Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:
a) (3x - 2)(4x + 5) = 0
<=> 3x - 2 = 0 hoặc 4x + 5 = 0
<=> 3x = 2 hoặc 4x = -5
<=> x = 2/3 hoặc x = -5/4
b) (2,3x - 6,9)(0,1x + 2) = 0
<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
<=> 2,3x = 6,9 hoặc 0,1x = -2
<=> x = 3 hoặc x = -20
c) (4x + 2)(x^2 + 1) = 0
<=> 4x + 2 = 0 hoặc x^2 + 1 # 0
<=> 4x = -2
<=> x = -2/4 = -1/2
d) (2x + 7)(x - 5)(5x + 1) = 0
<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
<=> 2x = -7 hoặc x = 5 hoặc 5x = -1
<=> x = -7/2 hoặc x = 5 hoặc x = -1/5
Bài 4
c) x(x - 2) + (x - 2)²
= (x - 2)(x + x - 2)
= (x - 2)(2x - 2)
= 2(x - 2)(x - 1)
d) 2x(x - y)² - 5(y - x)
= 2x(x - y)² + 5(x - y)
= (x - y)(2x + 5)
Bài 5
a) x² - 6x - 2xy + 12y
= (x² - 6x) - (2xy - 12y)
= x(x - 6) - y(x - 6)
= (x - 6)(x - y)
b) 10ax - 5ay - 2x + y
= (10ax - 5ay) - (2x - y)
= 5a(2x - y) - (2x - y)
= (2x - y)(5a - 1)
c) x⁴ + x³y - x - y
= (x⁴ + x³y) - (x + y)
= x³(x + y) - (x + y)
= (x + y)(x³ - 1)
= (x + y)(x - 1)(x² + x + 1)
d) x³ + 2x² - 4x - 8
= (x³ + 2x²) - (4x + 8)
= x²(x + 2) - 4(x + 2)
= (x + 2)(x² - 4)
= (x + 2)(x + 2)(x - 2)
= (x + 2)²(x - 2)
e) xy - 5x - y² + 5y
= (xy - 5x) - (y² - 5y)
= x(y - 5) - y(y - 5)
= (y - 5)(x - y)
f) ax - bx - 2cx - 2a + 2b + 4c
= (ax - bx - 2cx) - (2a - 2b - 4c)
= x(a - b - 2c) - 2(a - b - 2c)
= (a - b - 2c)(x - 2)
g) 5x²y + 5xy² - b²x - b²y
= (5x²y + 5xy²) - (b²x + b²y)
= 5xy(x + y) - b²(x + y)
= (x + y)(5xy - b²)
h) 4x³ - 4x² - 9x + 9
= (4x³ - 4x²) - (9x - 9)
= 4x²(x - 1) - 9(x - 1)
= (x - 1)(4x² - 9)
= (x - 1)(2x - 3)(2x + 3)
1) Ta có: \(x^2-4x+4=0\)
\(\Leftrightarrow\left(x-2\right)^2=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
Vậy: S={2}
\(a,=5\left(x-y\right)+a\left(x-y\right)=\left(5+a\right)\left(x-y\right)\\ b,=a\left(x+y\right)+b\left(x+y\right)=\left(a+b\right)\left(x+y\right)\\ c,=x\left(x+1\right)+a\left(x+1\right)=\left(x+a\right)\left(x+1\right)\\ d,Sửa:x^2y+xy^2-3x-3y=xy\left(x+y\right)-3\left(x+y\right)=\left(xy-3\right)\left(x+y\right)\\ e,=xy\left(x+1\right)-\left(x+1\right)=\left(xy-1\right)\left(x+1\right)\\ f,=x^2-4=\left(x-2\right)\left(x+2\right)\\ g,=\left(x+3\right)^2-y^2=\left(x-y+3\right)\left(x+y+3\right)\\ h,=\left(x+5\right)^2-y^2=\left(x-y+5\right)\left(x+y+5\right)\\ i,=\left(x-4\right)^2-24y^2=\left(x-2\sqrt{6}y-4\right)\left(x+2\sqrt{6}y+4\right)\)
Bài 1
a/ \(x\left(x^2+1\right)+2\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+1\right)=0\Rightarrow x=-2\)
b/
\(\Leftrightarrow x^3-6x^2+9x+5x^2-30x+45=0\)
\(\Leftrightarrow x\left(x-3\right)^2+5\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-3\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=3\end{matrix}\right.\)
1.
c/ \(\Leftrightarrow x^3+2x^2+2x+x^2+2x+2=0\)
\(\Leftrightarrow x\left(x^2+2x+2\right)+x^2+2x+2=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+2x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x^2+2x+2=0\left(vn\right)\end{matrix}\right.\)
d/
\(\Leftrightarrow x^4+x^3-2x^2-x^3-x^2+2x+4x^2+4x-8=0\)
\(\Leftrightarrow x^2\left(x^2+x-2\right)-x\left(x^2+x-2\right)+4\left(x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x^2-x+4\right)\left(x^2+x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+4=0\left(vn\right)\\x^2+x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
\(a.\left(x^2-2x+1\right)-4=0\\\Leftrightarrow \left(x-1\right)^2-2^2=0\\\Leftrightarrow \left(x-1-2\right)\left(x-1+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+1\right)=0\\\Leftrightarrow \left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{3;-1\right\}\)
\(b.x^2-x=-2x+2\\\Leftrightarrow x^2-x+2x-2=0\\\Leftrightarrow x\left(x-1\right)+2\left(x-1\right)=0\\\Leftrightarrow \left(x+2\right)\left(x-1\right)=0\\\Leftrightarrow \left[{}\begin{matrix}x+2=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{-2;1\right\}\)
\(c.4x^2+4x+1=x^2\\ \Leftrightarrow4\left(x^2+x+\frac{1}{4}\right)-x^2=0\\ \Leftrightarrow4\left(x+\frac{1}{2}\right)^2-x^2=0\\ \Leftrightarrow\left[2\left(x+\frac{1}{2}\right)-x\right]\left[2\left(x-\frac{1}{2}\right)+x\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}2\left(x+\frac{1}{2}\right)-x=0\\2\left(x+\frac{1}{2}\right)+x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x+1-x=0\\2x+1+x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\frac{1}{3}\end{matrix}\right.\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{-1;-\frac{1}{3}\right\}\)
a) x2y+xy+x+1= (x2y+xy)+(x+1)=xy(x+10+(x+1)=(x+1)(xy+1)
b) x2-(a+b)x+ab=x2-ax-bx+ab=(x2-ax)-(bx-ab)=x(x-a)-b(x-a)=(x-a)(x-b)
c) ax2+ay-bx2-by=(ax2+ay)-(bx2+by)=a(x2+y)-b(x2+y)=(a-b)(x2+y)
d) ax-2x-a2+2a=(ax-2x)-(a2-2a)=x(a-2)-a(a-2)=(a-2)(x-a)
e) 2x2+4ax+x+2a=(2x2+4ax)+(x+2a)=2x(x+2a)+(x+2a)=(x+2a)(2x+1)
f) x3+ax2+x+a=(x3+ax2)+(x+a)=x2(x+a)+(x+a)=(x2+1)(x+a)
1 ) \(x\left(a-b\right)+a-b=\left(x+1\right)\left(a-b\right)\)
2 ) \(2x\left(b-a\right)+a-b=2x\left(b-a\right)-\left(b-a\right)=\left(2x-1\right)\left(b-a\right)\)
3 ) \(-2x-2y+ax+ay=-2\left(x+y\right)+a\left(x+y\right)=\left(a-2\right)\left(x+y\right)\)
4 ) \(x^2-xy-2x+2y=x\left(x-y\right)-2\left(x-y\right)=\left(x-2\right)\left(x-y\right)\)
5 ) \(5x^2y+5xy^2+a^2x+a^2y\)
\(=5xy\left(x+y\right)+a^2\left(x+y\right)\)
\(=\left(5xy+a^2\right)\left(x+y\right)\)
6 ) \(2x^2-6xy+5x-15y\)
\(=2x\left(x-3y\right)+5\left(x-3y\right)\)
\(=\left(2x+5\right)\left(x-3y\right)\)
7 ) \(ax^2-3axy+bx-3by\)
\(=\left(ax^2+bx\right)-\left(3axy+3by\right)\)
\(=x\left(ax+b\right)-3y\left(ax+b\right)\)
\(=\left(x-3y\right)\left(ax+b\right)\)
8 ) \(x^2+4x-5x-20=0\)
\(\Leftrightarrow x\left(x+4\right)-5\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-4\end{matrix}\right.\)
9 ) \(x^2+10x-2x-20=0\)
\(\Leftrightarrow x\left(x+10\right)-2\left(x+10\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-10\end{matrix}\right.\)
10 ) \(x^2-6x-4x+24=0\)
\(\Leftrightarrow x\left(x-6\right)-4\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=6\end{matrix}\right.\)
:D