Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1
a) \(3x\left(4x^2-2x+3\right)\)
\(=3x.4x^2-3x.2x+3x.3\)
\(=12x^3-6x^2+9x\)
b) \(\left(2x+5\right)^2-4x^2\)
\(=\left[\left(2x+5\right)-4x\right]\left[\left(2x+5\right)+4x\right]\)
\(=\left(2x+5-4x\right)\left(2x+5+4x\right)\)
\(=\left(-2x+5\right)\left(6x+5\right)\)
c) \(\left(x-2\right)^2+\left(x-3\right)\left(x+3\right)\)
\(=\left(x^2-2.x.2+2^2\right)+\left(x^2-3^2\right)\)
\(=\left(x^2-4x+4\right)+\left(x^2-9\right)\)
Bài 2
a) \(6x^2y+18x\)
\(=6x\left(xy+3\right)\)
b) \(x^2-7x+3x-21\)
\(=\left(x^2-7x\right)+\left(3x-21\right)\)
\(=x\left(x-7\right)+3\left(x-7\right)\)
\(=\left(x-7\right)\left(x+3\right)\)
c) \(x^2-4y^2+2x+1\)
\(=\left(x^2+2x+1\right)-4y^2\)
\(=\left(x^2+2.x.1+1^2\right)-4y^2\)
\(=\left(x+1\right)^2-4y^2\)
\(=\left(x+1\right)^2-\left(2y\right)^2\)
\(=\left[\left(x+1\right)-2y\right]\left[\left(x+1\right)+2y\right]\)
\(=\left(x+1-2y\right)\left(x+1+2y\right)\)
d) \(x^2+3x-3y-y^2\)
\(=\left(x^2-y^2\right)+\left(3x-3y\right)\)
\(=\left(x-y\right)\left(x+y\right)+3\left(x-y\right)\)
\(=\left(x-y\right)\left[\left(x+y\right)+3\right]\)
\(=\left(x-y\right)\left(x+y+3\right)\)
Bài 3
a) \(\left(x+3\right)\left(x+2\right)-x\left(x+3\right)=10\)
\(\Rightarrow\left(x+3\right)\left[\left(x+2\right)-x\right]=10\)
\(\Rightarrow\left(x+3\right)\left(x+2-x\right)=10\)
\(\Rightarrow\left(x+3\right).2=10\)
\(\Rightarrow x+3=5\)
\(\Rightarrow x=2\)
b) \(\left(x+2\right)^2-\left(x-3\right)\left(x+3\right)=10\)
\(\Rightarrow\left(x^2+2.x.2+2^2\right)-\left(x^2-3^2\right)=10\)
\(\Rightarrow\left(x^2+4x+4\right)-\left(x^2-9\right)=10\)
\(\Rightarrow x^2+4x+4-x^2+9=10\)
\(\Rightarrow4x+13=10\)
\(\Rightarrow4x=-3\)
\(\Rightarrow x=-\frac{3}{4}\)
c) \(4x^2-25=0\)
\(\Rightarrow\left(2x\right)^2-5^2=0\)
\(\Rightarrow\left(2x-5\right)\left(2x+5\right)=0\)
\(\Rightarrow2x-5=0\) hoặc \(2x+5=0\)
\(\Rightarrow2x=5\) hoặc\(2x=-5\)
\(\Rightarrow x=\frac{5}{2}\) hoặc\(x=-\frac{5}{2}\)
d) \(2x\left(x+3\right)+x^2+3x=0\)
\(\Rightarrow2x\left(x+3\right)+x\left(x+3\right)=0\)
\(\Rightarrow\left(x+3\right)\left(2x+x\right)=0\)
\(\Rightarrow\left(x+3\right).3x=0\)
\(\Rightarrow x+3=0\) hoặc \(3x=0\)
\(\Rightarrow x=-3\) hoặc \(x=0\)
K MÌNH VỚI NHÉ
Bài 1 :
1) a2 - 4 + y ( a - 2 )
= ( a + 2 ) ( a - 2 ) + y ( a - 2 )
= ( a - 2 ) ( a + 2 + y )
2) ( x - 2 )2 - 9y2
= ( x - 2 - 3y ) ( x - 2 + 3y )
Bài 2 :
1) 3 ( x + 4 ) - 2x = 5
=> 3x + 12 - 2x = 5
=> x + 12 = 5
=> x = 5 - 12 = - 7
Vậy x = - 7
2) x ( x - 2 ) - x2 - 6 = 0
=> x2 - 2x - x2 - 6 = 0
=> - 2x - 6 = 0
=> 2x = - 6
=> x = \(-\frac{6}{2}=3\)
Vậy x = 3
3 ) x2 - 3x = 0
=> x ( x - 3 ) = 0
=> \(\orbr{\begin{cases}x=0\\x-3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
Vậy \(x\in\left\{0;3\right\}\)
4) 5 - 3 ( x - 6 ) = 4
=> 5 - 3x + 18 = 4
=> 3x = 5 + 18 - 4
=> 3x = 19
=> x = \(\frac{19}{3}\)
Vậy \(x=\frac{19}{3}\)
Mk làm bài 2 thui, bài 1 nhân ra rùi rút gọn đi là đc
a) \(5x^2-5y^2=5\left(x^2-y^2\right)=5\left(x-y\right)\left(x+y\right)\)
b) \(x^2-5xy+x-5y=x\left(x-5y\right)+\left(x-5y\right)=\left(x-5y\right)\left(x+1\right)\)
c) Phần này phải là \(x^2-y^2+4x+4y\)mới đúng, như vậy nó sẽ là :\(x^2-y^2+4x+4y=\left(x+y\right)\left(x-y\right)+4\left(x+y\right)=\left(x+y\right)\left(x-y+4\right)\)
d) \(x^2-2x-y^2-2y=\left(x^2-y^2\right)-\left(2x+2y\right)=\left(x+y\right)\left(x-y\right)-2\left(x+y\right)=\left(x+y\right)\left(x-y-2\right)\)
Chúc bạn hok tốt !
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 = \(-\frac{5}{7}\)
\(1,4x^2-y^2=\left(2x\right)^2-y^2=\left(2x-y\right)\left(2x+y\right)\)
\(2,9x^2-4y^2=\left(3x\right)^2-\left(2y\right)^2=\left(3x-2y\right)\left(3x+2y\right)\)
\(3,4x^2+y^2+4xy=\left(2x\right)^2+2.2x.y+y^2=\left(2x+y\right)^2\)
\(1,2x^2+8x=0\Rightarrow2x\left(x+4\right)=0\Rightarrow\orbr{\begin{cases}2x=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-4\end{cases}}\)
\(2,3\left(x-4\right)+x^2-4x=0\)
\(\Rightarrow3\left(x-4\right)+x\left(x-4\right)=0\)
\(\Rightarrow\left(3+x\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3+x=0\\x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=4\end{cases}}\)
\(3,3\left(x-2\right)=x^2-2x\)
\(\Rightarrow3\left(x-2\right)-x^2+2x=0\)
\(\Rightarrow3\left(x-2\right)-x\left(x-2\right)=0\)
\(\Rightarrow\left(3-x\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3-x=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}\)
\(4,x\left(x-2\right)-6\left(2-x\right)=0\)
\(\Rightarrow x\left(x-2\right)+6\left(x-2\right)=0\)
\(\Rightarrow\left(x+6\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+6=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-6\\x=2\end{cases}}\)
\(\left(-3x-2\right)^2+\left(3x+5\right)\left(5-3x\right)=-7\)
\(\Leftrightarrow9x^2+12x+4+15x-9x^2+25-15x=-7\)
\(\Leftrightarrow12x+36=0\Leftrightarrow x=-3\)
\(\left(x+2\right)\left(x^2+2x+2\right)-x\left(x-8\right)^2=\left(4x-3\right)\left(4x+3\right)\)
\(\Leftrightarrow x^3+2x^2+2x+2x^2+4x+4-x\left(x^2-16x+64\right)=16x^2-9\)
\(\Leftrightarrow x^3+4x^2+6x+4-x^3+16x^2-64=16x^2-9\)
\(\Leftrightarrow4x^2+6x-51=0\)
\(\cdot\Delta=6^2-4.4.\left(-51\right)=852\)
Vậy pt có 2 nghiệm phân biệt
\(x_1=\frac{-6+\sqrt{852}}{8}\);\(x_2=\frac{-6-\sqrt{852}}{8}\)
Bài 3:Ko phụ thuộc vào biết t nha
(3t+2)(2t+1)+(3-t)(t+2)=\(6t^2+3t+4t+2+3t+6-t^2-2t.\)
=\(5t^2+8t+8\) Vậy biểu thức phụ thuộc vào biến t
->đpcm sai.
1.a) \(\Leftrightarrow\) 3x+10-2x =0
\(\Leftrightarrow\text{ 3x-2x=-10}\)
\(\Leftrightarrow x=-10\)
b) coi lại có thiếu ngoặc ko nhé
cứ nhân vào dấu ngoặc rồi làm như thường