\right)\left[2\left(x+5\right)-1\right]=0\)
\(\left(x+5\right)\left(2x+10-1\right)=0\)
\(\left(x+5\right)\left(2x+9\right)=0\)
\(\Rightarrow x+5=0\) hoặc \(2x+9=0\)
*) \(x+5=0\)
\(x=-5\)
*) \(2x+9=0\)
\(2x=-9\)
\(x=-\dfrac{9}{2}\)
Vậy \(x=-5;x=-\dfrac{9}{2}\)
c) \(\left(x^2+1\right)\left(2x-1\right)+2x=1\)
\(\left(x^2+1\right)\left(2x-1\right)+2x-1=0\)
\(\left(x^2+1\right)\left(2x-1\right)+\left(2x-1\right)=0\)
\(\left(2x-1\right)\left(x^2+1+1\right)=0\)
\(\left(2x-1\right)\left(x^2+2\right)=0\)
\(\Rightarrow2x-1=0\) hoặc \(x^2+2=0\)
*) \(2x-1=0\)
\(2x=1\)
\(x=\dfrac{1}{2}\)
*) \(x^2+2=0\)
\(x^2=-2\) (vô lí)
Vậy \(x=\dfrac{1}{2}\)
d) Sửa đề:
\(\left(x^2+3\right)\left(x+1\right)+x=-1\)
\(\left(x^2+3\right)\left(x+1\right)+\left(x+1\right)=0\)
\(\left(x+1\right)\left(x^2+3+1\right)=0\)
\(\left(x+1\right)\left(x^2+4\right)=0\)
\(\Rightarrow x+1=0\) hoặc \(x^2+4=0\)
*) \(x+1=0\)
\(x=-1\)
*) \(x^2+4=0\)
\(x^2=-4\) (vô lí)
Vậy \(x=-1\)
Tìm x, biết :
a/ \(\dfrac{1}{3}x\left(x^2-4\right)=0\)
b/ \(x\left(x+5\right)=x+5\)
c/ \(x^3-\dfrac{1}{9}x=0\)
3)\(^2-\left(x+5\right)^2=0\)
e/ \(\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)
f/ \(x\left(2x-3\right)-6+4x=0\)
g/ \(2\left(3x-2\right)^2-9x^2+4=0\)
h/ \(x^2\left(x+1\right)+2x\left(x+1\right)=0\)
i/ \(4x^2+9x+5=0\)
a) \(\Rightarrow\dfrac{1}{3}x\left(x-2\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
b) \(\Rightarrow\left(x+5\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)
c) \(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
e) \(\Rightarrow\left(x+2\right)\left(x+2-x+2\right)=0\Rightarrow\left(x+2\right).4=0\Rightarrow x=-2\)
f) \(\Rightarrow x\left(2x-3\right)+2\left(2x-3\right)=0\Rightarrow\left(2x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)
g) \(\Rightarrow2\left(3x-2\right)^2-\left(3x-2\right)\left(3x+2\right)=0\Rightarrow\left(3x-2\right)\left(3x-6\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)
h) \(\Rightarrow x\left(x+1\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)
i) \(\Rightarrow4x\left(x+1\right)+5\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(4x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{5}{4}\end{matrix}\right.\)
BÀI 6 tìm x
1,\(2x\left(x-5\right)-\left(3x+2x^2\right)=0\) 2,\(x\left(5-2x\right)+2x\left(x-1\right)=13\)
3,\(2x^3\left(2x-3\right)-x^2\left(4x^2-6x+2\right)=0\) 4,\(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
5,\(6x^2-\left(2x-3\right)\left(3x+2\right)=1\) 6,\(2x\left(1-x\right)+5=9-2x^2\)
1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)
=>-13x=0
=>x=0
2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
=>3x=13
=>x=13/3
3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)
=>-2x^2=0
=>x=0
4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
=>-8x=6-14=-8
=>x=1
`1)2x(x-5)-(3x+2x^2)=0`
`<=>2x^2-10x-3x-2x^2=0`
`<=>-13x=0`
`<=>x=0`
___________________________________________________
`2)x(5-2x)+2x(x-1)=13`
`<=>5x-2x^2+2x^2-2x=13`
`<=>3x=13<=>x=13/3`
___________________________________________________
`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`
`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`
`<=>x=0`
___________________________________________________
`4)5x(x-1)-(x+2)(5x-7)=0`
`<=>5x^2-5x-5x^2+7x-10x+14=0`
`<=>-8x=-14`
`<=>x=7/4`
___________________________________________________
`5)6x^2-(2x-3)(3x+2)=1`
`<=>6x^2-6x^2-4x+9x+6=1`
`<=>5x=-5<=>x=-1`
___________________________________________________
`6)2x(1-x)+5=9-2x^2`
`<=>2x-2x^2+5=9-2x^2`
`<=>2x=4<=>x=2`
Tìm x , biết :
a. \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)
b. \(2x^3-50x=0\)
c.\(5x^2-4\left(x^2-2x+1\right)-5=0\)
d. \(x^3-x=0\)
e. \(27x^3-27x^2+9x-1=1\)
a) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)
\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)
\(\Leftrightarrow24x+25=15\)
\(\Leftrightarrow24x=-10\)
hay \(x=-\dfrac{5}{12}\)
b) Ta có: \(2x^3-50x=0\)
\(\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
c) Ta có: \(5x^2-4\left(x^2-2x+1\right)-5=0\)
\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)
\(\Leftrightarrow x^2+8x-9=0\)
\(\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=1\end{matrix}\right.\)
d) Ta có: \(x^3-x=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
e) Ta có: \(27x^3-27x^2+9x-1=1\)
\(\Leftrightarrow\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2-1^3=1\)
\(\Leftrightarrow\left(3x-1\right)^3=1\)
\(\Leftrightarrow3x-1=1\)
\(\Leftrightarrow3x=2\)
hay \(x=\dfrac{2}{3}\)
giúp mk với tứ tư mk phải nộp rùi
bài 1:
a, \(2x\left(3x^2-5x+3\right)\)
b, \(-2x\left(x^2+5x-3\right)\)
c, \(\dfrac{-1}{2}x\left(2x^3-4x+3\right)\)
bài 2:
a,\(\left(2x-1\right).\left(x^2-5-4\right)\)
b,\(-\left(5x-4\right).\left(2x+3\right)\)
c,\(\left(2x-y\right).\left(4x^2-2xy+y^2\right)\)
d,\(\left(3x-4\right).\left(x+4\right).\left(5-x\right).\left(2x^2+3x-1\right)\)
e,\(7\left(x-4\right)-\left(7x+3\right).\left(2x^2-x+4\right)\)
bài 3:
c/m rằng gtri của biểu thức ko phụ thuộc vào gtri của biến
a,\(x\left(3x+12\right)-\left(7x-20\right)+x^2\left(2x-3\right)-x\left(2x^2+5\right)\)
b,\(3\left(2x-1\right)-5\left(x-3\right)+6\left(3x-4\right)-19x\)
bài 4 :tìm x biết
a, \(3x+2\left(5-x\right)=0\)
b,\(x\left(2x-1\right).\left(x+5\right)-\left(2x^2+1\right).\left(x+4,5\right)=3,5\)
c,\(3x^2-3x\left(x-2\right)=36\)
d,\(\left(3x^2-x+1\right).\left(x-1\right)+x^2.\left(4-3x\right)=\dfrac{5}{2}\)
1,
a,\(2x\left(3x^2-5x+3\right)\)
\(=6x^3-10x^2+6x\)
b,\(-2x\left(x^2+5x-3\right)\)
\(=-2x^3-10x^2+6x\)
c,\(-\dfrac{1}{2}x\left(2x^3-4x+3\right)\)
\(=-x^4+2x^2-\dfrac{3}{2}x\)
Bài 2:
a) \(\left(2x-1\right)\left(x^2-5-4\right)\)
\(=\left(2x-1\right)\left(x^2-9\right)\)
\(=2x^3-18x-x^2+9\)
b) \(-\left(5x-4\right)\left(2x+3\right)\)
\(=-\left(10x^2+15x-8x-12\right)\)
\(=-10x^2-7x+12\)
c) \(\left(2x-y\right)\left(4x^2-2xy+y^2\right)\)
\(=8x^3-y^3\)
Bài 3: (chỉ cần kết quả ko chứa biến là ta có đpcm, nói chung bài này yêu cầu ta rút gọn)
a) \(x\left(3x+12\right)-\left(7x-20\right)+x^2\left(2x-3\right)-x\left(2x^2+5\right)\)
\(=3x^2+12x-7x+20+2x^3-3x^2-2x^3-5x\)
\(=20\)
b) \(3\left(2x-1\right)-5\left(x-3\right)+6\left(3x-4\right)-19x\)
\(=6x-3-5x+15+18x-24-19x\)
\(=-12\)
BT3: Tìm x
\(a,\left(x+2\right)^2-9=0\)
\(b,x^2-2x+1=25\)
\(c,\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)
\(d,\left(x-1\right)\left(x^2+x+1\right)+x\left(x+2\right)\left(2-x\right)=5\)
\(a,\left(x+2\right)^2-9=0\\ \Leftrightarrow\left(x+2-3\right)\left(x+2+3\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\\ Vậy\dfrac{ }{ }S=\left\{1;-5\right\}\)
\(b,x^2-2x+1=25\\ \Leftrightarrow\left(x-1\right)^2=25\\ \Leftrightarrow\left(x-1\right)^2-25=0\\ \Leftrightarrow\left(x-1-5\right)\left(x-1+5\right)=0\\ \Leftrightarrow\left(x-6\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ Vậy\dfrac{ }{ }S=\left\{6;-4\right\}\)
\(c,\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\\ \Leftrightarrow25x^2+10x+1-25x^2+9=30\\ \Leftrightarrow25x^2+10x-25x^2=30-1-9\\ \Leftrightarrow10x=20\\ \Leftrightarrow x=2\\ Vậy\dfrac{ }{ }S=\left\{2\right\}\)
\(d,\left(x-1\right)\left(x^2+x+1\right)+x\left(x+2\right)\left(2-x\right)=5\\ \Leftrightarrow x^3-1-x\left(x^2-4\right)=5\\ \Leftrightarrow x^3-1-x^3+4x=5\\ \Leftrightarrow x^3-x^3+4x=5+1\\ \Leftrightarrow4x=6\\ \Leftrightarrow x=\dfrac{3}{2}\\ Vậy\dfrac{ }{ }S=\left\{\dfrac{3}{2}\right\}\)
a: =>(x+2-3)(x+2+3)=0
=>(x-1)(x+5)=0
=>x=1 hoặc x=-5
b: =>(x-1)^2=25
=>x-1=5 hoặc x-1=-5
=>x=-4 hoặc x=6
c: =>25x^2+10x+1-25x^2+9=30
=>10x+10=30
=>x+1=3
=>x=2
d: =>x^3-1-x(x^2-4)=5
=>x^3-1-x^3+4x=5
=>4x=6
=>x=3/2
i, \(\left(x-1\right)\left(x+3\right)-\left(x-1\right)\left(2x+1\right)=0\)
k, \(\left(x+2\right)\left(x+1\right)-\left(x-3\right)\left(x+2\right)=0\)
l, \(\left(x-2\right)\left(x+3\right)=\left(x-2\right)\left(2x+5\right)\)
\(\left(x-1\right)\left(-x+2\right)=0\Leftrightarrow x=1;x=2\)
\(\left(x+2\right)\left(x+1-x+3\right)=0\Leftrightarrow x=-2\)
\(\left(x-2\right)\left(x+3\right)-\left(x-2\right)\left(2x+5\right)=0\Leftrightarrow\left(x-2\right)\left(-x-2\right)=0\Leftrightarrow x=-2;x=2\)
\(i,\left(x-1\right)\left(x+3\right)-\left(x-1\right)\left(2x+1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+3-2x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(-x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\\ k,\left(x+2\right)\left(x+1\right)-\left(x-3\right)\left(x+2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x+1-x+3\right)=0\\ \Leftrightarrow4\left(x+2\right)=0\\ \Leftrightarrow x+2=0\\ \Leftrightarrow x=-2\\ l,\left(x-2\right)\left(x+3\right)=\left(x-2\right)\left(2x+5\right)\\ \Leftrightarrow\left(x-2\right)\left(2x+5\right)-\left(x-2\right)\left(x+3\right)=0\\ \Leftrightarrow\left(x-2\right)\left(2x+5-x-3\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)