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Đề bài: Giải bất phương trình :
a) x2 + 5x + 6 ≥ 0
⇔x2+5x ≥ -6
⇔x(x+5) ≥ -6
⇔ x ≥ -6 hoặc x+5 ≥ -6
⇔x ≥ -6 hoặc x ≥ -11
b) x2 - 9x + 20 ≤ 0
⇔x(x-9) ≤ -20
⇔ x ≤ 20 hoặc x-9 ≤ - 20
⇔ x ≤ 20 hoặc x ≤ -11
Thấy đúng thì tick nha
a) \(x^2+5x+6\ge0\)
\(\Leftrightarrow x\left(x+5\right)\ge-6\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge-6\\x+5\ge-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge-6\\x\ge-11\end{matrix}\right.\)
\(\Leftrightarrow x\ge-6\)
b) \(x^2-9x+20\le0\)
\(\Leftrightarrow x\left(x-9\right)\le-20\)
\(\Leftrightarrow\left[{}\begin{matrix}x\le-20\\x-9\le-20\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\le-20\\x\le-11\end{matrix}\right.\)
\(\Leftrightarrow x\le-20\)
a) \(x^2+2xy+x+2y\)
\(=x\left(x+2y\right)+\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x+1\right)\)
b) \(7x^2-7xy-5x+5y\)
\(=7x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(7x-5\right)\)
c) \(x^2-6x+9-9y^2\)
\(=\left(x^2-6x+9\right)-9y^2\)
\(=\left(x-3\right)^2-\left(3y\right)^2\)
\(=\left(x-3-3y\right)\left(x-3+3y\right)\)
d) \(x^3-3x^2+3x-1+2\left(x^2-x\right)\)
\(=\left(x^3-1\right)-\left(3x^2-3x\right)+2\left(x^2-x\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)-3x\left(x-1\right)+2x\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1-3x+2x\right)\)
\(=\left(x-1\right)\left(x^2+1\right)\)
e) \(15\left(x-y\right)-25x+25y\)
\(=15\left(x-y\right)-25\left(x-y\right)\)
\(=\left(15-25\right)\left(x-y\right)\)
\(=-10\left(x-y\right)\)
f) \(12x^2-3xy+8xz-2yz\)
\(=3x\left(4x-y\right)+2z\left(4x-y\right)\)
\(=\left(4x-y\right)\left(3x+2z\right)\)
y) \(x^3+x^2y-x^2z-xyz\)
\(=x^2\left(x+y\right)-xz\left(x+y\right)\)
\(=x\left(x+y\right)\left(x-z\right)\)
lê thị hương giang e ko nghĩ câu F đề sai đâu ạ! Chị check giúp xem em có tính sai hay ko nha!
2/ Ta có:
\(F=\left(x^2-2xy+y^2\right)+4\left(x-y\right)+4+x^2+8x+16-20\)
\(=\left(x-y+2\right)^2+\left(x+4\right)^2-20\ge-20\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x=-4\\y=-2\end{matrix}\right.\)
Đề bài thế này chỉ có nước khai triển thôi ạ!
a) Khai triển ra, pt \(\Leftrightarrow-\left(5x^2-2x+1\right)=-\left(5x^2+x+22\right)\)
\(\Leftrightarrow-2x+1=x+22\Leftrightarrow3x=-21\Leftrightarrow x=-7\)
b) Khai triển ra, pt \(\Leftrightarrow x^3+3x^2+12x-9=x^3+3x^2+3x+1\)
\(\Leftrightarrow12x-9=3x+1\Leftrightarrow9x=10\Leftrightarrow x=\frac{10}{9}\) c) Cái này thì không cần khai triển đâu:v \(PT\Leftrightarrow\left(2x-1\right)^2-\left(2x+6\right)^2=5x-7\) Áp dụng hằng đẳng thức số 3 (nếu em nhớ không lầm) vào vế trái \(\Leftrightarrow-7\left(4x+5\right)=5x-7\Leftrightarrow x=-\frac{28}{33}\)
\(x^3-2x+y^3-2y=\left(x+y\right)\left(x^2-xy+y^2\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2-2\right)\)
\(x^2-2xy+y^2-16=\left(x-y\right)^2-16=\left(x-y-4\right)\left(x-y+4\right)\)
theo mình đề câu c là 6x2
\(x^3+6x^2+9x-xz^2=x\left(x^2-6x+9-z^2\right)\)
\(=\left(x-3-z\right)\left(x-3+z\right)\)
\(x^2-11x+30=x^2-5x-6x+30\)
\(=x\left(x-5\right)-6\left(x-5\right)=\left(x-5\right)\left(x-6\right)\)
\(4x^2-3x-1=4x^2-4x+x-1\)
\(=4x\left(x-1\right)+x-1=\left(4x+1\right)\left(x-1\right)\)
\(9x^2-7x-2=9x^2-9x+2x-2\)
\(=9x\left(x-1\right)+2\left(x-1\right)=\left(9x+2\right)\left(x-1\right)\)
\(\left(x^2+x\right)^2-2\left(x^2+x\right)-5=\left(x^2+x-1\right)^2-4\)
\(=\left(x^2+x-3\right)\left(x^2+x+1\right)\)
còn lại lát mình làm tiếp
Bài 1:
a, \(x^3-2x-y^3-2y=\left(x^3+y^3\right)-\left(2x+2y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)-2\left(x+y\right)=\left(x+y\right)\left(x^2-xy+y^2-2\right)\)
b, \(x^2-2xy+y^2-16=\left(x-y\right)^2-4^2=\left(x-y+4\right)\left(x-y-4\right)\)
c, \(x^3+6x^2+9x-xz^2=x\left(x^2+6x+9-z^2\right)\)
\(=x\left[\left(x+3\right)^2-z^2\right]=x\left(x+3+z\right)\left(x+3-z\right)\)
Mỗi bài mình sẽ làm một câu mẫu ạ
Bài 1:
a) \(x^3-2x+y^3-2y\)
\(=\left(x^3+y^3\right)-\left(2x+2y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2-2\right)\)
Bài 2:
a) \(x^2-11x+30\)
\(=x^2-5x-6x+30\)
\(=x\left(x-5\right)-6\left(x-5\right)\)
\(=\left(x-6\right)\left(x-5\right)\)
Bài 3:
a) \(x^2-5x+4=0\)
\(\Leftrightarrow x^2-4x-x+4=0\)
\(\Leftrightarrow x\left(x-4\right)-\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\)
Bài 2:
b: \(=4x^2-4x+x-1=\left(x-1\right)\left(4x+1\right)\)
c: \(=9x^2-9x+2x-2=\left(x-1\right)\left(9x+2\right)\)
e: Sửa đề: \(\left(x^2+3x+1\right)\left(x^2+3x+2\right)-2\)
\(=\left(x^2+3x\right)^2+3\left(x^2+3x\right)+2-2\)
\(=\left(x^2+3x\right)\left(x^2+3x+3\right)\)
\(=x\left(x+3\right)\left(x^2+3x+3\right)\)
ừ thì mình sẽ giúp bạn mà câu a bạn viết sai đề nha
1/a)\(2x^2+3x-5=2x^2-2x+5x-5=2x\left(x-1\right)+5\left(x-1\right)=\left(2x+5\right)\left(x-1\right)\)
b)\(4x^2-3x-1=4x^2-4x+x-1=4x\left(x-1\right)+\left(x-1\right)=\left(4x+1\right)\left(x-1\right)\)
c)Sai đề: \(3x^2+6xy+3y^2-3z^2\)
\(=3\left(x^2+2xy+y^2-z^2\right)\)
\(=3\left[\left(x+y\right)^2-z^2\right]\)
\(=3\left(x+y+z\right)\left(x+y-z\right)\)
d)Sai đề:\(x^3-2x^2y+xy^2-9x=x\left(x-2xy+y^2-9\right)=x\left[\left(x-y\right)^2-9\right]=x\left(x-y+3\right)\left(x-y-3\right)\)
e)\(2x-2y-x^2+2xy-y^2=2\left(x-y\right)-\left(x-y\right)^2=\left(x-y\right)\left(2-x+y\right)\)
f)Hình như sai đề đúng không?
\(x^3-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x+y+1\right)\left(x+y-1\right)\)
2/a.\(7x-6x^2-2=0\)
\(\Leftrightarrow-\left(6x^2-3x-4x+2\right)=0\)
\(\Leftrightarrow3x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(x-1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}3x-2=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\x=1\end{matrix}\right.\)
b.\(16x-5x^2-3=0\)
\(\Leftrightarrow-\left(5x^2-15x-x+3\right)=0\)
\(\Leftrightarrow5x\left(x-3\right)-\left(x-3\right)=0\)
\(\Leftrightarrow\left(5x-1\right)\left(x-3\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}5x-1=0\\x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\x=3\end{matrix}\right.\)
c.\(2x^2+3x-5=0\)
\(\Leftrightarrow2x^2-2x+5x-5=0\)
\(\Leftrightarrow2x\left(x-1\right)+5\left(x-1\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}2x+5=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}=-2,5\\x=1\end{matrix}\right.\)
a) x2 + 9x + 20
= x2 + 4x + 5x + 20
= x(x + 4) + 5(x + 4)
= (x + 5)(x + 4)
b) x2 + x - 12
= x2 + 4x - 3x - 12
= x(x + 4) - 3(x + 4)
= (x - 3)(x + 4)
c) 6x2 - 11x - 10
= 6x2 - 15x + 4x - 10
= 3x(2x - 5) + 2(2x - 5)
= (3x + 2)(2x - 5)