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\(a,x^2-5x\)
\(=x\left(x-5\right)\)
\(b,5x\left(x+5\right)+4x+20\)
\(=5x\left(x+5\right)+4\left(x+5\right)\)
\(=\left(5x+4\right)\left(x+5\right)\)
\(c,7x\left(2x-1\right)-4x+2\)
\(=7x\left(2x-1\right)-2\left(2x-1\right)\)
\(=\left(7x-2\right)-\left(2x-1\right)\)
\(d,x^2-16+2\left(x+4\right)\)
\(=x^2-16+2x+8\)
\(=x\left(x-2\right)-8\) ( Ý này thì k chắc lắm, sai thông cảm :)) )
\(e,x^2-10x+9\)
\(=x^2-x-9x+9\)
\(=x\left(x-1\right)-9\left(x-1\right)\)
\(=\left(x-9\right)\left(x-1\right)\)
\(f,\left(2x-1\right)^2-\left(x-3\right)^2=0\) ( mk đoán bài này là tìm x, sai thì bảo mk để mk sửa nhé )
\(\Rightarrow\left(2x-1\right)^2=\left(x-3\right)^2\)
\(\Leftrightarrow\pm\left(2x-1\right)=\pm\left(x-3\right)\)
\(\Rightarrow\hept{\begin{cases}2x-1=x-3\\-\left(2x-1\right)=-\left(x-3\right)\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}2x-1-x+3=0\\-2x+1-x+3=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x+2=0\\-3x+4=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\left(-2\right)\\x=\frac{4}{3}\end{cases}}\)
Vậy ...
1.x2-9
= (x-3)(x+3)
2. -2x2+2x+12
= -2x2+6x-4x+12
= -2x(x+2)+6(x+2)
= (x+2)(-2x+6)
4. -2x2+2x+24
= -2x2+8x-6x+24
= -2x(x+3)+8(x+3)
= (x+3)(-2x+8)
6. x2-5x+4
= x2-4x-x+4
= x(x-1) -4(x-1)
= (x-1)(x-4)
8. x2-7x+6
= x2-6x-x+6
= x(x-1)-6(x-1)
= (x-1)(x-6)
9. x2+5x+4
= x2+4x+x+4
= x(x+1)+4(x+1)
=(x+1)(x+4)
10. x2+7x+6
= x2 +x+6x+6
= x(x+1)+6(x+1)
= (x+6)(x+1)
K nhé
2) \(1-9x^2=\left(1-3x\right)\left(1+3x\right)\)
3) \(\frac{x^2}{9}-\frac{y^2}{16}=\left(\frac{x}{3}-\frac{y}{4}\right)\left(\frac{x}{3}+\frac{y}{4}\right)\)
4) \(a^4-b^4=\left(a^2-b^2\right)\left(a^2+b^2\right)=\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)\)
5) \(\left(a-b\right)^2-1=\left(a-b+1\right)\left(a-b-1\right)\)
6) \(4-\left(a-b\right)^2=\left(2-a+b\right)\left(2+a-b\right)\)
7) \(\left(x-y\right)^2-\left(m+n\right)^2=\left(x-y-m-n\right)\left(x-y+m+n\right)\)
8) \(\left(3x-2y\right)^2-\left(2x-3y\right)^2=\left(3x-2y-2x+3y\right)\left(3x-2y+2x-3y\right)\)
\(=\left[3\left(x+y\right)-2\left(x+y\right)\right]\left[3\left(x-y\right)+2\left(x-y\right)\right]=5\left(x+y\right)\left(x-y\right)\)
9) \(4x^2-12xy+9y^2=\left(2x-3y\right)^2\)
10) \(\left(x^4+2x^2+1\right)=\left(x^2+1\right)^2\)
11) \(\left(a^4+4-4x^2\right)=\left(a^2-2\right)^2\)
1 Điền vào chỗ chấm để có hằng đẳng thức thích hợp :
a) x2 - 4y2 = ........(x-2y)(x+2y)......................
b) 1/42 + 2xy + 4y2 =.............(1/4+2y)2..........................
c) 64x3 - 1 =..................(4x-1)(16x2+4x+1)................
d) 25 + 10y + y2 =........(5+y)2....................
a) \(\left(x+4\right)\left(x^2-4x+16\right)\)
\(x^3-4x^2+16x+4x^2-16x+64\)
\(=x^3+64\)
\(=x^3+4^3\)
\(=\left(x+4\right)\left(x^2-4x+16\right)\)
b) \(\left(\frac{1}{3}x+2y\right)\left(\frac{1}{9}x^2-\frac{2}{3}xy+4y^2\right)\)
\(=\frac{1}{27}x^3-\frac{2}{9}x^2y+\frac{4}{3}xy^2+\frac{2}{9}x^2y-\frac{4}{3}xy^2+8y^3\)
\(=\frac{1}{27}x^3+8y^3\)
\(=\left(\frac{1}{3}x\right)^3+\left(2y\right)^3\)
\(=\left(\frac{1}{3}x+2y\right)[\left(\frac{1}{3}x\right)^2-(\frac{1}{3}x.2y)+\left(2y\right)^2]\)
\(=\left(\frac{1}{3}x+2y\right)\left(\frac{1}{9}x^2-\frac{2}{3}xy+4y^2\right)\)
Câu c và d tương tự .
Sửa đề :Chứng minh hằng đẳng thức: ( x - y ) ( x4 + x3y + x2y2 + xy3 + y4 ) = x5 - y5
Ta có : ( x - y ) ( x4 + x3y + x2y2 + xy3 + y4 )
= x ( x4 + x3y + x2y2 + xy3 + y4 ) - y ( x4 + x3y + x2y2 + xy3 + y4 )
= x5 + x4y + x3y2 + x2y3 + xy4 - x4y - x3y2 - x2y3 - xy4 - y5
= x5 - y5
\(\implies\) ( x - y ) ( x4 + x3y + x2y2 + xy3 + y4 ) = x5 - y5 ( đpcm )
Mk chỉ lm 2 phần đầu thôi ,bn tham khảo nha!!!
\(a,\left(3x-1\right)^2-16=\left(3x-1+4\right)\left(3x-1-4\right)=\left(3x+3\right)\left(3x-5\right)=3\left(x+1\right)\left(3x-5\right)\)
\(b,\left(5x-4\right)^2-49x^2=\left(5x-4+7x\right)\left(5x-4-7x\right)\)
\(=\left(12x-4\right)\left(-2x-4\right)\)
\(=4\left(3x-1\right)\left(-2\right)\left(x+2\right)\)
\(=-8\left(3x-1\right)\left(x+2\right)\)
=.= hok tốt!!
\(\left(3x-1\right)^2-16\)
\(=\left(3x-1\right)^2-4^2\)
\(=\left(3x-1-4\right)\left(3x-1+4\right)\)
\(=\left(3x-5\right)\left(3x+3\right)\)
\(=3\left(x+1\right)\left(3x-5\right)\)
Trả lời:
Ta có: a + b + c = 0
<=> a + b = - c
=> ( a + b )3 = ( - c )3
<=> a3 + 3a2b + 3ab2 + b3 = - c3
<=> a3 + 3a2b + 3ab2 + b3 + c3 = 0
<=> a3 + 3ab ( a + b ) + b3 + c3 = 0
<=> a3 + 3ab ( - c ) + b3 + c3 = 0 (vì a + b = - c)
<=> a3 - 3abc + b3 + c3 = 0
<=> a3 + b3 + c3 = 3abc (đpcm)
Đặt B= 99 + 999+ .... + 999...9(100 số 9)= 10-1+102-1+103-1+...+10100-1
=(10+102+103+....+10100)+(-1-1-1-...-1) ( 100 số hạng -1)
=(10+102+103+....+10100)-100
Đặt A=10+102+103+....+10100
=>10A=102+103+....+10101
=>10A-A=10101-10
=>A=\(\frac{10^{101}-10}{9}\)
=>B=\(\frac{10^{101}-10}{9}-100=\frac{10^{101}-910}{9}\)(dpcm)
CM đẳng thức:
9 + 99 + 999+ .... + 999...9 = (10101 - 910) : 9
100 số 9