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a/ \(=x^3-2x^5\)
b/\(=5x^2+5-x^3-x\)
c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)
d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)
e/ \(=x^4-x^2+2x^3-2x\)
f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)
em 2k6, đọc phần lí thuyết r lm, nên có lỗi j sai mong mn thông cảm
bài 1,
a, \(3xy\left(4xy^2-5x^2y-4xy\right)\)
= \(3xy.4xy^2-3xy.5x^2y-3xy.4xy\)
=\(12x^2y^3-15x^3y^2-12x^2y^2\)
\(1,A=\left(3x+7\right)\left(2x+3\right)-\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\\ =6x^2+23x+21-2x-3-6x^2-23x+55\\ =73-2x\left(đề.sai\right)\\ B=x^4+x^3-x^2-2x^2-2x+2-x^4-x^3+3x^2+2x\\ =2\\ 2,\\ a,\Leftrightarrow30x^2+18x+3x-30x^2=7\\ \Leftrightarrow21x=7\Leftrightarrow x=\dfrac{1}{3}\\ b,\Leftrightarrow-63x^2+78x-15+63x^2+x-20=44\\ \Leftrightarrow79x=79\Leftrightarrow x=1\\ c,\Leftrightarrow\left(x+5\right)\left(x^2+3x+2\right)-x^3-8x^2=27\\ \Leftrightarrow x^3+3x^2+2x+5x^2+15x+10-x^3-8x^2=27\\ \Leftrightarrow17x=17\Leftrightarrow x=1\)
\(d,\Leftrightarrow7x-2x^2-3+x^2+x-6=-x^2-x+2\\ \Leftrightarrow9x=11\Leftrightarrow x=\dfrac{11}{9}\)
Bài 1
A= (x-2)(2x-1)-2x(x+3)=2x2-x-4x+2-2x2-6x=-11x+2
Bài 1:
a) \(A=\left(x-2\right)\left(2x-1\right)-2x\left(x+3\right)\)
\(A=2x^2-x-4x+2-2x^2-6x\)
\(A=-11x+2\)
b) \(B=\left(3x-2\right)\left(2x+1\right)-\left(6x-1\right)\left(x+2\right)\)
\(B=6x^2+3x-4x-2-6x^2-12x+x+2\)
\(B=-12x\)
c) \(C=6x\left(2x+3\right)-\left(4x-1\right)\left(3x-2\right)\)
\(C=12x^2+18x-12x^2+8x+3x-2\)
\(C=29x-2\)
d) \(D=\left(2x+3\right)\left(5x-2\right)+\left(x+4\right)\left(2x-1\right)-6x\left(2x-3\right)\)
\(D=10x^2-4x+15x-6+2x^2-x+8x-4-12x^2+18x\)
\(D=36x-10\)
A) \(\left(x-3\right)^2-\left(x+2\right)^2\)
\(=\left(x-3-x-2\right)\left(x-3+x+2\right)\)
\(=-5.\left(2x-1\right)\)
B) \(\left(4x^2+2xy+y^2\right)\left(2x-y\right)-\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)
\(=\left(2x\right)^3-y^3-\left[\left(2x\right)^3+y^3\right]\)
\(=8x^3-y^3-8x^3-y^3\)
\(=-2y^3\)
C) \(x^2+6x+8\)
\(=x^2+6x+9-1\)
\(=\left(x+3\right)^2-1\)
\(=\left(x+3-1\right)\left(x+3+1\right)\)
\(=\left(x+2\right)\left(x+4\right)\)
bài 3 A) \(x^2-16=0\)
\(\left(x-4\right)\left(x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-4=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
vậy \(\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
B) \(x^4-2x^3+10x^2-20x=0\)
\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\left(x^3+10x\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^3+10x=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x\left(x^2+10\right)=0\\x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
vậy \(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
Bài 4 :
Thay x=y+5 , ta có :
a ) ( y+5)*(y5+2)+y*(y-2)-2y*(y+5)+65
=(y+5)*(y+7)+y^2-2y-2y^2-10y+65
=y^2+7y+5y+35-y^2-2y-2y^2-10y+65
= 100
Bài 5 :
A = 15x-23y
B = 2x-3y
Ta có : A-B
= ( 15x -23y)-(2x-3y)
=15x-23y-2x-3y
=13x-26y
=13x*(x-2y) chia hết cho 13
=> Nếu A chia hết cho 13 thì B chia hết cho 13 và ngược lại
Bài 1:
a)
\((3x+2)(2x+9)-(x+2)(6x+1)=(x+1)-(x-6)\)
\(\Leftrightarrow (6x^2+27x+4x+18)-(6x^2+x+12x+2)=7\)
\(\Leftrightarrow 18x+16=7\Leftrightarrow 18x=-9\Leftrightarrow x=\frac{-1}{2}\)
b)
\(3(2x-1)(3x-1)-(2x-3)(9x-1)=0\)
\(\Leftrightarrow 3(6x^2-2x-3x+1)-(18x^2-2x-27x+3)=0\)
\(\Leftrightarrow 3(6x^2-5x+1)-(18x^2-29x+3)=0\)
\(\Leftrightarrow 14x=0\Leftrightarrow x=0\)
Bài 2:
\(M=(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)+x^2\)
\(=(x^2-ax-bx+ab)+(x^2-bx-cx+bc)+(x^2-cx-ax+ac)+x^2\)
\(=4x^2-2(a+b+c)x+(ab+bc+ac)\)
\(=(2x)^2-(a+b+c).2x+(ab+bc+ac)\)
\(=(a+b+c)^2-(a+b+c).(a+b+c)+(ab+bc+ac)\)
\(=(a+b+c)^2-(a+b+c)^2+(ab+bc+ac)=ab+bc+ac\)