Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(4x+1\right)\left(12x-1\right)\left(3x-2\right)\left(x+1\right)-4\) (Sửa đề)
\(=[\left(4x+1\right)\left(3x+2\right)][\left(12x-1\right)\left(x+1\right)]-4\)
\(=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)
Đặt \(12x^2+11x-1=n\)
\(=\left(n+3\right)n-4\)
\(=n^2+3n-4\)
\(=n^2-n+4n-4\)
\(=n\left(n-1\right)+4\left(n-1\right)\)
\(=\left(n-1\right)\left(n+4\right)\)
\(=\left(12x^2+11x-1-1\right)\left(12x^2+11x-1+4\right)\)
\(=\left(12x^2+11x-2\right)\left(12x^2+11x+3\right)\)
\(\left(3x+4\right)\left(x+1\right)\left(6x+7\right)^2=6\)
\(\Leftrightarrow\left(3x^2+7x+4\right)\left(36x^2+84x+49\right)=6\)(1)
Đặt \(\left(3x^2+7x+4\right)=n\)lúc đó (1):
\(\left(12n+1\right)n=6\)
\(\Rightarrow\hept{\begin{cases}n=0,75\\n=\frac{2}{3}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{-2}{3}\\x=\frac{-5}{3}\end{cases}}\)
Helpppppppppppppppppppppp!!!!
Trả lời:
Bài 1:
1, 4x2 - 25 + ( 2x + 7 )( 5 - 2x )
= ( 2x - 5 )( 2x + 5 ) - ( 2x + 7 )( 2x - 5 )
= ( 2x - 5 )( 2x + 5 - 2x - 7 )
= - 2 ( 2x - 5 )
2, 3 ( x + 4 ) - x2 - 4x
= 3 ( x + 4 ) - x ( x + 4 )
= ( x + 4 )( 3 - x )
3, 5x2 - 5y2 - 10x + 10y
= 5 ( x2 - y2 - 2x + 2y )
= 5 [ ( x2 - y2 ) - ( 2x - 2y ) ]
= 5 [ ( x - y )( x + y ) - 2 ( x - y ) ]
= 5 ( x - y )( x + y - 2 )
4, x2 - xy + x - y
= ( x2 - xy ) + ( x - y )
= x ( x - y ) + ( x - y )
= ( x - y )( x + 1 )
5, ax - bx - a2 + 2ab - b2
= ( ax - bx ) - ( a2 - 2ab + b2 )
= x ( a - b ) - ( a - b )2
= ( a - b )( x - a + b )
6, x2 + 4x - y2 + 4
= ( x2 + 4x + 4 ) - y2
= ( x + 2 )2 - y2
= ( x + 2 - y )( x + 2 + y )
mình cần câu trả lời gấp ạ, giúp với
\(x^2+2xy-8y^2+2xz+14yz-3z^2\)
\(=\left(x^2+y^2+z^2+2xy+2yz\right)+\left(-9x^2-12yz-4x^2\right)\)
\(=\left(x+y+z\right)^2-[\left(3x\right)2-2.3x2y+\left(2x\right)^2\)
\(=\left(x+y+z\right)^2-\left(3y-2x\right)^2\)
\(=\left(x+y+z-3y+2x\right)\left(x+y+z+3y-2x\right)\)
\(3x^2-22xy-4x+8y+7y^2+1\)
\(=3x^2-21xy-xy-3x-x+7y+y+7y^2+1\)
\(=\left(3x^2-21xy-3x\right)-\left(xy-7y^2-y\right)-\left(x-7y-1\right)\)
\(=3x\left(x-7y-1\right)-y\left(x-7y-1\right)-\left(x-7y-1\right)\)
\(=\left(x-7y-1\right)\left(3x-y-1\right)\)
ggiúp mình với, mình đang cần gấp lắm ạ
\(4x^2-25+\left(2x+7\right)\left(5-2x\right)\)
\(=\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)\)
\(=\left(2x-5\right)[2x+5-\left(2x+7\right)]\)
\(=\left(2x-5\right)\left(2x+5-2x-7\right)\)
\(=-2\left(2x-5\right)\)
\(3\left(x+4\right)-x^2-4x\)
\(=-x^2+3x+12-4x\)
\(=-\left(x^2+4x-3x-12\right)\)
\(=-\left(x-3\right)\left(x+4\right)\)
Ta có: \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ca\)
\(\Rightarrow\left(x+\frac{1}{6}y+3\right)^2=x^2+\left(\frac{y}{6}\right)^2+3^2+2.x.\frac{y}{6}+2.\frac{y}{6}.3+2.x.3\)
\(=x^2+\frac{y^2}{36}+9+\frac{xy}{3}+y+6x\)
d) \(\Leftrightarrow\left(x-2\right)^2=25\)
\(\Leftrightarrow x=7\)hoặc \(x=-3\)
e) \(\Leftrightarrow\left(5-2x-4\right)\left(5-2x+4\right)=0\)
\(\Leftrightarrow x=\frac{1}{2}\)hoặc \(x=\frac{9}{2}\)
d) \(x^2-4x+4=25\)
\(\Leftrightarrow\left(x-2\right)^2-5^2=0\)
\(\Leftrightarrow\left(x-2-5\right)\left(x-2+5\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\x+3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=7\\x=-3\end{cases}}\)
Vậy \(S=\left\{7;-3\right\}\)
b) \(\left(5-2x\right)^2-16=0\)
\(\Leftrightarrow\left(5-2x\right)^2-4^2=0\)
\(\Leftrightarrow\left(5-2x-4\right)\left(5-2x+4\right)=0\)
\(\Leftrightarrow\left(1-2x\right).\left(9-2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}1-2x=0\\9-2x=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-2x=-1\\-2x=-9\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{9}{2}\end{cases}}\)
Vậy \(S=\left\{\frac{1}{2};\frac{9}{2}\right\}\)