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Câu 1:
a: \(=6x^2+5x+1-6x^2-6x+x+1=2\)
b: \(=a^3+1-a^2+1=a^3-a^2+2\)
6.
\(A=x^2-2x+2\\ =x^2-2x+1+1\\ =\left(x-1\right)^2+1\\ \left(x-1\right)^2\ge0\forall x\\ \left(x-1\right)^2+1\ge1\forall x\)
Dấu "=" xảy ra khi \(\left(x-1\right)^2=0\Rightarrow x-1=0\Rightarrow x=1\)
Vậy \(Min_A=1\) khi \(x=1\)
5,
a,
\(x\left(x-2\right)+x-2=0\\ \left(x+1\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
b,
\(5x\left(x-3\right)-x+3=0\\ 5x\left(x-3\right)-\left(x-3\right)=0\\ \left(5x+1\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}5x+1=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{5}\\x=3\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=\dfrac{-1}{5}\\x=3\end{matrix}\right.\)
c,
\(3x\left(x-5\right)-\left(x-1\right)\left(2+3x\right)=30\\ 3x^2-15x-\left(2x-2+3x^2-3x\right)\\ 3x^2-15x-\left(3x^2-x-2\right)=30\\ 3x^2-15x-3x^2+x+2=30\\ -14x+2=30\\ -14x=28\\ x=-2\)
Vậy \(x=-2\)
d,
\(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\\ x^2+5x+6-\left(x^2+3x-10\right)=0\\ x^2+5x+6-x^2-3x+10\\ 8x+16=0\\ 8x=-16\\ x=-2\)
Vậy \(x=-2\)
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
^^
Phân tích đa thức thành nhân tử:
\(6xy+5x-5y-3x^2-3y^2\)
\(=-3x^2+6xy-3y^2+5x-5y\)
\(=-3\left(x^2-2xy+y^2\right)+5\left(x-y\right)\)
\(=-3\left(x-y\right)^2+5\left(x-y\right)\)
\(=\left(x-y\right)\left[-3\left(x-y\right)+5\right]\)
\(=\left(x-y\right)\left(-3x+3y+5\right)\)
Thực hiện phép tính:
a)\(\left(x^2+x-3\right)\left(x^2-x+3\right)\)
\(=\left[x^2+\left(x-3\right)\right]\left[x^2-\left(x-3\right)\right]\)
\(=\left(x^2\right)^2-\left(x-3\right)^2\)
\(=x^4-\left(x^2-6x+9\right)\)
\(=x^4-x^2+6x-9\)
b)\(\left(5x-1\right)\left(x+3\right)-\left(x-2\right)\left(5x-4\right)\)
\(=\left(5x^2+15x-x-3\right)-\left(5x^2-4x-10x+8\right)\)
\(=5x^2+15x-x-3-5x^2+4x+10x-8\)
\(=28x-11\)
Bài 2:
a: Ta có: \(x\left(2x-1\right)-2x+1=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)
Bài 2:
c: \(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
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