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18 tháng 6 2023
=>2x^2-4x+3x-6=6x-3x^2+10-5x
=>2x^2-x-6=-3x^2+x+10
=>5x^2-2x-16=0
=>x=2; x=-8/5
H9
HT.Phong (9A5)
CTVHS
10 tháng 8 2023
1) \(\left(x-3\right)^2-4=0\)
\(\Leftrightarrow\left(x-3-2\right)\left(x-3+2\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
2) \(x^2-2x=24\)
\(\Leftrightarrow x^2-2x-24=0\)
\(\Leftrightarrow x^2+4x-6x-24=0\)
\(\Leftrightarrow x\left(x+4\right)-6\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
LA
21 tháng 6 2017
b)(2x - 1)^2 - (2x + 5) (2x - 5 ) = 18
4x 2 -4x+1-4x 2+25=18
26-4x=18
4x=8
x=2
21 tháng 6 2017
a,27x-18=2x-3x^2
<=> 3x^2-2x+27-18x=0
<=> 3x^2-20x+27=0
\(\Delta\)= 20^2-4-12.27
tính \(\Delta\)rồi tìm x1 ,x2
HP
1 tháng 1 2017
a) thay x=-5 vào pt
=>-10-4m=6
=>m=-4
vậy m=-4 là giá trị cần tìm
b) tương tự
7 tháng 1 2018
a ) ( x - 2 )( x + 5 )
= x^2 + 5x - 2x + 10
= x^2 + 3x + 10
b ) 3x + 3y +ax + ay
= x( 3 + a ) + y( 3 + a )
= ( 3 + a )( x + y )
c ) ( x^2 + 2xy ) : ( x + 2y )
= [ x( x + 2y ) ] : ( x + 2y )
= x : 1
= x
d ) ( x - 2 )( x + 2 ) + ( x + 1 )^2 - 2x^2 = 0
x^2 + 2x - 2x - 4 + x^2 + x + x + 1 - 2x^2 = 0
x^2 - 4 + x^2 + 2x + 1 - 2x^2 = 0
2x^2 + 2x - 4 + 1 - 2x^2 = 0
2x - 3 = 0
2x = 0 + 3
2x = 3
x = 3 : 2
x = 3/2
KT
7 tháng 1 2018
a) \(\left(x-2\right)\left(x+5\right)\)
\(=x^2+5x-2x-10\)
\(=x^2+3x-10\)
b) \(3x+3y+ax+ay\)
\(=3\left(x+y\right)+a\left(x+y\right)\)
\(=\left(x+y\right)\left(3+a\right)\)
c) \(\left(x^2+2xy\right):\left(x+2y\right)\)
\(=\left[x\left(x+2y\right)\right]:\left(x+2y\right)\)
\(=x\)
d) \(\left(x-2\right)\left(x+2\right)+\left(x+1\right)^2-2x^2=0\)
\(\Leftrightarrow\)\(x^2-4+x^2+2x+1-2x^2=0\)
\(\Leftrightarrow\)\(2x-3=0\)
\(\Leftrightarrow\)\(2x=3\)
\(\Leftrightarrow\)\(x=\frac{3}{2}\)
Vậy....
26 tháng 9 2016
a ) \(9x^2-49=9\)
\(\Leftrightarrow9x^2=58\)
\(\Leftrightarrow x^2=29\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=29\\x=-29\end{array}\right.\)
Vậy ......................
b ) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-1\right)\left(x+1\right)-27=0\)
\(\Leftrightarrow\left(x^3+3^3\right)-x.\left(x^2-1^2\right)-27=0\)
\(\Leftrightarrow x^3+27-x^3+x-27=0\)
\(\Leftrightarrow x=0\)
c ) \(\left(x-1\right)\left(x+2\right)-x-2=0\)
\(\Leftrightarrow x^2+2x-x-2-x-2=0\)
\(\Leftrightarrow x^2-4=0\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-2\end{array}\right.\)
Vây .....................
M
1 tháng 4 2020
e, 3x(2-x) =15(x-2)
\(\Leftrightarrow3x\left(2-x\right)-15\left(x-2\right)=0\)
\(\Leftrightarrow-3x\left(x-2\right)-15\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(-3x-15\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\-3x-15=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
Vậy..
f, (x+5)(x+4)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x+5=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\x=-4\end{matrix}\right.\)
Vậy..
g, x(x+4)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
,h, (2x -4)(x-2)=0
\(\Leftrightarrow2\left(x-2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2-1\right)=0\)
\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
i, (x+1/5)(2x-3)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x+\frac{1}{5}=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{-1}{5}\\x=\frac{3}{2}\end{matrix}\right.\)
k, x²-4x=0
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
m, 4x²-1=0
\(\Leftrightarrow\left(2x\right)^2-1^2=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=1\\2x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\x=\frac{-1}{2}\end{matrix}\right.\)
n, x²-6x+9=0
\(\Leftrightarrow x^2-2.x.3+3^2=0\)
\(\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\)
<=> x=3
l, (3x-5)²-(x+4)²=0
\(\Leftrightarrow\left(3x-5-x-4\right)\left(3x-5+x+4\right)=0\)
\(\Leftrightarrow\left(2x-9\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-9=0\\4x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=9\\4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{9}{2}\\x=\frac{1}{4}\end{matrix}\right.\)
Vậy ..
o, 7x(x+2)-5(x+2)=0
\(\Leftrightarrow\left(x+2\right)\left(7x-5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\7x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\7x=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=\frac{5}{7}\end{matrix}\right.\)
Vậy....
p, 3x(2x-5)-4x+10=0
\(\Leftrightarrow3x\left(2x-5\right)-\left(4x-10\right)=0\)
\(\Leftrightarrow3x\left(2x-5\right)-2\left(2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-5=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=5\\3x=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\x=\frac{2}{3}\end{matrix}\right.\)
Vậy...
q, (2-2x)-x²+1=0
\(\Leftrightarrow2\left(1-x\right)-\left(x^2-1^2\right)=0\)
\(\Leftrightarrow2\left(1-x\right)-\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow2\left(1-x\right)+\left(1-x\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(2+x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
Vậy ....
r, x(1-3x)=5(1-3x)
\(\Leftrightarrow x\left(1-3x\right)-5\left(1-3x\right)=0\)
\(\Leftrightarrow\left(1-3x\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-3x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3x=-1\\x=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{3}\\x=5\end{matrix}\right.\)
s, 2x-3/4+x+1/6=3
\(\Leftrightarrow x-\frac{7}{12}=3\Leftrightarrow x=3+\frac{7}{12}=\frac{43}{12}\)
S
31 tháng 8 2017
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
\(\left(2x-1-x+5\right)\left(2x-1+x-5\right)=0\)
\(\left[{}\begin{matrix}x+4=0\\3x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=2\end{matrix}\right.\)
\(\Rightarrow\left(2x-1-x+5\right)\left(2x-1+x-5\right)=0\\ \Rightarrow\left(x+4\right)\left(3x-6\right)=0\\ \Rightarrow3\left(x-2\right)\left(x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)