help me.....
9x2=4(2x-3)2
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\(P=\frac{x^3-y^3+z^3+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
Ta có:Tử số: \(x^3-y^3+z^3+3xyz\)
\(=\left(x+z\right)^3-3zx\left(x+z\right)-y^3+3xyz\)
\(=\left(x+z-y\right)\left[\left(x+z\right)^2+y\left(x+z\right)+y^2\right]-3xz\left(x+z-y\right)\)
\(=\left(x-y+z\right)\left(x^2+y^2+z^2+xy+yz-zx\right)\)
Mẫu số: \(\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2\)
\(=x^2+2xy+y^2+y^2+2yz+z^2+z^2-2zx+x^2\)
\(=2\left(x^2+y^2+z^2+xy+yz-zx\right)\)
Suy ra \(P=\frac{x-y+z}{2}\)
\(\frac{x^3-y^3}{2y}.\left[\frac{2y}{4-2y-2x+xy}+\frac{2xy+4y}{\left(x-y\right)\left(x^2-4\right)}\right]\)
\(=\frac{x^3-y^3}{2y}.\left[\frac{2y}{\left(x-2\right)\left(y-2\right)}+\frac{2y\left(x+2\right)}{\left(x-y\right)\left(x-2\right)\left(x+2\right)}\right]\)
\(=\left(x^3-y^3\right).\left[\frac{x-y}{\left(x-y\right)\left(x-2\right)\left(y-2\right)}+\frac{y-2}{\left(x-y\right)\left(x-2\right)\left(y-2\right)}\right]\)
\(=\frac{\left(x-y\right)\left(x^2+xy+y^2\right)\left(x-2\right)}{\left(x-y\right)\left(x-2\right)\left(y-2\right)}\)
\(=\frac{x^2+xy+y^2}{y-2}\)
a) 4x2 - 12x - 7 = 4x2 - 12x + 9 - 16 = ( 4x2 - 12x + 9 ) - 16 = ( 2x - 3 )2 - 42 = ( 2x - 3 - 4 )( 2x - 3 + 4 ) = ( 2x - 7 )( 2x + 1 )
b) x2 + 6x - 1 ( nghiệm vô tỉ )
c) x2 - 4x + 2 ( nghiệm vô tỉ )
d) x2 + 4x - 12 = x2 - 2x + 6x - 12 = x( x - 2 ) + 6( x - 2 ) = ( x - 2 )( x + 6 )
e) x2 - 4x - 32 = ( x2 - 4x + 4 ) - 36 = ( x - 2 )2 - 62 = ( x - 2 - 6 )( x - 2 + 6 ) = ( x - 8 )( x + 4 )
f) x2 + 2x - 1 ( nghiệm vô tỉ )
g) x4 - 5x2 = x2( x2 - 5 )
h) Đặt t = x + 2y - 3
<=> t2 - 4t + 4
= ( t - 2 )2
= ( x + 2y - 3 - 2 )2
= ( x + 2y - 5 )2
i) x3( x2 + 1 )2 - 49x
= x[ x2( x2 + 1 )2 - 49 ]
= x[ ( x3 + x )2 - 72 ]
= x( x3 + x - 7 )( x3 + x + 7 )
k) x3 + y3 + z3 - 3xyz
= ( x3 + y3 ) + z3 - 3xyz
= ( x + y )3 - 3xy( x + y ) + z3 - 3xyz
= [ ( x + y )3 + z3 ] - [ 3xy( x + y ) + 3xyz ]
= ( x + y + z )[ ( x + y )2 - ( x + y )z + z2 ] - 3xy( x + y + z )
= ( x + y + z )( x2 + 2xy + y2 - xz - yz + z2 - 3xy )
= ( x + y + z )( x2 + y2 + z2 - xy - yz - xz )
\(4x^4-21x^2y^2+4y^2=\left(2x^2\right)^2-2.2x^2.2y^2+\left(2y^2\right)^2-13x^2y^2\)
\(=\left(2x^2-2y^2\right)^2-\left(\sqrt{13}xy\right)^2\)
\(=\left(2x^2-\sqrt{13}xy-2y^2\right)\left(2x^2+\sqrt{13}xy-2y^2\right)\)
Xem cách giải theo link: Câu hỏi của Trung Tính Hồ - Toán lớp 8 - Học toán với OnlineMath
Bài làm
9x2 = 4( 2x - 3 )2
<=> 9x2 - 22( 2x - 3 )2 = 0
<=> 9x2 - [ 2( 2x - 3 ) ]2 = 0
<=> ( 3x )2 - ( 4x - 6 )2 = 0
<=> ( 3x - 4x + 6 )( 3x + 4x - 6 ) = 0
<=> ( 6 - x )( 7x - 6 ) = 0
<=> \(\orbr{\begin{cases}6-x=0\\7x-6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=\frac{6}{7}\end{cases}}\)
Vậy x = 6 hoặc x = 6/7
\(9x^2=4\left(2x-3\right)^2\)
\(\left(3x\right)^2=\left(4x-6\right)^2\)
\(\left(3x\right)^2-\left(4x-6\right)^2=0\)
\(\left(3x-4x+6\right)\left(3x+4x-6\right)=0\)
\(\left(6-x\right)\left(7x-6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}6-x=0\\7x-6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=\frac{6.}{7}\end{cases}}\)