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T
27 tháng 6 2017
Bài 1:
a) \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)-3x^2=42\) (1)
\(\Leftrightarrow x^3+9x^2+27x+27-x\left(9x^2+6x+1\right)+8x^3+1-3x^2=42\)
\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1-3x^2=42\)
\(\Leftrightarrow26x+28=42\)
\(\Leftrightarrow26x=42-28\)
\(\Leftrightarrow26x=14\)
\(\Leftrightarrow x=\dfrac{7}{13}\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{7}{13}\right\}\)
LL
12 tháng 7 2017
1b) \(5x\left(x+3\right)^2-5\left(x+1\right)^3+15\left(x+2\right)\left(x-2\right)=5\Leftrightarrow5x\left(x^2+6x+9\right)-5\left(x^3+3x^2+3x+1\right)+15\left(x^2-4\right)=5\Leftrightarrow30x-65=5\Leftrightarrow30x=70\Leftrightarrow x=\dfrac{7}{3}\)
26 tháng 10 2020
Bài 1
1) 4x - x2 - 4 = 0
⇔ -( x2 - 4x + 4 ) = 0
⇔ -( x - 2 )2 = 0
⇔ x - 2 = 0
⇔ x = 2
2) 4( x - 1 )2 - ( 5 - 2x )2 = 0
⇔ 22( x - 1 )2 - ( 5 - 2x )2 = 0
⇔ ( 2x - 2 )2 - ( 5 - 2x ) = 0
⇔ ( 2x - 2 - 5 + 2x )( 2x - 2 + 5 - 2x ) = 0
⇔ ( 4x - 7 ).3 = 0
⇔ 4x - 7 = 0
⇔ x = 7/4
3) 9( x - 2 )2 - 4( 3 - x )2 = 0
⇔ 32( x - 2 )2 - 22( x - 3 )2 = 0
⇔ ( 3x - 6 )2 - ( 2x - 6 )2 = 0
⇔ ( 3x - 6 - 2x + 6 )( 3x - 6 + 2x - 6 ) = 0
⇔ x( 5x - 12 ) = 0
⇔ x = 0 hoặc 5x - 12 = 0
⇔ x = 0 hoặc x = 12/5
4) x2 - 6x + 5 = 0
⇔ x2 - 5x - x + 5 = 0
⇔ x( x - 5 ) - ( x - 5 ) = 0
⇔ ( x - 5 )( x - 1 ) = 0
⇔ x - 5 = 0 hoặc x - 1 = 0
⇔ x = 5 hoặc x = 1
26 tháng 10 2020
Bài 2.
1) x2 - z2 + y2 - 2xy
= ( x2 - 2xy + y2 ) - z2
= ( x - y )2 - z2
= ( x - y - z )( x - y + z )
2) a3 - ay - a2x + xy
= ( a3 - a2x ) - ( ay - xy )
= a2( a - x ) - y( a - x )
= ( a - x )( a2 - y )
3) 2xy + 3z + 6y + xz
= ( 2xy + 6y ) + ( xz + 3z )
= 2y( x + 3 ) + z( x + 3 )
= ( x + 3 )( 2y + z )
4) x2 + 2xz + 2xy + 4yz
= ( x2 + 2xy ) + ( 2xz + 4yz )
= x( x + 2y ) + 2z( x + 2y )
= ( x + 2y )( x + 2z )
5) ( x + y + z )3 - x3 - y3 - z3
= x3 + y3 + z3 + 3( x + y )( y + z )( x + z ) - x3 - y3 - z3
= 3( x + y )( y + z )( x + z )
M
30 tháng 10 2019
Câu 1 : Tìm x :
1. \(A=x^2+4x-2\)
\(A=x^2+2.x.2+2^2-2^2-2\)
\(A=\left(x^2+4x+2^2\right)-4-2\)
\(A=\left(x+2\right)^2-6\)
\(\left(x+2\right)^2-6\ge-6\)
MIn A= -6 khi \(\left(x+2\right)^2=0\)
=> \(x+2=0hayx=-2\)
Vậy x=2
những câu tiếp theo làm tg tự như thế nhé
30 tháng 10 2019
Câu 1:
a) Ta có: \(A=x^2+4x-2\)
\(=x^2+4x+4-6\)
\(=\left(x+2\right)^2-6\)
Ta có: \(\left(x+2\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+2\right)^2-6\ge-6\forall x\)
Dấu '=' xảy ra khi
\(\left(x+2\right)^2=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\)
Vậy: x=-2
b) Ta có: \(B=2x^2-4x+3\)
\(=2\left(x^2-2x+\frac{3}{2}\right)\)
\(=2\left(x^2-2\cdot x\cdot1+1+\frac{1}{2}\right)\)
\(=2\left[\left(x^2-2x\cdot1+1\right)+\frac{1}{2}\right]\)
\(=2\left[\left(x-1\right)^2+\frac{1}{2}\right]\)
\(=2\left(x-1\right)^2+1\)
Ta có: \(\left(x-1\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-1\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-1\right)^2+1\ge1\forall x\)
Dấu '=' xảy ra khi
\(2\left(x-1\right)^2=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)
Vậy: x=1
c) Ta có: \(C=x^2+y^2-4x+2y+5\)
\(=x^2-4x+4+y^2+2y+1\)
\(=\left(x^2-4x+4\right)+\left(y^2+2y+1\right)\)
\(=\left(x-2\right)^2+\left(y+1\right)^2\)
Ta có: \(\left(x-2\right)^2\ge0\forall x\)
\(\left(y+1\right)^2\ge0\forall y\)
Do đó: \(\left(x-2\right)^2+\left(y+1\right)^2\ge0\forall x,y\)
Dấu '=' xảy ra khi
\(\left\{{}\begin{matrix}\left(x-2\right)^2=0\\\left(y+1\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
Vậy: x=2 và y=-1
Câu 2:
a) Ta có: \(A=-x^2+6x+5\)
\(=-\left(x^2-6x-5\right)\)
\(=-\left(x^2-6x+9-14\right)\)
\(=-\left[\left(x^2-6x+9\right)-14\right]\)
\(=-\left[\left(x-3\right)^2-14\right]\)
\(=-\left(x-3\right)^2+14\)
Ta có: \(\left(x-3\right)^2\ge0\forall x\)
\(\Rightarrow-\left(x-3\right)^2\le0\forall x\)
\(\Leftrightarrow-\left(x-3\right)^2+14\le14\forall x\)
Dấu '=' xảy ra khi
\(-\left(x-3\right)^2=0\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy: GTLN của đa thức \(A=-x^2+6x+5\) là 14 khi x=3
b) Ta có: \(B=-4x^2-9y^2-4x+6y+3\)
\(=-\left(4x^2+9y^2+4x-6y-3\right)\)
\(=-\left(4x^2+4x+1+9y^2-6y+1-5\right)\)
\(=-\left[\left(4x^2+4x+1\right)+\left(9y^2-6y+1\right)-5\right]\)
\(=-\left[\left(2x+1\right)^2+\left(3y-1\right)^2-5\right]\)
\(=-\left(2x+1\right)^2-\left(3y-1\right)^2+5\)
Ta có: \(\left(2x+1\right)^2\ge0\forall x\)
\(\Rightarrow-\left(2x+1\right)^2\le0\forall x\)(1)
Ta có: \(\left(3y-1\right)^2\ge0\forall y\)
\(\Rightarrow-\left(3y-1\right)^2\le0\forall y\)(2)
Từ (1) và (2) suy ra
\(-\left(2x+1\right)^2-\left(3y-1\right)^2\le0\forall x,y\)
\(\Rightarrow-\left(2x+1\right)^2-\left(3y-1\right)^2+5\le5\forall x,y\)
Dấu '=' xảy ra khi
\(\left\{{}\begin{matrix}-\left(2x+1\right)^2=0\\-\left(3y-1\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x+1\right)^2=0\\\left(3y-1\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x+1=0\\3y-1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x=-1\\3y=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{-1}{2}\\y=\frac{1}{3}\end{matrix}\right.\)
Vậy: GTLN của đa thức \(B=-4x^2-9y^2-4x+6y+3\) là 5 khi và chỉ khi \(x=\frac{-1}{2}\) và \(y=\frac{1}{3}\)
Câu 3:
a) Ta có: \(x^2+y^2-2x+4y+5=0\)
\(\Rightarrow x^2-2x+1+y^2+4y+4=0\)
\(\Rightarrow\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\)
\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
Vậy: x=1 và y=-2
b) Ta có: \(5x^2+9y^2-12xy-6x+9=0\)
\(\Rightarrow x^2+4x^2+9y^2-12xy-6x+9=0\)
\(\Rightarrow\left(4x^2+12xy+9y^2\right)+\left(x^2-6x+9\right)=0\)
\(\Rightarrow\left(2x+3y\right)^2+\left(x-3\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x+3y\right)^2=0\\\left(x-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+3y=0\\x-3=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2\cdot3+3y=0\\x=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}6+3y=0\\x=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3y=-6\\x=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-2\\x=3\end{matrix}\right.\)
Vậy: x=3 và y=-2
20 tháng 5 2022
b: \(=\left(a+b\right)^2+2c\left(a+b\right)+c^2+\left(a+b\right)^2-2c\left(a+b\right)+c^2-4c^2\)
\(=2\left(a+b\right)^2-2c^2\)
\(=2\left(a+b+c\right)\left(a+b-c\right)\)
c: \(\left(a+b\right)^3-\left(a-b\right)^3\)
\(=a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2+b^3\)
\(=6a^2b+2b^3\)
\(=2b\left(3ab+1\right)\)
e: \(\left(x+y\right)^3-x^3-y^3\)
\(=x^3+3x^2y+3xy^2+y^3-x^3-y^3\)
\(=3x^2y+3xy^2=3xy\left(x+y\right)\)
19 tháng 10 2017
Bài 1 :
a ) \(3\left(x-y\right)^2-2\left(x+y\right)^2-\left(x-y\right)\left(x+y\right)\)
\(=3\left(x^2-2xy+y^2\right)-2\left(x^2+2xy+y^2\right)-\left(x^2-y^2\right)\)
\(=3x^2-6xy+3y^2-2x^2-4xy-2y^2-x^2+y^2\)
\(\)\(=2y^2-10xy\)
Câu b tương tự
Bài 2 :
a ) \(x^2-9+\left(x-3\right)^2\)
\(=\left(x-3\right)\left(x+3\right)+\left(x-3\right)^2\)
\(=\left(x-3\right)\left(x+3+x-3\right)\)
\(=2x\left(x-3\right)\)
b ) \(x^3-4x^2+4x-xy^2\)
\(=x\left(x^2-4x+4-y^2\right)\)
\(=x\left[\left(x-2\right)^2-y^2\right]\)
\(=x\left(x-2-y\right)\left(x-2+y\right)\)
c ) \(x^3-4x^2+12x-27\)
\(=x^3-9x^2+5x^2+27x-15x-3^3\)
\(=\left(x^3-9x^2+27x-3^3\right)+\left(5x-15x\right)\)
\(=\left(x-3\right)^3+5\left(x-3\right)\)
\(=\left(x-3\right)\left[\left(x-3\right)^2+5\right]\)
\(=\left(x-3\right)\left(x^2-6x+14\right)\)
d ) \(3x^2-7x-10\)
\(=3x^2+3x-10x-10\)
\(3x\left(x+1\right)-10x\left(x+1\right)\)
\(=-7x\left(x+1\right)\)
18 tháng 7 2019
2 .tìm x
a , x ( x + 2 ) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
b, x ( x-5 )= 5 -x
<=> x ( x-5 ) + x - 5 = 0
<=> x (x-5) + ( x-5)= 0
<=> (x-5)(x+1 )=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-5=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
c) ( x + 1 ) ( 6x2 + 2x ) + ( x - 1 ) ( 6x2 + 2x ) = 0
\(\Leftrightarrow\) ( 6x2 + 2x ) \([\)(x+1)(x-1)\(]\)=0
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}2x\left(3x+1\right)=0\\x^{2^{ }}-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=0\\3x+1=0\\x^2-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\frac{-1}{3}\\x=1\end{matrix}\right.\)
18 tháng 7 2019
1 ,a) 2a ( x - y ) - ( y - x ) = 2ax - 2ay - y + x
= x ( 2a + 1 ) - y ( 2a + 1 )
= ( 2a + 1 ) ( x - y )
b) a2 ( x - y ) - ( y - x ) = a2x - a2y - y + x
= x ( a2+ 1 ) - y ( a2 +1 )
= ( a2+1 ) - (x-y )
c) x ( x - y ) + y ( y - x ) - 3 ( x - y ) = x 2 - xy -+ y 2 - xy - 3x + 3y
= x2 - 2xy + y2 -3x + 3y
= (x-y)2 -3 ( x - y )
= ( x-y ) ( x-y+3)
9 tháng 8 2017
a) 4x2-8x=0
(2x)2-2.2.2x+4-4=0
(2x-2)2 =4
2x-2=2
2x =4
x=2
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