Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{4}{x+2}+\frac{-3}{x-2}+\frac{12}{x^2-4}.\)
\(=\frac{4\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{3\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{12}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{4x-8-3x-6+12}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x-4}{x^2-4}\)
\(\frac{4}{x+2}+\frac{\left(-2\right)}{x-2}+\frac{12}{x^2-4}\)
\(=\frac{4\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{12}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{4\left(x-2\right)-3\left(x+2\right)+12}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x-2}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{1}{x+2}\)
Đặt \(g\left(x\right)=f\left(x\right)+h\left(x\right)\left(1\right)\)trong đó \(h\left(x\right)=ax^2+bx+c\left(2\right)\)
Tìm \(a,b,c\)sao cho \(g\left(1\right)=g\left(2\right)=g\left(3\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}g\left(1\right)=f\left(1\right)+h\left(1\right)=0\\g\left(2\right)=f\left(2\right)+h\left(2\right)=0\\g\left(3\right)=f\left(3\right)+h\left(3\right)=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}h\left(1\right)=-5\\h\left(2\right)=-11\\h\left(3\right)=-21\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a+b+c=-5\\4a+2b+c=-11\\9a+3b+c=-21\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a+b+c=-5\\3a+b=-6\\5a+b=-10\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a=-2\\b=0\\c=-3\end{cases}}\)Thay vào (2) ta được:
\(h\left(x\right)=4x-3\)
Vì \(g\left(1\right)=g\left(2\right)=g\left(3\right)=0\)mà g(x) bậc 4 có hệ số cao nhất là 1 nên ta có
\(g\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-x_0\right)\)
Từ \(\left(1\right)\Rightarrow f\left(x\right)=g\left(x\right)-h\left(x\right)\)
\(=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-x_0\right)+4x-3\)
\(f\left(-1\right)=\left(-1-1\right)\left(-1-2\right)\left(-1-3\right)\left(-1-x_0\right)+4.\left(-1\right)-3\)
\(=-24\left(-1-x_0\right)-7\)
\(f\left(5\right)=\left(5-1\right)\left(5-2\right)\left(5-3\right)\left(5-x_0\right)+4.5-3\)
\(=24\left(5-x_0\right)+17\)
\(\Rightarrow f\left(-1\right)+f\left(5\right)\)\(=-24\left(-1-x_0\right)-7+24\left(5-x_0\right)+17\)
\(=24+24x_0+120-24x_0+10\)
\(=154\)
\(x^5-x^4+3x^3+3x^2-x+1=0\)
\(\Leftrightarrow x^5+x^4-2x^4-2x^3+5x^3+5x^2-2x^2-2x+x+1=0\)
\(\Leftrightarrow x^4\left(x+1\right)-2x^3\left(x+1\right)+5x^2\left(x+1\right)-2x\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^4-2x^3+5x^2-2x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x^4-2x^3+5x^2-2x+1=0\left(#\right)\end{cases}}\)
\(\Leftrightarrow x=-1\)(vì biểu thức # vô nghiệm) (cái này bạn tự cm)
vậy....
4(x-1)2=x2
<=>2(x-1)=x hoặc 2(x-1)=-x
<=>2x-2=x hoặc 2x-2=-x
<=>x=2 hoặc x=2/3
Vậy trung bình cộng các giá trị x là: (2+2/3):2=4/3
\(5-\left(6-x\right)=4\left(3-2x\right)\)
\(5-6+x=12-8x\)
\(-1+x=12-8x\)
\(x-1=12-8x\)
\(12+1=8x+1\)
\(8x=13-1\)
\(x=12:8\)
\(x=\dfrac{12}{8}=\dfrac{3}{2}\)
\(PT\Leftrightarrow5-6+x=12-8x\)
\(\Leftrightarrow9x=13\)
\(\Leftrightarrow x=\dfrac{13}{9}\)
Vậy: \(S=\left\{\dfrac{13}{9}\right\}\)
a) Ta có: \(6x^4-9x^3\)
\(=3x^3\cdot2x-3x^3\cdot3\)
\(=3x^3\left(2x-3\right)\)
b) Ta có: \(x^2y^2z+xy^2z^2+x^2yz^2\)
\(=xyz\cdot\left(xy+yz+xz\right)\)
c) Ta có: \(2x\left(x+3\right)+2\left(x+3\right)\)
\(=2\cdot\left(x+3\right)\cdot x+2\cdot\left(x+3\right)\cdot1\)
\(=2\left(x+3\right)\left(x+1\right)\)
d) Ta có: \(\left(x+5\right)^2-3\left(x+5\right)\)
\(=\left(x+5\right)\left(x+5-3\right)\)
\(=\left(x+5\right)\left(x+2\right)\)
e) Ta có: \(2x\left(x-3\right)-\left(x-3\right)^2\)
\(=\left(x-3\right)\left(2x-x+3\right)\)
\(=\left(x-3\right)\left(x+3\right)\)
a, 6x4 - 9x3 = 3x3 (2x-3x) = 3x3 (-x) = -3x4
b, x2y2z + xy2z2 + x2yz2 = xyz (xy+yz+xz)
c, 2x (x+3) + 2 (x+3) = (x+3) (2x+2) = (x+3) 2 (x+1)
d, (x+5)2 - 3 (x+5) = (x+5) (x+5-3) = (x+5) (x+2)
e, 2x (x-3) - (x-3)2 = (x-3) [2x-(x-3)] = (x-3) (2x-x+3) = (x-3) (x+3) = x2 - 9
Tự làm á! Đúng sai thì chịu
Đề là phân tích thành nhân tử hay rút gọn?
phân tích thành nhân tử