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a) \(\frac{3x+5}{2\left(x-1\right)}+\frac{4}{x-2}=\frac{\left(3x+5\right)\left(x-2\right)+4\cdot2\left(x-1\right)}{2\left(x-1\right)\left(x-2\right)}=\frac{3x^2-6x+5x-10+8x-8}{2\left(x-1\right)\left(x-2\right)}\)
\(=\frac{3x^2+7x-18}{2\left(x-1\right)\left(x-2\right)}\)
b) \(\frac{2x^2+1}{4x^2-2x}+\frac{3-3x}{1-2x}+\frac{3}{2x}=\frac{2x^2+1+4x\left(3-3x\right)+2\cdot3\left(1-2x\right)}{4x\left(1-2x\right)}=\frac{2x^2+1+12-12x+6-12x}{4x\left(1-2x\right)}\)\(=\frac{2x^2-24x+19}{4x\left(1-2x\right)}\)
Đề này... bạn xem lại đi. Chứ thế này thì dùng máy tính cũng không làm nổi T-T
bài 1.
a.\(\left(x+4\right)\left(x^2-4x+16\right)=x^3-4^3=x^3-64\)
b.\(\left(x^2-\frac{1}{3}\right)\left(x^4+\frac{1}{3}x^2+\frac{1}{9}\right)=\left(x^2\right)^3-\left(\frac{1}{3}\right)^3=x^6-\frac{1}{27}\)
bài 2.
a.\(892^2+892.216+108^2=892^2+2.892.108+108^2\)
\(=\left(892+108\right)^2=1000^2=1_{ }000_{ }000\)
b.\(36^2+26^2-52.36=36^2+26^2-2.26.36=\left(36-26\right)^2=10^2=100\)
a) \(\frac{x+3}{x}-\frac{x}{x-3}+\frac{9}{x^2-3x}=\frac{x+3}{x}-\frac{x}{x-3}+\frac{9}{x\left(x-3\right)}\)
\(=\frac{\left(x+3\right)\left(x-3\right)}{x\left(x-3\right)}-\frac{x.x}{x\left(x-3\right)}+\frac{9}{x\left(x-3\right)}\)
\(=\frac{x^2-3x+3x-9-x^2+9}{x\left(x-3\right)}=\frac{0}{x\left(x-3\right)}=0\)
b) \(\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{-10x+8}{9x^2-4}\)
\(=\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{-10+8}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\frac{1\left(3x+2\right)}{\left(3x-2\right)\left(3x+2\right)}-\frac{4\left(3x-2\right)}{\left(3x+2\right)\left(3x-2\right)}-\frac{-10x+8}{\left(3x-2\right)\left(3x+2\right)}\)
\(\frac{3x+2-12x+2+10x-8}{\left(3x-2\right)\left(3x+2\right)}=\frac{x-4}{\left(3x-2\right)\left(3+2\right)}\)
c) \(\frac{4a^2-3a+5}{a^3-1}-\frac{1-2a}{a^2+a+1}-\frac{6}{a-1}\)
\(=\frac{4a^2-3a+5}{\left(a-1\right)\left(a^2+a+1\right)}+\frac{2a-1}{a^2+a+1}-\frac{6}{a-1}\)
\(=\frac{4a^2-3a+5}{\left(a-1\right)\left(a^2+a+1\right)}+\frac{\left(2a-1\right)\left(a-1\right)}{\left(a-1\right)\left(a^2+a+1\right)}-\frac{6\left(a^2+a+1\right)}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\frac{4a^2-3a+5+2a^2-2a-a+1-6a^2-6a-6}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\frac{-12}{\left(a-1\right)\left(a^2+a+1\right)}\)
d) \(\frac{x+9y}{x^2-9y^2}-\frac{3y}{x^2+3xy}=\frac{x+9y}{\left(x-3y\right)\left(x+3y\right)}-\frac{3y}{x\left(x+3y\right)}=\frac{x\left(x+9y\right)}{x\left(x-3y\right)\left(x+3y\right)}-\frac{3y\left(x-3y\right)}{x\left(x-3y\right)\left(x+3y\right)}\)
\(=\frac{x^2+9xy-3xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}=\frac{x^2-6xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}=\frac{\left(x-3y\right)^2}{x\left(x-3y\right)\left(x+3y\right)}=\frac{x-3y}{x\left(x+3y\right)}\)
e) \(\frac{3x+2}{x^2-2x+1}-\frac{6}{x^2-1}-\frac{3x-2}{x^2+2x+1}\)
\(=\frac{3x-2}{\left(x-1\right)^2}-\frac{6}{\left(x-1\right)\left(x+1\right)}-\frac{3x-2}{\left(x+1\right)^2}\)
\(=\frac{\left(3x+2\right)\left(x+1\right)^2}{\left(x-1\right)^2\left(x+1\right)^2}-\frac{6\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x-1\right)\left(x+1\right)}-\frac{\left(3x-2\right)\left(x-1\right)^2}{\left(x+1\right)^2\left(x-1\right)^2}\)
\(=\frac{3x^3+6x^2+3x+2x^2+4x+2-6x^2+6-3x^3+6x^2-3x+2x^2-4x+2}{\left(x-1\right)^2\left(x+1\right)^2}\)
\(=\frac{8x^2+10}{\left(x-1\right)^2\left(x+1\right)^2}\)
f) \(\frac{5}{a+1}-\frac{10}{a-\left(a^2+1\right)}-\frac{15}{a^3+1}=\frac{5a^2}{a^3+1}+\frac{10}{a^3+1}-\frac{15}{a^3+1}\)
\(=\frac{5a^2+10-15}{a^3+1}=\frac{5a^2-5}{a^3+1}\)
a) \(\sqrt{4,9.1350.0,6}=\frac{7\sqrt{10}}{10}.15\sqrt{6}.\frac{\sqrt{15}}{5}=63\)
b) \(\sqrt{12,5}.\sqrt{0,2}.\sqrt{0,1}=\frac{5\sqrt{2}}{2}.\frac{\sqrt{5}}{5}.\frac{\sqrt{10}}{10}=\frac{1}{2}\)
c) \(\sqrt{\frac{484}{169}}=\frac{22}{13}\)
d) \(\sqrt{\frac{2}{288}}=\sqrt{\frac{1}{144}}=\frac{1}{12}\)
e) \(\frac{\sqrt{2^5}}{\sqrt{2^3}}=\sqrt{2^2}=2\)
\(\frac{1}{X-1}-\frac{X^3-X}{X^2+1}\left(\frac{1}{X^2-2X+1}+\frac{1}{1-X^2}\right)\)
=\(\frac{1}{X-1}-\frac{X^3-X}{X^2+1}.\frac{X+1+X-1}{\left(X-1\right)^2\left(X+1\right)}\)
=\(\frac{1}{X-1}-\frac{X\left(X^2-1\right)}{X^2+1}.\frac{2X}{\left(X-1\right)^2\left(X+1\right)}\)
\(\frac{3x^2+3x+3}{4x+4}\): \(\frac{9x^3-9}{2x^2-2}\)= \(\frac{3\left(x^2+x+1\right)}{4\left(x+1\right)}\): \(\frac{9\left(x^3-1\right)}{2\left(x^2-1\right)}\)
= \(\frac{3\left(x^2+x+1\right)}{4\left(x+1\right)}\). \(\frac{2\left(x-1\right)\left(x+1\right)}{9\left(x-1\right)\left(x^2+x+1\right)}\)= \(\frac{1}{6}\)
1. Ta có:
\(\frac{1}{x}+\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+2013\right)\left(x+2014\right)}\)
\(=\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+2013}-\frac{1}{x+2014}\)
\(=\frac{2}{x}-\frac{1}{x+2014}\)
\(=\frac{2\left(x+2014\right)}{x\left(x+2014\right)}-\frac{x}{x\left(x+2014\right)}\)
\(=\frac{2x+4028-x}{x\left(x+2014\right)}=\frac{x+4028}{x\left(x+2014\right)}\)
2a) ĐKXĐ: x \(\ne\)1 và x \(\ne\)-1
b) Ta có: A = \(\frac{x^2-2x+1}{x-1}+\frac{x^2+2x+1}{x+1}-3\)
A = \(\frac{\left(x-1\right)^2}{x-1}+\frac{\left(x+1\right)^2}{x+1}-3\)
A = \(x-1+x+1-3\)
A = \(2x-3\)
c) Với x = 3 => A = 2.3 - 3 = 3
c) Ta có: A = -2
=> 2x - 3 = -2
=> 2x = -2 + 3 = 1
=> x= 1/2
Xét: \(x^5+y^5=\left(x^2+y^2\right)\left(x^3+y^3\right)-\left(x^2y^3+x^3y^2\right)\)
\(=\left(x^2+y^2\right)\left(x^3+y^3\right)-x^2y^2\left(x+y\right)\)
Thay vào biểu thức:
\(A=\left(a^2+\frac{1}{a^2}\right)\left(a^3+\frac{1}{a^3}\right)-\frac{1}{a^2}.a^2\left(\frac{1}{a}+a\right)\)
Ta có:
\(a^2+\frac{1}{a^2}=\left(a+\frac{1}{a}\right)^2-2=9-2=7\)
\(a^3+\frac{1}{a^3}=\left(a+\frac{1}{a}\right)^3-3.a.\frac{1}{a}\left(a+\frac{1}{a}\right)=27-9=18\)
Thay vào biểu thức tính ez ^^