\(\dfrac{x^2}{6}=\dfrac{24}{25}\)
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a) ĐKXĐ: \(x\ne-5\)
\(\Leftrightarrow7x-7=6x+30\\ \Leftrightarrow x=37\)
b) \(\Leftrightarrow25x^2=144\\ \Leftrightarrow x^2=\dfrac{144}{25}\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{12}{5}\\x=-\dfrac{12}{5}\end{matrix}\right.\)
Bài 2:
a: -2*(-27)=54
6*9=54
=>Hai phân số này bằng nhau
b: -1/-5=1/5=5/25<>4/25
Bài 3:
a: =>16/x=-4/5
=>x=-20
b: =>(x+7)/15=-2/3
=>x+7=-10
=>x=-17
a. \(\dfrac{6}{2x+1}=\dfrac{2}{7}\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{21}\Rightarrow2x+1=21\)
\(\Rightarrow2x=21-1=20\Rightarrow x=\dfrac{20}{2}=10\)
Vậy x = 10
b. \(\dfrac{24}{7x-3}=\dfrac{-4}{25}\Rightarrow\dfrac{24}{7x-3}=\dfrac{24}{150}\Rightarrow7x-3=150\)
\(\Rightarrow7x=150+3=153\Rightarrow x=\dfrac{153}{7}\)
Vậy \(x=\dfrac{153}{7}\)
c. \(\dfrac{4}{x-6}=\dfrac{-12}{18}\Rightarrow-12\cdot\left(x-6\right)=4\cdot18=72\)
\(\Rightarrow x-6=\dfrac{72}{-12}=-6\Rightarrow x=-6+6=0\)
\(\dfrac{y}{24}=\dfrac{-12}{18}\Rightarrow y=\dfrac{-12\cdot24}{18}=-16\)
Vậy x = 0 ; y = -16
\(6.-\dfrac{5}{18}=-\dfrac{5}{3}\)
\(-\dfrac{13}{24}+-\dfrac{5}{21}.\dfrac{7}{25}=-\dfrac{13}{24}+-\dfrac{1}{15}=-\dfrac{73}{120}\)
\(-\dfrac{5}{13}-\dfrac{3}{26}.-\dfrac{2}{3}=-\dfrac{5}{13}-\left(-\dfrac{1}{13}\right)=-\dfrac{4}{13}\)
\(\left(\dfrac{2}{3}+-\dfrac{7}{6}\right).\left(\dfrac{3}{8}+-\dfrac{3}{24}\right)=-\dfrac{1}{2}.\dfrac{1}{4}=\dfrac{1}{8}\)
Kết quả lần lượt của các phép tính là :
\(\dfrac{-5}{3}\) , \(\dfrac{-73}{120}\) , \(\dfrac{-4}{13}\) , \(\dfrac{-1}{8}\)
24:
\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
\(\Leftrightarrow\left(x+2\right)\left(x+6\right)=8\left(x+6\right)-8\left(x+2\right)\)
\(\Leftrightarrow x^2+8x+12=8x+48-8x-16=32\)
=>(x+10)(x-2)=0
=>x=-10 hoặc x=2
25: \(\Leftrightarrow\dfrac{\left(x+1\right)^2+1}{x+1}+\dfrac{\left(x+4\right)^2+4}{x+4}=\dfrac{\left(x+2\right)^2+2}{x+2}+\dfrac{\left(x+3\right)^2+3}{x+3}\)
\(\Leftrightarrow x+1+\dfrac{1}{x+1}+x+4+\dfrac{4}{x+4}=x+2+\dfrac{2}{x+2}+x+3+\dfrac{3}{x+3}\)
\(\Leftrightarrow\dfrac{1}{x+1}+\dfrac{4}{x+4}=\dfrac{2}{x+2}+\dfrac{3}{x+3}\)
\(\Leftrightarrow x+5=0\)
hay x=-5
a: \(\Leftrightarrow x^2=900\)
=>x=30 hoặc x=-30
b: \(\Leftrightarrow\dfrac{2}{3}:\left(-0.1x\right)=\dfrac{4}{3}:\dfrac{-2}{25}=-\dfrac{4}{3}\cdot\dfrac{25}{2}=-\dfrac{100}{6}=\dfrac{-50}{3}\)
=>0,1x=2/3:50/3=2/3x3/50=1/25
=>1/10x=1/25
hay x=1/25:1/10=10/25=2/5
d: \(\Leftrightarrow x^2=\dfrac{144}{25}\)
=>x=12/5 hoặc x=-12/5
\(a,\dfrac{x-1}{x+5}=\dfrac{6}{7}\\ \Leftrightarrow\left(x-1\right).7=6\left(x+5\right)\\ \Rightarrow7x-7=6x+30\\ \Rightarrow7x-6x=7+30\\ \Rightarrow x=37\)
Vậy \(x=37\)
\(b,\dfrac{x^2}{6}=\dfrac{24}{25}\\ \Leftrightarrow x^2.25=24.6\\ \Rightarrow x^2.5^2=144\\ \Rightarrow\left(5x\right)^2=144\\ \Rightarrow\left(5x\right)^2=\left(\pm12\right)^2\\ \Rightarrow\left\{{}\begin{matrix}5x=12\\5x=-12\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{12}{5}\\x=-\dfrac{12}{5}\end{matrix}\right.\)
Vậy \(x=\pm\dfrac{12}{5}\)
\(\dfrac{x^2}{6}=\dfrac{24}{25}\)
\(\Rightarrow x^2=\dfrac{24.6}{25}=\dfrac{144}{25}\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{12}{5}\\x=-\dfrac{12}{5}\end{matrix}\right.\)