Bài 6 : Tìm x, biết :
a, ( 2x-5)^2 - ( 5+2x)^2 = 0
b, 27x^3 - 54x^2 + 36x = 8
c, ( x^3 + 8 ) - ( x+2) ( x-4) = 0
d, x^6 - 1 = 0
Giúp mk vs ạ mk đang cần gấp
\(1,\)
\(2x\left(x-3\right)-\left(3-x\right)=0\)
\(\Leftrightarrow2x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=3\end{cases}}\)
\(2,\)
\(3x\left(x+5\right)-6\left(x+5\right)=0\)
\(\Leftrightarrow\left(3x-6\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\x+5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)
\(3,\)
\(x^4-x^2=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
\(4,\)
\(x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(5,\)
\(x\left(x+6\right)-10\left(x-6\right)=0\)
\(\Leftrightarrow x^2+6x-10x+60=0\)
\(\Leftrightarrow x^2-4x+60=0\)
\(\Leftrightarrow x^2-4x+4+56=0\)
\(\Leftrightarrow\left(x-2\right)^2=-56\)(Vô lý)
=> Phương trình vô nghiệm
Bài 9 : Tìm x, biết :
a, ( x-2 ) ( x-3 ) + ( x-2 ) - 1 = 0
b, ( x+2 )^2 - 2x ( 2x + 3 ) = ( x+1 )^2
c, 6x^3 + x^2 = 2x
d, x^8 - x^5 + x^2 - x + 1 = 0
Giúp mk vs ạ mk đang cần gấp ạ
Bài 9 : Tìm x, biết :
a, (x - 2)(x - 3) + (x - 2) - 1 = 0
\(\Leftrightarrow\left(x-2\right)\left(x-3+1\right)-1=0\)
\(\Leftrightarrow\left(x-2\right)^2-1=0\)
\(\Leftrightarrow\left(x-2+1\right)\left(x-2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
Vậy x ={1; 3}
b, (x + 2)2 - 2x(2x + 3) = (x + 1)2
\(\Leftrightarrow\left(x+2\right)^2-\left(x+1\right)^2-2x\left(2x+3\right)=0\)
\(\Leftrightarrow\left(x+2+x+1\right)\left(x+2-x-1\right)-2x\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3-2x\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(1-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\1-2x=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{3}{2}\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy \(x=\left\{-\frac{3}{2};\frac{1}{2}\right\}\)
c, 6x3 + x2 = 2x
\(\Leftrightarrow6x^3+x^2-2x=0\)
\(\Leftrightarrow x\left(6x^2+x-2\right)=0\)
\(\Leftrightarrow x\left(6x^2+4x-3x-2\right)=0\)
\(\Leftrightarrow x\left[2x\left(3x+2\right)-\left(3x+2\right)\right]=0\)
\(\Leftrightarrow x\left(3x+2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3x+2=0\\2x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\frac{2}{3}\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy \(x=\left\{0;-\frac{2}{3};\frac{1}{2}\right\}\)
Bài 1: Tính giá trị của biểu thức
a, x3+12x2+48x+64 tại x=6
b, (2x-3)(2x+3)-(x+5)2-(x-1)(x+2) với x=-7/3
c, x2+y2 biết x+y=-8; xy=15
Bài 2: Tìm x, biết
a, 36x2-49=0
b, (x+3)2+(x+2)(x-2) 2(x-1)2=7
c, (x+3)(x2-3x+9)-x(x-1)(x+1)-27=0
Cả 2 bài đều liên quan đến 7 hđt, mk đang cần gấp ai giúp mk vs rồi mk tick cho ạh
Bài 4 : Phân tích các đa thức sau thành nhân tử :
a, x^3 + 6x^2 + 12x + 8
b, x^3 - 3x^2 + 3x - 1
c, 1 - 9x + 27x^2 - 27x^3
d, x^3 + 3/2x^2 + 3/4x + 1/8
e, 27x^3 - 54x^2y + 36xy^2 - 8y^3
Giúp mk vs ạ mk đang cần gấp ạ
Bài 4:
a) Ta có: \(x^3+6x^2+12x+8\)
\(=x^3+2x^2+4x^2+8x+4x+8\)
\(=x^2\left(x+2\right)+4x\left(x+2\right)+4\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+4x+4\right)\)
\(=\left(x+2\right)^3\)
b) Ta có: \(x^3-3x^2+3x-1\)
\(=x^3-x^2-2x^2+2x+x-1\)
\(=x^2\left(x-1\right)-2x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-2x+1\right)\)
\(=\left(x-1\right)^3\)
c) Ta có: \(1-9x+27x^2-27x^3\)
\(=1-3x-6x+18x^2+9x^2-27x^3\)
\(=\left(1-3x\right)-6x\left(1-3x\right)+9x^2\left(1-3x\right)\)
\(=\left(1-3x\right)\left(1-6x+9x^2\right)\)
\(=\left(1-3x\right)^3\)
d) Ta có: \(x^3+\frac{3}{2}x^2+\frac{3}{4}x+\frac{1}{8}\)
\(=x^3+3\cdot x^2\cdot\frac{1}{2}+3\cdot x\cdot\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3\)
\(=\left(x+\frac{1}{2}\right)^3\)
e) Ta có: \(27x^3-54x^2y+36xy^2-8y^3\)
\(=\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot2y+3\cdot3x\cdot\left(2y\right)^2-\left(2y\right)^3\)
\(=\left(3x-2y\right)^3\)
Bài 5 : Tìm x, biết :
a, x^2 ( x-5 ) + 5 - x = 0
b, 3x^4 - 9x^3 = -9x^2 + 27x
c, x^2 ( x+8 ) + x^2 = -8x
d, ( x+3 ) ( x^2 - 3x + 5 ) = x^2 + 3x
Giúp mk vs ạ mk đang cần gấp
a/ \(x^2\left(x-5\right)+5-x=0\)
\(\Leftrightarrow x^2\left(x-5\right)-\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=5\end{matrix}\right.\)
Vậy...
b/ \(3x^4-9x^3=-9x^2+27x\)
\(\Leftrightarrow3x^4-9x^3+9x^2-27x=0\)
\(\Leftrightarrow3x^3\left(x-3\right)+9x\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(3x^3+9x\right)=0\)
\(\Leftrightarrow3x\left(x-3\right)\left(x^2+3\right)=0\)
Vì \(x^2+3>0\forall x\)
\(\Leftrightarrow3x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Vậy..
c/ \(x^2\left(x+8\right)+x^2=-8x\)
\(\Leftrightarrow x^2\left(x+8\right)+x^2+8x=0\)
\(\Leftrightarrow x^2\left(x+8\right)+x\left(x+8\right)=0\)
\(\Leftrightarrow x\left(x+8\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+8=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\\x=-1\end{matrix}\right.\)
Vậy...
d/ \(\left(x+3\right)\left(x^2-3x+5\right)=x^2+3x\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+5\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left[\left(x-2\right)^2+1\right]=0\)
Vì \(\left(x-2\right)^2+1>0\forall x\)
\(\Leftrightarrow x+3=0\Leftrightarrow x=-3\)
Vậy..
\(a,x^2\left(x-5\right)+5-x=0\\ \Leftrightarrow x^2\left(x-5\right)-\left(x-5\right)=0\\ \Leftrightarrow\left(x^2-1\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=5\end{matrix}\right.\)
\(b,3x^4-9x^3=-9x^2+27x\\ \Leftrightarrow3x^4-9x^3+9x^2-27x=0\\ \Leftrightarrow3x^3\left(x-3\right)+9x\left(x-3\right)=0\\ \Leftrightarrow3x\left(x-3\right)\left(x^2+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
do \(x^2+2>0\)
\(c,x^2\left(x+8\right)+x^2=-8x\\ \Leftrightarrow x^2\left(x+8\right)+x^2+8x=0\\ \Leftrightarrow x^2\left(x+8\right)+x\left(x+8\right)=0\\ \Leftrightarrow x\left(x+1\right)\left(x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-8\end{matrix}\right.\)
\(d,\left(x+3\right)\left(x^2-3x+5\right)=x^2+3x\\ \Leftrightarrow\left(x+3\right)\left(x^2-3x+5\right)-x^2-3x=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-3x+5\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-3x+5-x\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x^2-4x+5=0\end{matrix}\right.\)
\(\Delta=b^2-4ac=\left(-4\right)^2-4\cdot1\cdot5=16-20=-4< 0\)\(\rightarrow\)\(x^2-4x+5\) vô nghiệm
Vậy \(x=-3\)
Tìm x, biết :
a, 8x ( x- 2017) - 2x + 4034 = 0
b, x/2 + x^2/8 = 0
c, 4 - x = 2(x-4)^2
d, ( x^2 + 1) ( x-2) + 2x = 4
Giúp mk vs ạ mk đang cần gấp
Bài 2 :Tìm x, biết
a, 2-x = 2 ( x-2)^3
b, ( x-1,5)^6 + 2 ( 1,5-x)^2 = 0
c, 2x^3 + 3x^2 + 3 + 2x = 0
d, x^2 (x+1) - x(x+1) + x(x-1) = 0
f, x^3 - 4x - 14x (x-2) = 0
Giúp mk vs ạ mk đang cần gấp
1. Tìm x biết:
a,5x.( x+3/4)=0
b, x+7/ 2010+ x+6/2011 =x+5 / 2012 + x + 4 /2013
2. Tìm x sao cho:
a, 2x-3 có giá trị dương
b, 13-5x có giá trị âm
c, x+3/2x-1 > 0
d, x+7/x+3 < 1
Giúp mk vs nha mk đang cần gấp. Mk tick cho. Thank nhìu
a) 5x.(x+3/4) = 0
=> x = 0
x+3/4 = 0 => x = -3/4
b) \(\frac{x+7}{2010}+\frac{x+6}{2011}=\frac{x+5}{2012}+\frac{x+4}{2013}.\)
\(\Rightarrow\frac{x+7}{2010}+\frac{x+6}{2011}-\frac{x+5}{2012}-\frac{x+4}{2013}=0\)
\(\frac{x+7}{2010}+1+\frac{x+6}{2011}+1-\frac{x+5}{2012}-1-\frac{x+4}{2013}-1=0\)
\(\left(\frac{x+7}{2010}+1\right)+\left(\frac{x+6}{2011}+1\right)-\left(\frac{x+5}{2012}+1\right)-\left(\frac{x+4}{2013}+1\right)=0\)
\(\frac{x+2017}{2010}+\frac{x+2017}{2011}-\frac{x+2017}{2012}-\frac{x+2017}{2013}=0\)
\(\left(x+2017\right).\left(\frac{1}{2010}+\frac{1}{2011}-\frac{1}{2012}-\frac{1}{2013}\right)=0\)
=> x + 2017 = 0
x = -2017
a) để 2x - 3 > 0
=> 2x > 3
x > 3/2
b) 13-5x < 0
=> 5x < 13
x < 13/5
c) \(\frac{x+3}{2x-1}>0\)
=> x + 3 > 0
x > -3
d) \(\frac{x+7}{x+3}=\frac{x+3+4}{x+3}=1+\frac{4}{x+3}\)
Để x+7/x+3 < 1
=> 1 + 4/x+3 < 1
=> 4/x+3 < 0
=> không tìm được x thỏa mãn điều kiện
ai giúp mih luôn với ạ mih cần gấp
tìm x
ý a (2x-5)^2-(5+2x)^2=0
y b 27x^3-54x^2+36x=8
ai giúp mih mih cảm ơn nhiều
b) \(27x^3-54x^2+36x=8\)
\(\Rightarrow27x^3-54x^2+36x-8=0\)
\(\Rightarrow\left(3x\right)^3-3.\left(3x\right)^2.2+3.3x.2^2-2^3=0\)
\(\Rightarrow\left(3x-2\right)^3=0\)
\(\Rightarrow3x-2=0\)
\(\Rightarrow3x=2\)
\(\Rightarrow x=\dfrac{2}{3}\)
(2x-5)^2-(5+2x)^2=0
<=>(2x-5-5-2x)(2x-5+5+2x)=0
<=>(-10).(4x)=0
<=>(-40x)=0
<=>x =0
27x^3-54x^2+36x=8
<=>27x^3-54x^2+36x-8=0
<=>(3x-2)^3=0
<=>3x-2=0
<=>3x=2
<=>x=2/3
a,\(\left(2x-5\right)^2-\left(5+2x\right)^2=0\)
\(\Rightarrow\left(2x-5\right)^2+\left(2x-5\right)^2=0\)
Do \(\left(2x-5\right)^2\ge0\)
\(\Rightarrow\left(2x-5\right)^2=0\)
\(\Rightarrow2x-5=0\)
\(\Rightarrow x=\dfrac{5}{2}\)
@Yukru giúp câu b với!