tìm x:
1.1/ x2(x-3)+3x=1
1.2/ 9(x-3)2=(x2-9)2
Tìm x biết:
d) (x-2)3-(x-3).(x2+3x+9)+6.(x+1)2=15
e) (x-1)3+(2-x).(4+2x+x2)+3x.(x+2)=17
f) (3x+3)2-18x=36+(x-3).(x2+3x+9)
Giải chi tiết giúp mình nha.Cảm ơn.
\(d,\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=15\\ \Leftrightarrow24x=-10\Leftrightarrow x=-\dfrac{5}{12}\\ e,\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=17\\ \Leftrightarrow9x=10\Leftrightarrow x=\dfrac{10}{9}\\ f,\Leftrightarrow9x^2+18x+9-18x=36+x^3-27\\ \Leftrightarrow x^3-9x^2=0\Leftrightarrow x^2\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\)
Bài 1: Tìm x:
1) (x-3)3 -( x-3)(x2+ 3x+9) +6( x+1)2+ 3x2 = -33
2) (X-3)( X2+ 3X+9) - X(X-2)( 2+X) = 1
3) (X+2)(X2 - 2X+4) – X(X-3)(X+3) = 26
a: Ta có: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2+3x^2=-33\)
\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+6x^2+12x+1+3x^2=-33\)
\(\Leftrightarrow39x=-34\)
hay \(x=-\dfrac{34}{39}\)
b: Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x-2\right)\left(x+2\right)=1\)
\(\Leftrightarrow x^3-27-x^3+4x=1\)
\(\Leftrightarrow4x=28\)
hay x=7
c: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x-3\right)\left(x+3\right)=26\)
\(\Leftrightarrow x^3+8-x^3+9x=26\)
\(\Leftrightarrow x=2\)
Tìm x,biết:
a)6x2-(2x+5).(3x-2)=-12
b)(x+3).(x2-3x+9)-x.(x2+2)=12-5x
c)x2-25=6x-9
\(a,\Leftrightarrow6x^2-6x^2-11x+10=-12\\ \Leftrightarrow-11x=-22\\ \Leftrightarrow x=2\\ b,\Leftrightarrow x^3+27-x^3-2x=12-5x\\ \Leftrightarrow3x=-15\\ \Leftrightarrow x=-5\\ c,\Leftrightarrow x^2-6x-16=0\\ \Leftrightarrow\left(x-8\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
a: ta có: \(6x^2-\left(2x+5\right)\left(3x-2\right)=-12\)
\(\Leftrightarrow6x^2-6x^2+4x-15x+10=-12\)
\(\Leftrightarrow-11x=-22\)
hay x=2
b: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2+2\right)=12-5x\)
\(\Leftrightarrow x^3+27-x^3-2x+5x=12\)
\(\Leftrightarrow x=-5\)
Tìm x :
a) x (3x + 1) + (x -1)2 - (2x + 1)(2x -1) = 0
b) (x + 1)3 + (2 - x)3 - 9(x - 3)(x+3) = 0
c) (x - 1)3 - (x + 3)(x2 - 3x + 9) + 3x2 = 25
d) (x + 2)3 - ( x +1)(x2 - x + 1) - 6(x - 1)2 = 23
e) (x + 3)(x2 - 3x + 9) - x(x - 2)(x+2) + 11 = 0
f) x(x - 3) - x + 3 = 0
Lời giải:
a. $x(3x+1)+(x-1)^2-(2x+1)(2x-1)=0$
$\Leftrightarrow (3x^2+x)+(x^2-2x+1)-(4x^2-1)=0$
$\Leftrightarrow 3x^2+x+x^2-2x+1-4x^2+1=0$
$\Leftrightarrow (3x^2+x^2-4x^2)+(x-2x)+(1+1)=0$
$\Leftrightarrow -x+2=0$
$\Leftrightarrow x=2$
b.
$(x+1)^3+(2-x)^3-9(x-3)(x+3)=0$
$\Leftrightarrow [(x+1)+(2-x)][(x+1)^2-(x+1)(2-x)+(2-x)^2]-9(x-3)(x+3)=0$
$\Leftrightarrow 3[x^2+2x+1-(x-x^2+2)+(x^2-4x+4)]-9(x-3)(x+3)=0$
$\Leftrightarrow 3(3x^2-3x+3)-9(x^2-9)=0$
$\Leftrightarrow 9(x^2-x+1)-9(x^2-9)=0$
$\Leftrightarrow 9(x^2-x+1-x^2+9)=0$
$\Leftrightarrow 9(-x+10)=0$
$\Leftrightarrow -x+10=0\Leftrightarrow x=10$
c.
$(x-1)^3-(x+3)(x^2-3x+9)+3x^2=25$
$\Leftrightarrow (x^3-3x^2+3x-1)-(x^3+3^3)+3x^2=25$
$\Leftrightarrow x^3-3x^2+3x-1-x^3-27+3x^2=25$
$\Leftrightarrow (x^3-x^3)+(-3x^2+3x^2)+3x-28=25$
$\Leftrightarrow 3x-28=25$
$\Leftrightarrow x=\frac{53}{3}$
d.
$(x+2)^3-(x+1)(x^2-x+1)-6(x-1)^2=23$
$\Leftrightarrow (x^3+6x^2+12x+8)-(x^3+1)-6(x^2-2x+1)=23$
$\Leftrightarrow x^3+6x^2+12x+8-x^3-1-6x^2+12x-6=23$
$\Leftrightarrow (x^3-x^3)+(6x^2-6x^2)+(12x+12x)+(8-1-6)=23$
$\Leftrightarrow 24x+1=23$
$\Leftrgihtarrow 24x=22$
$\Leftrightarrow x=\frac{11}{12}$
e.
$(x+3)(x^2-3x+9)-x(x-2)(x+2)+11=0$
$\Leftrightarrow x^3+3^3-x(x^2-4)+11=0$
$\Leftrightarrow x^3+27-x^3+4x+11=0$
$\Leftrightarrow (x^3-x^3)+4x+(27+11)=0$
$\Leftrightarrow 4x+38=0$
$\Leftrightarrow x=\frac{-19}{2}$
f.
$x(x-3)-x+3=0$
$\Leftrightarrow x(x-3)-(x-3)=0$
$\Leftrightarrow (x-3)(x-1)=0$
$\Leftrightarrow x-3=0$ hoặc $x-1=0$
$\Leftrightarrow x=3$ hoặc $x=1$
Bài 3.Tìm x biết
b) 3x( x2 + 4) –2x2 -8 =0
a) x2+5x =0
c) (x-3) (x2 + 3x +9) +x( x+2) (2- x) =36
b) \(\Leftrightarrow3x^3+12x-2x^2-8=0\\ \Leftrightarrow\left(3x^3-2x^2\right)+\left(12x-8\right)=0\\ \Leftrightarrow x^2\left(3x-2\right)+4\left(3x-2\right)=0\\ \Leftrightarrow\left(x^2+4\right)\left(3x-2\right)=0\)
Vì \(x^2+4>0\Rightarrow3x-2=0\Rightarrow x=\dfrac{2}{3}\)
c) \(x^2+5x=0\\ \Leftrightarrow x\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
d) \(\Leftrightarrow x^3-27+x\left(4-x^2\right)=36\\ \Leftrightarrow x^3+4x-x^3=63\\ \Leftrightarrow4x=63\\ \Leftrightarrow x=\dfrac{63}{4}\)
b) 3x(x\(^3\) +12x-2x\(^2\)-8=0
3x(x\(^2\)+4)-2(x\(^2\)+4)=0
(x\(^2\)+4)(3x-2)=0
\(\Leftrightarrow\left[{}\begin{matrix}X^2+4=0\\3X-2=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x\in Z\\X=\dfrac{2}{3}\end{matrix}\right.\)
a) x\(^2\)+5x=0
x(x+5)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
c)(x-3)(x\(^2\)+3x+9)+x(x+2)(2-x)=36
x\(^3\)-27+x(x+2)(2-x)=36
4x-27=36
4x=36+27
4x=63
x=\(\dfrac{63}{4}\)
Bài 5. Tìm x , biết rằng: a) x(x + 5)(x – 5) – (x + 2)(x2 – 2x + 4) = 3
b) (x – 3)3 – (x – 3)(x2 + 3x + 9) + 9(x + 1)2 = 15
c) (x+5)(x2 –5x +25) – (x – 7) = x3
d) (x+2)(x2 – 2x + 4) – x(x2 + 2) = 4
`a) x(x + 5)(x – 5) – (x + 2)(x^2 – 2x + 4) = 3`
`<=>x(x^2-25)-(x^3-8)=3`
`<=>x^3-25x-x^3+8=3`
`<=>-25x=-5`
`<=>x=1/5`
`b) (x – 3)^3 – (x – 3)(x^2 + 3x + 9) + 9(x + 1)^2 = 15`
`<=>x^3-9x^2+27x-27-(x^3-27)+9(x^2+2x+1)=15`
`<=>-9x^2+27x+9x^2+18x+9=15`
`<=>45x+9=15`
`<=>45x=6`
`<=>x=6/45=2/15`
`c) (x+5)(x^2 –5x +25) – (x – 7) = x^3`
`<=>x^3-125-x+7=x^3`
`<=>x^3-x-118=x^3`
`<=>-x-118=0`
`<=>-x=118<=>x=-118`
`d) (x+2)(x^2 – 2x + 4) – x(x^2 + 2) = 4 `
`<=>x^3+8-x^3-2x=4`
`<=>8-2x=4`
`<=>2x=4<=>x=2`
Tìm x biết:
a) ( x – 3 ) 3 – ( x – 3 ) ( x 2 + 3 x + 9 ) + 9 ( x + 1 ) 2 = 15;
b) x(x – 5)(x + 5) – (x + 2)( x 2 - 2x + 4) = 3.
a) Rút gọn VT = 45x + 8. Từ đó tìm được x = 2 15 .
b) Rút gọn VT = -25x – 8. Từ đó tìm được x = − 11 25 .
Bài 2: (2 điểm) Tìm x, biết:
a) (3x + 4)2 – (3x – 1)(3x + 1) = 49
b) x2 – 4x + 4 = 9(x – 2)
c) x2 – 25 = 3x - 15
d) (x – 1)3 + 3(x + 1)2 = (x2 – 2x + 4)(x + 2)
a) \(\Rightarrow9x^2+24x+16-9x^2+1=49\)
\(\Rightarrow24x=32\Rightarrow x=\dfrac{4}{3}\)
b) \(\Rightarrow x^2-13x+22=0\)
\(\Rightarrow\left(x-11\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=11\\x=2\end{matrix}\right.\)
c) \(\Rightarrow x^2-3x-10=0\)
\(\Rightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
Tìm x biết :
a) (x+2)2+(x-3)2 = 2x(x+7)
b) (x+3)(x2-3x+9) = x(x2+4)-1
a, (x+2)2+(x-3)2=2x(x+7)
x.2+2.2+x.2+(-3).2-2x=8
2x+4+2x-6-2x=8
(2x+2x)+(4-6)=8
4x-2=8
4x=8+2
4x=10
X=10:4
X=5/2
a) (x+2)2+(x-3)2 = 2x(x+7)
⇒x2+4x+4+x2-6x+9=2x2+14x
⇒x2+4x+4+x2-6x+9-2x2-14x=0
⇒ -16x+13=0
⇒ x=\(\dfrac{13}{16}\)
b) (x+3)(x2-3x+9) = x(x2+4)-1
⇒x(x2-3x+9)+3(x2-3x+9)=x3+4x-1
⇒x3-3x2+9x+3x2-9x+27-x3-4x+1=0
⇒-4x+28=0
⇒x=7
a: Ta có: \(\left(x+2\right)^2+\left(x-3\right)^2=2x\left(x+7\right)\)
\(\Leftrightarrow x^2+4x+4+x^2-6x+9-2x^2-14x=0\)
\(\Leftrightarrow-16x=-13\)
hay \(x=\dfrac{13}{16}\)
b: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)=x\left(x^2+4\right)-1\)
\(\Leftrightarrow x^3+27-x^3-4x=-1\)
\(\Leftrightarrow-4x=-28\)
hay x=7
Tìm x, biết:
a) x 2 (x - 5) + 5 - x = 0; b) 3 x 4 - 9 x 3 = -9 x 2 + 27x;
c) x 2 (x + 8) + x 2 = -8x; d) (x + 3)( x 2 -3x + 5) = x 2 + 3x.