Tìm x: x^3-3x^2-4x+12=0
Mình cảm ơn nhiều ạ
Tìm x: a) 2(x+3) - x^2-3x=0
b) 4x^2-25-(2x-5)(2x+7)=0
Mình cảm ơn nhiều ạ
\(a)2\left(x+3\right)-x^2-3x=0\)
\(\Leftrightarrow2x+6-x^2-3x=0\)
\(\Leftrightarrow-x^2+2x-3x+6=0\)
\(\Leftrightarrow-x^2-x+6=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow x^2+3x-2x-6=0\)
\(\Leftrightarrow x\left(x+3\right)-2\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow x+3=0\) hoặc \(x-2=0\)
\(\Leftrightarrow x=-3\) hoặc \(x=2\)
\(b)4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)
\(=\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)
\(=\left(2x-5\right)\left(2x+5-2x-7\right)=0\)
\(=-2\left(2x-5\right)=0\)
\(=2x-5=0\Leftrightarrow x=\frac{5}{2}\)
Cảm ơn bạn rất nhiều
Tìm x
a) 3*(x-3)+4x-4=1
b) 6x*(3x+5)-2x*(9x-2)+(17-x)*(x-1)+x*(x-18)=0
Mng giúp mk bài này nka ạ, cảm ơn nhiều
mọi người ơi giúp em với ạ ! em cảm ơn mọi người nhiều lắm ạ
3x(2 - x) - 5 = 4– ( 3x2 + 2)
(5-3x)(4x+1) = (2x + 1)(3x – 5)
a: =>6x-3x^2-5=4-3x^2-2
=>6x-5=2
=>6x=7
=>x=7/6
b: =>20x+5-12x^2-3x=6x^2-10x+3x-5
=>-12x^2+17x+5-6x^2+7x+5=0
=>-18x^2+24x+10=0
=>x=5/3 hoặc x=-1/3
Bài 1 ;Tìm x biết
1) (2x+1 )^3 - (2x+1)(4x^2-2x+1)-3(2x-1)=15
2) x(x-4)(x+4)-(x-5)(x^2 + 5x +25)=13
Bài 2 : Cmr các giá trị của biểu thức sau không thuộc vào giá ttij của biến :
A= (x+5)(x^2-5x+25)-(2x+1)^3-28x^3+3x(-11x+5)
B = (3x+2)^3 - 18x(3x+2)+(x-1)^3-28^3+3x(x-1)
C= (4x-1)(16x^2+4x+1)-(4x+1)^3+12(4x+1)^3+12(4x+1)-15
Tính giúp mình ạ ! Cảm ơn các cậu rất nhieeufuuuu :3
Bài 1.
1) ( 2x + 1 )3 - ( 2x + 1 )( 4x2 - 2x + 1 ) - 3( 2x - 1 ) = 15
<=> 8x3 + 12x2 + 6x + 1 - [ ( 2x )3 - 13 ] - 6x + 3 = 15
<=> 8x3 + 12x2 + 4 - 8x3 + 1 = 15
<=> 12x2 + 15 = 15
<=> 12x2 = 0
<=> x = 0
2) x( x - 4 )( x + 4 ) - ( x - 5 )( x2 + 5x + 25 ) = 13
<=> x( x2 - 16 ) - ( x3 - 53 ) = 13
<=> x3 - 16x - x3 + 125 = 13
<=> 125 - 16x = 13
<=> 16x = 112
<=> x = 7
Bài 2.
A = ( x + 5 )( x2 - 5x + 25 ) - ( 2x + 1 )3 - 28x3 + 3x( -11x + 5 )
= x3 + 53 - ( 8x3 + 12x2 + 6x + 1 ) - 28x3 - 33x2 + 15x
= -27x3 + 125 - 8x3 - 12x2 - 6x - 1 - 33x2 + 15x
= -33x3 - 45x2 + 9x + 124 ( có phụ thuộc vào biến )
B = ( 3x + 2 )3 - 18x( 3x + 2 ) + ( x - 1 )3 - 28x3 + 3x( x - 1 )
= 27x3 + 54x2 + 36x + 8 - 54x2 - 36x + x3 - 3x2 + 3x - 1 - 28x3 + 3x2 - 3x
= 7 ( đpcm )
C = ( 4x - 1 )( 16x2 + 4x + 1 ) - ( 4x + 1 )3 + 12( 4x + 1 )3 + 12( 4x + 1 ) - 15
= ( 4x )3 - 13 - [ ( 4x + 1 )3 - 12( 4x + 1 )3 - 12( 4x + 1 ) ] - 15
= 64x3 - 1 - ( 4x + 1 )[ ( 4x + 1 )2 - 12( 4x + 1 )2 - 12 ] - 15
= 64x3 - 16 - ( 4x + 1 )[ 16x2 + 8x + 1 - 12( 16x2 + 8x + 1 ) - 12 ]
= 64x3 - 16 - ( 4x + 1 )( 16x2 + 8x - 11 - 192x2 - 96x - 12 )
= 64x3 - 16 - ( 4x + 1 )( -176x2 - 88x - 23 )
= 64x3 - 16 - ( -704x3 - 528x2 - 180x - 23 )
= 64x3 - 16 + 704x3 + 528x2 + 180x + 23
= 768x3 + 528x2 + 180x + 7 ( có phụ thuộc vào biến )
Bài 1:Tìm x,biết:
a) 3/4x - 1/4 = 2 ( x - 3 ) + 1/4x
b) 30%x - x + 5/6 = 1/3
c) ( 2x - 1) . ( 3x + 12 ) = 0
d) | 2x - 1/2 | = 3/4
CÁC BẠN LÀM GIÚP TUI NHA TUI CẢM ƠN NHIỀU.
a) \(\frac{3}{4}x-\frac{1}{4}=2\left(x-3\right)+\frac{1}{4}x\)
\(\frac{3}{4}x-\frac{1}{4}=2x-6+\frac{1}{4}x\)
\(\frac{3}{4}x-2x-\frac{1}{4}x=\frac{1}{4}-6\)
\(x\left(\frac{3}{4}-2-\frac{1}{4}\right)=-\frac{23}{4}\)
\(-\frac{3}{2}x=-\frac{23}{4}\)
\(x=-\frac{23}{4}\div\left(-\frac{3}{2}\right)\)
\(x=\frac{23}{6}\)
b) \(30\%-x+\frac{5}{6}=\frac{1}{3}\)
\(\frac{3}{10}-x+\frac{5}{6}=\frac{1}{3}\)
\(\frac{3}{10}-x=\frac{1}{3}-\frac{5}{6}\)
\(\frac{3}{10}-x=-\frac{1}{2}\)
\(x=\frac{3}{10}-\left(-\frac{1}{2}\right)\)
\(x=\frac{4}{5}\)
c) \(\left(2x-1\right).\left(3x+12\right)=0\)
\(\hept{\begin{cases}2x-1=0\\3x+12=0\end{cases}\Rightarrow\hept{\begin{cases}2x=1\\3x=-12\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1}{2}\\x=-4\end{cases}}}\)
Vậy \(x\in\left\{\frac{1}{2};-4\right\}\)
bài 1 tìm x
a)3x(x-3)+4x-12=0
b)(x+1)(x^2-x+1)-x^3+2x-=17
c)(x-3)(x+5)+(x-1)^2-6x^4y^2:3x^2y^2=15x
giúp mik vs nhanh ak cảm ơn nhìu!
a: \(3x\left(x-3\right)+4x-12=0\)
=>\(3x\left(x-3\right)+\left(4x-12\right)=0\)
=>\(3x\left(x-3\right)+4\left(x-3\right)=0\)
=>\(\left(x-3\right)\left(3x+4\right)=0\)
=>\(\left[{}\begin{matrix}x-3=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{4}{3}\end{matrix}\right.\)
b: Sửa đề:\(\left(x+1\right)\left(x^2-x+1\right)-x^3+2x=17\)
\(\Leftrightarrow x^3+1-x^3+2x=17\)
=>2x+1=17
=>2x=17-1=16
=>\(x=\dfrac{16}{2}=8\)
c: \(\left(x-3\right)\left(x+5\right)+\left(x-1\right)^2-6x^4y^2:3x^2y^2=15x\)
=>\(x^2+2x-15+x^2-2x+1-2x^2=15x\)
=>\(15x=-14\)
=>\(x=-\dfrac{14}{15}\)
Tìm x,y,z biết :
a, x/3 = y/5 ; 2x + 4y = 28
b, 4x = 5y ; 3x - 2y = 35
c, x/-3 = y/-7 ; 2x + 4y = 68
d, x/2 = y/-3 =z/4 ; 4x - 3y - 2z = 16
giúp mình vs ạ , mình cần gấp ,cảm ơn ạ !
Áp dụng t/c dãy tỉ số bằng nhau:
a.
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{2x}{6}=\dfrac{4y}{20}=\dfrac{2x+4y}{6+20}=\dfrac{28}{26}=\dfrac{14}{13}\)
\(\Rightarrow\left\{{}\begin{matrix}x=3.\dfrac{14}{13}=\dfrac{52}{13}\\y=5.\dfrac{14}{13}=\dfrac{70}{13}\end{matrix}\right.\)
(Em có nhầm đề 26 thành 28 ko nhỉ, số xấu quá)
b.
\(4x=5y\Rightarrow\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{3x}{15}=\dfrac{-2y}{-8}=\dfrac{3x-2y}{15-8}=\dfrac{35}{7}=5\)
\(\Rightarrow\left\{{}\begin{matrix}x=5.5=25\\y=4.2=20\end{matrix}\right.\)
c.
\(\dfrac{x}{-3}=\dfrac{y}{-7}=\dfrac{2x}{-6}=\dfrac{4y}{-28}=\dfrac{2x+4y}{-6-28}=\dfrac{68}{-34}=-2\)
\(\Rightarrow\left\{{}\begin{matrix}x=-3.\left(-2\right)=6\\y=-7.\left(-2\right)=14\end{matrix}\right.\)
d.
\(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{4}=\dfrac{4x}{8}=\dfrac{-3y}{9}=\dfrac{-2z}{-8}=\dfrac{4x-3y-2z}{8+9-8}=\dfrac{16}{9}\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.\dfrac{16}{9}=\dfrac{32}{9}\\y=-3.\dfrac{16}{9}=-\dfrac{48}{9}\\z=4.\dfrac{16}{9}=\dfrac{64}{9}\end{matrix}\right.\)
tìm x, biết :
a, x/6 = 8/3
b, 5/x = 4/9
c, x + 3/ - 4 = 5/20
d, 7/ 3 + 4x = -2/9
f, 3/ x - 1 = x - 1/ 27
giúp e vs ạ , e cần gấp ạ , e cảm ơn nhiều ạ
a: \(\dfrac{x}{6}=\dfrac{8}{3}\)
=>\(x=6\cdot\dfrac{8}{3}=\dfrac{6}{3}\cdot8=8\cdot2=16\)
b: \(\dfrac{5}{x}=\dfrac{4}{9}\)
=>\(x=\dfrac{5\cdot9}{4}=\dfrac{45}{4}\)
c: \(\dfrac{x+3}{-4}=\dfrac{5}{20}\)
=>\(x+3=\dfrac{-4\cdot5}{20}=-1\)
=>x=-1-3=-4
d: \(\dfrac{7}{3+4x}=\dfrac{-2}{9}\)
=>\(4x+3=\dfrac{9\cdot7}{-2}=-\dfrac{63}{2}\)
=>\(4x=-\dfrac{63}{2}-3=-\dfrac{69}{2}\)
=>\(x=-\dfrac{69}{8}\)
f: ĐKXĐ: x<>1
\(\dfrac{3}{x-1}=\dfrac{x-1}{27}\)
=>\(\left(x-1\right)^2=3\cdot27=81\)
=>\(\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=10\left(nhận\right)\\x=-8\left(nhận\right)\end{matrix}\right.\)
((2)/(3)x-(1)/(2))*((x)/(4)-(1)/(8))=0
mình cần gấp mn giúp mình vs cảm ơn mn