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a) Kết quả M = 0. Chú ý: nhân tử chung là 2f - 5 = 0.
b) Kết quả N = 300000.
c) Kết quả p = 0. Chú ý: nhân tử x 2 + y -1 = 0.
d) Kết quả Q = 280. Chú ý: Q = (x - y)[ ( x - y ) 2 - xy].
Bài 6:
a) \(x\left(x-2\right)+x-2=0\)
\(\Leftrightarrow x\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
b) \(5x\left(x-3\right)-x+3=0\)
\(\Leftrightarrow5x\left(x-3\right)-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)
c) \(3x\left(x-5\right)-\left(x-1\right)\left(2+3x\right)=30\)
\(\Leftrightarrow3x^2-15x-2x-3x^2+2+3x=30\)
\(\Leftrightarrow-14x+2=30\)
\(\Leftrightarrow-14x=28\)
\(\Leftrightarrow x=-2\)
d) \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow x^2+3x+2x+6-x^2-5x+2x+10=0\)
\(\Leftrightarrow2x+16=0\)
\(\Leftrightarrow2x=-16\)
\(\Leftrightarrow x=-8\)
Bài 1:
a) \(8\left(x-2\right)-2\left(3x-4\right)=2\)
\(\Rightarrow2\left[4\left(x-2\right)-\left(3x-4\right)\right]=2\)
\(\Rightarrow4\left(x-2\right)-3x+4=0\)
\(\Rightarrow4x-8-3x+4=0\)
\(\Rightarrow x-4=0\)
\(\Rightarrow x=4\)
b) \(10\left(3x-2\right)-3\left(5x+2\right)+5\left(11-4x\right)=25\)
\(\Rightarrow5\left[2\left(3x-2\right)+11-4x\right]-3\left(5x+2\right)=25\)
\(\Rightarrow5\left(6x-4+11-4x\right)-3\left(5x+2\right)=25\)
\(\Rightarrow5\left(2x+7\right)-3\left(5x+2\right)=25\)
\(\Rightarrow10x+35-15x-6=25\)
\(\Rightarrow-5x+29=25\)
\(\Rightarrow-5x=25-29\)
\(\Rightarrow-5x=-4\)
\(\Rightarrow x=\dfrac{4}{5}\)
c) \(2x\left(x+1\right)-x^2\left(x+2\right)+x^3-x+4=0\)
\(\Rightarrow2x^2+2x-x^3-2x^2+x^3-x+4=0\)
\(\Rightarrow x+4=0\)
\(\Rightarrow x=-4\)
d) \(4x\left(3x+2\right)-6x\left(2x+5\right)+21\left(x-1\right)=0\)
\(\Rightarrow12x^2+8x-12x^2-30x+21x-21=0\)
\(\Rightarrow-x-21=0\)
\(\Rightarrow x=-21\)
Bài 2:
a) \(P=\left(4x^2-3y\right)2y-\left(3x^2-4y\right)3y\)
\(P=8x^2y-6y^2-9x^2y+12y^2\)
\(P=-x^2y+6y^2\)
Thay x = -1 ; y = 2 vào P ta được
\(P=-\left(-1\right)^2.2+6.2^2\)
\(P=-2+24=22\)
b) \(Q=4x^2\left(5x-3y\right)-x^2\left(4x+y\right)\)
\(Q=20x^3-12x^2y-4x^3-x^2y\)
\(Q=16x^3-13x^2y\)
Thay x = -1 ; y = 2 vào Q ta được
\(Q=16\left(-1\right)^3-13\left(-1\right)^2.2\)
\(Q=-16-26\)
\(Q=-42\)
c) \(H=x\left(x^3-y\right)+x^2\left(y-x^2\right)-y\left(x^2-3x\right)\)
\(H=x^4-xy+x^2y-x^4-x^2y+3xy\)
\(H=2xy\)
Thay x = 1/4 ; y = 2012 vào H ta được
\(H=2.\dfrac{1}{4}.2012\)
\(H=1006\)
1.a)\(8\left(x-2\right)-2\left(3x-4\right)=2\)
\(\Leftrightarrow8x-16-6x+8=2\)
\(\Leftrightarrow2x-8=2\Leftrightarrow2x=10\Leftrightarrow x=5\)
b)\(10\left(3x-2\right)-3\left(5x+2\right)+5\left(11-4x\right)=25\)
\(\Leftrightarrow30x-20-15x-6+55-20x=25\)
\(\Leftrightarrow-5x+29=25\Leftrightarrow-5x=-4\Leftrightarrow x=\dfrac{4}{5}=0,8\)
\(c)2x\left(x+1\right)-x^2\left(x+2\right)+x^3-x+4=0\)
\(\Leftrightarrow2x^2+2x-x^3-2x^2+x^3-x+4=0\)
\(\Leftrightarrow x+4=0\Leftrightarrow x=-4\)
\(d)4x\left(3x+2\right)-6x\left(2x+5\right)+21\left(x-1\right)=0\)
\(\Leftrightarrow12x^2+8x-12x^2-30x+21x-21=0\)
\(\Leftrightarrow-x-21=0\Leftrightarrow-x=21\Leftrightarrow x=-21\)
2.
a)\(P=\left(4x^2-3y\right)2y-\left(3x^2-4y\right)3y\)
\(\Leftrightarrow8x^2y-6y^2-9x^2y-12y^2\)
\(\Leftrightarrow x^2y-18y^2\)
tại x=-1 , y=2
ta có:\(x^2y-18y^2=\left(-1\right)^2.2-18.2^2=2-72=-70\)
vậy \(P=\left(4x^2-3y\right)2y-\left(3x^2-4y\right)3y=-70\) tại x=-1,y=2
b)\(Q=4x^2\left(5x-3y\right)-x^2\left(4x+y\right)\)
\(\Leftrightarrow20x^3-12x^2y-4x^3-x^2y\)
\(\Leftrightarrow17x^3-13x^2y\)
tại x=-1,y=2
ta có:\(17x^3-13x^2y=17\left(-1\right)^3-13\left(-1\right)^2.2=-17-26=-43\)
vậy \(Q=4x^2\left(5x-3y\right)-x^2\left(4x+y\right)=-43\)
c)\(H=x\left(x^3-y\right)+x^2\left(y-x^2\right)-y\left(x^2-3x\right)\)
\(\Leftrightarrow x^4-xy+x^2y-x^3-x^2y+3xy\)
\(\Leftrightarrow x^4+2xy-x^3\)
tại x=1/4 và y=2012
ta có:\(x^4+2xy-x^3=\left(\dfrac{1}{4}\right)^4+2.\dfrac{1}{4}.2012-\left(\dfrac{1}{4}\right)^3\approx1006\)
Bài 4:
a: \(\Leftrightarrow x^3-3x^2+3x-1-x^3-27+3x^2-12=2\)
\(\Leftrightarrow3x-40=2\)
=>3x=42
hay x=14
b: \(\Leftrightarrow x^3+8-x^3-2x=0\)
=>-2x+8=0
=>-2x=-8
hay x=4
c: \(x\left(x-2\right)+\left(x-2\right)=0\)
=>(x-2)(x+1)=0
=>x=2 hoặc x=-1
d: \(5x\left(x-3\right)-x+3=0\)
=>5x(x-3)-(x-3)=0
=>(x-3)(5x-1)=0
=>x=3 hoặc x=1/5
e: \(3x\left(x-5\right)-\left(x-1\right)\left(3x+2\right)=30\)
\(\Leftrightarrow3x^2-15x-3x^2-2x+3x+2=30\)
=>-14x=28
hay x=-2
f: \(\Leftrightarrow\left(x+2\right)\left(x+30-x-5\right)=0\)
=>x+2=0
hay x=-2
bài 1.
a) (4x3 - 2)(2x3- x + \(\dfrac{5}{8}\))
= 8x6 - 4x4 + \(\dfrac{5}{2}\)x3 - 4x3 + 2x - \(\dfrac{5}{4}\)
b) (x2y2 - xy + y)(x - y)
= x3y2 - x2y + xy - x2y3 + xy2 - y2
c) (x + 2y)(x2 - 2xy + y2)
= x3 + 8y3
d) (7x - 3)(7x + 3) + (2x - 3)2
= 49x2 - 9 + 4x2 - 12x + 9
= 53x2 - 12x
Bài 2.
a) 4(3x - 1) - 2(5 - 3x) = 24
12x - 4 - 10 + 6x - 24 = 0
18x - 38 = 0
\(\Rightarrow\) 18x = 38
\(\Rightarrow\) x = \(\dfrac{19}{9}\)
b) 4x2 - 9 = 0
\(\Rightarrow\) 4x2 = 9
\(\Rightarrow\) x2 = \(\dfrac{9}{4}\)
\(\Rightarrow\) x = \(\pm\dfrac{3}{2}\)
vậy x = 3/2 hoặc x = -3/2
c) x3 - 25x = 0
x(x2 - 25) = 0
x(x - 5)(x + 5) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-5=0\\x+5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
d) (x2 + 4)2 - 16x2 = 0
(x2 + 4 - 4x)(x2 + 4 + 4x) = 0
\(\Rightarrow\) (x - 2)2.(x + 2)2 = 0
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)^2=0\\\left(x+2\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Bài 3.
a) x(x + y) + y(x + y)
Ta có:
x(x + y) + y(x + y)
= (x + y)(x + y)
= (x + y)2
Thay x = 2004 và y = -2003 vào biểu thức đại số ta có:
[2004 + (-2003)]2 = 12
= 1
b) x2 + xy - xz - yz
Ta có:
x2 + xy - xz - yz
= (x2 + xy) - (xz + yz)
= x(x + y) - z(x + y)
= (x - z)(x + y)
Thay x= 6,5; y = 3,5 và z = 37,5 vào biểu thức đại số, ta có:
(6,5 - 37,5)(6,5 + 3,5)
= -31 . 10
= -310
c) x2 - 6xy + 9y2
ta có:
x2 - 6xy + 9y2
= (x - 3y)2
Thay x = 14 và y = -2 vào biểu thức đại số, ta có:
[14 - (-2)]2 = (14 + 2)2
= 162 = 256
Nhớ tik mik nhé không lần sau mik ko giúp đâu
có j ko hỉu cứ bình luận ở dưới