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a) \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}+\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)
\(=\dfrac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
b) \(\dfrac{\left(a^2-\left(b+c\right)^2\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a^2+c^2-2ac-b^2\right)}\)
\(=\dfrac{\left(a-b-c\right)\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(\left(a-c\right)^2-b^2\right)}\)
\(=\dfrac{\left(a-c-b\right)\left(a-c+b\right)}{\left(a-c-b\right)\left(a-c+b\right)}=1\)
c) \(\dfrac{x-1}{x^3}-\dfrac{x+1}{x^3-x^2}+\dfrac{3}{x^3-2x^2+x}\)
\(=\dfrac{x-1}{x^3}-\dfrac{x+1}{x^2\left(x-1\right)}+\dfrac{3}{x\left(x-1\right)^2}\)
\(=\dfrac{\left(x-1\right)^3-x\left(x+1\right)\left(x-1\right)+3x^2}{x^3\left(x-1\right)^2}\)
\(=\dfrac{x^3-3x^2+3x-1-x^3+x+3x^2}{x^3\left(x-1\right)^2}\)
\(=\dfrac{4x-1}{x^3\left(x-1\right)^2}\)
d) \(\left(\dfrac{x^2-y^2}{xy}-\dfrac{1}{x+y}\left(\dfrac{x^2}{y}-\dfrac{y^2}{x}\right)\right):\dfrac{x-y}{x}\)
\(=\left(\dfrac{\left(x-y\right)\left(x+y\right)}{xy}-\dfrac{1}{x+y}.\dfrac{x^3-y^3}{xy}\right):\dfrac{x-y}{x}\)
\(=\left(\dfrac{\left(x-y\right)\left(x+y\right)}{xy}-\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{xy\left(x+y\right)}\right):\dfrac{x-y}{x}\)
\(=\dfrac{\left(x-y\right)\left(x^2+2xy+y^2-x^2-xy-y^2\right)}{xy\left(x+y\right)}.\dfrac{x}{x-y}\)
\(=\dfrac{x}{x+y}\)
\(1.\)
\(a.\)
\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=x-1\)
\(b.\)
\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)
\(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2y}{\left(x-y\right)}\)
Tương tự các câu còn lại
a: \(=\dfrac{x+1}{x+2}\cdot\dfrac{x+3}{x+2}\cdot\dfrac{x+1}{x+3}=\dfrac{\left(x+1\right)^2}{\left(x+2\right)^2}\)
b: \(=\dfrac{x+1}{x+2}:\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x+3\right)^2}\)
\(=\dfrac{x+1}{x+2}\cdot\dfrac{\left(x+3\right)^2}{\left(x+1\right)\left(x+2\right)}=\dfrac{\left(x+3\right)^2}{\left(x+2\right)^2}\)
c: \(=\dfrac{\left(x+3\right)\left(x-1\right)-\left(2x-1\right)\left(x+1\right)-\left(x-3\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2+2x-3-2x^2-2x+x+1-x+3}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x^2+1}{\left(x-1\right)\left(x+1\right)}=-1\)
1 ) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=-15\)
\(\Leftrightarrow x^3-6x^2+12x-8-\left(x^3-27\right)+6\left(x^2+2x+1\right)=-15\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=-15\)
\(\Leftrightarrow\left(x^3-x^3\right)+\left(6x^2-6x^2\right)+\left(12x+12x\right)+\left(27+6-8\right)=-15\)
\(\Leftrightarrow24x+25=-15\)
\(\Leftrightarrow24x=-40\)
\(\Leftrightarrow x=-\dfrac{5}{3}\)
Vậy \(x=-\dfrac{5}{3}\)
a) PT \(\Leftrightarrow\dfrac{x^2-x+2}{\left(x-1\right)^3}=\dfrac{A+B\left(x-1\right)+C\left(x-1\right)^2}{\left(x-1\right)^3}\)
\(\Leftrightarrow x^2-x+2=A+Bx-B+Cx^2-2Cx+C\)
\(\Leftrightarrow x^2-x+2=Cx^2+x\left(B-2C\right)+\left(A+C-B\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}C=1\\B-2C=-1\\A+C-B=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}A=2\\B=1\\C=1\end{matrix}\right.\)
b: \(\Leftrightarrow\dfrac{x^2+2x-1}{\left(x-1\right)\left(x^2+1\right)}=\dfrac{A\cdot x^2+A+\left(Bx+C\right)\left(x-1\right)}{\left(x^2+1\right)\left(x-1\right)}\)
\(\Leftrightarrow x^2\cdot A+A+x^2\cdot B-x\cdot B+x\cdot C-C=x^2+2x-1\)
\(\Leftrightarrow x^2\left(A+B\right)+x\left(-B+C\right)+A-C=x^2+2x-1\)
=>A+B=1; -B+C=2; A-C=-1
=>A+C=3; A-C=-1; A+B=1
=>A=1; C=2; B=1-A=0
a, \(\dfrac{x^2-x}{x-2}+\dfrac{4-3x}{x-2}\)
\(=\dfrac{x^2-x+4-3x}{x-2}=\dfrac{x^2-4x+4}{x-2}\)
c) \(\dfrac{2}{x^2-9}+\dfrac{1}{x+3}\)
Ta có: \(\dfrac{1}{x+3}=\dfrac{1\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-3}{x^2-9}\)
\(\Rightarrow\dfrac{2}{x^2-9}+\dfrac{1}{x+3}=\dfrac{2}{x^2-9}+\dfrac{x-3}{x^2-9}=\dfrac{2+x-3}{x^2-9}=\dfrac{x-1}{x^2-9}\)
bai 1
a) \(\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|2,15\right|\)
\(\left|x+\dfrac{4}{15}\right|-3,75=-2,,15\)
\(\left|x+\dfrac{4}{15}\right|=-2,15+3,75=1,6\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{4}{15}=1,6\\x+\dfrac{4}{15}=-1,6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{28}{15}\end{matrix}\right.\)
Vậy ....
b) \(\left|\dfrac{5}{3}x\right|=\left|-\dfrac{1}{6}\right|\)
\(\left|\dfrac{5}{3}x\right|=\dfrac{1}{6}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{5}{3}x=-\dfrac{1}{6}\\\dfrac{5}{3}x=\dfrac{1}{6}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{10}\\x=\dfrac{1}{10}\end{matrix}\right.\)
c) \(\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|-\dfrac{3}{4}=\left|-\dfrac{3}{4}\right|\)
\(\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|-\dfrac{3}{4}=\dfrac{3}{4}\)
\(\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|=\dfrac{3}{2}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x-\dfrac{3}{4}=\dfrac{3}{2}\\\dfrac{3}{4}x-\dfrac{3}{4}=-\dfrac{3}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\-1\end{matrix}\right.\)
bai 2
a) \(\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|=\dfrac{1}{4}-\left|y\right|\)
\(\left|\dfrac{1}{6}+x\right|=\dfrac{1}{4}-\left|y\right|\) (*)
với mọi x ta luôn có \(\left|\dfrac{1}{6}+x\right|\ge0\)
\(\Rightarrow\dfrac{1}{4}-\left|y\right|\ge0\)
\(\Rightarrow\left|y\right|\le\dfrac{1}{4}\) \(\Rightarrow\dfrac{1}{4}-\left|y\right|=\left|\dfrac{1}{4}-y\right|\)
Nên từ * \(\Rightarrow\left|\dfrac{1}{6}+x\right|=\left|\dfrac{1}{4}-y\right|\)
\(\Rightarrow\left|\dfrac{1}{6}+x\right|-\left|\dfrac{1}{4}-y\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{6}+x=0\\\dfrac{1}{4}-y=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{6}\\y=\dfrac{1}{4}\end{matrix}\right.\)
b) \(\left|x-y\right|+\left|y+25\right|=0\)
với mọi x, y tao luôn có \(\left\{{}\begin{matrix}\left|x-y\right|\ge0\\\left|y+25\right|\ge0\end{matrix}\right.\)
mà \(\left|x-y\right|+\left|y+25\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x-y\right|=0\\\left|y+25\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=y\\y=-25\end{matrix}\right.\Rightarrow}\left\{{}\begin{matrix}x=-25\\y=-25\end{matrix}\right.\)
Bài 2:
a: \(A=\left|5x+1\right|-\dfrac{3}{8}>=-\dfrac{3}{8}\)
Dấu '=' xảy ra khi x=-1/5
b: \(B=\left|-\dfrac{1}{6}x+2\right|+0.25>=0.25\)
Dấu '=' xảy ra khi x=12
Bài 3:
a: \(A=2018-\left|x+2019\right|< =2018\)
Dấu '=' xảy ra khi x=-2019
b: \(=-10-\left|2x-\dfrac{1}{1009}\right|< =-10\)
Dấu '=' xảy ra khi x=1/2018
Từ x=\(\dfrac{1}{2}\)a+\(\dfrac{1}{2}\)b+\(\dfrac{1}{2}\)c=\(\dfrac{1}{2}\).(a+b+c)\(\Rightarrow\)2x=(a+b+c)
M=(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)+x\(^2\)
= x\(^2\)-xb-ax+ab+x\(^2\)-xc-bx+bc+x\(^2\)-ax-cx+ac+x\(^2\)
= 4x\(^2\)-2ac-2bx-2cx+ab+bc+ac
= 4x\(^2\)-2x(a+b+c)+ab+bc+ca
Thay 2x=a+b+c,ta được:
M= 4x\(^2\)-2x.2c+ab+bc+ca
M= 4x\(^2\)-4x\(^2\)+ab+bc+ca
M= ab+bc+ca