Phân thức đại số - Hỏi đáp

\(B=\frac{x^2+10x+20}{x^2+6x+9}=\frac{(x^2+6x+9)+4(x+3)-1}{x^2+6x+9}\)

\(=1+\frac{4(x+3)}{x^2+6x+9}-\frac{1}{x^2+6x+9}=1+\frac{4(x+3)}{(x+3)^2}-\frac{1}{(x+3)^2}\)

\(=1+\frac{4}{(x+3)}-\frac{1}{(x+3)^2}\)

Đặt \(\frac{1}{x+3}=a\Rightarrow B=1+4a-a^2=5-(a^2-4a+4)\)

\(=5-(a-2)^2\leq 5\)

Vậy \(B_{\max}=5\Leftrightarrow a=2\Leftrightarrow x=-\frac{5}{2}\)

\(\dfrac{x^2+y^2}{xy}=\dfrac{5}{2}\Leftrightarrow2x^2+2y^2-5xy=0\)

\(\Leftrightarrow2x^2+2y^2-4xy-xy=0\)

\(\Leftrightarrow\left(2x^2-xy\right)-\left(4xy-2y^2\right)=0\)

\(\Leftrightarrow x\left(2x-y\right)-2y\left(2x-y\right)=0\)

\(\Leftrightarrow\left(x-2y\right)\left(2x-y\right)=0\)

Ta có: \(x>y>0\Leftrightarrow x+x>y+0\Leftrightarrow2x>y\Leftrightarrow2x-y>0\)

Vậy \(x-2y=0\Leftrightarrow x=2y\)

\(E=\dfrac{3x+2y}{2x-3y}=\dfrac{6y+2y}{4y-3y}=\dfrac{8y}{y}=8\)

a) \(B=\dfrac{x^2+2x}{2x+10}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)

\(B=\dfrac{x^2+2x}{2\left(x+5\right)}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)

( ĐKXĐ : \(x\ne0,x\ne-5\) )

\(B=\dfrac{\left(x^2+2x\right).x}{2x\left(x+5\right)}+\dfrac{\left(x-5\right).2\left(x+5\right)}{2x\left(x+5\right)}+\dfrac{50-5x}{2x\left(x+5\right)}\)

\(B=\dfrac{x^3+2x^2+2x^2+10x-10x-50+50-5x}{2x\left(x+5\right)}\)

\(B=\dfrac{x^3+4x^2-5x}{2x\left(x+5\right)}\)

\(B=\dfrac{x^3-x^2+5x^2-5x}{2x\left(x+5\right)}\)

\(B=\dfrac{x^2\left(x-1\right)+5x\left(x-1\right)}{2x\left(x+5\right)}=\dfrac{\left(x-1\right)\left(x+5\right)x}{2x\left(x+5\right)}\)

\(B=\dfrac{x-1}{2}\)

Câu b và c dễ vì đã có kết quả rút gọn rồi :)

câu 1:

thực hiện phép chia được thương là 2x²+2x+1

ta có: 2x²+2x+1=2x²+2x+\(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)

=2(x²+x+\(\dfrac{1}{4}\))+\(\dfrac{1}{2}\)=2(x+\(\dfrac{1}{2}\))²+\(\dfrac{1}{2}\)

vì 2(x+\(\dfrac{1}{2}\))²\(\ge\)0 nên 2(x+\(\dfrac{1}{2}\))²+\(\dfrac{1}{2}\)\(\ge\)\(\dfrac{1}{2}\)

dấu = xảy ra khi x=\(\dfrac{-1}{2}\)

câu 2:

\(\dfrac{\left(x+2\right)P}{x-2}=\dfrac{\left(x-1\right)Q}{x^2-4}\Leftrightarrow\left(x^2-4\right)\left(x+2\right)P=\left(x-2\right)\left(x-1\right)Q\Leftrightarrow\left(x+2\right)^2P=\left(x-1\right)Q\)

\(\Leftrightarrow\)P=x-1

Q=(x+2)²

\(A=\dfrac{4}{4x^2-4x+7}\\ =\dfrac{4}{\left(2x\right)^2-2.\left(2x\right).1+1+6}=\dfrac{4}{\left(2x-1\right)^2+6}\\ \le\dfrac{4}{6}=\dfrac{2}{3}\)

Max A = 2/3 khi x=1/2

Ủa tao nhớ thầy cho là giá trị lớn nhất mà phương :<. Đổi đề thành lớn nhất rồi tao làm cho, ahihi:))

\(\left(x-4\right)^2-2\left(x-4\right)\left(x+5\right)+\left(x+5\right)^2\)

\(\left[\left(x-4\right)-\left(x+5\right)\right]^2=\left(-9\right)^2=81\)

\(\left(x-4\right)^2-2\left(x-4\right)\left(x+5\right)+\left(x+5\right)^2\)

\(=\left[x-4-\left(x+5\right)\right]^2\)

\(=\left[x-4-x-5\right]^2\)

\(=\left(-9\right)^2=9^2=81\)

1/

\(\dfrac{\left(x-y\right)^3-3xy\left(x+y\right)+y^3}{x-6y}\)

\(=\dfrac{x^3-3x^2y+3xy^2-y^3-3x^2y-3xy^2+y^3}{x-6y}\)

\(=\dfrac{x^3-6x^2y}{x-6y}\)

\(=\dfrac{x^2\left(x-6y\right)}{x-6y}\)

\(=x^2\)

\(2\)/

\(\dfrac{x^2+y^2+z^2-2xy+2xz-2yz}{x^2-2xy+y^2-z^2}\)

\(=\dfrac{\left(x-y+z^{ }\right)^2}{\left(x-y\right)^2-z^2}\)

\(=\dfrac{\left(x-y+z\right)^2}{\left(x-y-z\right)\left(x-y+z\right)}\)

\(=\dfrac{x-y+z}{x-y-z}\)

3/

\(\dfrac{\left(n+1\right)!}{n!\left(n+2\right)}\)

\(=\dfrac{n!\left(n+1\right)}{n!\left(n+2\right)}\)

\(=\dfrac{n+1}{n+2}\)

4/

\(\dfrac{n!}{\left(n+1\right)!-n!}\)

\(=\dfrac{n!}{n!\left(n+1\right)-n!}\)

\(=\dfrac{n!}{n!\left[\left(n+1\right)-1\right]}\)

\(=\dfrac{n!}{n!.n}\)

\(=\dfrac{1}{n}\)

5/

\(\dfrac{\left(n+1\right)!-\left(n+2\right)!}{\left(n+1\right)!+\left(n+2\right)!}\)

\(=\dfrac{\left(n+1\right)!-\left(n+1\right)!\left(n+2\right)}{\left(n+1\right)!+\left(n+1\right)!\left(n+2\right)}\)

\(=\dfrac{\left(n+1\right)!\left(-n-1\right)}{\left(n+1\right)!\left(n+3\right)}\)

\(=\dfrac{-n-1}{n+3}\)

a, (x³-11x+5-3x²) : (x-5)= (x³-3x²-11x+5) : (x-5)

Tính :

b, (4x⁴-5x²-3-3x³+9x) : (x²-3)= (4x⁴-3x³-5x²+9x-3) : (x²-3)

Tính:

Thay x=-y-z

suy ra x^2-y^2-z^2=(-y-z)^2-y^2-z^2=2yz

Vậy M=x^+y^+z^3/2xyz(quy đồng)

Tiếp tuc thay: x=-y-z

M=(-y-z)^3+y^3+z^3/2(-y-z)yz

M=-3x^2y-3xy^2/-2x^y-2xy^2=3/2

Lời giải:

Đặt \(f(x)=x^{2016}+x^{2015}+x^{200}+x^2=(x^2-1)Q(x)+ax+b\) trong đó, $Q(x)$ là đa thức thương, $ax+b$ là đa thức dư

Ta có:

\(f(1)=1+1+1+1=(1^2-1)Q(1)+a+b\)

\(\Leftrightarrow 4=a+b(1)\)

\(f(-1)=1+(-1)+1+1=[(-1)^2-1]Q(-1)-a+b\)

\(\Leftrightarrow 2=-a+b(2)\)

Từ \((1);(2)\Rightarrow a=1; b=3\)

Vậy đa thức dư là $x+3$