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1
a,\(\left(2x+1\right)\left(3x+1\right)-\left(6x-1\right)\left(x+1\right)\)
=\(6x^2+2x+3x+1-\left(6x^2+6x-x-1\right)\)
\(=6x^2+5x+1-6x^2-6x+x+1\)
\(=2\)
c,\(\left(a+1\right)\left(a^2-a+1\right)+\left(a+1\right)\left(a-1\right)\)
\(=\left(a^3+1\right)+\left(a^2-1\right)\)
\(=a^3+1+a^2-1\)
\(=a^3+a^2\)
2,
a,\(4ab+a^2-3a-12b\)
\(=\left(4ab-12b\right)+\left(a^2-3a\right)\)
\(=4b\left(a-3\right)+a\left(a-3\right)\)
\(=\left(4b+a\right)\left(a-3\right)\)
b,\(x^3+3x^2+3x+1-27y^3\)
\(=\left(x+1\right)^3-\left(3y\right)^3\)
\(=\left(x+1-3y\right)\left[\left(x+1\right)^2+\left(x+1\right).3y+\left(3y\right)^2\right]\)
\(=\left(x+1-3y\right)\left(x^2+2x+1+3xy+3y+9y^2\right)\)
4
a,\(2004^2-16\)
\(=2004^2-4^2\)
\(=\left(2004-4\right)\left(2004+4\right)\)
\(=2000.2008\)
\(=4016000\)
b,\(892^2+892.216+108^2\)
\(=\left(892+108\right)^2\)
\(=1000^2=1000000\)
c,\(10,2.9,8-9,8.0,2+10,2^2-10,2.0,2\)
\(=9,8\left(10,2-0,2\right)+10,2\left(10,2-0,2\right)\)
\(=9,8.10+10,2.10\)
\(=98+102\)
\(=200\)
d,\(36^2+26^2-52.36\)
=\(\left(36-26\right)^2\)
\(=10^2=100\)
3)\(A=-x^2+2x-3\)
\(\Leftrightarrow A=-x^2+2x-1-2\)
\(\Leftrightarrow A=-\left(x^2-2x+1\right)-2\)
\(\Leftrightarrow A=-\left(x-1\right)^2-2\)
Vậy GTLN của A=-2 khi x=1
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
Bài 1:
\(3a.\left(2a^2-ab\right)=6a^3-3a^2b\)
\(\left(4-7b^2\right).\left(2a+5b\right)=8a+20b-14ab^2-35b^3\)
Bài 2:
\(2x^2-6x+xy-3y=2x.\left(x-3\right)+y.\left(x-3\right)=\left(x-3\right).\left(2x+y\right)\)
Bài 3: Tại x = 3/2, y =1/3 thì Q = 67/9
Bài 4:
\(\left(\frac{1}{x+1}+\frac{2x}{1-x^2}\right).\left(\frac{1}{x-1}\right)\) \(\frac{1}{\left(x+1\right).\left(x-1\right)}+\frac{2x}{\left(1-x^2\right).\left(x-1\right)}=\frac{x-1}{\left(x+1\right).\left(x-1\right)^2}+\frac{-2x}{\left(x-1\right)^2.\left(x+1\right)}\)
= \(\frac{x-1-2x}{\left(x+1\right).\left(x-1\right)^2}=\frac{-\left(x+1\right)}{\left(x+1\right).\left(x-1\right)^2}=\frac{-1}{\left(x-1\right)^2}\)
mk chỉ phân tích thôi bạn tự chia nha!
a, \(16x^4-81=(4x^2)^2-9^2=(4x^2-9)(4x^2+9)\)
\(=[(2x)^2-3^2](4x^2+9)\)
\(=(2x+3)(2x-3)(4x^2+9)\)
b, \(x^3-3x^2+3x-1=(x-1)^3\)
\(x^2-2x+1=(x-1)^2\)
c, \(18x^5+9x^4+3x^3+6x^2+3x+1=(18x^5+9x^4+3x^3)+(6x^2+3x+1)\)
\(=(6x^2+3x+1)(3x^3+1)\)
câu c bạn đánh sai 1 dấu phép toán kìa!!!!
\(\left(x-3\right)\left(x-1\right)-3\left(x-3\right)\)
\(=\left(x-3\right)\left(x-1-3\right)\)
\(=\left(x-3\right)\left(x-4\right)\)
\(\left(x-1\right)\left(2x+1\right)+3\left(x-1\right)\left(x+2\right)\left(2x+1\right)\)
\(=\left(x-1\right)\left(2x+1\right)\left(1+3x+6\right)\)
\(=\left(x-1\right)\left(2x+1\right)\left(3x+7\right)\)
Câu 1:
a: \(=6x^2+5x+1-6x^2-6x+x+1=2\)
b: \(=a^3+1-a^2+1=a^3-a^2+2\)