Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(1.\)
\(a.\)
\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=x-1\)
\(b.\)
\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)
\(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2y}{\left(x-y\right)}\)
Tương tự các câu còn lại
1)
a) \(\dfrac{5x}{10}=\dfrac{x}{2}\)
b) \(\dfrac{4xy}{2y}=2x\left(y\ne0\right)\)
c) \(\dfrac{21x^2y^3}{6xy}=\dfrac{7xy^2}{2}\left(xy\ne0\right)\)
d) \(\dfrac{2x+2y}{4}=\dfrac{2\left(x+y\right)}{4}=\dfrac{x+y}{2}\)
e) \(\dfrac{5x-5y}{3x-3y}=\dfrac{5\left(x-y\right)}{3\left(x-y\right)}=\dfrac{5}{3}\left(x\ne y\right)\)
f) \(\dfrac{-15x\left(x-y\right)}{3\left(y-x\right)}=-5x\dfrac{x-y}{y-x}=-5x\dfrac{x-y}{-\left(x-y\right)}\)
\(=-5x.\left(-1\right)=5x\left(x\ne y\right)\)
2)
a) Nhớ ghi ĐK vào nhá, lười quá :V\(\dfrac{x^2-16}{4x-x^2}=-\dfrac{\left(x-4\right)\left(x+4\right)}{x^2-4x}=\dfrac{\left(x-4\right)\left(x+4\right)}{x\left(x-4\right)}=\dfrac{x+4}{x}\)
b) \(\dfrac{x^2+4x+3}{2x+6}=\dfrac{x^2+3x+x+3}{2\left(x+3\right)}=\dfrac{x\left(x+3\right)+\left(x+3\right)}{2\left(x+3\right)}\)
\(=\dfrac{\left(x+3\right)\left(x+1\right)}{2\left(x+3\right)}=\dfrac{x+1}{2}\)
c) \(\dfrac{15x\left(x+3\right)^3}{5y\left(x+y\right)^2}=\dfrac{3x\left(x+3\right)^3}{y\left(x+y\right)^2}\) ( câu này có gì đó sai sai )
d) \(\dfrac{5\left(x-y\right)-3\left(y-x\right)}{10\left(x-y\right)}=\dfrac{5\left(x-y\right)+3\left(x-y\right)}{10\left(x-y\right)}\)
\(=\dfrac{8\left(x-y\right)}{10\left(x-y\right)}=\dfrac{8}{10}=\dfrac{4}{5}\)
e) \(\dfrac{2x+2y+5x+5y}{2x+2y-5x-5y}=\dfrac{2\left(x+y\right)+5\left(x+y\right)}{2\left(x+y\right)-5\left(x+y\right)}\)
\(=\dfrac{7\left(x+y\right)}{-3\left(x+y\right)}=-\dfrac{7}{3}\)
a: \(=\dfrac{1}{x-y}-\dfrac{3xy}{\left(x-y\right)\left(x^2+xy+y^2\right)}+\dfrac{x-y}{x^2+xy+y^2}\)
\(=\dfrac{x^2+xy+y^2-3xy+x^2-2xy+y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2x^2-4xy+2y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{2\left(x-y\right)}{x^2+xy+y^2}\)
d: \(=\dfrac{x^3-1}{x-1}-\dfrac{x^2-1}{x+1}\)
\(=x^2+x+1-x+1=x^2+2\)
b) \(x^2+2\sqrt{3}x-6=0\)
\(\Leftrightarrow\) \(x^2+2\sqrt{3}x+3-9=0\)
\(\Leftrightarrow\) \(\left(x+\sqrt{3}\right)^2-9=0\)
\(\Leftrightarrow\) \(\left(x+\sqrt{3}-3\right).\left(x+\sqrt{3}+3\right)=0\)
\(\Leftrightarrow\) \(\left[\begin{array}{} x+\sqrt{3}-3=0 \\ x+\sqrt{3}+3=0 \end{array} \right.\)\(\Leftrightarrow\) \(\left[\begin{array}{} x= 3-\sqrt{3} \\ x= -3-\sqrt{3} \end{array} \right.\)
Vậy phương trình có tập nghiệm là S={\(3-\sqrt{3};-3-\sqrt{3}\)}
Câu 1:
\(\text{a) }\dfrac{x^2-xy}{3xy-3y^2}=\dfrac{x\left(x-y\right)}{3y\left(x-y\right)}=\dfrac{x}{3y}\)
\(\text{b) }\dfrac{2ax^2-4ax+2a}{5b-5bx^2}\\ =\dfrac{2a\left(x^2-2x+1\right)}{5b\left(1-x^2\right)}\\ =\dfrac{2a\left(x-1\right)^2}{5b\left(1-x\right)\left(1+x\right)}\\ =-\dfrac{2a\left(x-1\right)^2}{5b\left(x-1\right)\left(1+x\right)}\\ =-\dfrac{2a\left(x-1\right)}{5b\left(x+1\right)}\\ =-\dfrac{2ax-2a}{5bx+5b}\)
\(\text{c) }\dfrac{4x^2-4xy}{5x^3-5x^2y}=\dfrac{4x\left(x-y\right)}{5x^2\left(x-y\right)}=\dfrac{4}{5x}\)
\(\text{d) }\dfrac{\left(x+y\right)^2-z^2}{x+y+z}=\dfrac{\left(x+y+z\right)\left(x+y-z\right)}{x+y+z}=x+y-z\)
\(\text{e) }\dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6}\\ =\dfrac{\left(x^3+y^3\right)^2}{x\left(x^6-y^6\right)}\\ =\dfrac{\left(x^3+y^3\right)^2}{x\left(x^3-y^3\right)\left(x+y\right)^3}\\ =\dfrac{x^3+y^3}{x\left(x^3-y^3\right)}\\ =\dfrac{x^3+y^3}{x^4-xy^3}\)
Câu 3:
\(\text{ a) }\dfrac{\left(a+b\right)^2-c^2}{a+b+c}=\dfrac{\left(a+b+c\right)\left(a+b-c\right)}{a+b+c}=a+b-c\)
\(\text{b) }\dfrac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}\\ =\dfrac{\left(a^2+2ab+b^2\right)-c^2}{\left(a^2+2ac+c^2\right)-b^2}\\ =\dfrac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}\\ =\dfrac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+c+b\right)\left(a+c-b\right)}\\ =\dfrac{a+b-c}{a-b+c}\)
\(\text{c) }\dfrac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}\\ =\dfrac{2x^3-x^2-6x^2+3x-15x+45}{3x^3-10x^2-9x^2+3x+30x-9}\\ =\dfrac{\left(2x^3-x^2-15x\right)-\left(6x^2-3x-45\right)}{\left(3x^3-10x^2+3x\right)-\left(9x^2-30x+9\right)}\\ =\dfrac{x\left(2x^2-x-15\right)-3\left(2x^2-x-15\right)}{x\left(3x^2-10x+3\right)-3\left(3x^2-10x+3\right)}\\ =\dfrac{\left(x-3\right)\left(2x^2-x-15\right)}{\left(x-3\right)\left(3x^2-10x+3\right)}\\ =\dfrac{\left(x-3\right)\left(2x^2-6x+5x-15\right)}{\left(x-3\right)\left(3x^2-9x-x+3\right)}\\ =\dfrac{\left(x-3\right)\left[\left(2x^2-6x\right)+\left(5x-15\right)\right]}{\left(x-3\right)\left[\left(3x^2-9x\right)-\left(x-3\right)\right]}\\ =\dfrac{\left(x-3\right)\left[x\left(x-3\right)+5\left(x-3\right)\right]}{\left(x-3\right)\left[3x\left(x-3\right)-\left(x-3\right)\right]}\\ =\dfrac{\left(x-3\right)\left(x-3\right)\left(x+5\right)}{\left(x-3\right)\left(x-3\right)\left(3x-1\right)}\\ =\dfrac{x+5}{3x-1}\)
a: \(\left[\dfrac{1}{2}x^2\left(2x-1\right)^m-\dfrac{1}{2}x^{m+2}\right]:\dfrac{1}{2}x^2=0\)
\(\Leftrightarrow\left(2x-1\right)^m-x^m=0\)
\(\Leftrightarrow\left(2x-1\right)^m=x^m\)
=>2x-1=x
=>x=1
b: \(\left(2x-3\right)^8=\left(2x-3\right)^6\)
\(\Leftrightarrow\left(2x-3\right)^6\cdot\left(2x-4\right)\left(2x-2\right)=0\)
hay \(x\in\left\{\dfrac{3}{2};2;1\right\}\)
c: \(\Leftrightarrow4x^2-4x+1+y^2-\dfrac{2}{3}y+\dfrac{1}{9}+\dfrac{6}{9}=0\)
\(\Leftrightarrow\left(2x-1\right)^2+\left(y-\dfrac{1}{3}\right)^2+\dfrac{6}{9}=0\)(vô lý)
a: \(=\dfrac{4x^2+4x+1-\left(4x^2-4x+1\right)}{\left(2x-1\right)\left(2x+1\right)}\cdot\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{8x}{2x+1}\cdot\dfrac{5}{4x}=\dfrac{10}{2x+1}\)
c: \(=\dfrac{1}{x-1}-\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\left(\dfrac{x+1-x+1}{\left(x-1\right)^2\cdot\left(x+1\right)}\right)\)
\(=\dfrac{1}{x-1}-\dfrac{x}{x^2+1}\cdot\dfrac{2}{\left(x-1\right)}=\dfrac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}=\dfrac{x-1}{x^2+1}\)