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HL
24 tháng 9 2017
Dài dữ trời :V Về sau gửi từng bài một thôi, nhìn hoa mắt quá @@
B1: Phân tích thành nhân tử:
a) \(6x^2+9x=3x\left(2x+3\right)\)
b) \(4x^2+8x=4x\left(x+2\right)\)
c) \(5x^2+10x=5x\left(x+2\right)\)
d) \(2x^2-8x=2x\left(x-4\right)\)
e) \(5x-15y=5\left(x-3y\right)\)
f) \(x\left(x^2-1\right)+3\left(x^2-1\right)=\left(x^2-1\right)\left(x+3\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+3\right)\)
g) \(x^2-2x+1-4y^2=\left(x-1\right)^2-4y^2\)
\(=\left(x-1-2y\right)\left(x-1+2y\right)\)
h) \(x^2-100=\left(x-10\right)\left(x+10\right)\)
i) \(9x^2-18x+9=\left(3x-3\right)^2\)
k) \(x^3-8=\left(x-2\right)\left(x^2+2x+4\right)\)
l) \(x^2+6xy^2+9y^4=\left(x+3y\right)^2\)
m) \(4xy-4x^2-y^2=-\left(4x^2-4xy+y^2\right)\)
\(=-\left(2x-y\right)^2\)
n) \(\left(x-15\right)^2-16=\left(x-15-16\right)\left(x-15+16\right)\)
\(=\left(x-31\right)\left(x+1\right)\)
o) \(25-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3+x\right)\)
\(=\left(2+x\right)\left(8+x\right)\)
p) \(\left(7x-4\right)^2-\left(2x+1\right)^2\)
\(=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)\)
\(=\left(5x-5\right)\left(9x-3\right)\)
24 tháng 9 2017
Bài 1 :
a ) \(6x^2+9x=3x\left(x+3\right)\)
b ) \(4x^2+8x=4x\left(x+2\right)\)
c ) \(5x^2+10x=5x\left(x+2\right)\)
d ) \(2x^2-8x=2x\left(x-4\right)\)
e ) \(5x-15y=5\left(x-3y\right)\)
f ) \(x\left(x^2-1\right)+3\left(x^2-1\right)=\left(x^2-1\right)\left(x+3\right)\)
g ) \(x^2-2x+1-4y^2=\left(x-1\right)^2-\left(2y\right)^2=\left(x-1-2y\right)\left(x-1+2y\right)\)
h ) \(x^2-100=x^2-10^2=\left(x-10\right)\left(x+10\right)\)
i ) \(9x^2-18x+9=\left(3x-3\right)^2\)
k ) \(x^3-8=\left(x-2\right)\left(x^2+2x+2^2\right)\)
l ) \(x^2+6xy^2+9y^4=\left(x+3y^2\right)^2\)
m ) \(4xy-4x^2-y^2=-\left(2x-y\right)^2\)
n ) \(\left(x-15\right)^2=x^2-30x+15^2\)
o ) \(25-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3-x\right)=\left(2+x\right)\left(8-x\right)\)
p ) \(\left(7x-4\right)^2-\left(2x+1\right)^2=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)=\left(5x-5\right)\left(9x-3\right)\)
Bài 2 :
a ) \(3x^3-6x^2+3x^2y-6xy=3x\left(x^2-2x+xy-2y\right)\)
b ) \(x^2-2x+xy-2y=x\left(x-2\right)+y\left(x-2\right)=\left(x-2\right)\left(x+y\right)\)
c ) \(2x+x^2-2y-2xy=......................\)
d ) \(x^2-2xy+y^2-9=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\)
e ) \(x^2+y^2-2xy-4=\left(x-y\right)^2-2^2=\left(x-y-2\right)\left(x-y+2\right)\)
f )\(2xy-x^2-y^2+9=-\left(x-y\right)^2+9=3^2-\left(x-y\right)^2=\left(3-x+y\right)\left(3+x-y\right)\)
17 tháng 7 2018
Lần sau đăng thì chia thành nhiều câu hỏi nhé
\(16^2-9.\left(x+1\right)^2=0\)
\(16^2-\text{ }\left[3.\left(x+1\right)\right]^2=0\)
\(\left[16-3.\left(x+1\right)\right].\left[16+3\left(x+1\right)\right]=0\)
\(\left[16-3x-3\right]\left[16+3x+3\right]=0\)
\(\left[13-3x\right].\left[19+3x\right]=0\)
\(\Rightarrow\orbr{\begin{cases}13-3x=0\\19+3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=13\\3x=-19\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{13}{3}\\x=-\frac{19}{3}\end{cases}}}\)
KL:..............................
22 tháng 8 2018
a) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left(2x\right)^2-5^2-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left(-2\right).\left(2x-5\right)=0\)
\(\Leftrightarrow2x-5=0\)
\(\Leftrightarrow x=\dfrac{5}{2}\)
DN
22 tháng 8 2018
a,\(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Rightarrow\left(4x^2-25\right)-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Rightarrow\left(2x-5\right)^2-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Rightarrow\left(2x-5\right)\left(2x-5-2x-7\right)=0\)
\(\Rightarrow\left(2x-5\right)\left(-12\right)=0\)
\(\Rightarrow2x-5=0\)
\(\Rightarrow2x=5\)
\(\Rightarrow x=\dfrac{5}{2}\)
\(b,2x^3+3x^2+2x+3=0\)
\(\Rightarrow\left(2x^3+2x\right)+\left(3x^2+3\right)=0\)
\(\Rightarrow2x\left(x^2+1\right)+3\left(x^2+1\right)=0\)
\(\Rightarrow\left(2x+3\right)\left(x^2+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x+3=0\\x^2+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=-3\\x^2=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=1\end{matrix}\right.\)
\(c,x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Rightarrow\left(x^3+27\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Rightarrow\left(x+3\right)^3+\left(x+3\right)\left(x-9\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2+9+x-9\right)=0\)
\(\Rightarrow\left(x+3\right).x^3=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x^3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=0\end{matrix}\right.\)
\(d,x^2\left(x+7\right)-4\left(x+7\right)=0\)
\(\Rightarrow\left(x^2-4\right)\left(x+7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2-4=0\\x+7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2=4\\x=-7\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\)
5 tháng 11 2017
1.
a. x2 - 2x + 1 = 0
x2 - 2x*1 + 12 = 0
(x-1)2 = 0
............( tới đây tui bí rùi tự suy nghĩ rùi lm tiếp ik)
5 tháng 11 2017
1, Tìm x biết:
a, x2 - 2x +1 = 0
(x-1)2 = 0
x-1 = 0
x = 1. Vậy ...
b, ( 5x + 1)2 - (5x - 3) ( 5x + 3) = 30
25x2 +10x + 1 - (25x2 -9) = 30
25x2 +10x + 1 - 25x2 +9 = 30
10x + 10 =30
10(x+1) = 30
x+1 =3
x = 2. vậy ...
c, ( x - 1) ( x2 + x + 1) - x ( x +2 ) ( x - 2) = 5
(x3 - 1) - x(x2 -4) = 5
x3 - 1 - x3 + 4x = 5
4x - 1 = 5
4x = 6
x = \(\dfrac{3}{2}\) .vậy ...
d, ( x - 2)3 - ( x - 3) ( x2 + 3x + 9 ) + 6 ( x + 1)2 = 15
x3 - 6x2 + 12x - 8 - (x3 - 27) + 6 (x2 + 2x +1) =15
x3 - 6x2 + 12x - 8 - x3 + 27 + 6x2 + 12x +6 =15
24x + 25 = 15
24x = -10
x = \(\dfrac{-5}{12}\) vậy ...
ND
8 tháng 2 2018
a. \(9\left(x+2\right)-3\left(x+2\right)=0\)
\(\Leftrightarrow9x+18-3x-6=0\)
\(\Leftrightarrow6x+12=0\)
\(\Leftrightarrow x=-2\)
e. \(\left(2x-1\right)^2-45=0\)
\(\Leftrightarrow4x^2-2x+1-45=0\)
\(\Leftrightarrow4x^2-2x-44=0\)
Đến đó tự giải tiếp nha!
c. \(2\left(2x-5\right)-3x=0\)
\(\Leftrightarrow4x-10-3x=0\)
\(\Leftrightarrow x-10=0\)
\(\Leftrightarrow x=10\)
g. \(2x^2-6x=0\)
\(\Leftrightarrow2x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
JS
26 tháng 3 2020
b, \(\left(4x+2\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow4x+2=0\) (Vì \(x^2+1>0\forall x\))
\(\Leftrightarrow x=\frac{-1}{2}\)
Vậy phương trình có nghiệm \(x=\frac{-1}{2}.\)
c, \(\left(x^2-4\right)+\left(x-2\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2+3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(5-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\5-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
Vậy phương trình có tập nghiệm \(S=\left\{2;5\right\}\).
d, \(3x^2+2x-1=0\)
\(\Leftrightarrow3x^2+3x-x-1=0\)
\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm \(S=\left\{-1;\frac{1}{3}\right\}\).
TN
2 tháng 7 2017
\(a,x^4+2x^3+x^2=\left(x^2+x\right)^2\)
\(b,x^2+5x-6=x^2-x+6x-6=x\left(x-1\right)+6\left(x-1\right)\)\(=\left(x-1\right)\left(x+6\right)\)
\(c,5x\left(x-1\right)=x-1\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\Leftrightarrow\left(5x-1\right)\left(x-1\right)=0\)\(x^4+8x=x\left(x^3+8\right)=x\left(x+2\right)\left(x^2-2x+4\right)\) \(e,x^2+x-6=x^2+3x-2x-6=x\left(x+3\right)-2\left(x+3\right)=\left(x-2\right)\left(x+3\right)\)\(f,x^2-2x-3=x^2-3x+x-3=x\left(x-3\right)+\left(x-3\right)=\left(x+1\right)\left(x-3\right)\)\(h,2x^2+5x-3=0\Leftrightarrow2x^2-6x+x-3=0\Leftrightarrow2x\left(x-3\right)+\left(x-3\right)=0\Leftrightarrow\left(2x+1\right)\left(x-3\right)=0\)
DN
5 tháng 9 2018
\(a,x^2-2x=0\)
\(\Rightarrow x\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
Vậy ...
\(b,\left(5-2x\right)^2-16=0\)
\(\Rightarrow\left(5-2x\right)^2=16\)
\(\Rightarrow\left(5-2x\right)^2=4^2\)
\(\Rightarrow5-2x=\pm4\)
\(\Rightarrow\left[{}\begin{matrix}5-2x=4\\5-2x=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=1\\2x=9\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{2}{9}\end{matrix}\right.\)
Vậy ...
\(c,x\left(x+3\right)-x^2-11=0\)
\(\Rightarrow x^2+3x-x^2-11=0\)
\(\Rightarrow3x-11=0\)
\(\Rightarrow3x=11\)
\(\Rightarrow x=\dfrac{11}{3}\)
Vậy ...
\(x^3-x^2-2x+2=0\)
\(x^2\left(x-1\right)-2\left(x-1\right)=0\)
\(\left(x-1\right)\left(x^2-2\right)=0\)
\(\orbr{\begin{cases}x^2-2=0\\x-1=0\end{cases}}\)
\(\orbr{\begin{cases}x^2=2\\x=1\end{cases}}\)
\(\orbr{\begin{cases}x=\pm\sqrt{2}\\x=1\end{cases}}\)
\(x^3-x^2-2x+2=0\)
\(\Leftrightarrow\left(x^3-x^2\right)-\left(2x-2\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x^2-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x^2=2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\pm\sqrt{2}\end{cases}}\)
Vậy \(x=1\)hoặc \(x=\pm\sqrt{2}\)