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BB
1
5 tháng 4 2022
a: \(\Leftrightarrow\left(2-x\right)\left(x-3\right)+\left(x-1\right)\left(x+3\right)=-4x\)
\(\Leftrightarrow2x-6-x^2+3x+x^2+3x-x-3=-4x\)
=>7x-9=-4x
=>11x=9
hay x=9/11
b: \(\Leftrightarrow\left(5-x\right)\left(x-4\right)+\left(x+2\right)\left(x+4\right)=-3x\)
\(\Leftrightarrow5x-20-x^2+4x+x^2+6x+8=-3x\)
=>15x-12=-3x
=>18x=12
hay x=2/3
12 tháng 6 2023
2:
=>x^3-1-2x^3-4x^6+4x^6+4x=6
=>-x^3+4x-7=0
=>x=-2,59
4: =>8x-24x^2+2-6x+24x^2-60x-4x+10=-50
=>-62x+12=-50
=>x=1
MN
21 tháng 1 2020
a. \(\left(x^2-2x+1\right)-3x\left(x-1\right)=0\)
\(\Leftrightarrow x^2-2x+1-3x^2+3x=0\)
\(\Leftrightarrow-2x^2+x+1=0\)
\(\Leftrightarrow-2x^2+2x-x+1=0\)
\(\Leftrightarrow-2x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow-\left(2x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}\)
Vậy \(x\in\left\{-\frac{1}{2};1\right\}\)
b. \(4\left(7x-3\right)-\left(7x^2-3x\right)=0\)
\(\Leftrightarrow4\left(7x-3\right)-x\left(7x-3\right)=0\)
\(\Leftrightarrow\left(4-x\right)\left(7x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}4-x=0\\7x-3=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=\frac{3}{7}\end{cases}}\)
Vậy \(x\in\left\{4;\frac{3}{7}\right\}\)
c.\(\left(5-x\right)\left(2+3x\right)=4-9x^2\)
\(\Leftrightarrow\left(5-x\right)\left(2+3x\right)=\left(2-3x\right)\left(2+3x\right)\)
\(\Leftrightarrow\left(2+3x\right)\left(5-x-2+3x\right)=0\)
\(\Leftrightarrow\left(2+3x\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2+3x=0\\2x+3=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{2}{3}\\x=-\frac{3}{2}\end{cases}}\)
Vậy \(x\in\left\{-\frac{2}{3};-\frac{3}{2}\right\}\)
d. \(7-\left(2x+4\right)=-\left(x+4\right)\)
\(\Leftrightarrow7-2x-4=-x-4\)
\(\Leftrightarrow7-4+4=-x+2x\)
\(\Leftrightarrow7=x\)
Vậy x = 7
e. \(\left(x-1\right)-\left(2x-1\right)=9\)
\(\Leftrightarrow x-1-2x+1=9\)
\(\Leftrightarrow-x=9\)
\(\Leftrightarrow x=-9\)
g. \(x^3+x^2+x+1=0\)
\(\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x+1=0\end{cases}}\)Mà : \(x^2+1\ge1>0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Vậy x = -1
23 tháng 7 2018
\(A=x\left(x+2\right)\left(x+4\right)\left(x+6\right)-9 \)
\(A=x\left(x+6\right)\left(x+2\right)\left(x+4\right)-9\)
\(A=\left(x^2+6x\right)\left(x^2+6x+8\right)-9\)
Đặt \(x^2+6x+4=t\)
Ta được: \(A=\left(t-4\right)\left(t+4\right)-9\)
\(A=t^2-25\)
\(A=\left(t+5\right)\left(t-5\right)\)
\(A=\left(x^2+6x+9\right)\left(x^2+6x-1\right)\)
\(A=\left(x+3\right)^2\left(x^2+6x-1\right)\)
23 tháng 7 2018
\(B=\left(x^2-3x\right)^2+5x^2-15x+6\)
\(B=\left(x^2-3x\right)^2+5\left(x^2-3x\right)+6\)
\(B=\left(x^2-3x\right)\left(x^2-3x+5\right)+6\)
Đặt \(x^2-3x=a\)
Ta được: \(B=a\left(a+5\right)+6\)
\(B=a^2+5a+6\)
\(B=a^2+2a+3a+6\)
\(B=a\left(a+2\right)+3\left(a+2\right)\)
\(B=\left(a+2\right)\left(a+3\right)\)
\(B=\left(x^2-3x+2\right)\left(x^2-3x+3\right)\)
\(B=\left(x^2-x-2x+2\right)\left(x^2+3x+3\right)\)
\(B=\left[x\left(x-1\right)-2\left(x-1\right)\right]\left(x^2+3x+3\right)\)
\(B=\left(x-1\right)\left(x-2\right)\left(x^2+3x+3\right)\)
11 tháng 12 2017
a, (x-2)^2 - (x-3)(x+3)=6
x^2-4x+4-(x^2-9)=6
x^2-4x+4-x^2+9=6
(x^2-x^2)-4x+13=6
-4x=-7
x=1,75
b, 4(x-3)^2 - (2x-1)(2x+1)=10
4(x^2-6x+9)-(4x^2-1)=10
4x^2-24x+36-4x^2+1=10
-24x+37=10
x=9/8
c,(x-4)^2 - (x+2)(x-2)=6
x^2-8x+16-(x^2-4)=6
x^2-8x+16-x^2+4=6
-8x+20=6
x=7/4
d, 9(x+1)^2 - (3x-2)(3x+2)=10
9(x^2+2x+1)-(9x^2-4)=10
9x^2+18x+9-9x^2+4=10
18x+13=10
x=-1/6
D
1 tháng 7 2018
\(a,\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)
\(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)
\(-4x+13=6\)
\(-4x=6-13\)
\(-4x=-7\)
\(x=\frac{-7}{-4}\)
\(x=\frac{7}{4}\)
Vậy \(x=\frac{7}{4}\)
\(b,4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)
\(4\left(x^2-6x+9\right)-\left(4x^2-1\right)=10\)
\(4x^2-24x+36-4x^2+1=10\)
\(-24x+37=10\)
\(x=\frac{9}{8}\)
Vậy \(x=\frac{9}{8}\)
\(c,\left(x-4\right)^2-\left(x+2\right)\left(x-2\right)=6\)
\(x^2-8x+16-\left(x^2-4\right)=6\)
\(x^2-8x+16-x^2+4=6\)
\(-8x+20=6\)
\(x=\frac{7}{4}\)
Vậy \(x=\frac{7}{4}\)
\(d,9\left(x+1\right)^2-\left(3x-2\right)\left(3x+2\right)=10\)
\(9\left(x^2+2x+1\right)-\left(9x^2-4\right)=10\)
\(9x^2+18x+9-9x^2+4=10\)
\(18x+13=10\)
\(x=\frac{-1}{6}\)
Vậy \(x=\frac{-1}{6}\)
SM
19 tháng 9 2018
Bài 1 :
1) 4x2 - y2 = ( 2x + y ) ( 2x - y )
2) 9x2 - 4y2 = ( 3x - 2y ) ( 3x + 2y )
3) 4x2 + y2 + 4xy = ( 2x + y )2
Bài 2:
1) 2x2 + 8x = 0
=> 2x ( x + 4 ) = 0
=> \(\orbr{\begin{cases}2x=0\\x+4=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=-4\end{cases}}\)
2) 3 ( x - 4 ) + x2 - 4x = 0
=> 3 ( x - 4 ) + x ( x - 4 ) = 0
=> ( x - 4 ) ( 3 + x ) = 0
=> \(\orbr{\begin{cases}x-4=0\\3+x=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=4\\x=-3\end{cases}}\)
3) 3 ( x - 2 ) = x2 - 2x
=> 3 ( x - 2 ) - x2 + 2x = 0
=> 3 ( x - 2 ) - x ( x - 2 ) = 0
=> ( x - 2 ) ( 3 - x ) = 0
=> \(\orbr{\begin{cases}x-2=0\\3-x=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
4) x ( x - 2 ) - 6 ( 2 - x ) = 0
=> x ( x - 2 ) + 6 ( x - 2 ) = 0
=> ( x - 2 ) ( x + 6 ) = 0
=> \(\orbr{\begin{cases}x-2=0\\x+6=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=2\\x=-6\end{cases}}\)
5) 2x ( x + 5 ) = x2 + 5x
=> 2x ( x + 5 ) - x2 - 5x = 0
=> 2x ( x + 5 ) - x ( x + 5 ) = 0
=> ( x + 5 ) ( 2x - x ) = 0
=> \(\orbr{\begin{cases}x+5=0\\2x-x=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-5\\x=0\end{cases}}\)
6 ) ( x - 2 )2 - x ( x + 3 ) = 9
=> x2 - 4x + 4 - x2 - 3x = 9
=> - 7x + 4 = 9
=> - 7x = 5
=> x = \(-\frac{5}{7}\)
D
19 tháng 9 2018
\(1,4x^2-y^2=\left(2x\right)^2-y^2=\left(2x-y\right)\left(2x+y\right)\)
\(2,9x^2-4y^2=\left(3x\right)^2-\left(2y\right)^2=\left(3x-2y\right)\left(3x+2y\right)\)
\(3,4x^2+y^2+4xy=\left(2x\right)^2+2.2x.y+y^2=\left(2x+y\right)^2\)
\(1,2x^2+8x=0\Rightarrow2x\left(x+4\right)=0\Rightarrow\orbr{\begin{cases}2x=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-4\end{cases}}\)
\(2,3\left(x-4\right)+x^2-4x=0\)
\(\Rightarrow3\left(x-4\right)+x\left(x-4\right)=0\)
\(\Rightarrow\left(3+x\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3+x=0\\x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=4\end{cases}}\)
\(3,3\left(x-2\right)=x^2-2x\)
\(\Rightarrow3\left(x-2\right)-x^2+2x=0\)
\(\Rightarrow3\left(x-2\right)-x\left(x-2\right)=0\)
\(\Rightarrow\left(3-x\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3-x=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}\)
\(4,x\left(x-2\right)-6\left(2-x\right)=0\)
\(\Rightarrow x\left(x-2\right)+6\left(x-2\right)=0\)
\(\Rightarrow\left(x+6\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+6=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-6\\x=2\end{cases}}\)
9 tháng 7 2018
x2 -32 = 6x -x2 -18 +3x
(x -3 )(x+3 ) =9x -x2 - 9
x2 +3x-3x- 9 =9x - x2 -9
x2 - x 2 -9x = -9 +9
-9 x = 0
=> x = 0
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) \(x^2-2x=24\)
\(\Leftrightarrow x^2-2x-24=0\)
\(\Leftrightarrow x^2+4x-6x-24=0\)
\(\Leftrightarrow x\left(x+4\right)-6\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
\(x\left(x+2\right)\left(x+4\right)\left(x+6\right)=9\)
\(\Leftrightarrow x\left(x+2\right)\left(x+4\right)\left(x+6\right)-9=0\)
Đặt \(A=x\left(x+2\right)\left(x+4\right)\left(x+6\right)-9\)
\(A=x\left(x+6\right)\left(x+2\right)\left(x+4\right)-9\)
\(=\left(x^2+6x\right)\left(x^2+6x+8\right)-9\)(1)
Đặt \(a=x^2+6x\)
\(\Rightarrow\left(1\right)=a\left(a+8\right)-9=a^2+8a-9\)
\(=\left(a+4\right)^2-25=\left(a+4-5\right)\left(a+4+5\right)\)
\(=\left(a-1\right)\left(a+9\right)=\left(x^2+6x-1\right)\left(x^2+6x+9\right)\)
\(=\left(x^2+6x-1\right)\left(x+3\right)^2\)
\(pt\Leftrightarrow\left(x^2+6x-1\right)\left(x+3\right)^2=0\)
\(TH1:x^2+6x-1=0\)
\(\Leftrightarrow\left(x+3\right)^2-10=0\)
\(\Leftrightarrow\left(x+3\right)^2=10\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=\sqrt{10}\\x+3=-\sqrt{10}\end{cases}}\Leftrightarrow x=\pm\sqrt{10}+3\)
\(TH2:\left(x+3\right)^2=0\Leftrightarrow x+3=0\Leftrightarrow x=-3\)
Vậy \(x\in\left\{\pm\sqrt{10}+3;-3\right\}\)