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\(a.\dfrac{3x-2}{5}+\dfrac{x-1}{9}=\dfrac{14x-3}{15}-\dfrac{2x+1}{9}\\ \Leftrightarrow\dfrac{27x-18}{45}+\dfrac{5x-5}{45}=\dfrac{42x-9}{45}-\dfrac{10x+5}{45}\\ \Rightarrow27x-18+5x-5=42x-9-10x-5\\ \Leftrightarrow32x-23=32x-14\\ \Leftrightarrow0x=9\\ \Rightarrow Phươngtrìnhvônghiệm\\ \Rightarrow S=\phi\)
\(b.\dfrac{x+3}{2}-\dfrac{2-x}{3}-1=\dfrac{x+5}{6}\\ \Leftrightarrow\dfrac{3x-9}{6}-\dfrac{4-2x}{6}-\dfrac{6}{6}=\dfrac{x+5}{6}\\ \Rightarrow3x-9-4+2x-6=x+5\\ \Leftrightarrow5x-19=x+5\\ \Leftrightarrow4x=24\\ \Rightarrow x=6\\ \Rightarrow S=\left\{6\right\}\)
\(c.\dfrac{x+5}{2010}+\dfrac{x+4}{2011}+\dfrac{x+3}{2012}+\dfrac{x+2}{2013}=-4\\ \Leftrightarrow\dfrac{x+5}{2010}+1+\dfrac{x+4}{2011}+1+\dfrac{x+3}{2012}+1+\dfrac{x+2}{2013}+1=-4+4\\ \Rightarrow\dfrac{2015+x}{2010}+\dfrac{2015+x}{2011}+\dfrac{2015+x}{2012}+\dfrac{2015+x}{2013}=0\\ \Leftrightarrow\left(2015+x\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\right)=0\)
Do \(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}>0\)
nên \(2015+x=0\Rightarrow x=-2015\)
Câu d tương tự...thêm rồi chuyển vế sang :v
a) \(85^2+75^2+65^2+55^2-45^2-35^2-25^2-15^2\)
\(=\left(85^2-15^2\right)+\left(75^2-25^2\right)+\left(65^2-35^2\right)+\left(55^2-45^2\right)\)
\(=\left(85-15\right)\left(85+15\right)+\left(75-25\right)\left(75+25\right)+\left(65-35\right)\left(65+35\right)+\left(55-45\right)\left(55+45\right)\)
\(=70.100+50.100+30.100+10.100\)
\(=7000+5000+3000+1000\)
\(=16000\)
b) \(\frac{135^2+130.135+65^2}{135^2-65^2}\)
\(=\frac{135^2+2.60.135+65^2}{135^2-65^2}\)
\(=\frac{\left(135+65\right)^2}{\left(135-65\right)^2}\)
\(=\frac{200^2}{70^2}\) \(=\frac{200}{70}=\frac{20}{7}\)
a.\(\dfrac{5\left(x-3\right)}{4\left(x+1\right)}\) : \(\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x+1\right)^2}\)
= \(\dfrac{5\left(x-3\right)}{4\left(x+1\right)}\). \(\dfrac{\left(x+1\right)^2}{\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{5\left(x+1\right)}{4\left(x+3\right)}\)
b. \(\dfrac{6\left(x+8\right)}{7\left(x-1\right)}\). \(\dfrac{\left(x-1\right)^2}{\left(x-8\right)\left(x+8\right)}\)
= \(\dfrac{6\left(x-1\right)}{7\left(x-8\right)}\)
c.Tương tự hai câu trên nka!!
d. (\(\dfrac{1}{x\left(x+1\right)}\)-\(\dfrac{2-x}{x+1}\)).(\(\dfrac{x}{x-1}\))
=( \(\dfrac{1}{x\left(x+1\right)}\)-\(\dfrac{2x-x^2}{x\left(x+1\right)}\)). ....
= \(\dfrac{\left(1-x\right)^2}{x\left(x+1\right)}\). ...
= \(\dfrac{x-1}{x+1}\)
a)\(\frac{x}{5}+\frac{2x+1}{3}=\frac{x-5}{15}\)
\(\frac{3x}{15}+\frac{10x+5}{15}=\frac{x-5}{15}\)
\(3x+10x+5=x-5\)
\(13x+5-x+5=0\)
\(12x=-10\)
\(x=-\frac{5}{6}\)
a. 25 - \(x^2\) = (5-x) (5+x)
b) -196 + 4\(x^2\) = 196 - 4\(x^2\) = (14- 2x) (14+2x)
c)\(5^4-81x^4\) = \(\left[\left(5^2\right)^2\right]-\left[\left(81x^2\right)^2\right]\) = (\(\left(5^2-81x^2\right)\left(5^2+81x^2\right)\)
\(a,25-e=\left(5-\sqrt{e}\right)\left(5+\sqrt{e}\right)\)
\(b,-196+g=-\left(196-g\right)=-\left(14-\sqrt{g}\right)\left(14+\sqrt{g}\right)\)
\(c,2^6-47^2=\left(2^3\right)^2-47^2=\left(2^3-47\right)\left(2^3+47\right)\)
\(d,5^4-81x^4=\left(5^2\right)^2-\left(9x^2\right)^2=\left(5^2-9x^2\right)\left(5^2+9x^2\right)=\left(25-9x^2\right)\left(25+9x^2\right)\)
\(i,\dfrac{25}{16}-9y^2=\left(\dfrac{5}{4}-3y\right)\left(\dfrac{5}{4}+3y\right)\)
Bài 12:
a: \(=\left(xy+1+x+y\right)\left(xy+1-x-y\right)\)
\(=\left[x\left(y+1\right)+\left(y+1\right)\right]\left[x\left(y-1\right)-\left(y-1\right)\right]\)
\(=\left(x+1\right)\left(x-1\right)\left(y+1\right)\left(y-1\right)\)
b: \(=\left(x+y-x+y\right)\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)\)
\(=2y\cdot\left(3x^2+y^2\right)\)
c: \(=3y^2\left(x^4+x^3+x+1\right)\)
\(=3y^2\left[x^3\left(x+1\right)+\left(x+1\right)\right]\)
\(=3y^2\left(x+1\right)^2\left(x^2-x+1\right)\)
b: \(\Leftrightarrow4x^2-8x+4=x^2+2x+1+3\left(x^2+x-6\right)\)
\(\Leftrightarrow3x^2-10x+3=3x^2+3x-18\)
=>-13x=-21
hay x=21/13
c: \(\Leftrightarrow\left(\dfrac{x-90}{10}-1\right)+\left(\dfrac{x-76}{12}-2\right)+\left(\dfrac{x-58}{14}-3\right)+\left(\dfrac{x-36}{16}-4\right)+\left(\dfrac{x-15}{17}-5\right)=0\)
=>x-100=0
hay x=100
a) Ta có: